Solutions Manual Symmetry, Broken Symmetry, and Topology in Modern Physics Mike Guidry and Yang

Solutions Manual Symmetry, Broken Symmetry, and Topology in Modern Physics Mike Guidry and Yang

Solutions Manual Symmetry, Broken Symmetry, and Topology in Modern Physics Mike Guidry and Yang

Solutions Manual Symmetry, Broken Symmetry, and Topology in Modern Physics Mike Guidry and Yang

Student Solutions Manual Symmetry, Broken Symmetry, and Topology in Modern Physics Mike Guidry and Yang Sun This document gives the solutions for 172 of the 344 problems found at the ends of chapters (those marked with a ***) for the first edition of Symmetry, Broken Symmetry, and Topology in Modern Physics: A First Course by Mike Guidry and Yang Sun (Cambridge University Press, 2022). Unless otherwise indicated, literature references, equation numbers, figure references, table references, and chapter or section numbers refer to the print version of that book. The occasional reference to figures or tables contained specifically in this document (rather than the book) will be indicated explicitly by a tag “[this document]”.

Solutions Manual Symmetry, Broken Symmetry, and Topology in Modern Physics Mike Guidry and Yang

Solutions Manual Symmetry, Broken Symmetry, and Topology in Modern Physics Mike Guidry and Yang

Solutions Manual Symmetry, Broken Symmetry, and Topology in Modern Physics Mike Guidry and Yang

1

Introduction No problems for this introductory chapter.

1

Solutions Manual Symmetry, Broken Symmetry, and Topology in Modern Physics Mike Guidry and Yang

2

Some Properties of Groups 2.1 Suppose that there are two group identities, e and e′. Then e = e · e′ = e′. Thus there is only one group identity. Suppose that a−1 and a¯ are both inverses of the group element a.

Then using that a · a−1 and a · a¯ must both be equal to the group identity, a¯ = a¯ · (a · a−1) = (a¯ · a)· a−1 = ea−1 = a−1.

Thus each group element has a unique inverse. From the properties of the inverse and identity, (a · b)−1 · (a · b) = e. Multiply this expression from the right by b−1 and use b · b−1 = e to give (a · b)−1 · a = b−1.

Multiply from the right by a−1 and use a · a−1 = e to give (a · b)−1 = b−1 · a−1.

Check this result against entries in Table 2.2: Let a = (12) and b = (23). Then from the table a−1 = (12) and b−1 = (23), and also from the table a · b = (12)· (23) = (321).

Thus from the table (a · b)−1 = (321)−1 = (123). From the formula derived above and the table (a · b)−1 = b−1a−1 = (23)· (12) = (123).

Hence for this example (a · b)−1 = b−1a−1, as asserted.

2.2 Define the permutations ()=

123 123

(12) =

123 213

(23) =

123 132

(13) =

123 321

(123) =

123 231

(321) =

123 312

so that, for example, (12){abc} → {bac}. Carrying out all possible combinations, the multiplication table for A · B is A\B e

(123) (321)

e e (12) (23) (13) (123) (321)

(12) (12) e (123) (321) (23) (13)

(23) (23) (321) e (123) (13) (12)

(13) (13) (123) (321) e (12) (23)

(123) (123) (13) (12) (23) (321) e

(321) (321) (23) (13) (12) e (123)

2

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Some Properties of Groups

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where e is the identity, e ≡ ( ). This is equivalent to Table 2.2. From the multiplication table this is a group, because all group postulates are satisfied. The identity and any of the three transpositions of two objects are closed under multiplication, so these form 2-element subgroups isomorphic to S2, S2 ∼ {e,(12)} ∼ {e,(13)} ∼ {e,(23)}.

The set of even permutations A3 = {e,(123),(321)} is closed under multiplication, so this too is a subgroup. The multiplication table for C3 can be deduced from geometry.

The operators 1, c3, c2, and c3, rotate by 0, 2π , 4π , and 2π = 0, respectively, so the C3 3

3

3

3

multiplication table and isomorphism with A3 are c2 c2

e

c2

c2

c2 e

A3 e (123) (321)

→→

e e (123) (321)

(123) (123) (321) e

(321) (321) e (123)

if we invoke the correspondence e →→ e

c3 →→ (123)

c32 →→ (321)

between entries in the two multiplication tables.

2.4 Consider the multiplication table for G = {e, a, b}, where e is the identity. We must have a · a = b, since • If a · a = a, then a = e and a group can’t have two identities.

• If a · a = e, then the elements {e, a} alone close a group of order two.

Then since a · a = b, it follows that a · b = b · a = e and b · b = a are the only consistent choices. Therefore the multiplication table for the finite group of order three is unique and given by e a b

e a b e a b . a b e b e a

We conclude that there is only one independent group of order three. Note, for example, that this multiplication table is isomorphic to that for the cyclic group C3 if we identify a = c3 and b = c32 (see Example 2.18 and Problem 2.2).

2.5 From Example 2.15, the multiplication table for the cyclic group C4 is C4 e a a2 a3

e e a a2 a3

a a a2 a3 e

a2 a2 a3 e a

a3 a3 e a a2

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Some Properties of Groups

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For the subgroup H = {e, a2} eee−1 = e a2e(a2)

−1

−1

−1

= a2ea2 = e

ea2e−1 = a2 a2a2(a2)

aea−1 = aa−1 = e

= a2a2a2 = a2

a3e(a3) = a3ea = e

aa2a−1 = aa2a3 = a2 a3a2(a3)

−1

= a3a2a = a2,

where the cyclic condition a4 = e has been used and the inverses have been inferred from the multiplication table and the requirement xx−1 = e. Therefore H is identical to all of its conjugate subgroups H′ = {ghg−1 ; g ∈ G, h ∈ H}, and is an invariant subgroup (Section 2.12). In fact, since all cyclic groups are abelian, it is an abelian invariant subgroup.

2.6 An invariant subgroup consists of whole classes. For S3 there are three classes: {e}

{(12), (23), (13)}

{(123), (321)},

as may be verified by similarity transforms [see Eq. (2.21)]. Therefore, {e, (123), (321)} (the alternating subgroup) is invariant because it consists of two whole classes. The subgroup {e, (12)} cannot be invariant because e is a whole class but (12) is only part of a class.

2.8 (a) a ∼ a requires that there be some group element g such that a = gag−1. The choice g = e obviously satisfies this. (b) If a ∼ b, then a = gbg−1 where g ∈ G. Multiply from the left by g−1 and from the right by g to give g−1ag = b, which implies that b ∼ a. (c) If a ∼ b and b ∼ c, then there are elements p, q ∈ G such that a = pbp−1

b = qcq−1.

Therefore, a = pbp−1 = pqcq−1p−1 = pqc(pq)−1, where we have used that the reciprocal of a matrix product is formed by taking reciprocals of the factors in reverse order: (ABCD)−1 = D−1C−1B−1A−1.

Then, since pq must be a group element, a ∼ c. Thus class conjugation is an equivalence relation, as defined in Box 2.5.

2.11 The multiplication table of D2 is given in Problem 2.9. It has abelian invariant subgroups {e, a}, {e, b}, and {e, c}. Let’s form the left cosets of H = {e, a}: e{e, a} = {e, a} = H b{e, a} = {b, c}

a{e, a} = {a, e} = {e, a} = H c{e, a} = {c, b} = {b, c}.

Therefore, G/H = D2/H = {e, a} + {b, c}.

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Some Properties of Groups

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Now let’s construct the multiplication table for G/H. Let E = H = {e, a}

M = {b, c} = b{e, a}.

From the multiplication table and the coset multiplication law of Eq. (2.27), pH · qH = (pq)H we find that 2

E · E = e{e, a}e{e, a} = e {e, a} = E, E · M = e{e, a}b{e, a} = eb{e, a} = b{e, a} = {b, c} = M, M · E = b{e, a}e{e, a} = be{e, a} = b{e, a} = M, M · M = b{e, a}b{e, a} = bb{e, a} = e{e, a} = {e, a} = E, so the multiplication table of the quotient group is G/H E M

E E M

M M E

where E = H. This is the multiplication table for the group C2 ∼ Z2.

2.15 The representations Γ(4) and Γ(5) are equivalent (they have the same characters), so they are related by a similarity transform. You can verify that Γ(5) is converted to Γ(4) if each matrix is similarity transformed using [180] 1 √0 2 S= 1 3 −1 . √ 1 − 3 −1 Thus it is sufficient to examine one of Γ(4) or Γ(5). From Eq. (2.32), n χ∗(i)χ(i), aν = 1 i ν NG ∑ i

where the sum is over classes, and χν (i) and χ(i) are the characters of a given class i in the irrep ν and in the reducible representation U (g), respectively. Thus, for Γ(4) a1 = a2 = a3 =

1 6 1 6 1 6

[(1)(1)(3) + (2)(1)(0) + (3)(1)(1)] = 1, [(1)(1)(3) + (2)(1)(0) + (3)(−1)(1)] = 0, [(1)(2)(3) + (2)(−1)(0) + (3)(0)(1)] = 1,

so the Γ(4) and Γ(5) representations of S3 displayed in Fig. 2.4 are reducible, with the irrep content Γ(1) ⊕ Γ(3).

2.16 From Eq. (2.27), define coset multiplication by pHqH = (pq)H. Then,

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Some Properties of Groups

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1. For H an invariant subgroup, this is a consistent definition and leads to closure because Hg = gH and pHqH = pHHq = pHq = pqH, where we have used that under this multiplication law eHeH = eeH = H, so HH = H. Therefore, the product of two cosets under coset multiplication is itself a coset, since if p ∈ G and q ∈ G, then pq ∈ G. 2. Coset multiplication is associative since the original group multiplication is associative: (pH) (qH)(rH) = (pH)(qrHH) = p(qr)HHH = (pq)rHHH = (pH)(qH) (rH).

3. The coset E ≡ eH = H acts as an identity since, by the coset multiplication law, eHqH = eqH = qH. 4. For each coset pH there is a unique inverse p−1H, since (pH)(p−1H) = (pp−1)H = eH = E. Therefore, the cosets form a group under the coset multiplication law if H is an invariant subgroup.

2.17 If giH and g jH have an element in common, then gih1 = g jh2 for some elements h1, h2 ∈ H, which implies that g i g −j 1 = h2h1−1.

But h2h1−1 is an element of H by the group property and, by the rearrangement lemma (Section 6.2.5), H g i g −j 1 = H and thus Hgi = Hg j . Thus, two cosets are identical if a common element exists, so different cosets are completely disjoint.

2.18 (a) Let the order of G be n, the order of H be m, with the index of H in G being ℓ, with n = mℓ. Then trivially, if n is a prime number either m = 1 or m = n, so there are no proper subgroups. (b) Let G be of order n, where n is a prime number, and choose g∈ G but not equal to the identity e. Then if we take successive products of g with itself k times, gk = e by the group property, for some integer k. This means that n , g, g2, . . . , gk−1, e is a cyclic subgroup. But from part (a) G can’t a proper subgroup because it is of prime order by hypothesis. Thus the cyclic subgroup can only be the full group and generally groups of prime order n are isomorphic to cyclic groups Cn. 2.20 Consider the function set f1(x) = x

f2(x) = −x

f3(x) =

1 x

f4(x) = −

1 x

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Some Properties of Groups

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under the binary operation of substituting one function in another. For example, f1 · f1 ≡ f1( f1) = x = f1 f1 · f2 ≡ f1( f2) = −x = f2 1 1 f1 · f3 ≡ f1( f3) = = f3 f1 · f4 ≡ f1( f4) = − = f4. x x Thus, f1(x) plays the role of an identity. Likewise f2 · f1 = −x = f2 f ·f =− = f 1 2 3 4 x

f2 · f2 = −(−x) = x = f1 f ·f =− − = =f .

1 1 2 4 3 x x

Continuing in this manner we find the multiplication table f1 f2 f3 f4

f1 f1 f2 f3 f4

f2 f2 f1 f4 f3

f3 f3 f4 f1 f2

f4 f4 f3 f2 f1

where e ≡ f1 is the identity. This is a group, since 1. The set is closed under the multiplication operation of substitution.

2. There is an identity ( f1).

3. Each element has a unique inverse (the identity f1 appears exactly once in each row and column). In fact, each element is its own inverse.

4. The substitution operation is associative.

Since the multiplication table is symmetric about the diagonal, the group is abelian. As shown in Problem 2.19, the matrices a=

1 0 0 1

b=

1 0 0 −1

c=

−1 0 0 1

d=

−1 0 0 −1

have the same multiplication table under the binary operation of matrix multiplication if we identify a ↔ f1 = e, b ↔ f2, c ↔ f3, and d ↔ f4.

2.26 This problem is based on a discussion in Gilmore [72]. Parameterize the complex numbers of unit modulus by C = x + iy = cosφ + i sinφ.

Expand in a power series 1 1 1 1 C = cosφ + i sinφ = 1 − φ 2 + φ 4 − . . . + i φ − φ 3 + φ 5 − . . . .

2 4!

3!

5!

Comparing term by term with the exponential power series 1 i 1 eiφ = 1 + iφ − φ 2 − φ 3 + φ 4 − . . .

2 3!

4! we see that eiφ = cosφ + i sinφ

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Some Properties of Groups

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and eiφ is a faithful mapping for the complex numbers of unit modulus if we restrict −π ≤ φ ≤ π. These form a group under multiplication since ′

1. Closure: eiφ eiθ = ei(φ+θ) = eiφ with φ ′ = φ + θ.

2. Associativity: eiφ (eiθ eiα ) = (eiφ eiθ )eiα = ei(φ+θ+α).

3. Existence of inverse: e−iφ eiφ = 1.

4. Identity: e0eiφ = eiφ e0 = eiφ .

Now consider the product (cosφ + i sinφ)(cosθ + i sinθ) = eiφ eiθ = ei(φ+θ).

Multiplying out the left side and rewriting the right side using eiα = cosα + i sinα gives cosφ cosθ − sinφ sinθ + i(cosφ sinθ + sinφ cosθ) = cos(φ + θ) + i sin(φ + θ).

Equating separately the real and imaginary parts of the two sides then gives the two trigonometric identies cos(φ + θ) = cosφ cosθ − sinφ sinθ sin(φ + θ) = cosφ sinθ + sinφ cosθ that we sought to prove. This result is a simple example of a general property: all the special functions of mathematical physics are representations of some group and standard identities can be obtained by appropriate operations on representations of the group. See Gilmore [72] for further discussion.

2.27 This problem is adapted from an example in Sternberg [185]. (a) Divide the integers up into four equivalence classes: e ≡ {0, 4, −4, 8, −8,...}

a ≡ {1, 5, −3, 9, −7,...}

b ≡ {2, 6, −2, 10, −6,...}

c ≡ {3, 7, −1, 11, −5,...}

and construct a multiplication table of these classes under addition modulo 4. That is, add two integers and subtract an integer multiple of 4 to bring it in the range −4 to +4. Consider the product ac. Examples are 3 + 1 mod 4 = 0 ∈ e

5 + 7 mod 4 = 0 ∈ e

1 + 7 mod 4 = 0 ∈ e.

Therefore ac = ca = e. Likewise, consider the product ab. For example, 1 + 2 mod 4 = 3 ∈ c

5 + 6 mod 4 = 3 ∈ c

− 3 + 6 mod 4 = 3 ∈ c.

Therefore, ab = ba = c. Carrying out the other possible products gives the multiplication table Z4 e a b c e e a b c a a b c e b b c e a c c e a b

Solutions Manual Symmetry, Broken Symmetry, and Topology in Modern Physics Mike Guidry and Yang

Some Properties of Groups

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indicating that this is a group and it is abelian. (b) Consider the matrices e=

1 0 0 1

0 −1 1 0

a=

b=

−1 0

0 −1

=

0 −1

c=

0 −1

1 .

0

Matrix multiplication gives results like ab =

0 −1 1 0

−1 0

0 −1

bc =

−1 0

0 −1

1 0 = 0 1

0 −1

1 =c 0 −1 = a.

0

Carrying out all possibilities, we find that these matrices have the same multiplication table as Z4. Clearly e plays the role of the identity and the matrix inverse defines the inverse of an element. For example a−1 =

−1

0 —1 1 0

=

0 −1

1 = c, 0

since this leads to aa−1 =

0 −1 1 0

0 −1

1 1 0 = = e.

0 0 1

Thus, these matrices are a representation of Z4. (c) Consider following the rotation operations in a plane e = rotation by an integer multiple of 2π, a = rotation counterclockwise by π2 , b = rotation counterclockwise by π, c = rotation counterclockwise by 3π2 .

We can multiply these by implementing first one rotation and then another. For example, • ab = rotate counterclockwise by π + π2 = 3π2 = c.

5π π • bc = rotate counterclockwise by π + 3π 2 = 2 = 2 = a.

Carrying out all such multiplication we find the same multiplication table as for Z4 above.

Thus the cyclic group of four-fold rotations in the plane C4 is homomorphic to Z4. (d) The group ZN can be represented by the elements zi = e2πin/N

(n = 0, 1,... N − 1).

For Z4 we have N = 4 and n = 0, 1, 2, 3, giving the elements e ≡ z0 = e 0 = 1

π

a ≡ z1 = e2πi/4 = e 2 i

b ≡ z2 = e4πi/4 = eiπ

3

πi

c ≡ z3 = e 2 .

Taking products gives results like ac = e( 2 + 2 π)i = e2πi = e π

3

ab = e( 2 +π)i = e 2πi = c. π

3

Solutions Manual Symmetry, Broken Symmetry, and Topology in Modern Physics Mike Guidry and Yang

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Some Properties of Groups

Forming all such products gives once again the multiplication table for Z4 given above.

2.30 (a) For an arbitrary group element ga and group identity e, ga−1e ga = ga−1ga = e.

Thus e is in a class of its own. (b) Suppose that for group elements ga and gb, gc−1gbgc ≡ gd

−→

gb = gcgdgc−1

gc−1gagc ≡ gd

−→

ga = gcgdgc−1.

Multiply gd = gc−1gbgc from the left by gc and from the right by gc−1 to give gcgdgc−1 = gcgc−1gbgcgc−1 = gb.

Thus, combining with ga = gcgdgc−1 from above, gb = gcgdgc−1 = ga.

Hence the conjugate classes of gb and ga must be identical.

2.32 Suppose a direct product group G = A × B, with elements g = ab ∈ G, where a ∈ A and b ∈ B, and where the elements of A commute with the elements of B. Consider an arbitrary element ai of A. Then gaig−1 = abai(ab)−1 = abaib−1a−1 = abb−1aia−1 = aaia−1 ∈ A, where we’ve used in line 3 that the elements of A and B commute. Thus A is an invariant subgroup of G = A × B. Likewise, for an arbitrary element bi of B, gbig−1 = abbi(ab)−1 = abbib−1a−1 = aa−1bbib−1 = bbib−1 ∈ B, and B also is an invariant subgroup of G = A × B. 2.35 We follow a discussion in Elliott and Dawber [56]. Define ψ′(r) = ψ G−j 1 r . Then from Eq. (2.9), T (Gi)T (Gj)ψ(r) = T (Gi)ψ G−j 1 r = T (Gi)ψ′(r) = ψ ′ Gi−1 r = ψ G−j 1Gi−1 r = ψ (GiGj)−1r = T (GiGj)ψ(r).

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Some Properties of Groups

Thus T (Gi)T (Gj) = T (GiGj) and T (G) preserves the group multiplication law (2.8) for G and is a valid representation.

Solutions Manual Symmetry, Broken Symmetry, and Topology in Modern Physics Mike Guidry and Yang

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Introduction to Lie Groups 3.1 Do it in two steps. First use the highest-weight algorithm to show that 2 ⊗ 2 = 3 ⊕ 1.

Then take the product of that with the remaining2 to show that 2 ⊗ 2 ⊗ 2 = (3 ⊕ 1) ⊗ 2 = 4 ⊕ 2 ⊕ 2.

The relevant highest-weight decompositions are illustrated in the following diagrams:

2

2

= 3 -1

3

0

1 2

1

= 2

1

-1/2

2

= 4 -3/2

-1/2

1/ 2

1/ 2

2

3 /2

(see the example discussed in Section 3.3.4).

3.3 Evaluate the quantity U (α)−1U (β )−1U (α)U (β ) for two group elements U (α) = ei(α1X1+α2X2,+...)

U (β ) = ei(β1X1+β2X2,+...)

lying near the origin U (0) in the parameter space, where α = (α1, α2, . . .)

β = (β1, β2, . . .)

are two sets of angles parameterizing nearby points and X = (X1, X2, . . .) are the group generators. (This quantity measures the commutivity of the group because if the group elements commute, U (α)−1U (β )−1U (α)U (β ) = U (α)−1U (β )−1U (β )U (α) = U (α)−1U (α) = 1, and deviations of the product from unity indicate a failure to commute.) The product defined above must give a new element of the group, say U (γ), U (α)−1U (β )−1U (α)U (β ) = U (γ).

Substituting the exponential forms for all group elements and expanding in the vicinity of the origin for the exponentials on both sides of the equation gives 1 + ∑αkβℓ(XkXℓ − XℓXk) + . . . = 1 + ∑γmXm + . . . kℓ

m

12

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Introduction to Lie Groups

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Comparing term by term, consistency then requires that the Lie algebra [ Xk, Xℓ ] = ∑ckℓm Xm m

be satisfied, where [ Xk, Xℓ ] ≡ XkXℓ − XℓXk andkℓcm is an expansion coefficient. This is the form of the Lie algebra given in Box 3.2. It can be converted to the form in Eq. (3.3) by introducing new generators Xi → −iXi, as discussed in Box 3.2.

3.5 Using Eq. (3.7a) to change fcad → − facd and fcde → − fdce, the Jacobi identity (3.7b) may be written as fbcd fade − facd fbde = fabd fdce .

Now use (Ta)bc = −i fabc from Eq. (3.8) to write fbcd fade − facd fbde = −(Tb)cd (Ta)de + (Ta)cd (Tb)de = (TaTb)ce − (TbTa)ce = [ Ta, Tb ]ce, where the usual rules for matrix multiplication were used. Likewise, use Eq. (3.8) to write fabd fdce = i fabd (Td )ce.

Combining these results gives [ Ta, Tb ]ce = i fabd(Td)ce, or writing in matrix form and switching the dummy summation index from d to c, [ Ta, Tb ] = i fabcTc, which is Eq. (3.9).

3.6 The adjoint representation is generated by the structure constants; in particular, from Eq. (3.8) for SU(2) the adjoint representation is given in terms of three 3 × 3 matrices ta, with (ta)i j = −iεai j

+1 −1

if i jk cyclic if i jk anticyclic .

0

otherwise

εai j =

For example, if a = 1, (t1)23 = −iε123 = −i

(t1)32 = −iε132 = i,

and all other components (t1)i j are zero. Then, 0 0 t1 = 0 0 0 i

0 −i .

0

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Introduction to Lie Groups

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By a similar construction, 0 t2 = 0 −i

0 i 0 0 0 0

−i 0 0 0 .

0 0

0 t3 = i 0

Since this is a rank-1 algebra, we may diagonalize one generator. Let’s choose t3, and require by the usual eigenvalue methods that the characteristic (secular) equation, −λ i det(t3 − λ I) = det 0

−i −λ 0

0 0 −λ

= 0,

must be solved for the eigenvalues λ. This yields −λ 3 + λ = 0, which has solutions λ = 0, ±1. Therefore, we may choose the diagonal form with these eigenvalues on the diagonal, 1 0 0 t3′ = 0 0 0 .

0 0 −1 To transform t1 and t2 to this new basis in which t3 is diagonal, we need the matrix c that satisfies c−1t3c = t3′ . This matrix is formed from columns of eigenvectors of t3′ . Labeling the three eigenvectors by x, y, and z and solving −λ i 0

−i 0 x 0 −λ 0 y = 0 z 0 0 −λ

for each of the eigenvalues λ = 0, ±1 gives the eigenvectors x y z

−1

i 1 = √ 1 2 0

x y

= 0 z

0

0 1

x y z

−i 1 =√ 1 .

2 0 +1

i 0 −i

1 √0 0 2 1 0

Thus a suitable choice for c and its inverse are −i 1 c= √ 1 2 0

0 i 0 1 √ 2 0

−i 1 0

0 i 0 1 √ 2 0

1 1 c− = √ 2

which satisfy cc−1 =

1 2

i 0 −i

1 √0 1 0 0 0 2 = 0 1 0 .

0 0 1 1 0

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Then the matrices 0 −i 0 i 0 i , 2 0 −i 0

1

t1′ = c−1t1c = √ 1 t2′ = c−1t2c = √ 2 t3′ = c−1t3c =

0 −1 0

−1 0 1

1 0

0

0 1 , 0

0 0 0 , 0 0 −1

constitute a new basis for the adjoint representation in which t3 is diagonal. That this is an SU(2) representation may be checked by explicit multiplication. For example, t1′ ,t2′

1 = 2

0 i

−i 0 0 i 0 −i 0

1 0

0 −1 0

−1 0 1

0 0 1 − −1 0 0

−1 0 1

0 0 −i 0 1 i 0 i 0 0 −i 0

0

=i 0 0 0 0 0 −1

= it3′ ,

which satisfies the SU(2) algebra (3.13) since t1′ ,t2′ = iε123t3′ = it3′ .

3.7 More formal and mathematical proofs may be found in many books on group theory.

We give a proof here that is more heuristic. Consider the commutator [ Ta, M ] for some matrix M, where Ta is a generator, and assume that [ Ta, M ] = 0 for all a in an irreducible representation. Schur’s lemma then results from the following reasoning: 1. Since M commutes with all generators, none of the generators can connect states having different eigenvalues of M. 2. But irreducibility means that every state of the multiplet can be reached from any other state of the multiplet by application of some combination of generators.

3. Thus, there is a contradiction unless every member of the multiplet has the same eigenvalue with respect to M. 4. This means that the matrix M operating on the members of the irrep must be of the form M = λ 1, where 1 is the unit matrix.

Therefore M is necessarily a multiple of the unit matrix if it commutes with all generators of an irrep.

3.8 The commutators can be calculated by brute force. For example, ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ [M ,M ] = z −y x −z − x −z z −y 1 2 ∂y ∂z ∂z ∂x ∂z ∂x ∂y ∂z ∂ ∂ =y −x = M3. ∂x ∂y

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But more elegantly we may use Mi ≡ ε i jk x k

∂

x i,

∂xj

∂ = −δ i j ∂xj

εi jk εmnk = δim δ jn — δin δ jm

to give a general proof that [ Mi, Mj ] = εi jkMk

[ Mi, Nj ] = εi jkNk

[ Ni, Nj ] = εi jkMk,

which is the Lie algebra of SO(4). Then with the definitions 1 Ji ≡ (Mi + Ni) 2 only some algebra is required to show that [ Ji, Jj ] = εi jkJk

1 Ki ≡ (Mi − Ni), 2

[ Ki, Kj ] = εi jkKk

[ Ki, Jj ] = 0,

which is the Lie algebra of SU(2) × SU(2).

3.11 (a) Since (BAB−1)i j = Biℓ Aℓk B−k 1j and the trace is the sum over diagonal elements, Tr(BAB−1) = B iℓAℓk Bk−i1

= (B−1B)kℓAℓk = δkℓAℓk = Akk = Tr A,

where implied summation on repeated indices was assumed. (b) Expanding the exponentials in the product [82], ! ! ∞ ∞ 1 k A B 1 Bj e e = ∑ A × ∑ . k! k=0 j=0 j!

Changing to a summation variable m = k + j, this can be written ∞

∞

1

∞

1

∞

1

A kBm−k = ∑ A k B m−k . e e = ∑ ∑ k! (m − k)! m! ∑ k!(m − k)!

A B

m=0

m=0 k=0

m!

k=0

But from the binomial theorem m

m

(A + B) = ∑

m!

k

k=0 k!(m − k)!

implying that

∞ A B

e e = ∑ m=0

1 m!

m

m−k

A B

(A + B) = e

A+B

,

,

if we assume that A and B commute. If [ A, B ] /= 0, then the Baker–Campbell–Hausdorff formula quoted in Problem 3.15 must be used.

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(c) Let φi be the eigenvectors associated with the eigenvalues λi of A such that Aφi = λiφi.

Then it follows that ∞ 1 m A m λ ∞ 1 e φi = ∑ A φi = ∑ λi φi = e i φi m! m! m=0 m=0 and φi is an eigenvector of eA with eigenvalue eλi . (d) The determinant is the product of eigenvalues and the trace is the sum of eigenvalues.

Thus, from the result in part (c), det eA = eλ1 eλ2 . . . eλn = eλ1+λ2+...+λn = eTr A, for eigenvalues λn of a matrix A. (e) For a product of matrices A and B, Tr(AB) = (AB)ii = AikBki = BkiAik = (BA)kk = Tr(BA), where summation on repeated indices was assumed. (f) Matrix multiplication is associative and in part (e) we proved that Tr(XY ) = Tr(YX ).

Hence, Tr(ABC) = Tr (AB)C = Tr C(AB) = Tr(CAB) = Tr (CA)B = Tr B(CA) = Tr(BCA), and we have proved the cyclic property of the trace Tr(ABC) = Tr(BCA) = Tr(CAB) for a product of three matrices. This generalizes to invariance of the trace of a product of any number of matrices under cyclic permutation. (g) The hermitian conjugate of a matrix is the transpose of its complex conjugate. Therefore, (AB)† = (AB) ji

∗

= (AjkBki)∗ = A∗jkBk∗i = Bk∗iA∗jk

= B†ikA†k j = (B†A†)i j = B†A†.

Generalizing this, we find that the hermitian conjugate of a matrix product is the product of hermitian conjugates for each matrix in the product, in reverse order. (h) Let’s prove for a product of two matrices A and B. Let C = AB. Then A = ABB−1 = CB−1

B = A−1AB = A−1C.

Therefore we have that C = AB = (CB−1)(A−1B) = CB−1A−1C.

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But this equation can be true only if CB−1A−1 = 1, where 1 is the unit matrix, implying that B−1A−1 = C−1 = (AB)−1.

Therefore, we have shown that (AB)−1 = B−1A−1, which can be generalized to (ABCD · · · Z)−1 = Z−1 · · · D−1C−1B−1A−1 for an arbitrary number of matrices. (i) Utilizing ⟨a| A |b⟩∗ = ⟨b| A† |a⟩ and ∑|a⟩⟨a| = 1, a

∑⟨a| A |b⟩ ⟨a| A |b⟩ = ∑⟨b| A† |a⟩⟨a| A |b⟩ ∗

ab

ab

= ∑⟨b| A†A |b⟩ b

= Tr(A†A) = Tr(AA†), where the result of part (e) was used in the last step.

3.12 The product of two arbitrary matrices from the set is GG′ =

cosθ −sinθ

cosθ′ −sinθ ′

sinθ cosθ

sinθ′ cosθ′

=

cosθ cosθ′ − sinθ sinθ′ −sinθ cosθ′ − cosθ sinθ′

cosθ sinθ′ + sinθ cosθ′ −sinθ sinθ′ + cosθ cosθ′

=

cos(θ + θ ′) −sin(θ + θ ′)

=

sin(θ + θ ′ ) cos(θ + θ ′)

cosφ −sinφ

sinφ , cosφ

where φ ≡ θ +θ′. Thus the set is closed under matrix multiplication. Matrix multiplication is associative and since cosθ sinθ det = cos2 θ + sin2 θ = 1 = / 0 −sinθ cosθ the matrices are invertible and an inverse exists for each matrix. Finally an identity e exists: cos(θ = 0) −sin(θ = 0)

e≡ Therefore the matrices

G=

sin(θ = 0) 1 0 = . cos(θ = 0) 0 1

cosθ

sin θ

−sinθ

cosθ

(θ = 0 − 2π)

form a group under matrix multiplication. The group is obviously abelian since ′

′

GG = G G =

cos(θ + θ ′) −sin(θ + θ ′)

sin(θ + θ ′) cos(θ + θ ′)

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and the order of multiplication doesn’t matter. The law of composition for the matrix representation is clearly D(c) = D(a)D(b)

c ≡ a + b.

