1.1 Apply Tx = −x. Tx = −x Te x = e(−x) = e−x Here is another approach. Apply the operator to e x expressed in series form. ! x2 x3 x + + ... Te = T 1 + x + 2! 3! ! (−x)2 (−x)3 + + ... = 1−x+ 2! 3! ! x2 x3 = 1−x+ − + ... 2! 3! = e−x The operator has changed the sign of the odd terms.
2
FUNDAMENTAL CONCEPTS
1.2 Apply Tx = −x. T cos (x) = cos (−x) = cos (x) This result can also be understood by considering the series expansion of cos (x). x2 x4 x6 + − + ... 2! 4! 6! T cos