Therefore the group is continuous and analytical, so it is a one-parameter Lie group. In the (x1, x2) plane the matrices act as sin θ cosθ

x ′1 cosθ = ′ x2 −sinθ

x1 x1 cosθ + x2 sin θ = , x2 −x1 sinθ + x2 cosθ

which implies that x′2 + x′2 = (x1 cosθ + x2 sinθ)2 + (−x1 sinθ + x2 cosθ)2 1

2

= x2(cos2 θ + sin2 θ) + x2(sin2 θ + cos2 θ) 1

2

= x2 + x2.

1

2

Thus the group elements leave invariant the quadratic form x2 + x2.

1

2

3.19 By dimensional analysis, to convert σ given in natural units of MeV−2 to units of fm2 (where 1 fm = 10−13 cm), we must multiply by a combination of h¯ and c that has units of MeV2 fm2. This is just h¯ 2c2 = 3.89 × 104 MeV2 fm2 and σ=

h¯2c2

= (140)2 MeV2

3.89 × 104 MeV2 fm2 (140)2 MeV2

2

≃ 2 fm ≃ 20 mb

where mb stands for millibarns and 1 fm2 = 10 mb.

3.20 This problem is adapted from Ref. [88]. We have c = 1 = 3 × 108 ms−1, so 1 second is equal to 3 × 108 meters and 9.6 seconds is 9.6 × (3 × 108) = 2.88 × 109 meters. The meaning is clear when c is restored: the world record in the 100 meter dash is equal to the time for light to travel 2.88 × 109 meters, which is 9.6 seconds.

3.21 This problem is adapted from Ref. [85]. Mass and energy have the same units if h¯ = c = 1, so τ in these units has dimension energy−1. It must be multiplied by a combination of h¯ and c having units of energy × time to be dimensionally correct in normal units, so a factor of h¯ = 6.58 × 10−22 MeV s will do the job. Thus 137 (2307 MeV)2 −22 −19 τ= × 6.58 × 10 MeV s ≃ 2.9 × 10 s 3 4 (74.5 MeV) is the mean life for the decay.

3.22 This problem is adapted from Ref. [88]. Since c = 1 = 3 × 108 m s−1, one has 1 s = 3 × 108 m. Therefore, m2 1 m2 −17 1 J = 1 kg 2 = 1 kg = 1.1 × 10 kg.

8 2 s (3 × 10 m) Likewise, 1 atm = 105 N m−2 = 10

5 kg m−1

s2

5

= 10 kg

1 m−1

= 1.11 × 10 (3 × 108 m)2

−12

kg m−3,

where 1 s = 3 × 108 m was used again.

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Permutation Groups

4.2 Under the map {e,(123),(321)} → 1

{(12),(23),(31)} → −1

the S3 multiplication table A\B e

(123) (321)

e e (12) (23) (13) (123) (321)

(12) (12) e (123) (321) (23) (13)

(23) (23) (321) e (123) (13) (12)

(13) (13) (123) (321) e (12) (23)

(123) (123) (13) (12) (23) (321) e

(321) (321) (23) (13) (12) e (123)

becomes A\B +1

+1 +1

+1 +1

+1 +1

+1 +1 +1

+1 +1 +1

+1 +1 +1

+1 +1

+1 +1

+1 +1

+1 +1

which preserves the group multiplication.

4.4 Allowed diagrams for S4 are

The irreps of Sn are found by enumerating n-particle diagrams with n states available, and 20

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21

with each particle in a different state. The possibilities are

1 2 3 4

1 2 3 4

1 2 3 4

1 2 3 4

1 2 4 3

1 3 2 4

1 3 2 4 1 4 2 3

1 3 4 2

1D

3D

1 2 3 4

2D

3D

1D

where the dimensionalities follow from counting the number of diagrams for each box pattern. These dimensionalities may also be checked using Eq. (4.3). For example, n!

4!

Dim = = = 3, Πihi 4·2·1·1 which agrees with the result found above by counting diagrams. The total number of basis vectors is 12 + 32 + 22 = 32 + 12 = 24, which agrees with the 4! irreps expected for S4.

4.6 Use the method of Example 4.3. The two diagrams are 1 2 3 4

1 4 2 .

3

For the first diagram, labeling the four available states by a, b, c, and d, ψ1 = Yabcd = QP abcd, where Y is the Young operator, Q is the antisymmetrizing operator between particles 1 and 4, and P is the symmetrizing operator among particles 1, 2, and 3: Q = 1 − (14) P = 1 + (12) + (13) + (23) + (123) + (321).

Therefore, the wavefunction is ψ1 = [1 − (14)][1 + (12) + (13) + (23) + (123) + (321)]abcd = [1 − (14)](abcd + bacd + cbad + acbd + bcad + cabd).

Multiplying out the terms then gives ψ1 = abcd + bacd + cbad + acbd + bcad + cabd — dbca − dacb − dbac − dcba − dcab − dabc.

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A similar treatment of the second diagram gives ψ2 = Yabcd = QP abcd, where Q = 1 + (14) P = 1 − (12) − (13) − (23) + (123) + (321), which gives ψ2 = [1 + (14)][1 − (12) −(13) −(23) + (123) + (321)]abcd = abcd − bacd − cbad − acbd + bcad + cabd + dbca − bdca − cbda − dcba + bcda + cdba, for the properly symmetrized and antisymmetrized wavefunction.

4.7 For the outer product [21] × [21], following the rules given in Section 4.5.2, [21] × [21] =

× 1 1 =

1 1

2

2

⊕

1 1 2

⊕

1 ⊕ 1 1 2 2

1

2 1 1

1 1

⊕

1

⊕

⊕ 1

2

1

⊕ 1

2

2

where for example =

Dim

n!

6!

=

Π(hooks)

= 9.

5·4·2·1·2·1

Evaluating the other diagrams in the same way, the total dimensionality on the right side of the above equation is 9 + 5 + 5 + 16 + 16 + 10 + 9 + 10 = 80. We may check this using Eq. (4.9), which gives Dim

=

×

(3 + 3)!

(3!)(3!)

(2)(2) = 80.

This checks with the total dimensionality obtained above.

4.10 The character of a direct product is the product of characters. From the character table given in Problem 4.10, for the direct product ⊗ of S4 representations the class, order, and character of the direct product are given in the

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following table.

Class (i) Order ni Character χ(i)

e(1) A(8) 1 8 (2)(3) = 6 (−1)(0) = 0

B(3) 3 (2)(−1) = −2

C(6) D(6) 6 6 (0)(1) = 0 (0)(−1) = 0

Equation (2.32) may be used to find the number aν of each possible irrep ν in the direct product: ni ∗ a = χ (i)χ(i). ν

∑i nG ν

For S4 we have nG = 4! = 24. Then from the formula and the table above, a a

1 [(1)(3)(6) + 0 + (3)(−1)(−2) + 0 + 0] = 1 ! 24 1 = [(1)(3)(6) + 0 + (3)(−1)(−2) + 0 + 0] = 1 24 =

with all other aν = 0. Thus ⊗

=

⊕

.

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Electrons on Periodic Lattices 5.2 The result for T (σb) is worked out in Example 5.1. Applying the same approach to the transformation implemented by σa gives ! −cos( π ) −cos( π ) 3

T (σa) =

−cos( π6 )

6

cos(π3 )

=

√

− 12

− 23

— 23

1 2

√

!

and for the transformation σc, T (σ c) =

1

0

0 −1

.

These matrices are equivalent to those appearing in Eqs. (5.10)–(5.12). Now let us consider the rotations associated with C3v. From Eq. (5.13) and Fig. 5.11 we see that rotations R(α) in the plane through an angle α are implemented by a matrix T (R(α)) =

cosα sinα

−sinα cosα

.

Therefore, for the rotations c3 by 2π3 the transformation matrix is T (c3) =

! cos( 2π ) −sin( 2π ) 3

sin( 2π3 )

3

cos( 2π3 )

=

− 12

=

− 12

√ 3 2

√

− 23

!

— 12

and for the rotation c2 by 4π

T

3

3

c32

cos( 4π ) −sin( 4π ) 3 3 = 4π 4π sin( 3 ) cos( 3 )

!

√

— 23

√

3 2 — 12

! .

Finally, the matrix corresponding to the identity is equivalent to rotation by multiples of 2π and therefore is the unit matrix T (e) =

cos(0) sin(0)

−sin(0) 1 0 = .

0 1 cos(0)

24

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a

a

3

a

1

1

c3 1 c

.

Fig. 5.1

c2a

2

2 b

3

3

c

b

2

c

b

Products of D3 symmetry operations demonstrating that c2ac3 = c2b.

Collecting the previous results, we have a direct correspondence between the six symmetry operations of C3v and six matrices that take the form √ √ !

1 3!

1 3 −2 — 2 −2 2 √ √ T (σa) = T (σ b) = 1 3 1 — 23 2 2 2 √ !

— 21 — 23 1 0 T (c3) = √ T (σc) = 0 −1 3 − 21 2 √ !

1 3 −2 1 0 2 √ T c32 = .

T (e) = 0 1 — 23 − 12 in the basis implied by the coordinate system of Fig. 5.10. This is Eq. (5.14).

5.5 Labeling the axes bisecting the corners of the triangle by a, b, and c, a 3

1

c

2

b

there are six proper symmetry operations: 1. the identity e, 2. rotations c3 clockwise by 2π and c2 clockwise by 4π (which is equivalent to a counter3

3

3

clockwise rotation by 2π3 ), and 3. three rotations by π about the a, b, and c axes, denoted c2a, c2b, and c2c, respectively.

The multiplication table may be constructed by numbering the corners of the triangle and examining the effect of symmetry operations on the numbers, as illustrated in Fig. 5.1 [this document] for the product c2ac3. This leads to Table 5.1 [this document], which is seen to

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Table 5.1 A · B for D3 in Problem 5.5 A\B e c3 c23 c2a c2b c2c

e e c3 c23 c2a c2b c2c

c3 c3 c23 e c2b c2c c2a

c23 c23 e c3 c2c c2a c2b

c2a c2a c2c c2b e c23 c3

c2b c2b c2a c2c c3 e c23

c2c c2c c2b c2a c23 c3 e

a

a e

e



e

2

e



rotate by  around b to give new unit vectors

c

b

e'1 = - cos (/3)e1 + cos (/6)e2

c

b

e

e' e'2

e'2 = cos (/6)e1 + cos (/3)e2



cos (/6)e1 cos (/6)e2



e

2

- cos (/3)e1

Fig. 5.2



cos (/3)e2

b

e'

e'1

.



Basis vector transformation corresponding to rotation by π around the b axis for the D3 group. The e3 unit vector is perpendicular to the plane of the e1 and e2 unit vectors and directed upward from the plane of the page.

be isomorphic to the multiplication table for C3v given in Table 5.2 if we invoke the map e ↔ e c3 ↔ c3

c32 ↔ c32

c2a ↔ σa

c2b ↔ σb

c2c ↔ σc

between the elements of the two groups. The symmetry operations may be different physically for D3 and C3v but the two groups have the same multiplication table so they are isomorphic.

5.6 We may construct a 2-dimensional matrix representation by examining the effect of the symmetry operations on the unit vectors e1, e2, as illustrated in Figs. 5.2 and 5.3

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a

a e

e



e

e



2

rotate by  around c to give new unit vectors

c

2

e'1

b

c

b e

e'1 = - cos (/3)e1 - cos (/6)e2 - cos (/6)e1

e'2 = - cos (/6)e1 + cos (/3)e2 e'2

- cos (/6)e2



e2

cos (/3)e2



- cos (/3)e1

 

e'1

.

F i g . 5.3

Basis vector transformation corresponding to rotation by π around the c axis for the D3 group.

[this document] for the symmetry operations c2b and c2c, respectively. The approach is analogous to that used to construct a 2-dimensional matrix representation for C3v illustrated in Fig. 5.10, but in this example we employ a different choice of unit vectors. The matrices for c2b and c2c may be deduced directly from Figs. 5.2 and 5.3 [this document], the matrix for c2a follows by inspection since the only effect of this transformation is to change the sign of e1, the identity e is the unit matrix, and the two rotations are given by cosα −sinα sinα cosα with α = 2π for c3 and α = − 2π for c2. The corresponding matrix representation is R(α) =

3

3

T (c2b) =

0 T (c2a) = 10 −1 1

T (c3) =

− √2 3 2

3

√ 3! − 2 − 21

− 12 √ 3 2

T c23 =

√ !

3 2 1 2

− 21

√ 3

—2

T (c2c) = √

3

2 1

−2

!

− 12 √ — 23

T (e) =

√ !

— 23 1 2

1 0 , 0 1

which is the desired result.

5.8 The 2D representation worked out in Problem 5.6 is irreducible (proved in Problem

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Table 5.2 Characters for the group D3 Irrep\Class A1 A2 E

e(1) 1 1 2

c3(2) 1 1 −1

c2(3) 1 −1 0

Basis e3 (e1, e2)

5.7) and may be labeled E in the notation of Box 5.3. From the traces there are three classes: the identity, the two rotations about the principle axis e3, and the three rotations by π about the a, b, and c axes, with characters 2, −1, and 0, respectively. In addition to the 2D irrep of Problem 5.6 there will be a trivial 1D irrep corresponding to mapping all six symmetry operators to unity, T (c2a) = T (c2b) = T (c2c) = T (c3) = T (c23) = T (e) = +1, and having character +1 for all three classes. This 1D irrep may be labeled A1 in the notation of Box 5.3. Since the group D3 has three classes, there must be one additional 1D irrep so that the number of irreps is equal to the number of classes. The mapping T (c2a) = T (c2b) = T (c2c) = −1

T (e) = T (c3) = T (c23) = +1

satisfies the multiplication table and gives a third independent irrep of dimension one with characters −1 for the rotations by π and +1 for the rotations about the principle axis and the identity. This irrep may be labeled A2. Thus the character table is given by Table 5.2 [this document]. The three irreps of Table 5.2 [this document] have the following physical interpretation. As we have seen in the preceding discussion, the 2D representation E corresponds to transformations that mix the x and y coordinates (the unit vectors e1 and e2). The A2 representation corresponds to transformations that leave the unit vector e3 (z-axis) invariant under rotations about the primary axis but change its sign for the three c2 rotations. Finally, the A1 representation corresponds to complete invariance under all transformations.

5.10 The matrix representation of C3v given in Eq. (5.14) was constructed with respect to the particular coordinate system defined by the unit vectors e1 and e2 in Fig. 5.10.

Suppose that instead we wish to describe the problem in terms of the coordinate system (e′1, e′2) illustrated in Fig. 5.4 [this document]. The matrices for e, c3, and c32 in Eq. (5.14) were not derived with reference to a particular set of axes, so they should not change, but the three reflections were constructed assuming the axes (e1, e2), so we may expect the matrices corresponding to them to depend on coordinate system choice. We could proceed geometrically as in Fig. 5.10 but a more elegant approach is to use a rotation operator to perform a similarity transformation (Section 2.7) on the matrices derived in the original coordinate system. The 2-dimensional rotation matrix R(α) and its inverse R−1(α) are R(α) =

cosα sinα

−sinα cosα

R−1(α) =

cosα −sinα

sinα . cosα

where the inverse R−1(α) is found either by requiring R(α)R−1(α) = 1, or by noting that

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b

b −

e

e e

 

e' e

1

− 

 

c

.

Fig. 5.4

e'1

Rotate by  = -/6 to give new axes



a

c

a

Rotation from the (e1, e2) coordinate system to a new coordinate system (e′1, e′2).

the inverse rotation is obtained by substituting α → −α. A similarity transformation on a matrix Q by the matrix corresponding to a rotation by α is given by R−1(α)QR(α).

Specializing to α = − π6, the matrix T (σc) is similarity transformed by the rotation to π T ′(σc ) = R − 6 =

√ 3 2 1 2

π π π σc R − =R σc R − 6 6 6 √ ! !

1 1 3 1 −2 1 0 2 2 √ √ = √2 1 3 3 3 0 −1 — 2 2 2 2 −1

√

3 2 1

—2

! .

Carrying out analogous transformations for the other matrices in Eq. (5.14), √ ! √ !

1 1 3 3 − 2 −1 0 2√ 2 T ′(σa) = T ′(σb) = T ′(σc) = √2 , 3 0 1 — 23 − 12 — 12 2 √ ! √ !

1 3 1 3 − − ′ — T (c3) = √ 2 T ′(e) = 1 0 , 2 2 T ′ c23 = √2 3 3 1 1 −2 0 1 — − 2 2

2

which is Eq. (5.15).

5.11 The groups C3v and D3 have the same characters (see Table 5.4 of the book and Table 5.2 [this document]) but the basis functions corresponding to their irreps differ. For both groups x and y (unit vectors e1 and e2) transform together under the 2-dimensional E representation. However, the z coordinate (basis vector e3) transforms as the A1 irrep for C3v but as the A2 irrep for D3. This has to do with the difference in the symmetry operations contained in the 3-member class in each case.

1. For C3v these are reflections in the e1–e2 plane, which do not change a unit vector in the e3 direction.

2. But for D3 they are not reflections but are instead rotations about a 2-fold axis perpendicular to e3, and this changes e3 into −e3.

Therefore, z (e3) transforms according to the A1 representation of C3v, which has character +1 for the classes labeled σv, but it transforms as the A2 representation for D3, which has

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b

b

e

e e1 



 

 

c

e1

a



 



Reflect through a to give new unit vectors





a

a

c e

e' e

e'1 = − cos (/3)e1 − cos (/6)e2

e'2

e'2 = − cos (/6)e1 + cos (/3)e2

.

F i g . 5.5

   

a

e'1 The reflection σa acting on orthogonal unit vectors e1 and e2 (with e1 chosen coincident with the c axis) gives new unit vectors e′i and e′2. The lower diagram illustrates the geometrical relationship between the original unit vectors and the reflected ones. The two equations are equivalent to the matrix relation (5.11).

character −1 for the classes labeled c2. In both cases z2 transforms as the A1 representation, since then the sign does not matter.

5.12 By applying a geometrical analysis similar to that of Fig. 5.10 to Fig. 5.5 [this document], we find that the action of σa on the unit vectors displayed in Fig. 5.10 is given by the matrix equation √ ! ! ′ −cos(3 π ) −cos(6 π ) 1e −21 −2 3 1e 1e √ = = , π π e′ 2 σ e2 1 −cos( 6 ) cos( 3 ) e2 — 3 a 2

2

which is Eq. (5.11). Likewise, the action of σc is simply e′1 1 0 = e′ 2 σ 0 −1 c

e1 , e2

since the c reflection plane is parallel to e1 and perpendicular to e2, so that the action of σc is to leave e1 unchanged and to change only the sign of e2. This is Eq. (5.12).

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Electrons on Periodic Lattices

5.14 This solution follows an example in Tinkham [196]. Cyclic groups of order N are abelian with elements C1 = A, C2 = A2, C3 = A3, · · · , CN = AN = 1, where 1 is the group identity. In an irrep Γ, let Γ(A) = r, where r is just a number since the group is abelian.

Then from the group properties Γ(Cn) = Γ(A)n = rn, but the closure property requires that Γ(CN) = Γ(A)N = rN = 1, so rN = 1 and r is one of the Nth roots of unity, r = e2πip/N with p = 1, 2, 3,..., N. Thus the 1D irreps are Γ(p)(A) = e2πip/N

(p = 1, 2, 3,..., N),

which are labeled by the discrete index p.

5.15 From Problem 5.14, the irreps are 1D and given by Γ(p)(A) = e2πip/N

(p = 1, 2, 3,..., N).

From Eq. (2.9), if A represents a coordinate displacement by −a, then TAψ(x) = ψ(x + a) represents the action of a translation TA on wavefunctions. If the Hamiltonian is translationally invariant, all eigenfunctions must transform according to representations of the translation group. For example, utilizing Eq. (2.15) with ℓ = 1, TAψ(x)p = ψp(x + a) = Γ(p)(A)ψp(x) = e2πip/Nψp(x).

Let the total length of the 1D lattice with spacing a be L = aN, and define a crystal momentum k ≡ 2π ip/N. Then from above ψp(x + a) = e2πip/Nψp(x) = eikaψp(x), or relabeling with k instead of p, ψk(x + a) = eikaψk(x).

This is a 1D version of Eq. (5.8). If we define a function uk(x) by ψk(x) = uk(x)eikx, then from ψk(x + a) = eikaψk(x), uk(x + a)eik(x+a) = eikauk(x)eikx, implying that uk(x) is periodic: uk(x + a) = uk(x). Therefore, ψk(x) = uk(x)eikx

uk(x) = uk(x + a),

which is a 1D version of Eq. (5.9). This is easily generalized to 3D if we let a → a = (a1, a2, a3), where the ai define primitive translations in the three orthogonal directions.

Then, ψk(r + a) = eik·aψk(r).

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This also holds for any linear combinations of lattice vectors, T = n1a1 + n2a2 + n3a3, for integers ni. Thus, in 3D we have ψk(r + T ) = eik·T ψk(r), or equivalently ψk(r) = uk(r)eik·r

uk(r) = uk(r + T ),

which define Bloch’s theorem, given in Eqs. (5.8) or (5.9).

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The Rotation Group 6.1 Parameterize an arbitrary 2 × 2 matrix with real entries as S=

a

b

c

d

.

The orthogonality condition SST = 1 then implies that a c

b d

a c b d

a2 + b2 ca + db

=

ac + bd c2 + d2

=

1 0 .

0 1

This is equivalent to the constraint equations a2 + b2 = 1

c2 + d 2 = 1

ac + bd = 0.

The first equation can be satisfied by the choices a = cosθ and b = sinθ, and the second equation by the choices c = sinφ and d = cosφ. The third equation implies then that cosθ sin φ + sinθ cosφ = sin(θ + φ) = 0, and thus that θ = −φ. Therefore, S=

a c

b cosθ = d sinφ

sinθ cosφ

=

cos(−φ) sinφ

sin(−φ) cosφ

=

cosφ sinφ

−sinφ . cosφ

The requirement of unit determinant is satisfied by this parameterization since cos2 φ + sin2 φ = 1. Therefore, an arbitrary SO(2) matrix can be parameterized as Eq. (6.3) and thus interpreted as implementing rotation in a plane through an angle φ.

6.4 From Eq. (6.20) for a one-parameter group, ∫

dβ J(β ) f [g(β )] =

∫

dβ J(β ) f [g α(β , γ) ].

Now let us use the chain rule to change variables to α(β , γ) in the integral on the right side, ∫

dβ J(β ) f [g(β )] =

∫

dα

∂β ∂α

J(α) f [g(α)],

where γ is fixed. By comparing the two sides we identify J(α) =

∂α ∂β

−1

J(β ).

Because this is a Lie group, the behavior in the vicinity of the identity will be sufficient to determine the weighting factor, while the normalization can be determined by requiring that the group volume be normalized to unity. For SO(2) the result is almost trivial. The 33

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law of composition for rotation through an angle φ requires that φα = φβ + φγ so that for constant γ, dφβ = 1, dφα γ and normalization to unit volume gives J0 ≡

∫ 2π

−1

dφ

=

0

1 .

2π

This leads to Eq. (6.21), dgso(2) = J0

1 dφβ dφ = dφ, dφα γ 2π

for the invariant SO(2) integration measure. More generally for a group with multiple parameters, the partial derivative factor ∂β /∂/α become a Jacobian determinant for the change of variables. Thus the invariant group measure may be interpreted generally as a Jacobian that shifts a small standard volume from near the group identity to any point in the group space.

6.8 This is an example worked out in Elliott and Dawber [56]. Writing the sum in Eq. (6.41) out explicitly, χ(αj) = ei jα + e−i(− j+1)α + e−i(− j+2)α + . . . + e−i( j−2)α + e−i( j−1)α + e−i jα = e−i jα (1 + eiα + e2iα + . . . + e(2 j−2)iα + e(2 j−1)iα + e2 jiα ) 2j

inα

= e−i jα ∑ e

= e−i jα

n=0

=

sin( j + 21 )α sin 21 α

sin( j + 12 )α i jα e sin 21 α

,

where in the third line we have used the exponential sum formula 2j

∑ einα = n=0

sin( j + 21 )α sin 21 α

ei jα

(see http://mathworld.wolfram.com/ExponentialSumFormulas.html). Then expanding the sine functions in the numerator and denominator, sin( j + 1 )α ( j + 1 )α − 1 [( j + 1 )α]3 + . . . ( j) 2 2 3!

2 χα = ≃ , 1 1 1 3 + ... sin 21 α α − ( α) 2 3! 2 and taking the limit α → 0 of this expression gives ( j + 1 )α ( j) 2 Lim χα = = 2 j + 1.

1 α →0 α 2 But the limit α → 0 defines the identity, so the dimension of the SO(3) irrep labeled by j is the character of the identity, which is 2 j + 1.

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6.10 Expand the rotation operator in Eq. (6.52) to prove Eq. (6.55). Then let Xµ = J3 in Eq. (6.55) to give [ J3, T (LM)] = ∑ LM′ J3 |LM ⟩T (LM′) M′

= ∑ M δMM′ T (LM′) M′

= MT (LM), which is the first of Eqs. (6.57), with some notation change. Likewise, setting Xµ = J± in Eq. (6.55), [ J±, T (LM)] = ∑ LM′ J± |LM⟩ T (LM′) M′

√ = ∑ L(L + 1) − M(M ± 1) δM′,M±1 T (LM′) M′

=

√

L(L + 1) − M(M ± 1)T (LM),

which is the second of Eqs. (6.57) in different notation.

6.11 (a) The orbital angular momentum operators may be written (in h¯ = 1 units with implied summation on repeated indices) La = εabcrb pc.

Then we have [ La, rb ] = εacd[ rc pd, rb ] = εacd(rc pdrb − rbrc pd) = εacdrc[ pd, rb ] = iεacdrc.

But the structure constants generate the adjoint representation of the algebra (see Section 3.2.2), and (1)

[ La, r b ] = ⟨b| L a |c⟩ rc .

Therefore, the coordinate transforms as a vector (L = 1) under rotations. (b) By a similar proof as in part (a), the momentum is also an SO(3) vector.

6.12 Evaluate the Clebsch–Gordan coefficients in Eq. (6.67) and proceed by analogy with Eqs. (6.60)–(6.66). You should find T (AB)22 = A1B1,

1 T (AB)21 = √ (A0B1 + A1B0), 2

1 T (AB)20 = √ (A−1B1 + A1B−1 + 2A0B0), 6 for example.

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6.14 From the SO(3) Wigner–Eckart theorem (6.71), ⟨lm|YLM l′m′ =

∫

Yl∗m(θ , φ)YLM(θ, φ)Yl′m′ (θ, φ)d Ω

= l′ m′ LM l m⟩⟨l ||YL || l′⟩, and from the tensor properties of the spherical harmonics, YLM(θ, φ)Yl′ m′ (θ, φ) = ∑ kq

(2L + 1)(2l′ + 1) 4π(2k + 1)

1/2

× LM l′ m′ kq⟩ L 0 l′ 0 k 0⟩Ykq(θ, φ).

Inserting this expansion in the integral and using the orthonormality of the spherical harmonics, ∫

Yl∗m(θ , φ)Ykq(θ, φ) dΩ = δklδqm,

we obtain

s ⟨l ||YL || l′⟩ =

(2L + 1)(2l′ + 1) L 0 l′ 0 l 0⟩, 4π(2l + 1)

for the reduced matrix element of a spherical harmonic.

6.15 First let’s rewrite the Wigner–Eckart theorem in terms of 3J symbols rather than Clebsch–Gordan coefficients , so that we can use Table C.4. From Eqs. (6.71), (6.72), and (6.50), the SO(3) Wigner–Eckart theorem is given by ⟨JM| Tkq J′M′ = (−1)2k J′ M′ kq J M⟩⟨J || T k || ′J′⟩ √ ′ J′ k J = (−1)3k−J −M 2J + 1 ⟨αJ || T k ||α′J′⟩, M′ q −M and the inverse relation is ⟨J || T k || J′⟩ = (−1)2k ∑ J′ M′ kq J M⟩⟨JM| Tkq J′M′ M′q

√ ′ = (−1)3k−J −M 2J + 1 ∑ M′q

J′ M′

k J q −M

⟨JM| Tkq J′M′ .

From the definition of the quadrupole moment we may use the Wigner–Eckart theorem to write r 16π Q≡ ⟨ j, m = j| Q20 | j, m = j⟩ r 5 √ 16π J 2 J = (−1)6−2 j 2 j + 1 ⟨J || Q 2|| J⟩, 5 J 0 −J and therefore that the reduced matrix element for the quadrupole operator is r −1 5 (−1)2 j−6 j 2 j √ ⟨J || Q2 || J⟩ = Q. 16π 2 j + 1 j 0 −j

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On the other hand, evaluating ⟨ jm| Q20 | jm⟩ in an arbitrary state | jm⟩ using the Wigner– Eckartr theorem gives r √ 16π 16π j 2 j ⟨ jm| Q | jm⟩ = (−1)6− j−m 2 j + 1 ⟨ j || Q || j⟩ 2 20 m 0 −m 5 5 = (−1) j−m

j 2 j m 0 −m

= (−1) j−m

j j 2 m −m 0 3m2 — j( j + 1) = Q, j(2 j − 1)

j

2

j

j

0

−j

j j j −j

2 0

−1

Q −1

Q

where in the second step the explicit form of the reduced matrix element ⟨ j || Q2 || j⟩ from above was inserted, in the third step the symmetries listed in Section 6.3.8 were used to switch the columns of the 3J coefficients, and in the final step a

a

b −b

2 = (−1) a−b 0

3b2 − a(a + 1) a a 1 2a 3 2a 1 2a 1 1/2 ( + )( + )( + )( − )

from Table C.4 was used to evaluate the 3J coefficients.

6.17 This proof follows a discussion in Tung [199] (a) Associate the coordinate x = (x1, x2, x3) with a 2 ×2 matrix X through the map X = σixi, where the σi are the Pauli matrices of Eq. (3.11). Evaluating explicitly, X = σixi =

0 1 0 x1 + 1 0 i

−i 1 0 x3 x2 + x3 = 0 0 −1 x1 + ix2

x1 − ix2 −x3

.

The matrix X is hermitian and traceless, as it must be since the σi are hermitian and traceless. Evaluating the determinant of X gives det X = −(x12 + x22 + x32) = −|x|2. (b) Suppose U to be an arbitrary SU(2) matrix that induces a linear transformation X → X ′ = UXU −1 on X . The new matrix X ′ is hermitian and traceless since 1. The matrix X is hermitian and traceless by construction.

2. The trace is a matrix invariant that is not changed by a similarity transformation.1 3. The matrix U is a unitary element of SU(2) so U † = U −1. Thus UXU −1 = UXU † and X ′† = (UXU −1)† = (UXU †)† = (U †)†X †U † = UX †U † = UXU † = X ′, where the hermiticity (X = X †) of X has been used. (c) Since X ′ is a 2 × 2 hermitian and traceless matrix like X , it can be specified through the map X ′ = σix′i. Furthermore det X ′ = det X , since X and X ′ are related by a similarity transform, which can’t change the determinant (product of eigenvalues) of a matrix.

1 The eigenvalues of a matrix are invariant under similarity transforms. Thus the trace (sum of eigenvalues) and

the determinant (product of eigenvalues) for a matrix are also invariant.

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(d) From the results obtained above, det X ′ = |x′|2 = det X = |x|2

−→

|x′|2 = |x|2.

Since preservation of |x|2 is a defining characteristic of an SO(3) transformation in 3D euclidean space, we have shown that the SU(2) transformation X → X ′ = UXU −1 induces a corresponding SO(3) transformation in euclidean space. (e) Because the above equations are quadratic in the SU(2) element U , they are equally valid for ±U . Thus two SU(2) transformations are mapped to each SO(3) transformation and there is a 2:1 homomorphism between SU(2) and SO(3).

6.21 For the matrix R(φ) =

cosφ sinφ

−sinφ cosφ

T ≡ Tr R = 2 cosφ and D ≡ det R = 1. Then from Problem 14.14 the eigenvalues are q √ λ± = 21 T ± 14T 2 − D = cosφ ± cos2 φ − 1 = cosφ ± i sinφ = e±iφ , and the transformed matrix is diagonal with R′(φ) =

eiφ

0

.

0 e−iφ The corresponding basis vectors (eˆ−, eˆ+) are linear combinations of the original basis vectors (eˆ1, eˆ2) that are determined by cosφ sinφ

−sinφ cosφ

X1 X1 =λ .

X2 X2

Multiplying this out gives the equations X1 cosφ − X2 sinφ = λ X1

X1 sinφ + X2 cosφ = λ X2,

which may be used to define the new basis vectors. From the second equation with λ = eiφ = cosφ + i sinφ, X1 sinφ + X2 cosφ = X2 cosφ + iX2 sinφ, √ √ implying that X1 = iX2, which is satisfied by X1 = 1/ 2 and X2 = −i/ 2 so a suitable basis vector is 1 eˆ− = √ (eˆ1 − ieˆ2).

2 iφ − Likewise, choosing λ = e = cosφ − i sinφ implies that X1 = −iX2 and a suitable basis vector is 1 eˆ+ = √ (eˆ1 + ieˆ2).

2 We seek an operator C transforming between the eˆ1, eˆ2) basis and the (eˆ+, eˆ−) basis, implying that 1 eˆ1 + ieˆ2 eˆ1 C11 C12 eˆ1 C11eˆ1 + C12eˆ2 , C = = √ = eˆ2 C21 C22 eˆ2 2 eˆ1 − ieˆ2 C21eˆ1 + C22eˆ2

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which implies the constraints 1 i C11 = √ C12 = √ 2 2 and a transformation matrix

1 C21 = √ 2

i C22 = −√ .

2

1 1 i .

C= √ 2 1 −i We also need the inverse C−1 of C, such that CC−1 = 1. Generally, for a 2 × 2 matrix, 1 a b d −b A= −→ A −1 = . c d a det A −c Therefore we find that 1 1 i C = √ 1 −i 2

−1

C

1 1 1 =√ 2 −i i

CC

1 0 = 0 1 .

1 −i

1 i

−1

The similarity transform on R(φ) then gives explicitly R′(φ) = CR(φ)C−1 1 1 i = 2 1 −i = =

cosφ sinφ

cosφ + i sinφ 0 eiφ

0

0

e−iφ

−sinφ cosφ

0 cosφ − i sinφ

,

indicating that R(φ) is reducible to two 1-dimensional irreducible representations. Infinitesimally, 1 + iφ 0 eiφ 0 ≃ .

R′(φ) = 0 e−iφ 0 1 − iφ Comparing with R′(φ) ≃ 1 + iJφ, we see that the SO(2) generator J of Eq. (6.10) is J=

1

0

0 −1

,

in the new basis.

6.23 Note that d1/2 ⊗ d1/2 = d0 ⊕ d1, so the elements of the matrix d1 can be evaluated from direct products d1/2 ⊗ d1/2, where the elements of d1/2 have already been worked out in Eq. (6.33). In general, 1 = ⟨1M| d |1N⟩ dMN

=

∑ ∑ ⟨1M| m1m2⟩⟨m1m2| d |m3m4⟩⟨m3m4| 1N⟩

m1m2 m3m4

= ∑

∑ 21 m121 m2 1 M⟩ 21 m32 1 m4 1 N⟩⟨m1| d1/2 |m3⟩⟨m2| d1/2 |m4⟩,

m1m2 m3m4

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where from Eq. (6.33) 1/2

d

1/2

1/2

(β ) =

d 1/21/2

d1/2 −1/2

1/2

1/2

d −1/21/2

cos β

−sin β

β

β

2

=

2

sin 2

d−1/2 −1/2

! .

cos 2

Therefore, some elements of the matrix d1MN are 1/2

1/2

1 d11 =

1 1 1 1 2 2 2 2

11⟩ 12 12 12 12 11⟩ d 1/21/2d 1/21/2 = cos2

d110 =

1 1 1 1 2 2 2 2

11⟩

β 2

1/2 1/2 1 1 1 − 12 10⟩ d 1/21/2d 1/2 −1/2 + 2 2 2

= 12 (1 + cosβ )

1 − 12 12 12 2

1/2

1/2

10⟩ d 1/2 −1/2 d1/21/2

= √− 2 cos 2β sin β2 = − √1 sinβ 2

2

1

d1−1 =

1 1 1 1 2 2 2 2

11⟩ 21 − 21 21 − 12

d110 =

1 1 1 1 2 2 2 2

11⟩

= − √2 cos 2

β

β

1/2

1/2 1/2 1 1 1 − 12 10⟩ d 1/21/2d 1 1 + 2 2 2 /2 − /2

sin

2

1/2

1 − 1⟩ d 1/2 −1/2 d1/2 −1/2 = sin2 2

β 2

1 − 12 12 12 2

= 21 (1 − cosβ ) 1/2

1/2 d /2 −1/2 1/21/2

10⟩ d 1

= − √1 sinβ , 2

and so on. Evaluating all possibilities in the same way the final result is d111

d101 d1

d1 ′ = d1 mm

1 d1−1 d1 =

01

00

0−1

d1−11

d1−10

d1−1−1

1 (1 + cosβ ) 2 √1 sinβ

− √1 sinβ 2 cosβ

2 1 (1 − cosβ ) 2

√1 sinβ 2

1

2

(1 − cosβ ) − √1 sinβ .

2

2 (1 + cosβ ) 1

Changing the angle β to θ, this checks with Table C.1 of Appendix C upon using the symmetry j j (θ) d (θ) = (−1)k−md j (θ) = d mk

−k−m

km

to fill in the entries dmk (θ) not displayed explicitly in the table.

1

6.24 If y is a real number and x and z are positive real numbers, it is obvious that the matrices x y g= 0 z form a group under matrix multiplication. For example, gg′ =

x y 0 z

x′ y′ xx′ = ′ 0 z 0

xy′ + yz′ x′′ y′′ ≡ ′ 0 z′′ zz

which exhibits closure, since x and z are positive the determinant is non-zero and a unique inverse −1 x y 1/x −y/xz = , 0 z 0 1/z exists, the unit 2 × 2 matrix is the identity, and matrix multiplication is associative. Now let A=

∫

f (g)dg

dg ≡ dxdydz

g0 ≡

a b , 0 c

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where f (g) is an arbitrary function of the group elements. Then since g0g is a group element, ∫

f (g0g)d(g0g) =

∫

f (g′)dg′,

where we have used g′ ≡ g0g =

x y ax ay + bz x′ = ≡ 0 z 0 cz 0

a b 0 c

y′ , z′

which implies that x′ = ax

y′ = ay + bz

z′ = cz.

This is a Lie group, so we may investigate it for g0 near the origin, where g0 approaches the unit matrix with a, c → 1 and b → 0. Then x′ = ax

z′ = cz.

y′ ∼ ay

and for the volume element dx′dy′dz′ = (a2c)dxdydz.

Thus, dg′ = (a2c)dg /= dg and the integral with this measure is not invariant under left multiplication by a group element. However, if we define a new integration measure dgL ≡

dxdydz

,

x2z

then, proceeding as above, dg′L =

dx′dy′dz′ x′2z′

=

a2cdxdydz a2cx2z

=

dxdydz x2z

= dgL.

Hence, with this measure dgL = dg′L and the integration is invariant under left multiplication by a group element. Carrying out a similar proof after multiplying from the right by g0, you should find that dg′ = (ac2)dg /= dg, but defining a new measure dgR ≡

dxdydz xz2

you should find that dgR = dg′R and the measure is right invariant under group multiplication. In this example there are two invariant integration measures, depending on whether we require invariance under left or right multiplication. It can be shown that for the groups of most interest in physics (finite, compact, simple, and semisimple) the left and right invariant integration measures are identical.

6.26 Consider rotations about the z axis. From Eq. (6.27), OR f (x, y, z) = f R−1(x, y, z) = f (x + dx, y + dy, z) = f (x + ydφ, y − xdφ, z).

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Expanding in a Taylor series around f (x, y, z) ∂f

OR f (x, y, z) = f (x, y, z) +

dx +

∂f

dy + . . . ∂x ∂y ∂f ∂f = f (x, y, z) + y dφ − x dφ + . . . ∂x ∂y ∂ ∂ ≃ 1 + y dφ − x dφ f (x, y, z) ∂x ∂y

and we identify OR ≃ 1 + y

∂ ∂x

dφ − x

∂ ∂y

dφ.

But comparing with Eq. (6.6), for infinitesimal rotations about the z-axis OR ≃ 1 − iJzdφ.

Equating these two expressions for OR gives Jz = i y

∂ ∂ −x , ∂x ∂y

and by a similar analysis of rotations about the other two axes, ∂ ∂ ∂ ∂ J = −i z −x J = −i y −z , y x ∂x ∂z ∂z ∂y which confirms Eq. (6.28).

6.27 This proof may be found in Elliott and Dawber [56]. As for finite groups, we may use the characters of the SO(3) representations to find the coefficients cJ in the direct sum appearing on the right side of Eq. (6.44). The characters are invariants and they obey a relation analogous to Eq. (6.44): (j )

(j )

α

α

χ 1χ 2 =

(J)

∑ cJ χα .

J

Using (6.42) and assuming, to be definite, that j1 ≥ j2, we have (j )

(j )

1

2

χα χα

sin( j1 + 1 )α sin( j2 + 1 )α = =

2

2

sin2(α/2) −cos( j1 + j2 + 1)α + cos( j1 − j2)α

2 sin2(α/2) (j +j ) ( j −1/2) ( j2−1/2) = χα 1 2 + αχ 1 χα .

This procedure may now be repeated on the product of characters in the second term.

Iterating 2 j2 − 1 times we finally obtain [56] (j ) (j )

( j + j 2)

χα 1 χ α2 = χ α1 cj =

1 0

( j + j2−1)

+ χ α1

( j − j 2)

+ . . . χ α1

(J)

= ∑ cJ χ α ,

if ( j1 − j2) ≤ J ≤ ( j1 + j2) , otherwise

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where clearly the assumption j1 ≥ j2 was immaterial. Since all the cJ equal zero or one, the group SO(3) is simply reducible and we obtain for the SO(3) Clebsch–Gordan series D( j1) ⊗ D( j2) = D(J= j1 + j2) ⊕ D(J= j1+ j2−1) ⊕ . . . ⊕ D(| j1− j2|), which is Eq. (6.45).

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7

Classification of Lie Algebras

7.3 Simple roots are positive roots that can’t be written as a sum of other positive roots, and the number of simple roots equals the rank of the algebra. Thus there are two simple roots for each of these cases. (a) For θαβ = 30◦ the positive roots are displayed in Fig. 7.1 [this document; see also Fig. 7.4(a) in the book]. Notice that we can write for these vectors √ √ √ α4 + α6 = 0, 3 + 21 , − 23 = 12 , 23 = α3 √ √ √ α4 + 3α6 = 0, 3 + 3 2 1 , − 23 = 23 , − 23 = α5 √ √ α4 + 2α6 = 0, 3 + 2 21 , − 23 = (1, 0) = α1 √ √ √ 2α4 + 3α6 = 2 0, 3 + 3 21 , − 23 = 32 , 23 = α2.

Therefore the simple roots for θαβ = 30◦ (or 150◦) are √ α4 = 0, 3

α6 =

√ 1 , − 23 2

,

since all other vectors are linear combinations of these. This corresponds to the G2 root diagram displayed in Fig. 7.5(b). (b) For θαβ = 45◦ the positive roots are



(3/2, 3/2)

(3/2, − 3/2)

.

Fig. 7.1

Positive roots for the G2 root diagram of Problem 7.3.

44

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Classification of Lie Algebras

45

(0,1)

(1,1)

 (1,0)

(1,−1)

from which we see that α1 + α4 = (1, −1) + (0, 1) = (1, 0) = α2 α1 + 2α4 = (1, −1) + 2(0, 1) = (1, 1) = α3.

Therefore, the simple roots for θαβ = 45◦ (or 135◦) are α1 = (1, −1)

α4 = (0, 1),

since all other vectors are linear combinations of these. This corresponds to the B2 or SO(5) root diagram displayed in Fig. 7.5(c).

◦

◦

7.5 For A2 [SU(3)] the Dynkin diagram is , which is rank-2 since there are two circles. Thus there are two simple roots. The single line in the diagram indicates that the angle between simple roots is 120◦, so in the 2D weight space H2

120





H1

The simple roots α and β are of equal lengths because from Table 7.1, α ·α n′ = = 1. β ·β n The hyperplane perpendicular to α is the H2 axis, so another root β ′ is obtained by reflecting β in this plane H2



' 

H1

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Classification of Lie Algebras

46

H2

H1

.

Fig. 7.2

Final root diagram constructed in Problem 7.5.

Reflecting α through the plane perpendicular to β yields no new result but for each of the roots (β , β ′, α), their negatives (−β , −β , −α) are also roots, so the final diagram is displayed in Fig. 7.2 [this document].

7.6 From Section 7.2.2, let a Lie algebra be defined by [ Xi, Xj ] = ck X , ij k

where we use the Einstein convention of summing over repeated indices. Then the metric tensor is defined by gi j = g ji = cℓ ck . ik jℓ

From Problem 7.4, for the group SO(3) the operators Xi ≡ −iJi (where Ji is the usual angular momentum operator) obey the commutator [ Xi, Xj ] = ck X , or explicitly ij k

[ X1, X2 ] = X3

[ X2, X3 ] = X1

[ X3, X1 ] = X2.

Therefore, the non-vanishing cki j are c 12 = c 23 = c231 = 1 3

c132 = c321 = c213 = −1.

1

From this we can compute the components of the metric tensor: 3 2 g11 = c12 c13 + c213c312 = −2, 3 1 g22 = c21 c23 + c123c321 = −2, 1 2 g33 = c32 c31 + c231c132 = −2, 3 2 g12 = c12 c23 + c213c322 = 0, 3 2 g13 = c12 c33 + c213c332 = 0, 3 1 g23 = c21 c33 + c123c331 = 0,

and using gi j = g ji, SO(3)

gi j

=

−2

0

0

0 0

−2 0

0 −2

= −2δi j.

Since the metric tensor is negative definite (see Section 12.2), SO(3) is compact, and since

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Classification of Lie Algebras

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the determinant of the metric tensor det g = −8 is non-zero, SO(3) is semisimple. The Casimir operator is given by C = gi jXiXj, where the Xi are the generators. For SO(3) the generators can be taken as Xi = −iJi, [see Eq. (3.13)], and we obtain 1 2 2 2 C = − 2 X1 + X2 + X3 =— =

1 — J12 − J22 − J32 2

1 2 J + J22 + J23 2 1

for the SO(3) Casimir operator.

7.8 From Eqs. (7.14) and (7.3) βi = ⟨Hi| [ Eα , E−α ]⟩ 1 = Tr H i[ E α, E −α ] λ 1 = Tr(Hi Eα E−α ) − Tr(Hi E−α Eα ) λ 1 = Tr(E−α HiEα ) − Tr(E−α Eα Hi) λ 1 = Tr(E−α HiEα − E−αEαHi) λ 1 = Tr E−α [ Hi, Eα ] λ 1 = αiTr(E−αEα ) λ 1 = αiλ λ = αi, where the cyclic property of the trace was used in line four, Eq. (7.9) was used in line seven, and Eq. (7.12) was used in line eight.

7.10 In any representation Hi (E±α |mD⟩) = (HiE±α − E±αHi + E±αHi) |mD⟩ = ([ Hi, E±α ] + E±αHi) |mD⟩ = [ Hi, E±α ] |mD⟩ + miE±α |mD⟩ = ±αiE±α |mD⟩ + miE±α |mD⟩ = (m ± α)i E±α |mD⟩, where E±α Hi was added and subtracted in line one, Hi |mD⟩ = mi |mD⟩ was used in line three, and Eq. (7.9) was used in line four. So application of E±α to a state changes the

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Classification of Lie Algebras

48

weight quantum number of the state by ±α and E±α acts as a raising or lowering operator within the weight space.

◦ ◦ •

7.11 The Dynkin diagram is , which indicates that there are two simple roots ◦ with an angle of 150 between them (Table √7.3). From Table 7.1 and Eq. (7.31), the ratio of lengths between the two simple roots is 3. Let’s choose the longer simple root to be √ α1 = (0, 3). Then the√ preceding conditions are satisfied if we choose the second simple root to be α2 = ( 1 , − 3 ). From the simple roots 2 2 √ √ α1 = (0, 3) α2 = ( 1 , − 3 ), 2

2

and Eqs. (7.43) and (7.32)–(7.35), we have that α2 = 3 1

implying that n = −2

α2 = 1 2

α1 · α 2 α12

=1

α · α = − 3, 1 2 2 n′ = −2

α1 · α2 α22

= 3,

and that the alpha-string from Eq. (7.32) is {α1, α1 + α2, α1 + 2α2, α1 + 3α2} and the betastring from Eq. (7.34) is {α2, α1 + α2}. Therefore, the roots found to this point and their corresponding level defined through Eq. (7.37) are 1 2 3 4

α1, α2 α1 + α2 α1 + 2α2 α1 + 3α2

This exhausts the possible roots in levels 1–4: 1. The level-1 roots are unique since they are simple roots.

2. The level-2 roots are unique since possibilities like α1 − α2, 2α1, or 2α2 are forbidden

by Theorems 7.6 and 7.3.

3. The level-3 roots are unique because the other possibility, 2α1 + α2, was excluded from the beta-string above.

4. The level-4 roots are unique because the other possibility, 2α1 + 2α2 = 2(α1 + α2), is excluded by Theorem 7.6 because α1 + α2 is a root.

Next we ask whether there are any level-5 roots. By Theorem 7.9, such roots must be formed by the action of simple roots (α1 or α2) on the level-4 roots (the only possibility being α1 + 3α2). The possibility α2 + (α1 + 3α2) = α1 + 4α2 has already been excluded from the alpha-string above but α1 + (α1 + 3α2) = 2α1 + 3α2 needs to be checked. Let α ≡ α1 + 3α2. Then for a possible string α, α + α1, . . . , α + nα1, the value of n is [compare the alpha-string in Eqs. (7.32)–(7.33) above] n = −2

α · α1 −2(α1 · α1 + 3α1 · α2) = = 1.

2 3 α1

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Classification of Lie Algebras

Therefore α + α1 = 2α1 + 3α2 is a level-5 root. It is the only possible level-5 root since α1 + 4α2 was excluded earlier and possibilities like 3α1 + 2α2 can’t be obtained from the level-4 root by action of a simple root.

What about level-6 roots? If they exist, they must be obtained by the action of one of the simple roots on the level-5 root obtained above. The two possibilities are α1 + (2α1 + 3α2) = 3(α1 + α2) α2 + (2α1 + 3α2) = 2(α1 + 2α2).

But both of these are excluded by Theorem 7.3 because α1 + α2 and α1 + 2α2 are already known to be roots. Since there are no level-6 roots there can be no higher-level roots (because they would be formed by the action of simple roots on level-6 roots) and we have found all the positive roots. Using Theorem 7.1 to add the corresponding negative roots and employing the explicit values for the simple roots, we finally recover the root diagram of Fig. 7.4(a) for the algebra G2. Notice that this problem is essentially the inverse of Example 7.6, where the Dynkin diagram for G2was deduced from the simple roots.

Solutions Manual Symmetry, Broken Symmetry, and Topology in Modern Physics Mike Guidry and Yang

8

Unitary and Special Unitary Groups 8.1 This solution follows a discussion in Ref. [56]. For example, from the properties in Table 8.1 [V+,V− ] = 3Y + T3 ≡ 2V3

V3 = 3Y + 1 T3

→

2

4

2

[V3,V± ] = [ 43Y + 21 T3,V± ] = 43 [Y,V± ] + 21 [ T3,V± ] = ± 3 V± ± 1 V± 4

4

= ±V±.

Thus the operators {V±,V3} are generators of an SU(2) group: [V+,V− ] = 2V3

[V3,V± ] = ±V±.

In a similar way we can prove that {U±,U3} and {T±, T3} constitute two other independent sets of SU(2) generators. Since T3, U3, and V3 are SU(2) generators, their possible eigenvalues are 0, ±21 , ±1, ± 32 + . . . , and from Table 8.1 3 Y − T3 = 2U3 2

Y + T3 = 2V3

3

2

−→

Y = 23 (U3 + V3).

Since U3 and V3 are integer or half-integer, the possible values of Y are 0, ± 13 , ± 23 , ±1, ± 43, . . .

Hence if we label the weight space by T3 and Y , the points will be spaced at intervals of 1 integer in the Y direction and 1 integer in the T3 direction, as illustrated in Fig. 8.1 [this 3

2

document] . We may now use the algebra of the operators U±, V±, and T± to deduce their action in this weight space. For example, since [ T+, T− ] = 2T3

[ T3, T± ] = ±T±

[Y, T± ] = 0,

it follows that the action of T± on a basis vector |T3Y ⟩ where1 Tˆ3 |T3Y ⟩ = T3 |T3Y ⟩

Yˆ |T3Y ⟩ = Y |T3Y ⟩

is to increase or decrease T3 by one and to leave Y unchanged: Tˆ3 Tˆ± |T3Y ⟩ = ± Tˆ ±+ Tˆ±Tˆ3 |T3Y ⟩ = T±ˆ T3 ± 1 |T3Y ⟩ = T3 ± 1 Tˆ± |T3Y ⟩ , 1

We now use a hat to distinguish a (generally non-commuting) operator from its (generally commuting) eigenvalues, where necessary to avoid confusion.

50

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Unitary and Special Unitary Groups

51

Y

T3

.

Fig. 8.1

Possible sites for Y and T3 in Problem 8.1

while the action of Yˆ on this eigenvector is Yˆ Tˆ± |T3 Y ⟩ = Tˆ± Yˆ |T3 Y ⟩ = Y Tˆ± |T3 Y ⟩ , so Y us unchanged by Tˆ±. In summary, the action of Tˆ± is T±ˆ |T3 Y ⟩ ∼ |T3 ± 1,Y ⟩.

By similar proofs, Uˆ± |T3Y ⟩ ∼ T3 ∓21 ,Y ± 1

Vˆ± |T3Y ⟩ = T3 ±21 ,Y ± 1 .

Thus the action of Uˆ±, Vˆ±, and T±ˆ may be summarized as in Fig. 8.2 [this document], where the horizontal spacing between points is one unit in T3 and the vertical spacing is one unit in Y .

Since all basis vectors of an irrep must be accessible from a given basis vector using linear combinations of generators, the occupied sites in the SU(3) irreps are even more restricted than we have already concluded, owing to the restrictions on the actions of the operators Uˆ±, Vˆ±, and T±ˆ . In particular, only the lattice sites marked by a heavy dot in Fig. 8.3[this document] correspond to irreps (where in this plot the horizontal spacing between dots is 1 and the vertical spacing is 1 ), because only such points can be reached 2 3 from the other sites marked by heavy dots using the stepping operators Uˆ±, Vˆ±, and Tˆ± .

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Unitary and Special Unitary Groups

52

Y

.

Fig. 8.2

T3 Actions of the stepping operators U±, T±, and V± in Problem 8.1.

Y T3

.

Fig. 8.3

Possible occupied sites for Problem 8.1.

8.3 For the product (2, 1) ⊗ (3, 0) in Young diagrams

⊗

1 1 1

=

1

⊕

1 1

⊕

⊕

1 1 1

1 1 ⊕ 1

=

⊕ ⊕

⊕

1 1

⊕

1 1

1 1

⊕ ⊕

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Unitary and Special Unitary Groups

53

By the hooks rule, the dimensionalities are =

dim

3·4·5·6·7·8·2 = 48 2·3·4·5·7

3·4·5·6·7·2·3 = 42 2·3·5·6·2 3·4·5·6 = = 15 2·3·4 3·4·5·2 = = 15 4·2 3·4·2·3 = =6 2·3·2 3·4·5·6·2·3·4 = = 24.

3·4·5·2·3 =

dim dim dim dim dim

These sum to 150, as they should, since (2, 1) ⊗ (3, 0) = 15 ⊗ 10. In summary, (2, 1) ⊗ (3, 0) = (5, 1) ⊕ (3, 2) ⊕ (4, 0) ⊕ (2, 1) ⊕ (0, 2) ⊕ (1, 3) = 48 ⊕ 42 ⊕ 15 ⊕ 15 ⊕ 6 ⊕ 24 for the direct product of (2, 1) and (3, 0) in SU(3).

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9

SU(3) Flavor Symmetry 9.1 For the fundamental representation 3 of SU(3) the weights m(T3,Y ) are m(1) = 12 , 13

m(2) = − 12 , 13

m(3) = 0, − 23 .

For the 10-dimensional SU(3) representation (λ, µ) = (3, 0), the Young diagram is The weights are additive so for the allowed ten diagrams m

1 1 1

= m(1) + m(1) + m(1) =

3 ,1 2

,

m

1 1 2

= m(1) + m(1) + m(2) =

1 ,1 2

,

m

1 1 3

= m(1) + m(1) + m(3) = (1, 0),

m

1 2 2

= m(1) + m(2) + m(2) = − 21 , 1 ,

m

1 2 3

= m(1) + m(2) + m(3) = (0, 0),

m

2 2 2

= m(2) + m(2) + m(2) = − 23 , 1 ,

m

1 3 3

= m(1) + m(3) + m(3) = 21 , −1 ,

m

2 3 3

= m(2) + m(3) + m(3) = − 12, −1 ,

m

3 3 3

= m(3) + m(3) + m(3) = (0, −2),

m

2 2 3

= m(2) + m(2) + m(3) = (−1, 0).

.

Plotting these in a weight diagram for the 10, Y 1 0

10

−1 −2 −2

−1

T 0

1

2

3

Proceeding as in Section 9.3.2 for the nucleon octet, we identify the weights of the funda54

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SU(3) Flavor Symmetry

55

−

ddd ddu uud uuu

+



++

1

0

uds

dds

Y −1

ssd

* 

* −

uus

 Decuplet

ssu

* −

10 (B=1, JP = 3/2+) −2

* +

* 

−

sss 

−

.

Fig. 9.1

0

−

1

T3

0

1

T3

Quark content of the ∆ decuplet from Problem 9.1.

mental representation with the quark triplet (u, d, s), m(1) → u

m(2) → d

m(3) → s

Thus from the list of weights obtained above for the 10 we may construct Fig. 9.1 [this document].

9.4 We follow a solution given in Ref. [85]. The u, d, and s quantum numbers are given in Table 9.1, and Y = B + S and T3 are additive. Each baryon in Fig. 9.1 contains three quarks 1 B = 1 3+ 1 + , with the sum of Y and T3 for the three quarks giving Y and T3 for the 3 3 baryon. For example, ∆++ in Fig. 9.1 has T3 = 32 and Y = 1. From Table 9.1 and Y = B + S, this requires a uuu quark structure so that T3 = 1 + 1 + 1 = 3 Y = 1 + 1 + 1 = 1.

2

2

2

2

3

3

3

Likewise, ∆+ has T3 = 1 and Y = 1, so it must have a uud structure such that 2

T3 = 21 + 21 − 12 = 12

Y = 31 + 31 + 13 = 1.

Lower rows of the diagram result from replacing u or d quarks with s quarks, giving finally the quark content shown for the ∆ decuplet in Fig. 9.4. The properly-symmetrized flavor wavefunctions may be constructed using the methods of Section 4.4. The standard arrangements of Young tableau for the 10 are 1 1 1

1 1 2

1 1 3

1 2 2

1 2 3

1 3 3

2 3 3

3 3 3

2 2 2

2 2 3

where the numbers in boxes are state labels, (u, d, s) ≡ (1, 2, 3). The wavefunctions are completely symmetric since all diagrams have a single row of boxes. For example, 1 ψ 1 1 2 ≡ u u d = √ (uud + duu + udu), 3

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56

where for example uud ≡ ψu(a)ψu(b)ψd(c) for particles a, b, and c, and the wavefunction has been normalized. The weight vector is the sum of the weight vectors for the constituent quarks; for example, (T3,Y ) = ( 12 , 1) and this symmetrized wavefunction is associated with the ∆+ from Fig. 9.1. The other nine states of the 10 follow in the same way. For example, ψ

1 2 3

1 = √ (uds + sud + sdu + usd + dsu + dus), 6

which by comparison of weights corresponds to the Σ∗0 in Fig. 9.1. The states of such a multiplet can also be constructed starting from any one of the states using the stepping operators in Eq. (8.8). The same methods may be used for other SU(3) flavor multiplets, except that if there are degeneracies (for example, the center state of the 8), the degenerate solutions must be orthogonalized.

9.5 Utilizing the methods of Section 9.4.2, the SU(2) reduction of the 15-dimensional SU(3) representation (2, 1) is 3

SU(3):

3 3

3 3

3

3 3 3

⇓

⇓

⇓

⇓

3

2

4

3

⇓

SU(2): Dimension:

1

2

where boxes with numbers have been removed and columns with more than one row have been eliminated for the reduction to SU(2) representations. Check of dimensionalities: 3 + 2 + 4 + 3 + 1 + 2 = 15.

9.7 This solution is given in Ref. [85]. From Problem 9.11 the ground-state baryon wavefunction must have ψ( spin)· ψ( flavor) = symmetric.

A proton is uud and has spin 12, so two of the quarks must be in a spin-singlet state 1 √ (↑↓ − ↓↑), 2 which is antisymmetric under particle exchange. Hence the corresponding flavor function must be antisymmetric too, 1 √ (ud − du).

2 Adding a third quark u↑ then gives (u↑d↓ − u↓d↑ − d↑u↓ + d↓u↑)u↑, which is symmetric in the first two quarks, but not in the third. To get the fully symmetrized

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SU(3) Flavor Symmetry

wavefunction, take all cyclic permutations (three for each term) and normalize to give, 1 ψ(p↑) = √ 2u↑u↑d↓ + 2d↓u↑u↑ + 2u↑d↓u↑ − u↓d↑u↑ 18 −u↑u↓d↑ − d↑u↑u↓ − d↑u↓u↑ − u↑d↑u↓ − u↓u↑d↑ .

This is symmetric under simultaneous exchange of spin and flavor for any two quarks.

9.11 The flavor wavefunction of the 10 analyzed in Problem 9.4 is completely symmetric.

We also assume that the ground state orbital wavefunction corresponds to L = 0 and thus is symmetric, and that for the spin- 1 baryons in the 10 the spin wavefunction must be a 2 symmetric triplet |↑↑↑⟩ for the three quarks to yield a baryon spin of 3 . Thus for the spin- 3 2

2

baryons of the 10 the wavefunction has the symmetry ψ( space)· ψ( spin)· ψ( flavor) = symmetric.

But this is inconsistent with the Pauli principle, which requires the total baryonic wavefunction to be antisymmetric. This problem can be alleviated if we assume that there is an additional quark degree of freedom (let’s call it “color”) distinct from flavor that transforms under its own SU(3) symmetry. Then if the SU(3) color state has the symmetry (0, 0) = 1 =

= antisymmetric,

the total baryon wavefunction would be antisymmetric under exchange of all coordinates (space, spin, flavor, and color): ψ( space)· ψ( spin)· ψ( flavor) · ψ( color) = antisymmetric.

Applying the methods of Section 4.4 to the (0, 0) Young diagram, for the three quarks of a baryon, a color singlet is given by 1 (qqq)1 = √ (RGB − RBG + BRG − BGR + GBR − GRB), 6 where R, G, and B are quark color labels and an expression like RGB means that quark 1 is red (R), quark 2 is green (G), and quark 3 is blue (B). This color wavefunction is manifestly antisymmetric under exchange of any two particles.

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10

Harmonic Oscillators and SU(3) 10.1 Using Hψ = Eψ and the commutators Eq. (10.5), we may write H(aqψ) = (aqH − aq)ψ = aq(H − 1)ψ = aq(E − 1)ψ = (E − 1)(aqψ), and likewise, H(a†ψ) = (a†H + a†)ψ q

q

q

= aq(H + 1)ψ †

= (E + 1)(a†qψ).

Therefore, a†q operating on an oscillator wavefunction raises the oscillator energy by one unit and aq lowers it by one unit.

10.2 Generally for a commutator [ A + B,C + D ] = [ A,C ] + [ A, D ] + [ B,C ] + [ B, D ], so we can write 1 [ A˜i j, A˜kl ] = [ A,C ] + [ A, D ] + [ B,C ] + [ B, D ] 4 † B ≡ a j a†i C ≡ a†ka l D ≡ a l a†k.

A ≡ ai a j Then using the commutators in Eq. (10.3) we can evaluate [ A,C ] = ai †aj ak†al − ak†al ai†a j = ai†a j ak†al − a†k(δil + ai†a l )a j = ai†aj ak†al − a†k a j δil − ak†ai†al a j = ai†aj ak†al − a†k a j δil − ai†ak†a j al = ai†a j ak†al − a†ka j δil − a†i (a j a†k − δ jk)a l = ai†aj ak†al − a†ka jδil + ai†al δ jk − ai†a j a†ka l = ai†al δ jk − ak†a j δil.

Evaluating the other commutators in the same way gives [ A, D ] = ai †al δ jk − ak†a j δil

[ B,C ] = al ai†δ jk − a ja†kδil

[ B, D ] = al ai†δ jk − a ja†kδil.

58

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Thus collecting results and using the definition (10.7), 1 [ A˜i j, A˜kl ] = [ A,C ] + [ A, D ] + [ B,C ] + [ B, D ] 4 1 1 † † † † = (ai al + alai )δ jk − (akaj + a jak ) δil 2 2 = A˜il δ jk − Ã k jδil , which is Eq. (10.8).

10.5 First, from Eqs. (8.2) and (8.7)–(8.8) we may construct useful matrix representations of T±, U±, and V±: λ1

λ2

1 T± = 2 ± i 2 = 2

0 1

0 1 0 1 0 0 0 ± −1 0 0 .

0 0 0 0 0 0

Therefore we have, 0 1 0 T+ = 0 0 0 0 0 0

0 0 0 T− = 1 0 0 .

0 0 0

Likewise, for the U± operators, λ6

U± = 2

±i

λ7 2

=

1 2

0 0

0 0 0 0 1 ± 0 0 1 0 0

0 0 −1

0 0 U− = 0 0 0 1

0 0 .

0

0 1 , 0

and therefore, 0 0 0 U+ = 0 0 1 0 0 0 Likewise, for the V± operators, λ4

V± = 2

±i

λ5 2

=

0 0 0 1 1 0 0 0 ± 0 2 1 0 0 −1

0 1 0 0 , 0 0

and therefore, 0 0 1 V+ = 0 0 0 0 0 0

0 0 0 V− = 0 0 0 .

1 0 0

Finally, T3 =

λ3

1 2

= 0 2 0

01 −2 0

0 0 0

2 λ8 1 3 Y=√ = √ λ8 = 0 3 2 3 0 1

01 3

0

0 0 . − 23

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Now let’s evaluate the matrix elements of 1 (Aij)kl ≡ δik δ jl — δkl , 3 1 2 3 subject to the constraint A + A + A = 0. First consider the non-diagonal elements (k /= l).

1

For j = 2,

2

3

(A21)kl ≡ δ1kδ2l , which implies that the only non-vanishing element of (A12)kl is 1 when k( row) = 1 and l( column) = 2. Therefore,

0 1 0 0 0 0 0 0 0

= T+.

0 0 1 A1 = 0 0 0 0 0 0

= V+.

A1 = 2

Likewise, by similar reasoning 3

We can also work in reverse. For example, 0 0 0 3 U+ = 0 0 1 = δ2kδ3l = A2 0 0 0 0 0

0 0 0 2 U− = 0 0 0 = δ3kδ2l = A3, 0 1 0

0

0

V− = 0 0 0 = δ3kδ1l = A3 1 0 0

1

0 0

T− = 1 0 0 = δ2kδ1l = A2.

0 0 0 1

Now consider the diagonal elements.

1 (A33)kl = δ3kδ3l − δkl = 3

− 13 = 0 0

1 0 0 0 3 0 0 0 − 0 0 0 1 0

0 1 3

0

0 0 1

3

0 0 −3 1 0 = −Y. 2 0 3

Likewise, we can write 1 1 0 0 3 (A1)kl = δ1kδ1l − 3 δkl = 0 0 0 − 0 0 0 0 0

01

1 0 0 0 3 (A2)kl = δ2kδ2l − 3 δkl = 0 1 0 − 0 0 0 0 0

01

1

2

1

1

3

0 3

0

2 0 3 0 = 0 1 0 3

−13 0 0 = 0 1 0 3

01 0 −3 0 , 0 − 13 0

0

2

0 , − 13

3

0

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from which we have clearly 1 0 0 A1 − A2 = 0 −1 0 0 = 2T3.

0 0 0 1

2

Finally, note the condition 1

2

3

2 3

A1 + A2 + A 3 = 0 0

−13 01 0 −3 0 + 0 0 − 13 0

02 3

0

−13 0 0 + 0 0 − 13

01 0 − 3 0 = 0, 2 0 3

which reduces the nine Aji to the eight independent generators of SU(3).

10.7 From Eqs. (8.2) and (8.7)–(8.8), the explicit form for the matrices T±, T3, and Y are 0 1 0 T+ = 0 0 0 0 0 0 1 2

T3 = 0 0

0 0 0 T− = 1 0 0 , 0 0 0 1 3

0 0 − 21 0 0 0

Y= 0 0

0

0 0 .

3 0 − 23 1

In the oscillator model, the three-component column vector upon which the SU(3) matrices operate can be labeled by the number of quanta in the x, y, and z directions, respectively, Nx |ψ⟩ = NxNyNz = Ny .

Nz Comparing with the above matrix definitions for T3 and Y , we see that in terms of the oscillator creation and annihilation operators 1 † 1 ax ax − a†yay = (Nx − N y), 21 1 † 2 † Y= a a + a a − 2a†a — 2N ).

(Nx + Ny z z = z x x y y 3 3 By similar consideration, T3 =

T+ = ax†a

y

T− = ay†ax ,

since, for example, 0

1 0 0 1 0 0 0 1 = 0 0 0 0 0 0

corresponds to the annihilation of an oscillator quantum along the y axis and the creation of one along the x axis. Carrying out similar considerations for the other raising and lowering

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operators defined in Eq. (8.8) and Fig. 8.1, we have for the oscillator weight space 1 1 † T— = a†ya x T3 ≡ (Nx − Ny) Y ≡ (N x + Ny − 2Nz) T+ = axa y 2 3 V+ = ax†az V− = az†a x U+ = a†ya z U− = az†ay , where the operators T3 and Y label the weight space and the stepping operators T±, V±, and U± step through the oscillator SU(3) states of an irrep like those displayed in Fig. 10.3 by shifting oscillator quanta among the x, y, and z axes.

10.9 For the 2p and 1f orbitals there are ∑(2l + 1) = (2 · 1 + 1) + (2 · 3 + 1) = 10 orbital states, so the orbital symmetry is U(10) ⊃ SU(10). For SU(10) with two particles, computing the dimensionalities as in Example 8.4 gives for the possible representations =

Dim

110 10 11 N = = 55 = D 2 2 1 10

Dim

90 9 N = = 45, = = D 2 2 1

and the 55 + 45 = 100 states checks with the 102 = 100 possibilities expected from distributing 2 particles on 10 states. For the 2-particle SU(6) ⊃ SU(3) states in the N = 3 fp shell, each particle carries three oscillator quanta, so the 2-particle SU(3) Young diagrams will be the direct product of two 3-box diagrams, giving 2 × 3 = 6 boxes for each diagram.

Applying the algorithm for the direct product given in Section 8.9, and Eq. (8.20) for the dimensionality of SU(3) representations, ⊗ 1 1 1 =

1 1 1 ⊕

=

⊕

1 1 1

⊕ ⊕

1 1 1

⊕

1 1 1

⊕

= 28 ⊕ 10 ⊕ 27 ⊕ 35, and 28 + 10 + 27 + 35 = 100 sums to the expected 102 configurations. Since 28 + 27 = 55 and 10 + 35 = 45, the 28 and 27 of SU(3) must be associated with the symmetric 55 of SU(10), and the 35 and 10 of SU(3) must be associated with the SU(10) irrep 45. These results agree with the classification given in Table 3.III of Harvey [106].

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11

SU(3) Matrix Elements 11.1 Consider the SU(3) product 6 ⊗ 3 = 10 ⊕ 8 =

.

⊕

From the graphical procedure of Section 8.11, the occupied sites in the SU(3) weight space are those shown in the following figure, with circles indicating additional degeneracy.

Y 1

0 −1

10 + 8 −2 −2

−1

0

1

2

T3

Consider the 10 or (p, q) = (3, 0) representation that appears in the product. In schematic notation this is given by |10⟩ = ⟨6, 3| 10⟩ |6⟩ |3⟩, where ⟨6, 3| 10⟩ is the SU(3) Clebsch– Gordan coefficient. Using Young diagrams and the weight space diagram, the rightmost occupied site of the above diagram corresponds to the upper right state of the 10 and has the wavefunction (10)

1 1 1 = u1u1u1 ≡ ψ1

,

because only three u1 components can give the maximally stretched weights associated with this state. This gives the wavefunction in terms of the components ui of the fundamental representation 3 of Fig. 11.1(b). Comparing with Example 11.1, we may also express it as a product of 6 and 3 basis functions: (10) = (u1u1)u1 = ψ(6)u1.

1 1 1 = ψ1 1

Now consider 1

(10)

1 1 2 = √ (u1u1u2 + u1u2u1 + u2u1u1) ≡ ψ2

.

3 This state clearly corresponds to one of the states on the doubly-occupied site one unit to the left of ψ1(10) in the above figure, since it can be obtained by applying the lowering operator T− to ψ1(10) [remember that T− can operate on any of the three particles in u1u1u1, 63

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and note the action of T− on the 3 in Fig. 11.1(b): T−(u1) = u2]. Comparing with the previous example, this may be expressed in terms of the 6 and 3 bases as √ (6) 1 1 (10) (6) 1 1 2 =ψ = √ (u1u1u2 + u1u2u1 + u2u1u1) = √ (ψ1 u2 + 2 ψ2 u1).

2 3 3 Proceeding in analogous fashion yields the other eight states of the 10.

1 √ (6) 1 (10) (6) 1 2 2 =ψ = √ (u1u2u2 + u2u1u2 + u2u2u1) = √ ( 2 ψ2 u2 + ψ3 u1) 3 3 3 (10) (6) = ψ = u u u = ψ u 2 2 2 2 2 2 2 4 3 √ (6) 1 1 (6) (10) 1 1 3 =ψ √ √ = (u u u + u u u + u u u ) = (ψ u + 2ψ4 u1) 1 1 3 1 3 1 3 1 1 3 5 1 3 3 1 (10) 1 2 3 =ψ = √ (u1u2u3 + u1u3u2 + u2u1u3 + u2u3u1 + u3u1u2 + u3u2u1) 6 6 1 (6) (6) (6) = √ (ψ2 u3 + ψ4 u2 + ψ5 u1) 3 √ (6) 1 1 (10) (6) 2 2 3 =ψ = √ (u2u2u3 + u2u3u2 + u3u2u2) = √ (ψ3 u3 + 2ψ5 u2) 7 3 3 √ 1 1 (6) (10) (6) 1 3 3 =ψ √ √ = (u u u + u u u + u u u ) = ( 2ψ4 u3 + ψ6 u1) 1 3 3 3 1 3 3 3 1 8 3 3 √ 1 1 (6) (10) (6) 2 2 3 =ψ √ √ = (u u u + u u u + u u u ) = ( 2ψ u + ψ 2 3 3 3 2 3 3 3 2 3 9 5 6 u2) 3 3 (10) (6) = u3u3u3 = ψ u3.

3 3 3 =ψ 10

6

These symmetrized and normalized wavefunctions are displayed in Fig. 11.4. The corresponding SU(3) Clebsch–Gordan coefficients are the coefficients in these ten equations, which may be extracted by taking overlaps as in Example 11.1.

11.2 From the following figure, Y 1

0 −1

10 + 8 −2 −2

−1

0

1

2

T3

the upper right state of the octet (arrow) has the same weight as ψ2(10). Hence, we must (10). Write construct a wavefunction having this weight that is orthogonal to ψ 2 r 2 (6) 1 (6) (8) (6) (6) (10) ψ2 u1 ψ1 = aψ1 u2 + bψ2 u1, ψ2 = √ ψ1 u2 + 3 3

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with a and b to be determined by the orthogonality and normalization requirements D E (10) (8) ψ2 ψ1 = 0 a2 + b2 = 1.

√ Inserting the first equations in the first of these gives a = − 2 b; imposing the normalization requirement then gives 1 √ (6) (8) (6) ψ1 = √ 2 ψ1 u2 − ψ2 u1).

3 The remaining states of the 8 are 1 √ (6) (8) (6) ψ 2 = √ ψ2 u2 − 2 ψ3 u1 3 1 √ (6) (8) (6) 2 ψ1 u3 − ψ4 u1) ψ3 = √ 3 1 (8) (6) (6) (6) ψ 4 = √ 2ψ2 u3 − ψ5 u1 − ψ4 u2 6 1 √ (6) (8) (6) √ ψ5 = 2 ψ3 u3 − ψ5 u2 6 1 (8) (6) (6) ψ 6 = √ ψ4 u2 − ψ5 u1 2 1 √ (6) (8) (6) ψ 7 = √ ψ4 u3 − 2 ψ6 u1 3 1 √ (6) (8) (6) √ ψ8 = ψ5 u3 − 2 ψ6 u2 , 3 which were obtained by applying lowering operators to the state ψ1(8) and orthogonalizing where necessary.

11.3 The six states on the boundary of the product diagram in Fig. 11.2 clearly belong to the 8 and can be made in only one way each from the basis vectors (u1, u2, u3) of the 3 and (v1, v2, v3) of the 3: ψ

(8)

1 (8)

ψ

= u1v3

ψ

= u2v2

ψ

6

(8)

2 (8) 7

= u2v3

ψ

= u3v1

ψ

(8)

3 (8)

= u1v1 = u3v2.

8

Thus, the corresponding SU(3) Clebsch–Gordan coefficients are all unity for these states.

There are three combinations that give T3 = Y = 0 at the center site: u2v1, u1v2, and u3v3.

The three states at the center are the three independent linear combinations of these products that satisfy the required conditions, two belonging to the octet and one to the singlet.

For the octet we may write, (T = 1) :

ψ4(8) = a |u2v1⟩ + b |u1v2⟩ + c |u3v3⟩,

(T = 0) :

ψ5(8) = a′ |u2v1⟩ + b′ |u1v2⟩ + c′ |u3v3⟩,

and for the singlet, ψ(1) = a′′ |u2v1⟩ + b′′ |u1v2⟩ + c′′ |u3v3⟩.

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The unknown coefficients in these three equations may be determined by the following reasoning. First, ψ(8) must be part of a T = 1 multiplet containing ψ(8) and ψ(8), and we 4

may obtain it by applying T− = T−(1) + T−(2) to ψ(38): E (8) (8) ψ4 = T− ψ3 = T− |u1v1⟩ = |u2v1⟩ + |u1v2⟩.

3

6

Therefore, the normalized wavefunction is 1 (8) ψ4 = √ (u2v1 + u1v2).

2 Now consider the SU(3) singlet state ψ(1) at the center. Because it is a scalar under SU(3), any of the SU(3) raising and lowering operators must give zero when they act on it. Operate first on it with V−, V−ψ(1) = V−(a′′u2v1 + b′′u1v2 + c′′u3v3) = b′′u3v2 + c′′u3v2 = 0.

Hence, b′′ = −c′′. Operating now with T− gives T−ψ(1) = T−(a′′u2v1 + b′′u1v2 + c′′u3v3) = a′′u2v2 + b′′u2v2 = 0.

Hence, b′′ = −a′′ = −c′′ and we may choose 1 1 1 (1) √ √ ψ = −√ |u1v2⟩ + |u2v1⟩ + |u3v3⟩, 3 3 3 which is orthogonal to ψ(8). The final member of the octet, ψ(8), must be orthogonal to 4

5

both ψ4(8) and ψ(1). Utilizing the preceding results, the requirement of orthogonality with ψ4(8) gives

D

(8)

ψ5

E

(8)

ψ4

a′ b′ = √ ⟨u2v1| u2v1⟩ + √ ⟨u1v2| u1v2⟩ = 0 2 2 a′ + b′ = 0,

while the requirement of orthogonality with ψ(1) yields D E a′ b′ c′ (8) (1) ψ5 ψ = √ ⟨u2v1| u2v1⟩ − √ ⟨u1v2| u1v2⟩ + √ ⟨u3v3| u3v3⟩ = 0 3 3 3 a′ − b′ + c′ = 0.

These requirements, and the normalization condition (a′)2 + (b′)2 + (c′)2 = 1, are satisfied by the choice 1 1 2 ′ ′ ′ ′ ′ ′ a = −√ b = −a = √ c = b −a = √ , 6 6 6 and the final member of the octet may be written 1 1 2 (8) ψ5 = −√ |u2v1⟩ + √ |u1v2⟩ + √ |u3v3⟩.

6 6 6

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Let’s check that this final state is an isosinglet. Operating with T± gives 1 1 (8) T−ψ5 = −√ |u2v2⟩ + √ |u2v2⟩ + 0 = 0 6 6 1 1 (8) T+ψ5 = −√ |u1v1⟩ + √ |u1v1⟩ + 0 = 0, 6 6 and ψ5(8) is indeed a T = 0 state. On the other hand, 1 1 (8) (8) T−ψ4 = √ |u2v2⟩ + √ |u2v2⟩ ∝ ψ6 2 2 1 1 (8) (8) T+ψ4 = √ |u1v1⟩ + √ |u1v1⟩ ∝ ψ3 , 2 2 and ψ(8), ψ(8), and ψ(8) are components of a T = 1 triplet.

6

3

4

11.6 Consider the operator 1 1 11 T11 = −√ (Lx + iLy) = −√ L+ = L1 2 2 given in Eqs. (11.12) and (11.13). We wish to show that T1111 satisfies Eq. (11.10) for ν = 0, ±1. This was already proven in Example 11.3 for ν = 0. For ν = 1 we clearly have [ L1, T1111 ] = [ L1, L1 ] = 0 and that (using L1 ∝ L+) (λ µ)κ′L′(M + ν) L1 |(λ µ)κ11⟩ = 0, since L+ acting on the maximally stretched state |LM⟩ = |11⟩ must give zero. Thus, T 11 11 satisfies Eq. (11.10) for ν = 1. Finally consider that for ν = −1, √ [ L−1, T 11 ] = [ L−1, L1 ] = − 2 ⟨1 − 111| 10⟩ L0 = L0 = T 11, 11

10

where Eqs. (10.18) and (11.12), and ⟨1 − 111| 10⟩ = − √1 that

2

from Table C.3 were used, and

1 ′ ′ ′ ′ (λ µ)κ L (M − 1) L−1 |(λ µ)κ11⟩ = √ (λ µ)κ L (M − 1) L− |(λ µ)κ11⟩ 2 1 √ = √ 1(1 + 1) − M(M − 1)δκκ′ δLL′ 2 = δκκ′ δLL′ .

Thus Eq. (11.10) is satisfied by T 11 for ν = −1. Since we have now shown that T 11 satisfies 11

Eq. (11.10) for ν = 0, ±1, it is a vector operator.

11

11.8 Let us illustrate the construction of isoscalar factors by considering the states of the 8 in the SU(3) product 3 ⊗ 3 = 8 ⊕ 1. In Section 11.2 we constructed coupled wavefunctions for this product; the coupled products are summarized in Fig. 11.3 and the uncoupled bases ui and vi are summarized in Fig. 11.1(b) and Table 11.1. The isoscalar factors corresponding to the boundary of the octet in Fig. 11.3 are all unity because (1) the

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SU(3) Matrix Elements

corresponding SU(3) Clebsch–Gordan coefficients are unity and (2) the associated SU(2) Clebsch–Gordan coefficients are unity too, because each site on the boundary is a maximally stretched state of an SU(2) multiplet. (Remember that horizontal rows are SU(2) representations and that a maximally stretched state can be made with only one combination of the product basis).

Now, consider the center member of the 8 isotriplet [the wavefunction labeled T = 1 in Fig. 11.3]. It also must have f = 1 because the isoscalar factor has no dependence on the SU(2) weight quantum numbers; therefore, it is the same for all members of an isospin multiplet and we have established that the other two members of this T = 1 triplet (which lie on the boundary) must have f = 1. More formally, we note that this is a T = 1, T3 = 0, Y = 0 state and the SU(3) Clebsch–Gordan coefficient associated with the first term for ψ(48) may be expressed in the notation of Eq. (11.8) as 1 10⟩ = √ , 2 where we use the dimensionality to label the SU(3) representations. But the SU(2) Clebsch– Gordan coefficient is √1 (Table C.3) and solving for the isoscalar factor gives 2 √ 1/ 2 11 11 ⟨3 ; 3 − ||801⟩ = √ = 1, 32 32 1/ 2 3 31 12 − 21 ; 3 − 31 21 21 8, 010⟩ = ⟨313 12 ; 3 − 13 12 ||801⟩

1 − 111 2 222

as we had already deduced on more heuristic grounds. The second term in ψ4 (8) involves the same isoscalar factor and as a consistency check you may verify that it gives the same result.

We are left with the other center state of the 8, which is an isosinglet that is labeled T = 0 in Fig. 11.3. This is a T = T3 = Y = 0 state and the SU(3) Clebsch–Gordan coefficient associated with the first term in ψ5(8) may be factored into a product of an isoscalar factor and an SU(2) Clebsch–Gordan coefficient, 3 31 12 − 21 ; 3 − 31 21 21 8, 000⟩ = ⟨313 12 ; 3 − 13 12 ||800⟩

1 − 111 2 222

1 00⟩ = −√ .

6

The SU(2) Clebsch–Gordan coefficient is − √1 , so the isoscalar factor is 2 √ −1/ 6 1 1 1 11 √ =√ . ⟨3 3 2 ; 3 − 3 2 ||800⟩ = −1/ 2 3 The second term in ψ5(8) gives the same isoscalar factor as the first. Finally, the SU(3) Clebsch–Gordan coefficient associated with the third term may be expressed as 2 3 − 3200; 3 32 00 8, 000⟩ = ⟨3 − 320; 3 32 0||800⟩⟨0000| 00⟩ = √ .

6 The SU(2) Clebsch–Gordan coefficient is 1, so the isoscalar factor is r √ 2/ 6 2 2 2 ⟨3 − 0; 3 0||800⟩ = = .

3 3 1 3 We see in this simple example that most of the SU(3) Clebsch–Gordan coefficients are

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numerically equal to an SU(2) Clebsch–Gordan coefficient, and that the others are related to an SU(2) Clebsch–Gordan coefficient through an easily-determined factor. The isoscalar factors for other cases may be constructed in a similar way.

11.10 The spherical operators are expressed in terms of cartesian components by Eq. (11.23), 1 1 † † † † † a±1 = ∓√ (ax ± iay) a0 = az a±1 = ∓√ (ax ± iay ) a0 = az .

2 2 We may calculate the commutators of the spherical operators using the commutator identity [ a + b, c + d ] = [ a, c ] + [ a, d ] + [ b, c ] + [ b, d ].

For example, † [ a1 , a−1 ]=−

1

[ (ax + iay ),(ax† − iay†)] 2 1 = − [ a x, a†x] − i[ a x, a†y] + i[ a y, a†x ] + [ a y, a†y ] 2 = −1,

where we have used from Eq. (10.3), [ aa, a† ] = δi j, b

[ aa, a ] = [ a†, a† ] = 0, a

b

b

with the labels a, b referring to the cartesian axes x, y, z. As another example, 1 [ a−1 , a1 † ] = − [ (ax − iay ),(ax† + iay†)] 2 1 = − [ a x, a†x] + i[ a x, a†y ] − i[ a y, a†x ] + [ a y, a†y ] 2 = −1.

Proceeding in this way we find that in general [ ap, a† ] = (−1)pδp,−q q

[ ap, aq ] = [ a†, a† ] = 0 p

q

for p and q = (0, ±1).

11.12 From Eq. (11.27), −1 00 T00 = √ 3

H−

3 .

2

Let’s now use this result to construct the reduced matrix elements for the oscillator, starting from those of a†. Since ⟨H⟩ = ⟨NLM| H |NLM⟩ = N + 23 , 1 00 ⟨NLM| T00 |NLM⟩ = −√ 3

⟨H⟩ −

3 1 = −√ N. 2 3

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But using steps from the derivation of Eq. (11.26), we may also write 1 00 q † ⟨NLM| T00 |NLM⟩ = −√ ∑⟨NLM| (−1) aqa−q |NLM⟩ 3 q 1 = −√ ⟨NLM| aq† N ′L ′M ′ N′ L′ M′ (−1) qaq |NLM⟩ 3 ∑ q N′∑ L′M′ ′ ∗ 1 † ′ ′ = −√ ⟨NLM| aq† N ′L ′M ′ ⟨NLM| aq N L M 3 ∑ q N′∑ L′M′ 1 ′ ′ 2 † = −√ ⟨NLM| aq N − 1, L M . ∑ ∑ 3 q L′M′ Now apply the Wigner–Eckart theorem (11.15) to each matrix element on the right side: ⟨NLM| aq† N − 1, L′M′ = ∑⟨NL; N − 1, L′||ρ 31⟩ LML′M′ 1q⟩⟨N |a† |N − 1⟩, ρ

so that 1 00 ⟨NLM| T00 |NLM⟩ = −√ ∑ ⟨NL; N − 1, L||ρ 31⟩ 2 LML′ M ′ 1q⟩ 2 ⟨N |a† |N − 1⟩ 2 3 qL′M′ 1 † 2 = −√ ⟨N |a |N − 1⟩ , 3 and comparing with 00

⟨NLM| T00 |NLM⟩ = −√ 3

1 3 ⟨H⟩ − 2

1 = −√ N. 3

from above, we finally conclude that √ ⟨N |a† |N − 1⟩ = N. Let’s define for an SU(3) tensor TLλ2µ2 , 2

⟨(λ1µ1)L1 || T Lλ2µ2 || (λ3µ3⟩ ≡ (−1)L2−L1−L3 2

√ 2L1 + 1

⟨(λ2µ2)L2; (λ3µ3)L3||(λ1µ1)L1⟩⟨(λ1µ1) || T λ2 µ2 || (λ3µ3)⟩.

Then the magnetic substate dependence may be factored as ⟨(λ1µ1)L1M1| TLλ22Mµ22 |(λ3µ3)L3M3⟩ L L L λ2 µ2 1 2 3 L −M = (−1) 1 1 ⟨(λ1µ1)L1 || TL2 || (λ3µ3)L3⟩, M1 M2 M3 so it is convenient to tabulate the reduced matrix elements ⟨(λ1µ1)L1 || TL2λ2µ2 || (λ3µ3)L3⟩.

10 Since a†qtransforms as T 1q , and the states |NLM⟩ of the oscillator span the irrep (λ, µ) =

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(N, 0) of SU(3), †

10

⟨N + 1, L + 1 || a || NL⟩ = ⟨(N + 1, 0), L + 1 || T1 || (N, 0)L⟩ √ = (−1)2L 2L + 3 ⟨(10)1;(N0)L||(N + 1, 0)L + 1⟩ × ⟨N + 1 || a†|| N⟩ √ = − (L + 1)(N + L + 3), (see Wybourne [224], Section 20.4), which is the first equation (11.28). The corresponding reduced matrix element for a = T1 01 follows from the observation that ′

⟨NL || T λ µ || N′L′⟩ = (−1)L−L ⟨N′L′ || (T λ µ)† || NL⟩, from which we have

√ ⟨N − 1, L − 1 || a || NL⟩ = −⟨NL || a† || N − 1, L − 1⟩ = (N + L + 1)L,

which is the second equation (11.28).

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Non-Compact Groups 12.1 Define the momentum operator Pi and the angular momentum operator Lj by ∂ ∂ Pi = −i L j = −iε jkmxk . ∂ xi ∂ xm Then the commutator of momentum and angular momentum is ∂ ∂ , ε jkm xk ∂ xm ∂ xi ∂ ∂ ∂ ∂ xk −x k = −ε jkm ∂ xm ∂ xm ∂ xi ∂ xi ∂ xk ∂ ∂2 ∂2 — x + x k = −ε jkm ∂ x ∂ x k ∂ x m ∂ xi i m ∂ x m ∂ xi ∂ = −ε jim ∂ xm = iεi jmPm,

[ Pi , L j ] = −

where we have used the standard formula for the derivative of a product and ∂ xk ∂ ∂ ∂2 ∂2 = = ε jim = −εi jm. ∂x∂x ∂x ∂x ∂ x = δik ∂x ∂x i

i

m

i

m

m

i

Since the order of partial differentiation doesn’t matter, partial derivatives commute and ∂ ∂ ∂ ∂ − ∂ xi ∂ x j ∂ x j ∂ xi

[ Pi , Pj ] = −

= 0.

Finally we already know from Eq. (6.29) that the angular momentum algebra is [ Li, Lj ] = iεi jkLk. Thus P and L generate a lie group (E3) with the closed algebra given by Eq. (12.10). The group is non-abelian since there are non-zero commutators.

12.3 Define the operators P± through P± ≡ P1 ± iP2.

From (12.14), these have the commutation properties [ J, P+ ] = [ J, P1 ] + i[ J, P2 ] = iP2 + P1 = P+, [ J, P− ] = [ J, P1 ] − i[ J, P2 ] = −P1 + iP2 = −P−, which may be summarized as [ J, P± ] = ±P±.

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Then, for a state with J |m⟩ = m |m⟩, J P± |m⟩ = P±J + [ J, P± ] |m⟩ = (P±J ± P±) |m⟩ = P±(J ± 1) |m⟩ = (m ± 1) P± |m⟩ .

Thus P± act as raising and lowering operators for eigenvalues of J. 12.7 Substituting P1 → iX2, P2 → iX3, and J → iX1 in the E2 commutators (12.14) gives [ X1, X2 ] = X3

[ X1, X3 ] = −X2

[ X2, X3 ] = 0,

which is in the form [ Xi, Xj ] = c X of Eq. (7.18) with non-zero structure constants k

ij k

3 2 c 12 = c 31 = 1

2 c321 = c13 = −1.

Then from Eq. (7.19) the components of the E2 metric tensor are 3 2 g11 = c12 c13 + c213 c312 = −2 3 2 g12 = c12 c23 + c213c322 = 0

3 1 g22 = c21 c23 = 0 3 2 g13 = c12 c33 + c213c332 = 0

2 1 g33 = c31 c32 = 0, 3 1 g23 = c21 c33 = 0.

Using gi j = −gji then, the metric tensor is 2E

gi j =

−2 0 0 0 0 0 .

0 0 0

By the criteria in Section 7.2.2, det g = 0 (so g is singular with no inverse), and E2 isn’t semisimple (or simple, since simple algebras are necessarily semisimple). This is because it has an invariant subgroup that is abelian, corresponding to the translation subgroup consisting of (P1, P2) in Eq. (12.14). Since E2 isn’t semisimple, it can’t be written as a direct product of simple groups. As discussed in Section 12.4.3, E2 is a semidirect product of the 2D translation subgroup and the 1D subgroup generated by J of rotations in the plane. It is obviously non-compact, since the translations are unbounded.

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The Lorentz Group 13.2 For two 4-vectors A and B to be orthogonal their scalar product must vanish, so from Eq. (13.1) orthogonality requires ηµν Aµ Bν = 0. But lightlike vectors X obey the condition ηµν X µ X ν = 0, by definition. Thus a lightlike Minkowski vector is orthogonal to itself.

13.3 Evaluating the commutators of Eq. (13.20) explicitly [ Kx, Ky ] = −iJz

[ Ky, Kz ] = −iJx

[ Kz, Kx ] = −iJy,

[ Jx, Ky ] = iKz

[ Jy, Kz ] = iKx

[ Jz, Kx ] = iKy,

[ Jy, Kx ] = −iKz

[ Jz, Ky ] = −iKx

[ Jx, Kz ] = −iKy,

[ Ji, Jj ] = iεi jkJk

[ Jx, Kx ] = [ Jy, Ky ] = [ Jz, Kz ] = 0.

Then, for example, using the definitions (13.30) [ A x, A y ] =

1 4 1

([ Jx + iKx, Jy + iKy ])

= 4 ([ Jx, Jy ] + i[ Jx, Ky ] + i[ Kx, Jy ] − [ Kx, Ky ]) 1 = (2iJz − 2Kz) 4 = iAz. As a second example, [ B x, B y ] =

1 4 1

([ Jx, Jy ] − i[ Jx, Ky ] − i[ Kx, Jy ] − [ Kx, Ky ])

= (2iJz + 2Kz) 4 = iBz. Continuing in this manner we find [ Ax, Ay ] = iAz

[ Ay, Az ] = iAx

[ Bx, By ] = iBz [ By, Bz ] = iBx

[ Az, Ax ] = iAy,

[ Bz, Bx ] = iBy

[ Ai, B j ] = 0,

which corresponds to the SU(2) × SU(2) commutator algebra of Eq. (13.31).

13.6 Let σ µ ≡ (σ 0, σ 1, σ 2, σ 3), where σ 0 is the unit 2 × 2 matrix and the σ i are the three Pauli matrices of Eq. (3.11). Likewise define the Minkowski-space coordinate xµ = (x0, x1, x2, x3). Then using the metric (13.3) σµ = ηµν σ ν = (σ 0, −σ 1, −σ 2, −σ 3)

xµ = ηµν xν = (x0, −x1, −x2, −x3).

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From Eq. (13.26), inserting explicitly the unit matrix and the Pauli matrices (3.11) gives X = η µν xµ σν = xµ σµ = σ0x0 + σ1x1 + σ2x2 + σ3x3 =

1 0 0 0 1 1 0 x + x + 0 1 1 0 i

=

x0 + x3 x1 − ix2 x1 + ix2 x0 − x3

=

x0 − x3 −x1 + ix2 , −x1 − ix2 x0 + x3

−i 2 1 0 3 x + x 0 0 −1

which is Eq. (13.27). The transformation (13.29) preserves the hermitian character of X since if X is hermitian to begin with, (X ′)† = (AXA†)† = (A†)†X †A† = AX †A† = AXA† = X ′, where we have used that the hermitian conjugate of a matrix product is the product of hermitian conjugates for each matrix in reverse order [see Problem 3.11(g)], that (A†)† = A, and that X is assumed hermitian, X = X †.

13.7 Letting σ 0 equal the 2 × 2 unit matrix and σ i(i = 1, 2, 3) equal the Pauli matrices of Eq. (3.11), define σ µ ≡ (σ 0, σ 1, σ 2, σ 3)

σµ ≡ ηµν σ ν = (σ 0, −σ 1, −σ 2, −σ 3),

where ηµν = diag(1, −1, −1, −1) is given in Eq. (13.3). Starting from Eq. (13.26), X = xµ σµ , multiply both sides by σ ν and take the trace of both sides to give Tr(X σ ν ) = xµ Tr(σµ σ ν ), where as usual we are summing on the repeated index µ. Now if ν = 0 we have Tr(σµ σ 0) = Trσµ = 2δ µ0, since σ 0 is the unit 2 × 2 matrix and the Pauli matrices σi(i = 1, 2, 3) are traceless. Likewise, if ν = 1, 2, 3, Tr(σµ σ ν ) = 2δµν , using the properties of the Pauli matrices in Table 3.1. Thus, solving for xν gives 1 ν xν = Tr(X σ ).

2 This result can be checked by explicit multiplication. For example, using Eq. (13.27) and

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choosing ν = 2, 1 x2 = 2Tr(X σ 2) 1 x0 − x3 −x1 + ix2 0 −i = Tr 2 −x1 − ix2 x0 + x3 i 0 1 −ix1 − x2 −(x0 − x3)i = Tr (x0 + x3)i ix1 — x2 2 1 = (−2x2) = −x2, 2 which is correct since the Minkowski metric is η = diag(1, −1, −1, −1), which implies that x2 = η22x2 = −x2.

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Lorentz Covariant Fields 14.2 Using coshξ = γ from Eq. (13.14) and basic hyperbolic trigonometry identities r r r r ξ γ −1 ξ coshξ + 1 γ +1 2 ξ sinh = cosh −1 = . cosh = = 2 2 2 2 2 2 Inserting this in Eq. (14.20) and introducing the unit vector in the direction of the boost pˆ = p/p with p ≡ | p|, " # γ + 1 1/2 γ − 1 1/2 ψR( p) = + σ · p̂ ψR(0) 2 2 " # E + m 1/2 E − m 1/2 + σ · p̂ ψR(0) = 2m " 2m # p = √ E +m σ · pˆ ψR (0) +√ 2m(E + m) 2m(E + m) E +m+σ· p ψR(0), =√ 2m(E + m) where we’ve used γ = E/m in line two, multiplied by (E + m)1/2/(E + m)1/2 and used (E2 − m2)1/2 = p in line three, and used p pˆ = p in the last line. This result is Eq. (14.21).

14.3 Utilizing the momentum 4-vector (14.24) E p1 E µ p = p = p2 ≡ p2 p3

p0 p1

ν

p0 −p1

pµ = ηµν p = −p 2 , −p 3

p3

and the Weyl representation of the γ-matrices (14.28), γ

0 ≡

0 1 1 0

γ ≡

p0 0

γi pi =

i

σi

i −σ , 0

0 σi

−σi p i= 0

0

we may write γ0 p

0 =

0 1 0 p = 1 0 0 p0

1 0 m m= 0 1 0

0 σ·p

−σ · p 0

0 , m

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where σ · p = σi pi. Then for the operators in the Dirac equation (14.31) γ µ pµ − m = γ0 p0 + γi pi − m p0 0 + 0 σ·p

−σ · p m 0 − 0 0 m

=

0 p0

=

−m p0 + σ · p . p0 − σ · p −m

Thus, with the substitution (14.26) the Dirac equation in covariant notation (14.31) is equivalent to Eq. (14.25).

14.5 Solve Eqs. (14.21) and (14.22) for the unboosted spinors ψR(0) and ψL(0) and then equate them to give E +m+σ · p ψ ( p) = ψ ( p).

R L E+m−σ · p Simplify by multiplying the numerator and denominator by (E + m) + σ · p, utilizing the identity (σ · p)2 =

1 0 2 p, 0 1

proved in Problem 14.4, and making judicious use of the energy relation E2 = p2 + m2 in both the numerator and denominator. After a small amount of algebra E+σ · p ψ ( p) = ψ ( p) R L m which is the expression for ψR( p) in Eq. (14.23). The expression for ψL( p) in Eq. (14.23) can be obtained in a similar way.

14.6 The properties of the Pauli matrices σi are given in Table 3.1. In the Pauli–Dirac representation 1 0 0 σi γ = γi = .

0 0 −1 −σi 0 If this is a valid representation it must satisfy the anticommutator (14.27), { γµ , γν } ≡ γµ γν + γν γµ = 2ηµν, where the metric ηµν = diag(1, −1, −1, −1) is given by Eq. (13.3). Let’s check for i, j = 1, 2, 3: { γi, γ j } =

0 −σi

=−

σi 0

{ σ i, σ j } 0

0 −σ j

σj 0 + 0 −σ j

σj 0

0 −σi

σi 0

0 . { σ i, σ j }

But from Table 3.1, { σi, σ j } = 2δi j, which gives { γi, γ j } = 2ηi j,

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in accord with Eq. (14.27). Now let’s check γ0. { γ0, γ0 } = 2η00 and 0

σi

σi

0

{ γ , γi } = 0

+

0

−σi

−σi

0

= 0 = 2η .

0i

Thus the matrices of the Pauli–Dirac representation satisfy Eq. (14.27).

14.7 Utilizing σ µν = i [2γ µ , γν ], the current ψσ µν ψ transforms as ψS−1σ µν Sψ =

i

ψS−1(γµ γν — γ ν γ µ )Sψ 2 i i = ψS−1γµ γν Sψ − ψS−1γν γ µ Sψ 2 2 i i = ψS−1γµ SS−1γν Sψ − ψS−1γν SS−1γµ Sψ 2 2 i = Λµ λ Λ εν ψ (γλ γε − γ ε γ λ ) ψ 2 = Λµ λ Λε ν ψσ λ ε ψ,

where in line three we have inserting SS−1 = 1 in each term and in line four we have used Example 14.3 and the solution of Problem 14.13, to give S−1γµ SS−1γν S = S−1γµ S

S−1γν S = Λµ λγλ

Λ νε γε = Λµ λ Λ εν γλ γε .

From Eq. (13.10) this is the transformation law for a contravariant second-rank tensor, and by inspection it is antisymmetric under exchange of indices.

14.11 From Eq. (14.50), ψ† = (P+ψ)† + (P−ψ)† = ψ†P+ + ψ†P−, where we have used that (AB)† = B†A†, and that P† = P , since from Table 14.1 γ† = γ . ±

±

5

5

Then the adjoint is given by ψ ≡ ψ†γ0 = ψ†P+γ0 + ψ†P−γ0 = ψ†γ0P− + ψ†γ0P+ = ψP− + ψP+, where we have used that γ5 anticommutes with γ0. Therefore, from (14.50) and (14.52), ψψ = ψP−P+ψ + ψP−P−ψ + ψP+P+ψ + ψP+P−ψ = ψP−P−ψ + ψP+P+ψ = RL + LR,

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where the properties (14.48) have been used, for vector interactions, ψγ µ ψ = ψP−γµ P+ψ + ψP−γµ P−ψ + ψP+γµ P+ψ + ψP+γµ P−ψ = ψP−γµ P+ψ + ψγ µ P+P−ψ + ψγ µ P−P+ψ + ψP+γµ P−ψ = ψP−γµ P+ψ + ψP+γµ P−ψ = Rγµ R + Lγµ L, and for axial vector interactions ψγ µ γ5ψ = Rγµ γ5R + Lγµ γ5L, by a similar proof. Therefore, vector and axial vector interactions conserve chiral symmetry but fermion mass terms and scalar interactions involve ψψ, which mixes left-handed and right-handed components, thereby breaking chiral invariance.

14.14 This solution follows the solution for two problems given in Ref. [89]. (a) To find the eigenvalues of an n × n matrix A requires solving the characteristic equation det(A − λ I) = 0 for λ, with I the n × n unit matrix and det indicating the determinant. For the general 2 × 2 matrix A α β λ 0 α −λ β det(A − λ I) = det − = det = 0. γ δ −λ γ δ 0 λ Evaluation of the determinant yields a quadratic equation in the eigenvalues λ with solur √ tions α + δ ± (α + δ)2 − 4(αδ − β γ) 1 1 2 λ± = = T± T − D, 2 2 4 where we have defined D ≡ det A = αδ − β γ

T ≡ Tr A = α + δ ,

with Tr indicating the trace operation (sum of diagonal elements). (b) Obtaining the eigenvectors corresponding to the eigenvalues found in part (a) requires solving the original linear system AX = λ X with the eigenvalues λ inserted. Choose a basis 1 0

χ1 ≡

0 1

χ2 ≡

.

The states X will in general be linear combinations of these basis states X≡

X+

1

= aχ1 + bχ2 = a

0

+b

0

X−

=

1

a

.

b

Thus the linear system is AX =

α γ

β δ

a b

=λ

a b

,

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implying that αa + β b = λ a. Inserting λ = λ+ and choosing b = 1 gives a = β /(λ+ −α), and ignoring normalization the eigenvector associated with eigenvalue λ+ is β χ1 + χ2.

X+ = aχ1 + bχ2 = λ+ — α Likewise, inserting λ = λ− in αa + β b = λ a and choosing a = 1 gives b = (λ− − α)/β , which implies a second eigenvector 1 X− = χ1 + (λ− − α)χ2. β (c) Applying the results of part (a) to the neutrino mass matrix M =

0 m m M

that was given in Box 14.4 yields λ± =

1 1 4m2 M ± M 1+ 2 2 M 2

1/2

≃

1 1 2m2 M ± M 1+ 2 2 M 2

,

where in the last step the square root has been expanded assuming 4m2/M2 ≪ 1. Thus the neutrino mass eigenvalues are λ+ ≃ M ≃ 1015 GeV

λ− ≃

−m2

≃ − 10−2 eV, M where 2m2/M2 has been ignored relative to 1 and the eigenvalues have been evaluated assuming m = 100 GeV and M = 1015 GeV. Applying the general results of part (b) for the eigenvectors of the mass matrix M with the assumption m ≪ M gives m m X+ = νL + νR ≃ νR X− = νL − νR ≃ νL, M M where νL denotes the wavefunction of a pure left-handed neutrino and νR the wavefunction of a pure right-handed neutrino. Thus the eigenvector associated with λ− is approximately the left-handed neutrino of the Standard Model but with a tiny mass, and the eigenvector associated with λ+ is a very massive right-handed neutrino. These results constitute the simplest form of the seesaw mechanism described in Box 14.4.

14.15 From the solution of Problem 14.12, the chirality operator does not commute with the Dirac Hamiltonian for massive particles. Helicity is the projection of the intrinsic spin on the momentum axis. That is Lorentz invariant for a massless particle which moves always at lightspeed, but is not for a massive particle which moves at less than lightspeed because for the massive particle the helicity can be reversed by boosting to a frame moving at less than c but greater than the speed of the massive particle. The Dirac Hamiltonian H is from Eq. (14.44) H = α · p+ βm and the helicity operator h is given by Eq. (14.40), h=

1 σ · pˆ 0 .

2 0 σ · pˆ

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Therefore, the commutator of helicity with the Dirac Hamiltonian is [ h, H ] = hH − Hh. Explicit commutation of the matrices in the Pauli–Dirac representation (14.43) indicates that 1 σ 0 0 σ·p [h, α · p] = , = 0, 2 0 σ σ·p 0 where we have used α·p=

σ·p 0

0 σ·p

from the definition of α in Eq. (14.43). Likewise, from Eq. (14.43) β=

1 0 0 −1

and therefore [ h, β m ] = 0. Thus the helicity commutes with the Dirac Hamiltonian, [ h, H ] = 0. We conclude that for a massive fermion the chirality is Lorentz invariant but it is not conserved because it doesn’t commute with the Dirac Hamiltonian, and the helicity is conserved because it commutes with the Dirac Hamiltonian but it is not Lorentz invariant, so its value is frame dependent.

14.20 Consider a Lorentz boost along the x axis analogous to Eq. (13.16), p′0 = E′ = E coshξ + p1 sinhξ

p ′2 = p2

p′1 = E sinhξ + p1 coshξ

p′3 = p3.

The transformation between volume elements is given by d3 p′ = Jd3 p, where J is the Jacobian of the transformation between the p and p′ coordinates. Only one coordinate (p1) changes in the boost so the Jacobian matrix has a single entry ∂ p′1

∂E

= sinhξ + coshξ ∂ p1 ∂ p1 = =

p1 sinhξ + coshξ E p1 sinhξ + E coshξ

E′ = , E

E

where ∂ E/∂ p1 = p1/E has been deduced from the relativistic energy expression E = ( p2 + m2)1/2. Thus, ∂ p′1 3 3 ′ E′ 3 p = d p, d p = d E ∂ p1 or equivalently 3

d p d3 p′ = .

E′ E

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This suggests defining the invariant volume element in momentum space as D 3 p ≡ constant×

d3 p E

= constant×

d3 p p0

,

where the constant may be chosen for convenience.

14.21 Consider a massive free-particle Dirac spinor u(α) with a normalization E u(α)†( p)u(β)( p) = δ , m αβ where α and β are spinor indices. Free particles will propagate as plane-wave states of the form ψ(α)(x) = u(α)e−ip·x.

The probability of finding a free fermion in a box of volume V in some frame is then P=

∫

ρ dV = (ψ†ψ)V = (u†u)V,

where we’ve used that the density ρ is independent of the coordinates for plane-wave states. Rotations don’t change the volume element entering this formula but boosts do. If in the rest frame the volume is V0, in a boosted frame for which the particle has energy E the new volume will be V = (m/E)V0. Thus, we find that ∫

and the normalization

ρ dV = u†uV0

m = E

E m V0 m E

u(α)†( p)u(β)( p) =

= V 0,

E

δ , m αβ for massive-fermion Dirac spinors is unaltered by a change of Lorentz frame.

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Poincaré Invariance 15.1 The most general transformation of the Poincaré group is given by Eq. (15.1), x′µ = Λµν xν + bµ, which may be written as the matrix equation x → x′ = Λx + b. If we apply a Poincaré transformation and then follow it by a second Poincaré transformation [199], x′ → x′′ = Λ′x′ + b′ = Λ′(Λx + b) + b′ = (Λ′Λ)x + (Λ′b + b′) = (Λ′b + b′) + (Λ′Λ)x.

Denoting this sequence of operations by − x, x −−−→ x′ −−−′ −→ x′′ →−− ′ ′ −−− ′− ′− g(b,Λ)

g(b ,Λ )

g(Λ b+b ,Λ Λ)

it is seen that the Poincaré multiplication rule is g(b′Λ′)g(b, Λ) = g(Λ′b + b′, Λ′Λ), where Λ parameterizes a Lorentz transformation and b parameterizes a spacetime translation. This result is Eq. (15.2).

15.3 (a) In the Poincaré multiplication law (15.2), g(b′Λ′)g(b, Λ) = g(Λ′b + b′, Λ′Λ), set g(b, Λ) = g(b, I) ≡ T (b) (where I is the unit matrix) and g(b′, Λ′) = g(0, Λ) ≡ Λ to give ΛT (b) = g(Λb, Λ) = T (Λb)Λ, where in the last step we have used Eq. (15.4) to write, g(Λb, Λ) = T (Λb)Λ. This becomes ΛT (b)Λ−1 = T (Λb), upon multiplying from the right by Λ−1 and using ΛΛ−1 = 1. This result is Eq. (15.5). (b) From Eq. (15.4) g(b, Λ) = T (b)Λ, where T (b) ≡ g(b, I). Multiply this expression from the right by T (a)g(b, Λ)−1 to give g(b, Λ)T (a)g(b, Λ)−1 = T (b)ΛT (a)g(b, Λ)−1 = T (b)ΛT (a)Λ−1T (b)−1, 84

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where in the last step g(b, Λ)−1 = T (b)Λ

−1

= Λ−1T (b)−1

was used. But from part (a), ΛT (a)Λ−1 = T (Λa), giving g(b, Λ)T (a)g(b, Λ)−1 = T (b)T (Λa)T (b)−1 = T (Λa), where the last step has used that a translation commutes with any other translation. Thus, the translations give translations under conjugation with arbitrary group elements, implying that they form an invariant subgroup of the Poincaré group (see Section 2.12).

15.5 The action of an infinitesimal group element on a 4-vector xρ is [125] 1 1 µν e− 2 i εµν L xρ ≃ (1 − iεµν Lµν )xρ 2 1 = xρ + εµν (xµ ∂ ν − xν ∂ µ )xρ 2 1 = xρ + εµν xµ ηνρ − x ν η µ ρ 2 1 1 = xρ + εµν xµ ηνρ − εµν xν η µ ρ 2 2 1 1 ν µ = xρ + ενµ x η ρ − εµν xν η µ ρ 2 1 12 = xρ − εµν xν η µ − εµν xν ηµ ρ ρ 2 2 = xρ − εµν xν ηµ ρ = xρ − ερν xν = xρ − ε ρ ν x ν , where Eq. (15.8) was used in line two, in line three ∂ α xρ = ηα ρ = δ αρ was used, in line five the dummy summation indices were switched µ ↔ ν in the second term, and in line six εµν = −ενµ was used. This result is equivalent to Eq. (15.7). Thus, Eq. (15.8) defines generators of proper Lorentz transformations.

15.9 Beginning from Eq. (15.17) with the standard lightlike vector (15.26), 1 Wµ ≡ − εµνρσ Lνρ Pσ 2

p¯µ = (ω, 0, 0, ω),

assuming εµνρσ to be completely antisymmetric in its indices with ε0123 ≡ +1, setting Pσ = p¯σ , and utilizing from Eq. (15.27), Lµν = −Lνµ

1 Ji = εi jk Ljk 2

Ki = L0i,

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we may evaluate components of the Pauli–Lubanski pseudovector as follows.

1 1 νρ σ νρ 0 1 νρ 3 W0 = − ε0νρσ L P = − ε 0νρ0 L p¯ − ε 0νρ3 L p¯ 2 2 2 1 1 21 3 νρ 3 12 3 = − ε0νρ3 L p¯ = − ε0123 L p¯ + ε0213 L p¯ 2 2 1 1 = − L12 p¯3 − L21 p3 = (L12 p¯3 − L21 p¯3) 2 2 1 = ω(L12 − L21) = ωL12 = ωJ3, 2 where we have used ε0νρ0 = 0

ε0123 = +1

ε0213 = −1

L21 = −L12

p¯3 = −p¯3,

lowered the indices using Li j = Li j, and used that since Ji = 12εi jkLjk, J3 =

1

(ε123L12 + ε132L21) =

2 (see Problem 15.8). By a similar procedure,

1 2

(L12 − L21) = L12

1 1 νρ 3 νρ σ νρ 0 1 W1 = − ε1νρσ L p¯ = − ε 1νρ0 L p¯ − ε1νρ3 L p¯ 2 2 2 1 23 0 32 0 02 3 20 3 = − (ε1230 L p¯ + ε1320 L p¯ + ε1023 L p¯ + ε1203 L p¯ ) 2 1 = − (−L23 p¯0 + L32 p¯0 − L02 p¯3 + L20 p¯3) 2 1 = − (−L23 p¯0 + L32 p¯0 − L02 p¯3 − L02p¯3) 2 = −ω(L32 − L02) = ω(J1 + K2), where we’ve used L20 p¯3 = −L02 p¯3 = −L02 p¯3

p¯0 = p¯3 = ω

J1 = −L32

K2 = L02.

Applying similar steps for W2, 1 13 0 31 0 10 3 01 3 W2 = − (ε2130 L p¯ + ε2310 L p¯ + ε2103 L p¯ + ε2013 L p¯ ) 2 1 = − (L13 p¯0 − L31 p¯0 − L10 p¯3 + L01 p¯3) 2 1 = − (L13 p¯0 − L31 p¯0 − L10 p¯3 + L01 p¯3) 2 = −ω(L13 + L01) = ω(J2 − K1), and likewise for W3, W = − 1 (ε L12 p¯0 + ε L21 p¯0) = − 1 (−L12 p¯0 + L21 p¯0) 3 3120 3210 2 2 1 1 = − (−L12 p¯0 + L21 p¯0) = − ω(L21 − L12) = −ωL21 = ωJ3.

2 2

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Summarizing, for a massless particle propagating on the z-axis with E = ω (natural units) we have found W0 = ωJ3

W1 = ω(J1 + K2)

W2 = ω(J2 − K1)

W3 = ωJ3,

for the components of the Pauli–Lubanski pseudovector. This is the result quoted in Eq. (15.28).

15.12 This solution is adapted from a discussion in Jones [125]. For lightlike particles the little group is the euclidean group E2. Let us take a set of E2 generators (L1 = J1 + K2, L2 = J2 − K1, J3) with the commutation relations [ L1, L2 ] = 0

[ L1, J3 ] = −iL2

[ L2, J3 ] = iL1.

The standard lightlike vector is given by Eq. (15.28): p¯µ = ω(1, 0, 0, 1). One can write for the state p¯µ corresponding to the standard vector, P0 p¯µ = ω p¯µ

P1 p¯µ = 0

P2 p¯µ = 0

P3 p¯µ = ω p¯µ .

We then make the following observations: 1. The standard vector p¯µ = ω(1, 0, 0, 1) is manifestly invariant under rotations exp(−iJ3φ) about the z-axis.

2. For the commutator of L1 with P0, [ L1, P0 ] = [ J1 + K2, P0 ] = [ J1, P0 ] + [ K2, P0 ] = −iP2, where Eq. (15.15) was used. But the state p¯µ corresponding to the standard vec tor is, by construction, an eigenstate of P2 with zero momentum in the 2-direction, P2 p¯µ = 0.

Thus, effectively [ L1, P0 ] vanishes in this space since [ L1, P0 ] p¯µ = −iP2 p¯µ = 0.

3. Again using Eq. (15.15), the commutators of L1 with Pi may be written [ L1, Pi ] = [ J1, Pi ] + [ K2, Pi ] = −iεi1kPk − iP0δ2i and we find upon utilizing P0 p¯µ = ω p¯µ

P1 p¯µ = 0

P2 p¯µ = 0

P3 p¯µ = ω p¯µ .

that [ L1, P1 ] p¯µ = − iε11kPk p¯µ = 0

[ L1, P2 ] p¯µ = i(P3 − P0) p¯µ = 0

[ L1, P3 ] p¯µ = −iP2 p¯µ = 0.

4. Likewise, for the commutators of L2 with Pi [ L2, Pi ] = [ J2, Pi ] − [ K1, Pi ] = −iεi2kPk + iP0δ1i, which gives [ L2, P1 ] p¯µ = i(P0 − P3) p¯µ = 0

[ L2, P2 ] p¯µ = −i(P0δ12 − ε22kPk) p¯µ = 0

[ L2, P3 ] p¯µ = iP1 p¯µ = 0.

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Poincaré Invariance

5. Finally, for the commutator of L2 and P0, [ L2, P0 ] p¯µ = [ J2, P0 ] − [ K1, P0 ] p¯µ = iP1 p¯µ = 0.

Collecting these results, the generators of E2 leave the standard lightlike vector invariant.

It follows that they generate the Poincaré little group for massless particles. Notice that the relations we have used to prove these results follow both from the properties of the commutators for the little group E2 and from the nature of the standard vector in this space; for example the commutator [ L1, P0 ] does not vanish for the general case, but it does vanish when acting on the standard lightlike vector. These results are not dependent on a particular choice of the standard vector in this space. Any lightlike vector would do, because any such vector can be transformed to our choice of standard vector by Lorentz transformations, which do not mix the classes of vectors listed in Table 15.1.

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Gauge Invariance 16.3 This solution follows a discussion in Quigg [166]. Poincaré invariance requires that the action be unchanged by a spacetime transformation xµ → x′µ = xµ + aµ , where the infinitesimal displacement aµ is independent of xµ . The corresponding change in a scalar field Lagrangian density is δ L = L (x′) − L (x) = aµ ∂µ L .

Assuming the Lagrangian density to be independent of coordinates x (translational invariance) but to depend on the field φ and its derivatives ∂µ φ, we may also write δ L = ∂ L δφ + ∂ L δ(∂ φ) µ ∂φ ∂(∂µ φ)

δφ ≡ φ(x′) − φ(x) = aµ∂ φ, µ

δ(∂µ φ) ≡ ∂µ φ(x′) − ∂µφ(x) = aν ∂ν (∂µ φ).

But from the Euler–Lagrange equations (16.14), ∂ L /∂φ = ∂µ ∂ L /∂(∂µ φ) . From this and the above equations δ L = ∂ν

∂L ∂(∂ν φ)

aµ ∂ µ φ +

∂L µ ∂L µ a ∂ µ (∂νφ) = ∂ν a ∂ µφ, ∂(∂ ∂(∂ν φ) ν φ)

where the validity of the last step may be checked by taking the derivative of the product, remembering that aµ does not depend on the coordinates. Equating this expression with the first equation above and doing some rearranging (repeated indices are dummy indices) leads to ∂L ∂ ν φ − η µν L = 0, aν∂µ ∂(∂µ φ) where η µν is the metric tensor. This relation must be satisfied for arbitrary displacements aν , implying that ∂L ∂ ∂ ν φ − η µν L = 0. µ ∂(∂µ φ) Define the stress–energy–momentum tensor Θµν ≡

∂L

∂ ν φ − η µν L , ∂(∂µ φ)

for which it may be shown that Θ00 is the energy density and the Θ0k are momentum densities. This leads to a local conservation law ∂µ Θµν = 0 89

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for energy and momentum. The quantity Θµν is an example of a conserved Noether tensor; in this case it links 4-momentum conservation with translational invariance.

16.4 Obviously the Lagrangian density (16.10) is unchanged by the global U(1) rotations ψ(x) → eiα ψ(x) and ψ(x) → ψ(x)e−iα , L0 = ψ (i∂/ − m)ψ = ψe−iα (i∂/ − m)eiαψ = ψe−iα eiα (i∂/ − m)ψ = ψ (i∂/ − m)ψ, since α does not depend on the spacetime coordinates for a global U(1) transformation.

The Noether current associated with this symmetry is given by Eq. (16.20). Comparing U = eiα with a general U(1) group element U = eiJα indicates that the generator for the single Noether current can be taken to be J = 1 and ∂ L0 ∂ L0 Jµ = −i Jψ = −i (1)ψ = ψγµ ψ, ∂(∂ µ ψ) ∂(∂ µ ψ) where we have used ∂/ ≡ γµ ∂ µ and ∂ ψ(iγµ ∂ µ

∂ L0

m)ψ − −i = −i = ψγµ . ∂(∂ µ ψ) ∂(∂ µ ψ) The current Jµ is conserved, ∂ µ Jµ = 0, implying that the charge defined in Eq. (16.21) is a constant of motion.

16.5 This solution follows a problem worked in Ref. [85]. Express the two scalar fields in terms of a column vector, φ1 . φ2

φ≡ Then utilizing Eq. (6.3), φ ′ = Rφ =

cosα sin α

φ1 ≃ φ2

−sinα cosα

1 α

−α 1

φ1 , φ2

which implies that to order α under SO(2) rotations, φ2 → φ2 + αφ1.

φ1 → φ1 − αφ2

Substituting in Eq. (16.27), under infinitesimal SO(2) rotations the Lagrangian density transforms as L → L + O α2 and (16.26) is invariant under SO(2) transformations.

From above the infinitesimal SO(2) rotation operator is R≃

1 α

−α 0 = 1+ 1 α

−α 0

.

Upon comparison with Eq. (6.6), we conclude that the infinitesimal SO(2) generator L is L=

0

i

−i

0

.

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Inserting this generator in Eq. (16.20), the current associated with the SO(2) invariance of the Lagrangian density (16.26) is Jµ = −i

∂L ∂(∂ µ φ i)

Li jφ j = (∂µ φ1)φ2 − (∂µ φ2)φ1,

which is conserved by virtue of Noether’s theorem.

16.6 The Lagrangian density (16.30) L = (∂µ φ)†(∂ µ φ) − m2(φ†φ) − λ(φ † φ) 2 , is clearly invariant under the U(1) rotations φ → eiα φ and φ † → φ†e−iα . Inserting the infinitesimal transformations φ → φ + iα

φ † → φ † − iα

in Eq. (16.30) and utilizing Eq. (16.20), the conserved Noether current associated with the U(1) invariance is ∂L ∂L (−1)φ† = i (∂µ φ)φ † − (∂µ φ†)φ , Jµ = −i (1)φ − i µ ∂(∂ φ) ∂(∂µ φ†) which is equivalent to Eq. (16.28) upon inserting the definitions 1 1 † φ = √ (φ1 + iφ2) φ = √ (φ1 − iφ2).

2 2 from Eq. (16.29).

16.8 The solution of this problem may be found in many books. This solution follows a discussion in Ref. [88]. Consider paths in spacetime between the fixed points A and B and introduce the integral ∫ B

S=

L (xµ (σ), x˙µ (σ)) dσ,

A

where in the Lagrangian L xµ (σ), x˙µ (σ) the path is parameterized by σ and x˙µ ≡ dx µ /dσ.

For a small variation in the path xµ (σ) → xµ (σ) + δ xµ (σ), the change in the value of the integral is δS ≡

∫ B

δ Ldσ =

∫ B

A

∂L ∂L µ δ x˙µ (σ) + δ x (σ) dσ. µ ∂ x˙µ (σ) ∂ x (σ)

A

Integrate the first term by parts to give δS =

B ∂ L δ x (σ) µ + µ ∂ x˙ (σ) A

∫

B

A

— d dσ

∂L ∂ x˙µ (σ)

+

∂L ∂ xµ (σ)

δ x µ (σ)dσ.

But δ xµ (σ) vanishes at the fixed endpoints so the first term in this expression is zero and the variation in S is δS =

∫ B A

—

d dσ

∂L ∂ x˙µ (σ)

+

∂L ∂ xµ (σ)

δ x µ (σ)dσ.

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Paths that extremize the integral satisfy δ S = 0, which is true generally only if the expression inside the square brackets of the above integrand vanishes for arbitrary variations δ xµ (σ). Thus we obtain d — dσ

∂L ∂ x˙µ (σ)

+

∂L = 0, ∂ xµ (σ)

which is the Euler–Lagrange equation (16.14). We have shown that satisfying the variational condition δ S = 0 characterizing an extremal path is equivalent to requiring that L satisfy the Euler–Lagrange differential equation.

16.12 The gauge transformation is implemented by U (θ ) = exp(−iτ · θ ) ≃ 1 − iτkθ k.

Then from Eq. (16.51) we may write the gauge-transformed vector potential as A′µ = τ jA′ jµ, where i τj A′µj = U τj Aµj U −1 − U ∂µU −1 g i k = (1 − iτkθ )τ jA jµ(1 + iτ kθ k) − (1 − iτ θk k)∂ µ(1 + iτ kθ k). g For the first term, multiplying it out and omitting all terms of order θ 2, (1 − iτkθ k)τ j A j µ(1 + iτkθ k) = τ j A jµ + iAµj θ k(τ jτk − τkτ j) = τ j A j + iAj θ k[ τ j, τk ] µ

µ

= τ j Aµ − A µθ k f jklτl, j

j

where we’ve used [ τ j, τk ] = i f jklτl, with f jkl a structure constant for the group. For the second term, noting that ∂µ (1) = 0 and discarding terms of order θ 2, 1 i k k — (1 − iτkθ k)∂µ (1 + iτkθ ) = τk∂µ θ . g g Putting this together, 1 τ jA′µj = τ jA jµ + τ k∂µ θ k − Aµj θ k f jkl τl . g Multiplying from the left by τi and taking the trace of both sides gives 1 Tr(τ iτ j )A′µj = Tr(τ iτ j )Aµj + Tr( τi τk )∂µ θ k − Aµj θ k fjkl Tr(τ iτ l ), g which becomes, upon using the normalization Tr(τaτb) = 2δab and that f jki is antisymmetric under interchange of indices, 1 1 A′i = Ai + ∂ θ i + f A j θ k = Ai + ∂ θ i + f Al θ k, ik j µ ikl µ µ µ µ g µ g µ where in the last step the summation index j has been replaced with l. This is Eq. (16.52) if the index i is replaced with j.

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16.13 Suppose that the current Jµ has vanishing 4-divergence: ∂µ Jµ ≡ ∂0J0 + ∇ · J = ∂0J0 + ∂kJk = 0.

Then from Eqs. (16.23) and (16.21), Q˙ = ∂0

∫

d 3x J0 =

∫

d 3x ∂0J0 = −

∫

d 3x ∇ · J.

But the divergence theorem implies that for a 2D surface S bounding a volume V , ∫

d 3x ∇ · J =

V

∫

J · n ds, S

where n is an outward normal to S and ds is a surface differential element. Therefore, Q˙ = —

∫

J · n ds = 0,

S∞

since by hypothesis no currents flow on S at infinity. Hence, from Eq. (16.23), [ H, Q ] = Q˙ = 0 if ∂µ Jµ = 0, and Q is conserved.

16.14 The Lagrangian density L0 of Eq. (16.16) is invariant under the global isospin rotation (16.17) because αa and τa are independent of the spacetime coordinates. Therefore, Noether’s theorem implies that there is a conserved quantity associated with this continuous symmetry. Invariance of the Lagrangian density under the infinitesimal global isospin rotation (16.18) implies that ∂ L0 ∂ L0 δψ + δ(∂ ψ) = 0, δ L ≡ ∂ L0 δψ + ∂ L0 δ(∂ ψ) + 0

∂ψ

∂(∂ µ ψ)

µ

∂ψ

∂(∂ µ ψ)

µ

where from Eq. (16.18), i i δψ = τaαa ψ δ(∂µ ψ) = τaα a(∂µ ψ).

2 2 In addition, the Dirac equation (16.11) and the Lagrangian density (16.16) imply that ∂ L0 ∂ L0 = (i∂/ − m)ψ = 0 ∂(∂µ ψ) = 0. ∂ψ Collecting these results and utilizing a generalization of Eq. (16.14) gives i ∂ L0 i ∂ L0 a ∂ τa α ψ + τ α (∂ ψ) = 0, a a µ 2 µ ∂(∂µ ψ) 2 ∂(∂µ ψ) which is equivalent to ∂µ αa Jµ = 0 a

L 1 i ∂ 0 a Jµ ≡ − τ ψ = ψγ µ τ a ψ, 2 2 ∂(∂µ ψ) a

where Jµa is the isospin current. Since ∂µ αaJaµ = 0 must be satisfied for arbitrary constant αa, we conclude that the isospin current has vanishing 4-divergence, ∂ µ Jaµ = 0, and therefore is conserved.

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Spontaneous Symmetry Breaking 17.1 The condition U |0⟩ = |0⟩ implies that U |0⟩ = ei(L1θ1+L2θ2+...+LN θN ) |0⟩ = |0⟩, where U denotes a group element and Li a group generator. This can be satisfied for arbitrary θi only if each generator annihilates the vacuum state: Li |0⟩ = 0 (i = 1, 2,..., N).

Thus, spontaneous symmetry breaking corresponding to U |0⟩ /= |0⟩ implies that at least one generator fails to annihilate the vacuum (ground state).

17.3 From Eq. (17.14) a potential may be identified as 1 1 2 V = µ2 φ 21+ φ 22 + φ 32 + λ φ 21 + φ22 + φ32 .

2 4 From the first and second derivatives the potential has a Wigner-mode minimum when φ1 = φ2 = φ3 = 0 with µ2 > 0, or a Goldstone-mode minimum when r q −µ 2 2 2 2 ≡ v, |φ| = φ1 + φ2 + φ3 = λ with µ2 < 0. For the Goldstone mode the degenerate vacua lie on a sphere of radius v.

Choose a vacuum state ⟨φ⟩0 = v εˆ3, where εˆ3 is a unit vector. Because of the spontaneous symmetry breaking the ground state is no longer SO(3) invariant but it remains invariant under SO(2) rotations about the 3-axis.

Introducing a shifted field χ through φ3(x) = χ(x) + v, and rewriting the Lagrangian in terms of the physical fields φ1, φ2, and χ allows us to identify masses by inspection of the quadratic Lagrangian terms: a massive scalar with mχ = (2λ v2)1/2 and two Goldstone bosons with mφ 1 = mφ 2 = 0.

94

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The Higgs Mechanism 18.1 Choosing µ2 < 0 to break the symmetry spontaneously in Eq. (18.1), the absolute minima (degenerate vacua) occur at |φ|2 = −

µ2 v2 ≡ .

2λ 2

Choose a ground state v ⟨φ⟩0 = √ , 2 where v is real and positive, expand in polar coordinates, 1 1 iξ(x)/v φ(x) = √ v + η(x) e = √ v + η(x) + iξ (x) + . . . , 2 2 and substitute in Eq. (18.1) to give L ≃

1 µ 1 (∂µ η)(∂ µ η) + µ2η2 + (∂ ξ)(∂µ ξ) 2 2 1 1 + qvAµ(∂ µ ξ) + q2v2Aµ Aµ − Fµν F µν + . . . , 2 4

where higher-order terms were discarded. This looks like the Lagrangian density corresponding to three fields: η, ξ , and Aµ . The η field seems to have a mass [compare Eq. (16.4)] √ mη = −2µ2, as implied by the term µ2η2, the photon appears to have a mass mA = qv, as implied by the term 12 q2v2Aµ Aµ [compare Eq. (16.13)], and the ξ field seems to be massless. However, in manipulating gauge theories it is particularly important to keep track of the number of degrees of freedom because particular choices of gauge can obscure the physical content of a theory. Originally we had two degrees of freedom from a complex scalar field plus two degrees of freedom from the transverse components of the photon field, for a total of four degrees of freedom. After the spontaneous symmetry breaking we have one degree of freedom from the real scalar field η, one from the massless ξ field, and three for the massive vector field, for a total of five degrees of freedom. The seeming acquisition of a new degree of freedom is an illusion that can be eliminated by an 95

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appropriate choice of gauge. Implementing the local gauge transformation v + η(x) φ(x) → e−iξ(x)/v φ(x) = √ , 2 1 Aµ (x) → A µ(x) + ∂µ ξ(x) ≡ A′µ(x), qv and dropping the primes on Aµ and Fµν, the Lagrangian density takes the form 1 1 1 (∂µ η)(∂ µ η) + µ 2 η 2 + q2v2Aµ Aµ − Fµν F µν .

2 2 4 Now the particle spectrum exhibits the original four degrees of freedom, but re-distributed √ 2 among a scalar particle η with mass −2µ (one degree of freedom) and a massive vector field Aµ with mass qv (three degrees of freedom), while the ξ field has disappeared. This is the Higgs mechanism in action: spontaneous breaking of a local gauge symmetry has given the photon an effective mass, without producing massless Goldstone bosons.

L =

18.2 Let’s take the 4-divergence of both sides of Eq. (18.11). The right side gives ∂ν Jν and the left side gives ∂ν (∂µ ∂ µ )Aν + m2Aν − ∂ν(∂µ Aµ ) = ∂µ ∂ µ ∂ν Aν + m2∂ν Aν − ∂ν∂ν ∂µ Aµ = ∂µ ∂ µ ∂ν Aν + m2∂ν Aν − ∂µ∂ µ ∂ν Aν = m2∂ν Aν , where in the second line we have used that repeated indices are dummy indices that may be replaced with any indices not already in use. Equating the divergence of the right and left sides then gives m2∂ν Aν = ∂ν Jν = 0, where the last step follows from ∂ν jν = 0 in Eq. (14.13). Therefore, ∂ν Aν = 0.

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The Standard Model 19.2 From Eq. (16.10) an explicit Dirac mass term is of the form mψψ. Thus, decomposing ψ into right and left chiral components, ψ = R + L, we have ψψ = R¯R +R¯L + L¯R + L¯L = L¯R +R¯L, where we have used R¯R = L¯ L = 0 because, for example, 1 R = (1 + γ5)ψ 2

¯ 1 R = ψ(1 − γ5)ψ 2

[see Problem 14.8], so that 1 1 R¯R = ψ¯ (1 − 5γ )(1 + γ5 )ψ = ψ¯ (1 − γ52)ψ = 0, 4 4 where γ52 = 1 has been used. But in the standard electroweak theory R is an SU(2) singlet and L is an SU(2) doublet, so ψψ = L¯ R + R¯L cannot be an SU(2) invariant. Inserting Eq. (19.9) in Eq. (19.13) and carrying out the matrix multiplications gives Lf–s = −Ge e¯R(φ−νL + φ¯0eL) + (ν¯Lφ+ + e¯Lφ 0)eR .

Choosing unitary gauge (19.19), φ+ = φ− = 0

1 φ 0 = √ (v + η), 2

then gives Ge Ge Lf–s = −√ v(e¯ReL + e¯LeR) − √ η(e¯ReL + e¯LeR) 2 2 Gev Ge = − √ e¯e − √ ηe¯e, 2 2 where e¯e = e¯ReL + e¯LeR has been used. This is Eq. (19.20).

19.3 From Eq. (19.25) the Weinberg angle is defined by tanθ = sinθW = g′ , W cosθW g which implies that g′ = g

sinθW cosθW

g = g′

cosθW . sinθW

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Therefore we have that

s √

2 g2 + g′2 = g2 + g2 sin 2 θW cos θW s cos2 θW + sin2 θW =g cos2 θW g′ g = , = cosθW sinθW

where g = (cosθW/ sinθW)g′ was used in the last step. Thus, from the first expression in Eq. (19.23) −g′aµ + gb3µ Zµ0 ≡ √ g2 +√g′2 √ −sinθW g 2 + g ′2 a µ + cosθW g 2 + g ′2 b3 µ √ = g2 + g′2 = −sinθWaµ + cosθWb3µ and from the second expression in Eq. (19.23) gaµ + g′b3µ Aµ ≡ √ g2 +√g′2 √ cosθW g2 + g′2 aµ + sinθW g2 + g′2 b3 µ = √ 2 ′2 g +g = cosθWaµ + sinθWb3µ, which corresponds to Eq. (19.26).

19.6 By analogy with Example 4.3, the Young operator for the SU(3) color singlet is 1

Y 2

= 1 − (12) − (13) − (23) + (123) + (321).

3

This gives a normalized state ψ=Y

1 2 3

1 |rgb⟩ = √ (rgb − grb − bgr − rbg + brg + gbr), 6

where r ≡ 1, g ≡ 2, and b ≡ 3 are the three possible SU(3) color states. Using the methods of previous examples and assuming gluons to transform as the color SU(3) 8, the 2-gluon

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state GG has the color SU(3) decomposition GG = 8 ⊗ 8 = =

⊗ ⊕

⊕

⊕

⊕

⊕

= 27 ⊕ 10 ⊕ 1 ⊕ 8 ⊕ 8 ⊕ 10, which contains a color singlet (1). A 3-gluon state GGG has the color decomposition GGG = (GG) ⊗ G =

⊕

⊕

⊕

⊕

⊕

⊗

= (27 ⊗ 8) ⊕ (10 ⊗ 8) ⊕ (1 ⊗ 8) ⊕ (8 ⊗ 8) ⊕ (8 ⊗ 8) ⊕ (10⊗ 8), which we already know contains a color singlet since we saw above that 8 ⊗ 8 contains a singlet.

19.7 Two-quark SU(3) states allow symmetric and antisymmetric combinations: q⊗q = 3⊗3 =

⊕ symm.

antisymm.

= 3 ⊕ 6.

To determine the corresponding wavefunctions we must construct all possible standard arrangements of Young tableaux, as described in Section 4.2, which are Symmetric :

1 1

1 2

1 3

2 2

Antisymmetric :

1 2

1 3

2 3

3 3 ,

2 , 3

where boxes represent particles and numbers represent states (for this example there are two particles with three states available). A row of boxes is an injunction to symmetrize in the particle labels and a column of boxes implies antisymmetrization. After replacing the state numbers with color labels (1 = r, 2 = b, 3 = g) the color wavefunctions are Symmetric :

rr, rb + br, rg + gr, bb, gb + bg, gg,

Antisymmetric :

rb − br, rg − gr, bg − gb,

where we have omitted normalization factors. Now for SU(3), 3 ⊗ 3 = 8 ⊕1 and 6 ⊗3 = 10 ⊕8. Hence, to produce a qqq color singlet the qq in the antisymmetric 3 must combine with the remaining q (in a 3). Thus each pair of quarks is in an antisymmetric state and the non-color part of the two-quark wavefunction in a baryon must be symmetric.

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Dynamical Symmetry 20.2 Consider the spin operator Sy. From Eqs. (20.7a) and (3.11), Sy = ∑ ∑ σ ′ σy |σ⟩ c†τσ ′ m cτσ mk k

mk τσσ′

= ∑ ∑ ⟨↑|σy |↑⟩ c†τ↑mk cτ↑m + ⟨↑|σy |↓⟩ c†τ↑m cτ↓m k

mk τ=±

k

k

+⟨↓|σy |↑⟩ cτ†↓m cτ↑mk + ⟨↓|σy |↓⟩ c†τ↓m cτ↓mk k

=∑ mk

k

† c+↓m k + ic†+↓m c+↑m k — ic †−↑m c−↓m k + ic†−↓m c−↑m k −ic+↑m k k k k

= −iB12 + iB21 − iB34 + iB43, where in line two we have used the representations σy =

0 −i i 0

σ = |↑⟩ =

1 0

σ ′ = |↓⟩ =

0 1

to show that ⟨↑|σy |↑⟩ = 0

⟨↑|σy |↓⟩ = −i

⟨↓|σy |↑⟩ = i

⟨↓|σy |↓⟩ = 0,

and we’ve used equivalences between the indices a and (τ, σ) in the table of Fig. 20.5(a) to map to indices for Bab. For example, we find

∑c† mk

c

+↑mk +↓mk

= ∑c† c mk

1mk 2mk

= B12,

from Eq. (20.9) and the mappings (τ, σ) = (+, ↑) ↔ a = 1 and (τ, σ) = (+, ↓) ↔ a = 2. Proceeding as in the above example, we find that the spin operators (20.7a) may be expressed as Sx = B12 + B21 + B34 + B43, Sy = −i (B12 − B21 + B34 − B43), Sz = B11 − B22 + B33 − B44, the valley isospin operators (20.7b) as Tx = B13 + B31 + B24 + B42, Ty = −i (B13 − B31 + B24 − B42), Tz = B11 + B22 − B33 − B44, 100

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The Néel vector of Eq. (20.7c) as 1

Nx =

(B12 + B21 − B34 − B43), i Ny = − (B12 − B21 − B34 + B43), 2 1 Nz = (B11 − B22 − B33 + B44), 2 and the operators Παβ of Eq. (20.7d) as 2

Πxx =

1

(B14 + B41 + B23 + B32), i Πyx = − (B23 − B32 + B41 − B14), 2 1 Πzx = (B13 + B31 − B24 − B42), 2 i Πxy = − (B32 − B23 + B41 − B14), 2 1 Πyy = − (B41 + B14 − B23 − B32), 2 i Πzy = − (B31 − B13 − B42 + B24), 2 in terms of the multipole operators Bab defined in Eq. (20.9).

2

20.3 Net spin is measured by ⟨Sz⟩, which characterizes ferromagnetic order, the difference in charge between the A and B sublattices is measured by ⟨Tz⟩, which characterizes charge density wave order, and the difference in spins between the A and B sublattices is measured by ⟨Nz⟩, which characterizes antiferromagnetic order. The order parameters are [223] ⟨Sz⟩ = ⟨nˆ1⟩ − ⟨nˆ2⟩ + ⟨nˆ3⟩ − ⟨nˆ4⟩ ⟨Tz⟩ = ⟨nˆ1⟩ + ⟨nˆ2⟩ − ⟨nˆ3⟩ − ⟨nˆ4⟩ ⟨Nz⟩ = ⟨nˆ1⟩ − ⟨nˆ2⟩ + ⟨nˆ4⟩ − ⟨nˆ3⟩ The order parameters for the states in Fig. 20.6 worked out using these formulas are summarized in Table 20.1 [this document].

20.4 As a representative example, from Eq. (20.11) the S = 1, MS = 0, T = 0 pair is, A0010 = ∑ ∑ ∑ †

mk m1m2 n1n2

1

2

m1 21 m2 10⟩

1 n 1 2 1 2 n2

00⟩ c†m1n1mk cm2 n2−mk . †

But from the result of Problem 6.5 1

1

2 n1 2 n2 00⟩ =

(−1)1/2−n1 √ δn1,−n2 , 2

so that (−1)1/2−n1 √ 2 mk m1 m2 n1

A0010 = ∑ ∑ ∑ †

1

1 2 m1 2 m2

10⟩ c†m1 n1mk c†m2−n1−mk .

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Table 20.1 Order parameters for states in Fig. 20.6 State

⟨Sz⟩

⟨Tz⟩

⟨Nz⟩

|S⟩ = S† |0⟩

0

0

0

†

0

†

0

|D2⟩ = D2 |0⟩

0

2

0

|D−1⟩ = D−1 |0⟩

−2

0

0

†

|D1⟩ = D†1 |0⟩ † |D0⟩ = D0 |0⟩ 1 |Q+⟩ = 2 (|S⟩ + |D0⟩) 1 |Q−⟩ = 2 (|S⟩ −|D0⟩)

2

0

0

0

0

0

0

0

1

0

0

−1

The remaining Clebsch–Gordan coefficient vanishes unless m2 = −m1, so

(−1)1/2−n1 √ 2 mk m1n1

†10 A00 =∑ ∑

1

1 2 m1 2 − m1

10⟩ c†m 1 n 1 m k c†−m 1 −n 1 −m k .

Writing the four terms in the sum over m1 and n1 out explicitly for n1 = (− 1 , 1 ) and 2 2 m1 = (− 1 , 1 ) gives 2 2

†

1 † † 1 1 1 1 2 − 2 2 2 10⟩ c− 1 − 1 m c 1 1 −m k 2 m∑ 2 2 22 k k 1 † † 1 + √ ∑ 2 − 21 21 21 10⟩ c− 1 1 mk c 1 − 1 −mk 22 2 2 2 mk 1 — √ ∑ 12 12 12 − 21 10⟩ c †1 − 1 m c†− 1 1 −m 2 2 k 2 2 k 2 mk 1 † † 1 1 1 1 + √ ∑ 2 2 2 − 2 10⟩ c 1 1 mk c− 1 − 1 −mk .

22 2 2 2 mk

A0010= −√

Utilizing from Table C.3 that

1 1 1 1 2 − 2 2 2 10⟩ =

1 1 1 1 1 2 2 2 − 2 10⟩ = √

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we may write †

A 10 = − 00

1

1

∑c†— 1 − 1 mk c†1 1 −mk + 2 ∑c†— 1 1 mk c†1 − 1 −mk

2m 2 2 2 2 22 2 2 mk k 1 1 † † † † — ∑ c 1 − 1 m c− 1 1 −m + ∑ c 1 1 m c− 1 − 1 −m 2 k k 2 k k 22 2 2 mk 2 2 mk 2 2 1 1 1 = − ∑c† c† + ∑c† c† − 1 ∑c† c† + ∑c† c† 4mk 1−mk 2mk 3−mk 3mk 2−mk 2 mk 2 mk 2 mk 1mk 4−mk 2 mk 1 † 1 † 1 † 1 † = A14 + A23 + A23 + A14 2 2 2 2 † , = A†14 + A23

where in the second equation the mapping between spin and isospin quantum numbers and the label a in Fig. 20.5(a) has been used to replace labels (σ, τ) with the label a, we have performed manipulations such as 1 1 1 1 † † † † † † † — ∑c4mk c1−mk = ∑c1−mk c4mk = ∑ c1mk c4−mk = A14, 2 2 2 2 mk

−mk

mk

where the first equality is because independent fermion creation operators anticommute, the second equality is because mk is a dummy summation index that may be replaced with † another summation index, and the third equality employed the definition of Aab in Eq. (20.8). The other five possibilities may be determined in a completely analogous way, with the results √ √ † † † A 10 = A† + A† A 10 = 2 A† A 10 = 2 A† , 00

†01

A00

14

23

13

−10

A01 = 2 A†12

†01

10

†01

= A14† − A†23

√

√

24

A0−1 = 2 A†34 .

The hermitian conjugates of these give the six corresponding pair annihilation operators in coupled representation. These are the generators given in Eq. (20.12), up to a normalization. † 20.6 From Eq. (20.9) and using nˆi ≡ ∑mk cim c − 14 Ω, k imk

B =

—1 Ω = n , 1 ∑ 1mk 1mk 4 mk B = c† c — 1 Ω = n , c† c

11

22

B = 33

B = 44

∑ 2mk 2mk mk

c† c

2 4 —1 Ω = n ,

c† c

3 4 —1 Ω = n .

∑ 3mk 3mk mk

∑ 4mk 4mk mk

4

4

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Then, utilizing the results of Problem 20.2, Sz = B11 − B22 + B33 − B44 = n1 − n2 + n3 − n4, Tz = B11 + B22 − B33 − B44 = n1 + n2 − n3 − n4, 1 1 Nz = (B11 − B22 − B33 + B44) = (n1 − n2 − n3 + n4), 2 2 which may be compared with Eq. (20.15).

20.8 This is a straightforward solution of Eq. (20.18) using the mapping in Fig. 20.5(a) between the Bmj −ml defined in Eq. (20.19) and the Bab defined in Eq. (20.9). As an example, let’s compute P13. From Eq. (20.18), 3

P13 = ∑ (−1) 2 +ml

3

3

2 m j 2 ml

31⟩ Bmj −ml

mjml

= (−1)3/2−1/2 32 32 32 − 1 2 31⟩ B3/2,1/2 + (−1)3/2+1/2 32 12 32 12 31⟩ B1/2,−1/2 + (− 1)3/2+3/2 23 − 12 32 32 31⟩ B−1/2,−3/2 r r r 1 3 1 =− B12 + B23 — B34 , 5 5 5 where we have used the restriction that m j + ml = 1, the values of the Clebsch–Gordan coefficients r r 1 3 3 3 3 1 3 1 3 3 3 1 3 1 − 2 31⟩ = 2 − 2 2 2 31⟩ = 31⟩ = 2 2 2 2 2 2 2 5 5 and the mappings B3/2,1/2 ↔ B12

B1/2,−1/2 ↔ B23

B−1/2,−3/2 ↔ B34

inferred from the table in Fig. 20.5(a). Doing all in the same way yields r r 1 9 1 0 1 P0 = (B11 + B22 + B33 + B44) P0 = (B11 − B44) + (B22 − B33), 2 20 20 r r r r r r 4 B — 3 B 4 3 3 3 1 1 — P1 = − B12 P−1 = B21 + B32 + B43 , 23 34 10 10 10 10 10 10 1 1 2 2 P0 = (B11 − B22 + B44 − B33) P1 = √ (B34 − B12), 2 2 1 1 1 2 2 2 P−1 = √ (B21 − B43) P2 = −√ (B13 + B24) P−2 = −√ (B31 + B42), 2 2 2 r r r r r 1 9 1 3 1 3 3 P0 = (B11 − B 44 ) + (B33 − B22 ) P1 = − B12 + B23 − B34 , 20 20 5 5 5 r r r r 1 3 1 1 3 3 21 32 43 2 P−1 = 5 B − 5 B + 5 B P = 2 (B24 − B13 ), r 1 3 P3 = (B — B31 ) P33 = −B14 P−3 = B41, −2 2 42 with Bab defined in Eq. (20.9).

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Generalized Coherent States 21.4 Since the Lie algebra is equivalent to that of angular momentum SU(2), we expect that the states of the Hilbert space are labeled by two quantum number | jm⟩, with J2 | jm⟩ = j( j + 1) | jm⟩

J3 | jm⟩ = m | jm⟩,

where we assume h¯ = 1 units (see Appendix B) and where from Eq. (3.18) 1 2 2 J = (J+J− + J−J+) + J3 .

2 Let’s introduce a vacuum state |0⟩ corresponding to no fermions. The anticommutator algebra given in Problem 21.3 leads to the Pauli principle requirement a†a† |0⟩ = a†a†a† |0⟩ = a† . . . a† |0⟩ = 0, which means that only two states are permitted in the Hilbert space: • the zero-fermion vacuum state |0⟩, and • the one-fermion state a †|0⟩.

Thus, the Hilbert space has two degrees of freedom and may be described by the j = 21 two-dimensional representation of SU(2). This may be verified explicitly. Utilizing the commutators and SU(2) mapping found in Problem 21.3, 1 2 2 J = (J+J− + J−J+) + J3 2 1 1 1 3 † = (a †a + aa †) + a† a − = . a a− 2 4 2 2 Therefore, j( j + 1) = 3 , implying that j = 1 [doublet SU(2) representation] for all states 4

2

in the Hilbert space. Likewise, the action of J3 on these states is easily computed, 1 1 |0⟩ = n |0⟩ = — | 0⟩ 2 2 1 † 1 J3 (a† |0⟩) = na† |0⟩ = a†a − a |0⟩) = (a † |0⟩).

2 2 J3 |0⟩ = a†a −

We conclude that the Hilbert-space states may be labeled by the ( j, m) quantum numbers: 1 1 11 † 2 − 2 = |0⟩ and 2 2 = a |0⟩. The action of the J± operators on these states is then J+ 12 − 12 = a† |0⟩ = 12 12 J− 12 − 12 = a |0⟩ = 0

J+ 12 12 = a†a† |0⟩ = 0, J− 12 12 = aa† |0⟩ = 21 − 12 ,

which are the actions expected for SU(2) raising and lowering operators applied to states in the j = 21 irrep of SU(2) [see Eqs. (3.19)–(3.20)].

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21.5 We take as a reference state the SU(2) minimal-weight state |21 − 12⟩, which is the fermion vacuum. As shown in Problem 21.4, the actions of the generators on this state are J+ 21 − 12 = 12 12

J− 12 − 12 = 0

J3 21 − 12 = − 12 21 − 12 ,

so only J3 leaves the reference state invariant and the stability subgroup H is U(1), with the single generator J3 and group elements h = eiJ3 λ ∈ H. The coset space is then G/H = SU(2)/U(1).

21.6 The dynamical group is SU(2), with group elements g ∈ G of the form ∗

∗

g = ei(J3λ+J+ξ−J−ξ ) = ei[(a a−1/2)λ+ξ a −ξ a]. †

†

From Problem 21.5 the coset space is G/H = SU(2)/U(1), with elements h = eiJ3 λ of the stability subgroup H. By comparing these expressions for g and h, the condition that the cosets yield a unique decomposition g = Ωh requires a coset representative to take the form ∗

∗

Ω(ξ) = eξ J+−ξ J− = eξ a −ξ a, †

∗

since then Ωh = ei(J3 λ+J+ξ−J−ξ ) = g.

21.7 The coherent state is obtained from the action of the coset representative given in Problem 21.6, ∗

∗

Ω(ξ) = eξ J+−ξ J− = eξ a −ξ a, †

on the reference state 21 , − 21 of Problem 21.5, ∗

| j ξ⟩ = 21 ξ = Ω(ξ) 21 − 21 = e(ξa −ξ a) 12 − 12 . †

Since the first form of Ω(ξ) in the first equation above is identical to Eq. (21.28), we may appropriate the properties of the coset space already worked out for the SU(2) example of atoms interacting with a classical radiation field in Section 21.4. The topology of the coset space (and thus of the coherent-state space) is S2. Parameterizing the unit 2-sphere in terms of the angular variables θ and φ defined in Eq. (21.32) gives |ξ | = 21 θ, and from (21.31) θ θ eiφ sin 2 θ2 . θ −iφ sin cos e — 2 2 Therefore, the coherent states | j, ξ⟩ may be expressed in terms of two angular variables defined on S2 , Ω(ξ) =

1 ,ξ 2

cos

= 21 θ φ = e−iφ sin

θ 2

11 22

+ cos

θ 2

1 − 12 2

,

with the ranges of θ and φ given in Eq. (21.32).

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Restoring Symmetry by Projection ′

22.2 Solving δ ⟨ΨBCS| Hˆ ′ |ΨBCS⟩ = 0 with respect to the variational parameter λ gives D ′ ′ E ′ E ′ E, d D λ′ d nD λ ′ ′ Ψλ λ λ λ ˆ ˆ ˆ H = H Ψ − λ Ψ N Ψ Ψ Ψ BCS BCS BCS BCS BCS BCS dλ ′ dλ ′ D d D λ′ ′ ′ E ′ E d λ = ′ Ψ BCS Hˆ ΨλBCS − λ Nˆ ΨλBCS Ψ BCS dλ dλ ′ dE dN = −λ ′ dλ ′ dλ = 0, where we have used D λ′ ′ E λ =E Ψ BCS Hˆ ΨBCS

D ′ ′ E λ ΨλBCS Nˆ ΨBCS = N,

with E the energy and N the average particle number. Therefore, dE dλ

′

=λ

dN dλ

−→

′

λ=

dE dN

,

so the chemical potential λ measures the energy change accompanying a change in the particle number.

22.4 Utilizing Eqs. (22.29), the anticommutator between quasiparticle creation and annihilation operators is †

†

†

{ αk, αk′ } = αk αk′ + αk′ αk † = uk ck − v kc †−k u k′ ck†′ − vk′ c−k′ + uk′ ck†′ − vk′ c−k′ u kc k — vkc −k † † † = u u ′ { c , c } − u v ′ { c , c ′ } − u ′v { c † , c } + v v ′ { c , c ′ } k k

k

k′

k k

k

−k

k k

−k

k′

k k

−k

−k

= (ukuk′ + vkvk′ )δkk′ = δkk′ , where in going from line three to line four we have used the bare-particle anticommutation relations †

{ ci , c j } = δ i j

†

†

{ ci , c j } = { ci , c j } = 0

and in the last line u2 + v2 = 1 was used. Likewise, for the quasiparticle anticommutator k

k

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108

{ αk , αk′ }, { αk , αk′ } = αkαk′ + αk′ αk † † † = uk ck − vk c†−k uk ′ ck ′ − vk′ c−k uk ck − v kc −k ′ + u ′c ′ − v ′c ′ k k k −k † † = u u ′{ c , c ′ } − u v ′ { c , c ′ } − u ′v { c † , c ′ } + v v ′ { c , c† ′ } k k

k

k

k k

k

−k

k k

−k

k

k k

−k

−k

= −ukvk′ δk,−k′ − uk′ vkδk′,−k = 0, where in the last step we have used that k and k′ are by definition positive, so k = −k′ and k′ = −k cannot be satisfied. Likewise, { α†k, α†k′} = 0 by a similar proof, or by taking the hermitian conjugate of { αk , αk′ } = 0. Thus the Bogoliubov quasiparticles defined by the transformation (22.29) satisfy Eq. (22.30) and obey Fermi–Dirac statistics, just as the bare fermions do.

22.7 Consider ⟨x|Π xˆΠ |ψ⟩, where |ψ⟩ is a generic state. Note that in the following derivations we assume the operators to always act on the left state. ⟨x|Π xˆ Π|ψ⟩ = ⟨−x| xˆ Π|ψ⟩ = −x ⟨−x|Π|ψ⟩ = −x ⟨x|ψ⟩ = −⟨x| xˆ|ψ⟩ .

This result implies that Π xˆΠ = −xˆ, or equivalently (since Π−1 = Π), Π xˆ−xˆΠ = −2 xˆΠ /= 0. Therefore, Π does’t commute with xˆ (they anti-commute). Now consider 2

2

⟨x|Π xˆ Π|ψ⟩ = ⟨−x| xˆ Π|ψ⟩ = x2 ⟨−x|Π|ψ⟩ = x2 ⟨x|ψ⟩ = ⟨x| xˆ2 |ψ⟩, which indicates that Π xˆ2 Π = xˆ2, or equivalently Π xˆ—2 xˆ2 Π = 0, so the parity operator commutes with xˆ2.

22.8 The energy eigenfunctions of a quantum system with a central force (such as the hydrogen atom) are ψnlm(r, θ, φ) = Rnl(r)Ylm(θ, φ), which we abbreviate as |nlm⟩. The matrix element for electric dipole transitions is n′l′m′ r |nlm⟩.

When applying the parity operator to |nlm⟩ the radial function Rnl(r) is invariant under r → −r but the angular part Ylm is affected. The spherical harmonics Ylm have the property that if they are multiplied by r l the result is a homogeneous polynomial in the Cartesian

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Restoring Symmetry by Projection

coordinates of degree l. Since Cartesian coordinates change sign under a parity transformation, Ylm takes an extra (−1)l factor. We thus have Π|nlm⟩ = (−1)l |nlm⟩.

But r is odd under parity transformation, Π r Π = −r. Thus we obtain ′

n′l′m′ r |nlm⟩ = − n′l′m′ Π r Π|nlm⟩ = −(−1)l+l n′l′m′ r |nlm⟩, ′

which implies that the matrix element vanishes unless (−1)l+l = −1, or ∆l = l − l′ = odd, and an electric dipole transition can occur only between states that have opposite parities.

This selection rule is equivalent to a statement that an electric dipole transition between states with ∆l = even (including between states with the same l) is forbidden.

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Quantum Phase Transitions 23.2 Using the results of Problem 23.1, for classical thermal particles the associated “thermal pressure” PT is PT = nkT = nmvT2 , where the ideal gas equation of state P = nkT was used. Likewise, for quantum particles the associated “quantum pressure” PQM is h¯2 PQM ∼ n5/3 = nm m

h¯n1/3 m

!2 2 = nmvQM ,

where the Fermi–Dirac equation of state P ∼ (h¯ 2/m)n5/3 was used. Therefore, utilizing the velocities determined in Problem 23.1, the ratio of quantum pressure to thermal pressure is v2 h¯2n2/3 PQM , = QM = PT mkT v2T indicating that the pressure is dominated by quantum effects at high density and low temperature, and by classical thermal effects for high temperature and low density. As an aside, this result is of considerable importance in astrophysics, where stars having pressure dominated by thermal effects behave very differently those with pressure dominated by quantum effects. This difference in stellar equations of state is discussed more extensively in Ch. 3 of Ref. [89].

23.3 The Bose–Hubbard model Hamiltonian may be expressed as [163] ! t † (nˆ − n )2 − H = H +H = E (b†b + b b ) , pot

kin

C

∑ i

0

EC ∑ ⟨i j⟩

i

j

j i

where bi† creates a Cooper pair (see Box 32.2) on patch i and b i annihilates it, n0 is the average Cooper pair density, EC is the energy cost to convert an isolated Cooper pair into charge carriers in a conductor, t is the energy associated with a Cooper pair hopping between adjacent patches i and j, and ⟨i j⟩ indicates that the summation is only over patches i and j that are nearest neighbors. The two terms (potential and kinetic energies) express competing ground-state tendencies. If t/EC ≪ 1 the first (potential-energy) term dominates and the Cooper pairs are locked on the isolated patches (insulating behavior). Conversely, if t/EC ≫ 1 the second (kinetic-energy) term dominates and the Cooper pairs move freely from patch to patch (superconducting behavior). We may prove that these terms do not commute with each other as follows. Since the kinetic-energy term is restricted to hopping 110

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between adjacent sites i and i ± 1, we must evaluate h i h i h i † † bi†bi , b†i b i±1 + bi±1 b i = b†i b i , b†i b i±1 + bi †bi , bi±1 bi † = bi†bi±1 — bi±1 bi /=0,

where we have used the commutator identity [ AB,CD ] = A[ B,C ]D + [ A,C ]BD + CA[ B, D ] + C[ A, D ]B, assumed operators on different sites to commute, and used the boson commutators (10.3) for operators on the same site. Because this commutator is non-zero, in general the kinetic and potential energy terms in the Hamiltonian do not commute so the two tendencies of the Hamiltonian (insulating and superconducting behavior) are in fundamental tension at intermediate values of t/EC. In fact, the phase transition caused by varying t/EC exhibits quantum critical behavior. This problem is discussed more extensively in Ch. 13 of Ref. [163].

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Topology, Manifolds, and Metrics 24.1 This problem is an example given in Pires [164]. From the definition of a topology in Section 24.2.1, we must show that for the set M = {a, b, c, d, e} the subsets in } τ ≡ 0/ , M,{a},{c, d},{a, c, d},{b, c, d, e} , are closed under arbitrary unions and finite intersections (give members of τ under these operations). First consider the unions. Obviously M and 0/ are closed under union with the subsets of τ, while {a} ∪ {c, d} = {a, c, d} ∈ τ, {a} ∪ {a, c, d} = {a, c, d} ∈ τ, {a} ∪ {b, c, d, e} = {a, b, c, d, e} = M ∈ τ, {c, d} ∪ {a, c, d} = {a, c, d} ∈ τ, {c, d} ∪ {b, c, d, e} = {b, c, d, e} ∈ τ, {a, c, d} ∪ {b, c, d, e} = {a, b, c, d, e} = M ∈ τ.

Next let’s check the intersections. Obviously the intersections of 0/ and M with any subset are in τ, while {a} ∩ {c, d} = 0/ ∈ τ, {a} ∩ {a, c, d} = {a} ∈ τ, {a} ∩ {b, c, d, e} = 0/ ∈ τ, {c, d} ∩ {a, c, d} = {c, d} ∈ τ, {c, d} ∩ {b, c, d, e} = {c, d} ∈ τ, {a, c, d} ∩ {b, c, d, e} = {c, d} ∈ τ.

Since the subset τ contains 0/ and M and is closed under unions and intersections, it satisfies all the conditions of Section 24.2.1 and defines a topology on M. 24.3 This solution follows a discussion in Nakahara [152]. Take X = (−π 2, π2) and Y = R, and consider the map f : X → Y given by f (x) = tan x. But tan x is one to one on X and its inverse tan−1 x is one to one on R, so the interval (−π2, π2) and R are homeomorphic.

An interval is bounded if its bounds are real numbers. Thus (−π 2, π 2) is bounded but R isn’t since its bounds are ±∞. Since the homeomorphism connects topological spaces with different boundedness properties, this example shows that boundedness cannot be a topological invariant.

24.4 For the 2-torus the topology is T 2 = S1 × S1. From Table 24.1, π1(A × B) = π1(A) × π1(B)

π1(S1) = Z.

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113

Therefore, π1(T 2) = Z× Z. More generally, T d = S1 × S1 . . . S1 and ` ˛ ¸ x d f ac tors

π1 (T ) = Z × Z . . . Z ` ˛¸ x d

d f ac tors

for a d-dimensional torus.

24.5 Consider the open unit disk shown in Fig. 24.5(b) and specified by Y = {(x, y) : x2 + y2 < 1}. Take as on open covering the family of concentric open disks Sα = (x, y) : x2 + y2 < 1 −

1 1+α

,

with α = 1, 2,... Then ∪Sα ⊃ Y . Now if Y is compact, there is a finite subcovering S′ = {S1, S2, . . . , Sn} such that S1 ∪ S2 ∪ . . . ∪ Sn ⊃ Y. But no finite set S′ can cover Y because the set S′ has one Sα corresponding to the largest radius (< 1). We can always choose another open disk not in the set that has a radius < 1 but larger than that of the largest Sα . Therefore, Y cannot be covered by the finite set S′ and is non-compact. Comparing this result with Example 24.7, it is clearly the open character of the disk that is responsible for the non-compact space associated with this case.

24.7 This problem is based on a discussion in Wybourne [224]. From Probs. 2.9 and 2.11, the group D2 = {e, a, b, c} has a quotient group with respect to the abelian invariant subgroup H = {e, a}, D2/H = H + M = {e, a} + {b, c}, with a 2:1 map φ from D2 to D2/H given by :

e

b H

a

M. c

If we equip the space for D2 with a simple topology defined by the open sets } τ = 0/ , {e}, {a}, {b}, {c}, {e, a, b, c} , the space for D2 becomes a topological space. Furthermore, the above map endows the quotient space with a topology we may call the quotient topology in terms of the open sets } τ′ = 0/ , {H}, {M}, {H, M} , because the inverse map φ−1 for the set τ′ in D2/H maps to open sets τ in D2.

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Topological Solitons 25.1 The Lagrangian density is given by Eq. (25.1), from which ∂L ∂U (x) ∂L ∂φ =− =− . ∂φ ∂φ ∂(∂φ/∂ x) ∂x Inserting in the Euler–Lagrange equation (16.14) gives the equation of motion (25.2).

25.3 Multiplying both sides of Eq. (25.2) by dφ/dx and integrating over dx gives ∫ ∫ d 2φ dφ dU dφ dx, dx = 2 dx dx dφ dx which implies that 1 2

dφ dx

2

= U (φ),

where the integration constant is zero because both dφ/dx and U (φ) are assumed to vanish at spatial infinity. Therefore dφ ±√ = dx, 2U (φ) and integration gives Eq. (25.5), x = x0 ±

∫ φ(x)

dφ

, 2U (φ) where x0 is arbitrary because of translational invariance. φ(x0 )

√

25.5 This solution follows a discussion in Ref. [85]. We assume a φ 4 scalar field theory having a Lagrangian density given by Eq. (25.1) with a potential 2 1 U (φ) = λ φ 2 − m λ 4

2

,

given by Eq. (25.6), where λ and m2 are positive constants. The degenerate vacuum states √ occur at φ0 = ±m/ λ, as illustrated in Fig. 25.1 [this document]. Two nontrivial solutions are consistent with the boundary conditions, ( √ √ −m/ λ, +m/ λ φ(−∞), φ(+∞) = , √ √ +m/ λ, −m/ λ which are called the kink and antikink, respectively. From Eq. (25.5) the solution is r ∫ 2 φ(x) dφ . x − x0 = ± λ φ(x0) φ 2 − m2/λ 114

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115

U ( )

(b) Analog problem

-m-1/2 -m-1/2

.

Fig. 25.1

U ( )

+m-1/2

+m-1/2

(a) Field theory

Field theory and the mechanical analog problem for the kink soliton solution.

Integrating over φ, setting φ(x0) = 0, and inverting to solve for φ(x) leads to Eq. (25.7), m tanh √ (x − x0) , 2 λ where the solution with the positive sign is the kink and the one with the negative sign is the antikink. The energy density for time-independent solutions is 2 1 dφ m m4 + U (φ) = 2U (φ) = sech 4 √ (x − x0) , E (x) = 2λ 2 dx 2 where (25.6) and (25.7) have been used. The total energy is √ ∫ +∞ 2 2 m3 E= E (x)dx = 3 λ −∞ φ±(x) = ±√

m

for both the kink and antikink. The solutions and associated energy density are plotted in Fig. 25.4.

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Geometry and Gauge Theories 26.1 Choosing the spherical coordinates (x1, x2) = (S, φ) defined from the following figure, z S r C = 2r

 R



x

y

the metric tensor components are g11 = 1

g22 = R2 sin 2

x1 R

,

and insertion into the expression for the gaussian curvature ( " 1 1 ∂ g11 ∂ g22 ∂ g11 ∂ 2g22 ∂ 2g11 + K= — − + (∂ x1)2 (∂ x2)2 2g 11 ∂ x1 ∂ x1 ∂ x2 2g11 g22 " #) 2 1 ∂ g11 ∂ g22 ∂ g 22 + + , ∂ x1 2g 22 ∂ x2 ∂ x2

#

2

gives K = R−2, which is constant. Choosing the cylindrical coordinates (x1, x2) = (r, φ) gives the metric tensor components g11 =

R2 R2 − r2

2

g22 = r ,

and insertion into the equation for K yields K = R−2. Thus the Gaussian curvature is independent of the coordinates for a sphere and is unchanged by changing the parameterization of the 2D manifold.

26.2 From Box 26.1 the tangent vector for a coordinate curve can be written dxµ t = t µ eµ = eµ , dλ 116

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so that for the infinitesimal line interval ds = eµ (x)dxµ and ds2 = ds · ds = (eµ · eν)dxµ dxν .

But from Eq. (13.5) with the Minkowski metric ηµν replaced by a more general metric gµν , ds2 = gµν dxµ dxν .

Therefore, gµν = eµ · eν in holonomic coordinates.

26.4 This solution follows one given in Ref. [85]. Define a gauge transformation ψ′(x) = G(x)ψ(x). Then from ψ(x1) = Uγ (x0, x1)ψ(x0), ψ′(x1) = Uγ′ (x0, x1)ψ′(x0)

G(x1)ψ(x1) = Uγ′ (x0, x1)G(x0)ψ(x0)

G(x1)Uγ (x0, x1)ψ(x0) = Uγ′ (x0, x1)G(x0)ψ(x0).

Since ψ(x0) is arbitrary, Eq. (26.10) follows upon dropping ψ(x0) and multiplying from the right by G−1(x0).

26.7 Suppose a vector field V (λ) defined along a curve xµ (λ) parameterized by λ. Assuming that V (λ) = V µ (λ)eµ (λ), where eµ (λ) is a coordinate basis vector evaluated at the point on the curve labeled by λ, then from Eq. (26.1), ν dV dV µ µ ∂ eµ dx = eµ + V ∂ xν dλ . dλ dλ Expand the partial derivative factor in the vector basis, ∂ eµ = Γα e , µν α ν ∂x where Γαµν is an expansion coefficient. This gives ν µ dx α eµ + Γµν V dλ eα dλ dλ ν µ α dx µ dV eµ = V e + Γ µ αν dλ dλ µ ν dV µ α dx = + ΓανV eµ dλ dλ µ DV ≡ eµ , Dλ where in line two dummy summation indices in the second term were interchanged, α ↔ µ.

Thus ν DV µ dV µ µ α dx = + ΓανV , Dλ dλ dλ

dV

=

dV µ

which is Eq. (26.4) for the absolute derivative of a vector field along a curve parameterized by λ.

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Geometrical Phases 27.2 The shift in phase between trajectories 1 and 2 produced by the vector potential is given by Eq. (27.4), ∆δ =

∫

∫

e h¯

A · dr =

A· dr − 1

2

e h¯

I

A · d r.

Now apply Stokes’ theorem [Eq. (27.5) and Box 27.1] I

A · dr = C

∫

(∇ × A)· ndS ≡

S

∫

(∇ × A)· d S,

S

where S is the surface enclosed by the contour C and n is an outward normal to the surface.

From (27.4) and (27.5), we conclude that ∫

e (∇ × A)· d S h¯ S ∫ e = h¯ S B · d S e = Φ, h¯ where Φ is the magnetic flux through the solenoid. ∆δ =

27.5 The Hamiltonian is

µ

µ

(σx Bx + σyBy + σzB z), 2 2 where the σi are Pauli matrices. The scalar product is H=−

σ·B=

σ·B=−

−i 1 0 Bz By + Bz = 0 0 −1 Bx + iBy

1 0 0 Bx + 1 0 i

Bx − iBy , −Bz

which requires us to solve the eigenvalue problem Hψ =

−µ

Bz

Bx − iBy

2

Bx + iBy

−Bz

ψ = Eψ.

The energy eigenvalues are determined from Bz − E

Bx − iBy

Bx + iBy

−Bz − E

which has the solution (27.29), E±(B) = ±

= 0,

µq 2 µ B x+ B2y+ B2z = ± B, 2 2

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119

where B ≡ |B|. Defining the angles θ and φ through Bx = B sinθ cosφ

By = B sinθ sin φ

Bz = B cosθ,

the corresponding eigenvectors are given by θ cos e −iφ/2 2 |ψ+⟩ = θ sin eiφ/2 2

|ψ−⟩ =

θ −iφ/2 e 2 , cos θ eiφ/2 2

−sin

which is Eq. (27.30).

27.6 In this solution we follow Dittrich and Reuter [50]. The Berry phase may be expressed in the form given by Eqs. (27.26)-(27.27), which requires evaluation of the quantity ∇BH(B). For the Hamiltonian (27.28), µ σ ·B 2 µ ∂ ∂ ∂ =− i +j +k ∂Y ∂Z 2 ∂X = − µ (iσ + jσ + kσ ) x z z µ2 = σ.

2

∇BH(B) = ∇B −

(σxX + σ yY + σ zZ)

Let m = |ψ−⟩ and n = |ψ+⟩ in Eq. (27.27). Then we must evaluate ⟨mB| ∇BH |nB⟩ × ⟨nB| ∇BH |mB⟩ (En(B) − En(B))2 =

=

=

⟨ψ−| ∇BH |ψ+⟩ × ⟨ψ+| ∇BH |ψ−⟩ (E− — E+ )2 i

µ2 2 4(E− — E+ ) µ2 2 4(E− − E+)

⟨−|σx |+⟩

µ2 ⟨ψ−| σ |ψ+⟩ × ⟨ψ+| σ |ψ−⟩ 4 = (E− — E+)2 j k

⟨−|σy |+⟩

⟨−|σz |+⟩

y z ⟨+|σ x|−⟩ ⟨+|σ |−⟩ ⟨+|σ |−⟩

⟨−|σy |+⟩⟨+|σz |−⟩ − ⟨−|σz |+⟩⟨+|σy |−⟩ −⟨−|σx |+⟩⟨+|σz |−⟩ + ⟨−|σz |+⟩⟨+|σx |−⟩

.

⟨−|σx |+⟩⟨+|σy |−⟩ − ⟨−|σy |+⟩⟨+|σx |−⟩

Let us choose the coordinate system so that the z axis points along B, σz |±⟩ = ±|±⟩

σx |±⟩ = |∓⟩

σy |±⟩ = ±|∓⟩.

Then, for example, ⟨−|σx |+⟩⟨+|σy |−⟩ − ⟨−|σy |+⟩⟨+|σx |−⟩ = −i ⟨−| −⟩⟨+| +⟩ − i ⟨−| −⟩⟨+| +⟩ = −2i.

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120

Evaluating the other matrix elements in a similar way implies that ⟨mB| ∇BH |nB⟩⟨nB| ∇BH |mB⟩ =

µ2 4

0

0 . −2i

Then from Eq. (27.27), ⟨−| ∇BH |+⟩⟨+| ∇BH |−⟩ (E− − E+)2

Ω− (B) = i

2

=i =i

=−

µ (−B − B) 2 0 1 0 2 4B −2i 1

2B2

0

µ2

1

4

0 −2i

0 0 .

1

This result was obtained assuming a unit vector in the z direction, which is of the form B/B, so the general solution for arbitrary axes is 1 B Ω (B) = − .

— 2 B3 By a similar analysis, Ω+ (B) =

1 2

B B3

.

These are the results expected for a magnetic monopole of strength ± 12 at the origin of parameter space. From Eq. (27.26), the corresponding Berry phase is given by γm(C) = −

∫

d S · Ωm(B)

S

∫

1∫ = − d S · Ω± (B) =∓ 3 d S · B 2B S S ∫ ∫ 1 = ∓ 3 dφ dθ B2 sinθ n · B 2B ∫ B3 ∫ = ∓ 3 dφ sinθ dθ 2B = ∓π(1 − cosθ), where π(1 −cosθ) is half the solid angle enclosed by the path C as seen from the origin of parameter space, n is a unit vector perpendicular to the surface, and n · B = B because the monopole field is radial.

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27.7 The second derivative in Eq. (27.13) may be expanded as ∂2 ∂R

2

∂ ∂ ∂Φ(R) Φ(R) |n, R⟩ + |n, R⟩ ∂R ∂ R ∂R ∂Φ(R) ∂ ∂2 = |n, R⟩ + Φ(R) |n, R⟩ ∂R ∂R ∂ R2 ∂ 2Φ(R) ∂Φ(R) ∂ + |n, R⟩ |n, R⟩ + 2 ∂R ∂R ∂R

|Φ(R)⟩ |n, R⟩ =

.

Inserting this result and Eq. (27.11) into Eq. (27.13) and multiplying from the left by ⟨n, R| gives ∂ 2Φ(R) −h¯2 ∂2 n, R + ⟨n, R| H |ψ⟩ = Φ(R) n, R 2M ∂ R2 ∂ R2 ∂Φ(R) ∂ +2 n, R n, R + εn(R)Φ(R) ∂R ∂R = EΦ(R), which is Eq. (27.14).

27.9 Take the derivative of Eq. (27.18) to give ∂ ∂ exp |n˜,RR⟩ = ∂R ∂R

∫

i R An(R)d R R R |n, R A R⟩ h¯ R0 ∫

∫

i i R A (R)d R i R A (R)d R ∂ |n, R⟩ = AAn(R) exp n R R |n, R R⟩ + exp n R A h¯ R0 h¯ R0 A R ∂R R h¯ ∫ i ∂ i R R |n˜, R An(R)d |n, R R⟩. = AAn(R) R⟩ + exp A R R ∂R h¯ h¯ R 0 Multiply this expression from the left by ih¯ ⟨n˜, R| to give ∂ A (R) ≡ ih¯ n˜, R n˜, R n˜

∂R

∫ R

R An(R) R |n˜, R⟩ R + ih¯ n˜, R exp = −⟨n˜, R| R = −An(R) + An(R)

∂ A R R R i n, R A (R)d R n h¯ R0 ∂R

= 0, where in the second line we have used ∫

R ⟨n, R|R= ⟨n˜, R|exp

i R R RR AAn(R)d h¯ R0

evaluated from Eqs. (27.18) and (27.19). Thus, under the transformation ∫

R → |n˜, R |n, R⟩ R⟩ ≡ exp

i R An(R)d R R |n, R⟩, R A h¯ R0

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we have that

h¯ ∂

— An (R) →

h¯ ∂

i ∂R i ∂R and the Berry connection (vector potential) is eliminated from the Hamiltonian of the Schrödinger equation in favor of the additional ∫phase for the wavefunction, i R A R R γ exp(i n) = exp An(R)d R h¯ R0 given in Eq. (27.20).

27.12 The Berry connection is (h¯ = 1 units) An(R) ≡ i ⟨n, R| ∇R |n, R⟩.

Simplifying notation by letting λ stand for the parameters, we have for the Berry phase evaluated over a path from λ1 to λ2 ∫ λ2

γ(λ1, λ2) =

λ1

A(λ)dλ.

Under a local gauge transformation |uλ ⟩ → eiχ(λ) |uλ ⟩ ≡ |u˜λ ⟩ the Berry connection is transformed to (see Problem 27.10) dχ A˜(λ ) = A(λ) + . dλ Then the gauge-transformed Berry phase accumulated between endpoints λ1 and λ2 is γ˜(λ1, λ2) = =

∫ λ2 λ1 ∫ λ2

˜ A(λ)dλ A(λ)dλ +

∫ λ2 dχ

dλ dλ = γ(λ1, λ2) + χ(λ2) − χ(λ1). λ1

λ1

Thus γ(λ1, λ2) is gauge invariant if χ(λ2) − χ(λ1) = 2π m, where m is an integer, since γ is a phase. Evaluating on a closed path where λ1 = λ2 implies that |u˜λ =1⟩ = |u˜λ =2⟩, which requires γ˜λ = γλ + 2π m

(m = integer)

and the Berry phase is gauge invariant modulo 2π. That is, the Berry phase is gauge invariant when regarded as a phase angle. The Berry curvature is Ω = ∇ × A, so under a gauge transformation A → A − ∇χ

Ω → ∇ × (A − ∇χ).

But by a basic vector identity, ∇ × ∇χ = 0 for a acalar χ (see Section 14.1.1), so under a gauge transformation Ω → ∇ × A − ∇ × ∇χ = ∇ × A = Ω, and the Berry curvature Ω is gauge invariant.

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Topology of the Quantum Hall Effect 28.1 (a) Requiring that the forces in the y direction cancel, from Eq. (28.1) 1 F = −e E − v B = 0, y x z y c which implies a relationship 1 Ey = vxBz c between the transverse electric field and the perpendicular magnetic field. (b) The electron velocity v depends on the interactions of the electrons with the lattice.

This can be approximated by writing an equation of motion based on the Drude model [131, 163]1 dv v 1 F=m + = −e E + v × B , dt τ c where m is the effective mass of an electron and τ is the mean time between electron collisions in the sample. In steady state d v/dt = 0 and the equations of motion are vx = −

eτ m

Ex −

eτB mc

vy

vy = −

eτ m

Ey +

eτB mc

vx

vz = −

eτ m

Ez,

where B ≡ Bz = |B|. Defining the cyclotron frequency ωc by ωc ≡

eB

mc the electron equations of motion may be written eτ eτ vx = − Ex − ωcτvy vy = − Ey + ωcτvx m m

vz = −

eτ m

Ez.

(c) Inserting the results of part (b) into Ey = vxB/c from part (a) gives a relationship between the transverse electric field Ey and longitudinal electric field Ex, Ey =

B c

vx =

B eτ eBτ − Ex — ωc τvy = − E x, c m mc

where we have required that at equilibrium vy = 0.

1 The Drude model is an old pre-quantum model (proposed in the year 1900) of electron transport where the

microscopic behavior of electrons in a solid is treated classically and looks much like a pinball machine, with electrons bouncing sequentially off heavier, immobile positive ions. It fails to account fully for the quantum behavior of matter but is often used to get a very rough idea of electron transport properties in a solid.

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124

(d) You were given in the problem that if τ is the mean collision time for an electron, the current density may be approximated in the Drude model as e2ne

nee2τ

, m m where ne is the electron number density and σ is the conductance. In a classical Hall effect experiment one typically measures the transverse voltage VH and the longitudinal voltage VL = ExL, where L is the length of the device. If the sample is approximated as 2D with transverse width w [see Fig. 28.1(a)], the Hall voltage VH and the total current I are given by VH Ey = I = wjx . w The Hall resistance RH is then defined by R = VH = wEy = Ey = − B , j=

H

τE = σE

I

wjx

σ≡

jx

ecne

where results from parts (a)-(c) were used. Likewise, the longitudinal resistance RL is given by VL LEx L R = = = , L I wjx wσ where jx = σ Ex was used. Collecting results, VH E = I = wj R ≡ VH = − B y

w

x

H

I

ecne

R ≡ VL = L , L I wσ

which is Eq. (28.2).

28.2 We work in the Landau gauge (28.4), so that Ay = Bx and Ax = 0 (with the z dimension neglected, since our concern is with 2D electron gases). The Schrödinger equation is then " # h¯2 ieBx 2 2 ∂ + ∂y − ψ(x, y) = Eψ(x, y), — ∗ x h¯c 2m where m∗ is the effective mass, ∂ x2 ≡ ∂ 2/∂ x2, and so on. In this gauge the Hamiltonian does not depend on y so let’s write the wavefunction in the separable form ψ(x, y) = φ(x)eiky.

Inserting this into the Schrödinger equation we find that φ(x) obeys 2 h̄ ω c x −Λ2∂x 2 + −Λk φ(x) = Eφ(x), Λ 2

where the cyclotron frequency ωc and the magnetic length Λ are defined by Eq. (28.6), r eB h¯c ωc ≡ ∗ Λ≡ . mc eB But this is just the Schrödinger equation for a harmonic-oscillator with wavefunction 2 2 x ψnk(x, y) = Hn − Λk e−(x−xk ) /2Λ eiky, Λ

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where Hn is a Hermite polynomial and xk = Λ2k. The corresponding energy is εn = h¯ ωc n +

1 , 2

which is independent of k, depending only on the principle Landau quantum number n = 0, 1, 2,... We see that the wavefunction is localized in x near x = xk = Λ2k but extended in y (which is a consequence of our gauge choice).

28.6 From Eq. (27.24) using the toroidal coordinates in Fig. 28.7(a), Φλρ ≡

∫

∫λ

Ωµν dS =

λi

S

f

∫1

dλ 0

But the second term vanishes because ∫ λf λi

∫ 1

dλ 0

dρ(∂ρ Aλ ) =

∫ λf λi

dρ(∂λ Aρ − ∂ρ Aλ ). ∫ Aλ (1)

dλ

Aλ (0)

dAλ = 0,

since Aλ have the same value at ρ = 0 and ρ = 1 by the periodic gauge condition. Thus Φλρ ≡ = = = =

∫

Ωµν dS

S

∫ λ

f

λi

∫ 1

dλ

0

∫ λ

∂

f

λi ∫ λf

dλ dλ

λi ∫ λf λi

dρ

∂ Aρ ∂λ

∫ 1

∂λ 0 ∂γ ρ

dρ Aρ

∂λ

dγρ

= γρ (λf) − λρ(λi), where in line 3 we have used that the Berry phase evaluated for a cyclic path in ρ is γρ =

∫ 1

Aρ dρ, 0

from Eq. (27.22). On the torus the endpoints for λ are identified modulo 2π, so at the end of a cycle in λ the Berry phase will have evolved by Φλρ = γρ (λf) − λ ρ (λi) = 2π m, where m is an integer. But from Box 28.4 the Berry flux is related to the Chern number C by Φλρ ≡

∫

Ωµν dS = 2πC S

Comparing these results gives C = m, and the Chern number may be interpreted as the winding number of the Berry phase along ρ in one cycle over λ, as illustrated in Fig. 28.7.

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29

Topological Matter 29.2 Using the algorithm of Fig. 29.16,

=

(a)

=

=

(b)

The last step of solution (a) has invoked the rule that two braids that can be made to look the same just by pulling on the strands are equivalent.

29.4 Let U ≡ iσ2 so that Θ = UK . Then |α˜ ⟩ = Θ|α⟩ = UK |α⟩ = ∑ UK |a⟩⟨a| α⟩ a

= ∑⟨a| α⟩ UK |a⟩ = ∑⟨a| α⟩∗ U |a⟩ ∗

a

a

= ∑ ⟨α| a⟩U |a⟩, a

where we’ve used that K has no effect on a ket |a⟩. Likewise, for |β˜⟩, |β˜⟩ = Θ|β ⟩ = UK |β ⟩ = ∑ UK |b⟩⟨b| β ⟩ b

= ∑⟨b| β ⟩ UK |b⟩ = ∑⟨b| β ⟩∗ U |b⟩, ∗

b

b

so that ⟨β˜| = ∑⟨b| β ⟩⟨b|U †. b

Combining these results gives ⟨ β˜ | α̃ ⟩ = ∑⟨b| β ⟩⟨b|U †U |a⟩⟨α| a⟩ ab

= ∑⟨α| a⟩⟨a| β ⟩ a

= ⟨α| β ⟩ = ⟨β | α⟩∗ , 126

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where ⟨b|U †U |a⟩ = δab was used.

29.6 From the solution of Problem 14.14, ther eigenvalues of a 2 × 2 matrix H are given by λ± = T ± T − D, 12 1 2 4 where T is the trace of H and D is the determinant of H. For the matrix given in Eq. (29.4), H=

d3

d1 − id2

d1 + id2

−d3

d1 ≡ a0 sinx d2 ≡ a0 sin ky

d3 ≡ m + 2 − coskx − cos ky T = Tr H = 0 and D = det H = −(d2 + d2 + d2), so the eigenvalues are given by q

1

2

3

E± = ± d21+ d22 + d32 q = ± (m + 2 − coskx − cos ky)2 + a20(sin2 kx + sin2 ky).

If we set the off-diagonal matrix elements in H to zero to get the energies of the noninteracting bands, T = 0 and D = −d32, so q E±0 = ± d23= m + 2 − coskx − cos ky.

This may be plotted with standard plotting packages. For example, the GnuPlot script used to create the panels in Fig. 29.9 of the book is: # Two-band mixing eigenvalues # set key off to suppress legend which gets in way set key off left top vertical Right noreverse enhanced autotitles\ box linetype -1 linewidth 1.000 set samples 600 # Define linestyles set style line 1 lt rgb "#000000" lw 1 set style line 2 lt rgb "#ff0000" lw 1

# Define linestyle 1 # Define linestyle 2

# X-Y range for plot set xrange [-1.0*pi:1.0*pi] set yrange [-5.0:5.0] # change independent variable from default x to kx set dummy kx # Constants m = -0.5 # Controls phases a = 0.2 scale = 1.95 # Overall scale for plot ky = 0.0 # Set k_y to a constant for plots # Eigenvalue formulas t1(kx) = (m + 2.0 - cos(kx) - cos(ky))**2 t2(kx) = a**2 * ( sin(kx)**2 + sin(ky)**2 )

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E1(kx) = sqrt(t1(kx) + t2(kx)) Eun1(kx) = sqrt(t1(kx)) # Plot non-interacting bands (off-diagonal set to zero in H) plot scale*Eun1(kx) ls 2 replot -scale*Eun1(kx) ls 2 # Plot interacting bands replot scale*E1(kx) ls 1 replot -scale*E1(kx) ls 1 # Width and height of postscript figure in inches width = 3.0 height = 3.0 # Create a postscript file of the plot set out "bandInversion.eps" # Output file set terminal postscript eps size width, height enhanced color\ solid lw 2 "Arial" 32 replot # Plot to EPS file # Create a PNG file of the plot set out "bandInversion.png" # Assume 72 pixels/inch and make bitmap twice as # large for display resolution set terminal png transparent size 2*width*72, 2*height*72 \ lw 2 enhanced font ’Arial,28’ replot # Plot to PNG file

From the solution of Problem 14.14 the eigenvectors are |X+⟩ = N+ a |χ1⟩ + |χ2⟩ for the system H=

α

β

γ

δ

where we define a≡

|X−⟩ = N− |χ1⟩ + b |χ2⟩

|χ ⟩ =

1

|χ ⟩ =

0

1

0

2

1

β E+ − α

b≡

,

E− − α , β

and where E± are the eigenvalues found above, |χ1⟩ and |χ2⟩ are wavefunctions for the unmixed bands, and N± are normalizations. Requiring that ⟨X+| X+⟩ = ⟨X−| X−⟩ = 1 gives 1 1 N+ = √ N− = √ .

1 + a2 1 + b2 Thus the normalized eigenfunctions are |X+⟩ = A |χ1⟩ + B |χ2⟩

|X−⟩ = C |χ1⟩ + D |χ2⟩

where we define A≡ √

a 1 + a2

B≡ √

1 1 + a2

C≡ √

1 1 + b2

D≡ √

b 1 + b2

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with a≡

β E+ − α

b≡

E− − α β

and with α = d3 = m + 2 − coskx − cos ky

β = d1 − id2 = a0 sin kx − ia0 sin ky,

assuming the identification H=

α β = γ δ

3d

d1 + id2

1d

− id 2 −d3

d1 ≡ a0 sin kx d2 ≡ a0 sin ky d3 ≡ m + 2 − coskx − cos ky.

Figure 29.10 then results from plotting E± and |X±⟩. A gnuPlot script for the eigenvalues was given above and one that will do the job for the eigenfunctions in Fig. 29.10 is listed below. # Two-band mixing eigenvalues and eigenfunctions # set key off to suppress legend which gets in way set key off left top vertical Right noreverse enhanced autotitles\ box linetype -1 linewidth 1.000 set samples 600 # Define linestyles set style line 1 lt rgb "#000000" lw 1 set style line 2 lt rgb "#ff0000" lw 1 set style line 3 lt rgb "#0000ff" lw 1 set style line 4 lt rgb "#0000ff" lw 1 # X-Y range for plot set xrange [-1.0*pi:1.0*pi] set yrange [-0.25:2.5] # change independent variable from default x to kx set dummy kx # Constants m = -2.5 # Controls phases a0 = 0.2 scale = 1.95 # Overall scale for plot ky = 0.0 # Set k_y to a constant for plots # Eigenvalue formulas t1(kx) = m + 2.0 - cos(kx) - cos(ky) t2(kx) = a0**2 * ( sin(kx)**2 + sin(ky)**2 ) E1(kx) = sqrt(t1(kx)**2 + t2(kx)) Eun1(kx) = t1(kx) # No coupling # Wavefunctions alpha(kx) = t1(kx) beta(kx) = a0*sin(kx)

# Assumes ky = 0.0

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a(kx) = beta(kx)/(E1(kx) - alpha(kx)) b(kx) = (-E1(kx) - alpha(kx))/beta(kx) A(kx) = a(kx)/sqrt(1.0 + a(kx)**2) B(kx) = 1.0/sqrt(1.0 + a(kx)**2) C(kx) = 1.0/sqrt(1.0 + b(kx)**2) D(kx) = b(kx)/sqrt(1.0 + b(kx)**2) plot A(kx)**2 ls 2 replot B(kx)**2 ls 3 # Offset by offset from above so they # aren’t on top of each other offset = 1.25 replot (C(kx)**2 + offset) ls 2 replot (D(kx)**2 + offset) ls 3 ### ----- PLOT FILE OUTPUT ------------ ### # Width and height of postscript figure in inches width = 2.9 height = 2.0 # Create a postscript file of the plot set out "bandInversionWF.eps" # Output file set terminal postscript eps size width, height enhanced color\ solid lw 2 "Arial" 32 replot # Plot to EPS file # Create a PNG file of the plot set out "bandInversionWF.png" # Assume 72 pixels/inch and make bitmap twice as # large for display resolution set terminal png transparent size 2*width*72, 2*height*72 \ lw 2 enhanced font ’Arial,28’ replot # Plot to PNG file

29.7 Assume a time-reversal invariant Hamiltonian for a spin-21 system. Then if |ψ⟩ is an eigenstate with energy E, because of the time-reversal symmetry Θ|ψ⟩ is also an eigenstate with energy E. Now we check whether |ψ⟩ and Θ|ψ⟩ are independent states. Let |α⟩ ≡ |ψ⟩

|α˜ ⟩ ≡ Θ|α⟩ = Θ|ψ⟩

|β ⟩ ≡ Θ|ψ⟩

|β˜⟩ ≡ Θ|β ⟩ = Θ2 |ψ⟩ = −|ψ⟩,

where in the final expression we used that Θ2 = −1 for spin- 12 particles. Then using ⟨β˜ | α̃ ⟩ = ⟨α| β ⟩ from Problem 29.4, we have that ⟨ψ| Θψ⟩ = −⟨ψ| Θψ⟩, which can be true only if ⟨ψ| Θψ⟩ = 0. Thus |ψ⟩ and Θ|ψ⟩ are independent states since they are orthogonal, and if there is a state at energy E in a time-reversal invariant spin- 12system, it is at least two-fold degenerate (Kramers theorem).

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Angular Momentum Recoupling 30.3 The matrix element can be written using the Wigner–Eckart theorem (6.71) as

=√

1 J 0 M

J −M

⟨LSJM| Lz + 2Sz |LSJM⟩ = (−1)J−M M

J(J + 1)(2J + 1)

⟨LSJ || L || LSJ⟩ + 2⟨LSJ || S || LSJ⟩

⟨LSJ || L || LSJ⟩ + 2⟨LSJ || S || LSJ⟩ ,

where we have used the symmetries of the 3J symbol given in Section 6.3.8 and Table C.4 to evaluate (−1)J−M

J −M

1 J 0 M

J J M −M

= (−1)J−M

1 M =√ .

0 J(J + 1)(2J + 1)

To proceed we must evaluate the reduced matrix elements of L and S. The angular momentum operator L is a rank-1 tensor that operates on only the orbital part of the angular momentum state, so from Eq. (30.12), ⟨LSJ || L || LSJ⟩ = (−1)J+L+S+1(2J + 1)

L L 1 ⟨L || L || L⟩.

J J S

From the first equation above, ⟨LSJM| Lz |LSJM⟩ =

( — 1)J+L+S+1(2J + 1) M L L 1 ⟨L || L || L⟩, √ J J S J(J + 1)(2J + 1)

so we must evaluate the 6J symbol and the reduced matrix element ⟨L || L || L⟩. Using Eq. (30.8) and Table 30.1, L L 1 J

J

S

= (− 1)2L+2JW (LLJJ; 1S) = (−1)

L(L + 1) + J(J + 1) − S(S + 1)

2L+2J+L+J−S−1

√

.

4LJ(J + 1)(L + 1)(2L + 1)(2J + 1)

Inserting this in the above equation gives M √ ⟨LSJM| Lz |LSJM⟩ = 2J(J + 1) L(L + 1)(2L + 1) × J(J + 1) + L(L + 1) − S(S + 1) ⟨L || L || L⟩, 131

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so we are left to evaluate ⟨L || L || L⟩. This may be done by analogy with Example 6.9. From the Wigner–Eckart theorem (6.71), ML = ⟨LML| Lz |LML⟩ = (−1)L−ML

L −ML

1 L ⟨L || L || L⟩ 0 ML

L 1 ⟨L || L || L⟩ −ML 0 ML 2L−2ML √ ⟨L || L || L⟩ = (−1) L(L + 1)(2L + 1) ML =√ ⟨L || L || L⟩, L(L + 1)(2L + 1) where symmetries of the 3J symbol from Section 6.3.8 were invoked in the second line and where in the third line Table C.4 was used to evaluate the 3J symbol as = (−1)L−ML

L L ML −ML

1 0

L ML

= (−1)L−ML √

ML L(L + 1)(2L + 1)

.

Solving for the reduced matrix element then gives √ ⟨L || L || L⟩ = L(L + 1)(2L + 1).

Inserting this in the expression above for the matrix element of Lz then gives ⟨LSJM| L z |LSJM⟩ =

M J(J + 1) + L(L + 1) − S(S + 1) .

2J(J + 1)

By a similar procedure, since S is also a rank-1 tensor that operates on only the spin degree of freedom, from Eq. (30.12) we find ⟨LSJ || S || LSJ⟩ = (−1)J+L+S+1(2J + 1)

S S 1 ⟨S || S || S⟩.

J J L

Proceeding in the same way as for Lz above leads to √ ⟨S || S || S⟩ = S(S + 1)(2S + 1), so that 2 ⟨LSJM| S z |LSJM⟩ =

M J(J + 1)

S(S + 1) − L(L + 1) + J(J + 1) .

Combining the above results yields finally, M 2J(J + 1) + J(J + 1) − L(L + 1) + S(S + 1) 2J(J + 1) J(J + 1) + S(S + 1) − L(L + 1) = M 1+ 2J(J + 1) ≡ Mg,

⟨LSJM|(L z + 2S z ) |LSJM⟩ =

where the quantity in parentheses, g ≡ 1+

J(J + 1) + S(S + 1) − L(L + 1) , 2J(J + 1)

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is called the Landé g-factor.

30.5 Proceeding as in the derivation of Eq. (30.11), ⟨ j1 j2JM| Tkq(1) j1′ j2′ J′(M − q) = δ j2 j2′ ∑ ∑ ⟨ j1m1 j2m2| JM⟩ m1m2 m′

1

× j1′ m′1 j2m2 J′(M − q) ⟨ j1m1| Tkq(1) j1′ m′1 .

Introducing reduced matrix elements on both sides of the equation and exploiting the additivity of the magnetic quantum numbers in the Clebsch–Gordan coefficients gives an expression involving a sum over products of four Clebsch–Gordan coefficients, which may be replaced by a 6J coefficient, and the reduced matrix element is √ ′ ⟨ j1 j2J || T k(1) || j1′ j2′ J′⟩ = (−1)J +k+ j1+ j2 (2J + 1)(2J′ + 1) ×

j1 J′

j1′ J

k ⟨ j1 || T k(1) || j1′ ⟩δj2 j2′ . j2

This is Eq. (30.12). A specific physical example of applying this equation would be the matrix element of a spherical harmonic (which operates only on the orbital part of the wavefunction) between states in which the orbital angular momentum l and the spin angular momentum s have been coupled to a total angular momentum j. Problem 30.6 illustrates.

30.6 From Eq. (30.12),

√ √ ′ ⟨ls j || Y L || l′s j′⟩ = (−1) j+l −L−S (2l + 1) (2 j′ + 1) W ll′ j j′; LS ⟨l || Y L || l′⟩ s (2L + 1) √ ′ j j′ L = (−1) j−1/2 (2 j + 1)(2 j + 1) , 1 — 2 12 0 4π(2l + 1)

where in the second line we have used Eq. (6.74) for the reduced matrix element of the spherical harmonic between states of good l and the explicit Racah coefficient for s = 1 2 from Table 30.1. This result is Eq. (30.13).

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Nuclear Fermion Dynamical Symmetry 31.1 For a clear and concise overview of the quasispin model, see Ring and Schuck [171], Section 6.2. This solution follows that presentation. The commutators may be worked out explicitly. For example [ s(m), s(m) ] = a† a† a

a −a

= a† a† a

a −a

(1 − a† a )a†

= a† a† a

a −a

a† − a

= a† a† a

a − 1 + a† a

= a† a† a

a − 1 + a† a

+

−

a a† a†

m −m −m m

−m m m −m

m −m −m m

−m

m −m −m m

−m

m m

−m −m

a† a a†

−m m m −m

− a† a a a −m −m m −m m −m † † = a a a a − 1 + a † a — a† a a† a m −m −m m −m −m m −m −m m †

m −m −m m

m −m −m m

−m −m

+ a† (1 − a† a

−m −m

m

)a m

= am am + a−m am − 1 †

†

= 2s0(m).

Calculating the other commutators in a similar manner, we obtain (m)

(m)

(m)

(m)

[ s+ , s— ] = 2s0

(m)

(m)

[ s0 , s ± ] = ±s ± ,

which is an SU(2) algebra with s playing the role of a “quasispin”. Quasispin models are of physical relevance for any case in which there are fermion shells with a strong tendency for pairs of particles to couple to zero total angular momentum. Such models and their generalizations have found substantial application in nuclear, atomic, and condensed matter physics.

31.3 Defining the operators S+ ≡ ∑ a† a†

m −m =

m>0

∑ s+(m) m>0

S— ≡ ∑ ama−m = m>0

— ∑ s(m)

m>0

permits the Hamiltonian to be expressed as H = −G

∑ a†m′ a†— m′ a−mam

m,m′ >0

= −GS+S− (m′) (m)

= −G ∑ s+ s — . mm′

134

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But S, and S± are generators of SU(2), so they must obey the relation familiar from angular momentum, 1 2 2 S = (S+S− + S−S+) + S0, 2 where S2 is the total quasispin Casimir operator and 1 † (m) † S0 = ∑ (am am + a −m a−m — 1) = ∑ s0 .

2 m>0 m>0 Therefore, S+S− = 2S2 − 2S02 − S−S+ = S2 − S20 + S0, where [ S+, S− ] = 2S0 was used in the second step. This permits us to express the Hamiltonian entirely in terms of diagonal quasispin operators: H = −GS+S− = −G(S2 − S20 + S0).

Choosing a basis where S2 and S0 (and hence H) are diagonal, ⟨SS0| S 2 |SS0⟩ = S(S + 1).

Furthermore, the operator S0 may be written as 1 1 † † S0 = ∑ (am am + a −m a−m ) — ∑ 1.

2 m>0 2 m>0 The first term just counts the number of pairs of particles N in ±m orbitals in the shell, while the second term counts the total pair degeneracy Ω of the shell (twice the maximum number of particles it could hold). Therefore, 1 ⟨SS0| S0 |SS0⟩ = (N − Ω) 2 and the energy is given by 1 1 E = ⟨SS0 | H |SS0 ⟩ = −G S(S + 1) − (N − Ω) 2 + (N − Ω) .

4 2 31.4 Defining the seniority ν through 1 S = (Ω − ν), 2 permits the energy to be expressed as G 2 E(N, ν) = − ν − 2ν(Ω + 1) + 2N(Ω + 1) − N2 .

4 Successive quasispin multiplets differ in S by 1, so the difference in energy between adjacent multiplets is ∆E = E(S − 1) − E(S) = −G (S − 1)(S − 1 + 1) − S(S + 1) = −G S2 − S − S2 − S = 2GS

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and the spectrum is independent of the number of particles in the shell. Since −GS(S + 1) is an attractive contribution to the Hamiltonian (assuming G > 0), the ground state results from the full alignment of quasispins. Because S = ∑ s(m) m>0

and the length of each s(m) is 12, the ground-state quasispin Sg is 1 Sg = Ω.

2 The ground state is a seniority-zero state since νg = Ω − 2Sg = Ω − 2

Ω

= 0.

2 The first excited state will correspond to the second largest alignment of quasispins, 1 S1 = Ω − 1, 2 and since Ω ν1 = Ω − 2S1 = Ω − 2 + 2 = 2, 2 this is a seniority-two state.

31.7 For n1 = 8 fermions (N1 = 4 fermion pairs), the U(3) Young diagrams have 8 boxes, subject to the condition that no row is longer than the row above. The four Young diagrams meeting these requirements and the associated partitions [h1, h2, h3] are shown in Fig. 31.5.

The quantum numbers (λ, µ) specifying the allowed SU(3) irreps are given by Eq. (31.8) for 2N1 = 8, leading to the SU(3) irreps (λ µ) = {(8, 0),(4, 2),(0, 4),(2, 0)} The spectra associated with these SU(3) irreps may be evaluated using Eqs. (31.7)-(31.12). (a) Consider the (8, 0) irrep. From Eq. (31.10), K0 = 0

Km = 8

κ = 0.

Then from Eq. (31.9) the allowed values of J are J = 0, 2, 4, 6, 8, and Eq. (31.12) gives the Casimir energy C(8, 0) = 88. (b) Consider the (4, 2) SU(3) irrep. From From Eq. (31.10), K0 = 2

Km = 4

κ = 0, 2.

Thus this SU(3) irrep has two possible values of κ. For κ = 0, Eq. (31.9) gives the allowed values J = 0, 2, 4. For κ = 2, Eq. (31.9) gives the allowed values J = 2, 3, 4, 5, 6. For the (4, 2) irrep Eq. (31.12) gives the Casimir energy C(4, 2) = 46.

By a similar procedure we find for the (0, 4) irrep K0 = 0

Km = 4

κ =0

J = 0, 2, 4

C(0, 4) = 28

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and for the (2, 0) irrep K0 = 0

Km = 2

κ=0

J = 0, 2

C(0, 4) = 28

From Eq. (31.11), the states labeled by J are members of rotational bands with energies αJ(J + 1) and the different irreps are offset from each other by βC(λ, µ). This allows construction of Fig. 31.5 [where we have assumed the J = 0 state of the irrep (8, 0) to define the zero of the energy scale].

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Superconductivity and Superfluidity 32.1 That a weakly doped AF Mott insulator contains seeds that allow it to rearrange spontaneously into a superconductor follows directly from SU(4) symmetry, which requires that 1 2 2 2 2 Q + ∆ + Π = (1 − x ), 4 where ∆ is the singlet pair correlation, Π is the triplet pair correlation, and Q is the AF correlation [190]. But for the pure SU(4) ⊃ SO(4) AF solution, Q2 = 14 (1 − x)2. Thus, 1 2 2 2 2 ∆ + Π = (1 − x ) − Q 4 1 1 2 2 = (1 − x ) − (1 − x) 4 4 x(1 − x) = , 2 and even the AF Mott insulator limit has finite pairing correlations ∆2 + Π2 unless x vanishes identically. SU(4) symmetry implies that there is indeed a superconductor lurking in the AF Mott insulator ground state that can be summoned forth by a small amount of doping!

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Current Algebra 33.1 Let ta ≡ τa/2. Then from Eq. (33.4) the commutator with matrix indices displayed explicitly is (sum on repeated indices) [ Qa(t), Qb(t)] =

∫

d 3xd 3x′ itja kltb [ iq†(x)qj(x), qk†(x′)ql(x′)].

We must evaluate commutators of the form [ AB,CD ] where A, B,C, D are fermionic creation and annihilation operators. Expand the commutator using the identity [ AB,CD ] = [ A,C ]DB + C[ A, D ]B + A[ B,C ]D + AC[ B, D ] and replace commutators with anticommutators using [A, B] = 2AB − {A, B} to prove [ AB,CD ] = −AC{ D, B } + {C, A }DB + A{C, B }D −C{ A, D }B. Then substitute A ≡ q†(x) i

B ≡ q j(x)

C ≡ q†(x′)

D ≡ ql(x′)

k

and use the fermion anticommutators { q a (x,t), q †(x ′ ,t)} = δ ( 3)(x − x′)δab

{ q a(x,t), q b(x ′,t)} = { q †(x,t), q †(x ′,t)} = 0 a

b

b

to show that [ qi†(x)qj(x), qk†(x′)ql(x′)] =

δk jqi †(x)ql(x′) − δilqk†(x′)q j(x) δ(3)(x − x′).

Inserting this commutator in the integral for the charge commutator gives [ Qa(t), Qb(t)] = =

∫

∫

= = = = = =

∫ ∫ ∫ ∫

∫ ∫

d 3xd 3x′ ti aj tklb [ qi †(x)qj (x), qk†(x′)ql (x′)] d 3xd 3x′ itja kltb

† † ′ ′ ( 3) ′ δk j q i (x)ql(x ) − δilqk (x )q j(x) δ (x − x )

d 3xi jtakltb δk jiq†(x)ql(x) − δilqk †(x)qj(x) a b † d 3x iq†(x)t (x)ti ja kitb q j(x) i j jlt ql(x) − q k a b † d 3x iq†(x)t (x)tkib itja q j(x) i j jlt ql(x) − q k a b † b a d 3x iq†(x)t il l jt q j(x) − q il ltj q j(x) i (x)t

d 3x qi †(x) ilta ltjb − tilb tlaj q j(x) d 3x q†(x)[ta,tb ]q(x),

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Solutions Manual Symmetry, Broken Symmetry, and Topology in Modern Physics Mike Guidry and Yang

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where in step five we have switched the order of ta and tb in the second term (permitted ij

ki

since these are matrix entries and thus just numbers), in step six we have changed dummy (repeated) indices several times in both terms so that their indices match (permitted because a repeated summation index may be replaced by any index not already in use), and in the final step we have used that the matrix elements of the commutator are [t a ,t b ]i j = ilta ltbj − tilb tlaj .

But for the Pauli matrices 1 a 1 b τ, τ 2 2

[ta,tb ] = Therefore we obtain [ Qa , Q b ] =

= iεabc

τc .

2

∫

d 3x q†(x)[ta,tb ]q(x) ∫ τc 3 † = iεabc d x q (x) q(x) 2 = iεabcQc

for the vector current commutator, which is the first of Eqs. (33.6).

33.3 The Dirac parity operation applied to a wavefunction q(x,t) gives Pq(x,t)P−1 = γ0q(−x,t), up to a phase. Applying this operation to Qi(t) defined in Eq. (33.4), ∫

1 d3x Pq†(x)τi q(x)P−1 21 ∫ = d3x Pq†(x)P−1τ Pq(x)P−1 i 2 ∫ 1 = (γ 0 )2 d 3xq † (−x)τiq(−x) 2 = Qi(t),

PQi (t)P−1 =

where we have used (γ0)2 = 1 and changed integration variables to x′ = −x. In a similar way, applying a parity transformation to Eq. (33.5), ∫

1 d3x Pq†(x)γ5 τi q(x)P−1 2∫ 1 = d3x Pq†(x)P−1γ 5τi Pq(x)P−1 ∫ 21 = − (γ0)2 d3xq†(−x)γ τ q(−x) 5 i 2 = −Qi5,

PQi5 (t)P−1 =

where we have used that { γ5, γ µ } = 0, the same integration change of variable as above,

Solutions Manual Symmetry, Broken Symmetry, and Topology in Modern Physics Mike Guidry and Yang

Current Algebra

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and that (γ0)2 = 1. Therefore, from Eq. (33.7) 1 1 PQi P−1 − PQi5 P−1 = (Qi + Qi5) = QRi , 2 2 1 1 −1 i −1 −1 PQR P = (PQ iP + PQi5 P ) = (Qi − Qi5 ) = QLi .

2 2 Thus the right-handed and left-handed currents are related by parity transformations.

PQLi P−1 =

Solutions Manual Symmetry, Broken Symmetry, and Topology in Modern Physics Mike Guidry and Yang

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Grand Unified Theories 34.1 If ψ annihilates a particle, the charge conjugate field ψc = CψT annihilates an antiparticle. Using the projection operators of Section 14.5.4, 1 c ψc (1 − γ5)ψ L= 2 1 T = (1 − γ5)Cψ 2 1 T = C [ ψ(1 − γ5)] 2 = C (ψ R )T .

Therefore, the charge conjugate of a right-handed field is left-handed.

34.3 Choose T a = T b = T c = Q, where Q is the charge operator, and evaluate 1 abc Tr {Ta(R), Tb(R)} T c(R) = d A(R), 2 for the 5 and 10 representations of Eq. (34.2). For the 5, Tr Q3 5 = 3Q(d¯)3 + Q(νe)3 + Q(e−)3 =3 1 3 8 =− , 9

3

+ 03 + ( —1)3

while for the 10, Tr Q3 (10) = 3Q(u)3 + 3Q(d)3 + 3Q(u¯)3 + Q(e+)3 =3

2 3

8 =+ , 9

3

+3 −

1 3

3

+3 −

2 3

3

+ (1)3

where Table 9.1 was used for the quark charges. Thus A(5) + A(10) = 0 and the SU(5) representation 5 ⊕ 10 is free of anomalies because finite anomalies from the 5 and 10 cancel exactly.

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Solutions Manual Symmetry, Broken Symmetry, and Topology in Modern Physics Mike Guidry and Yang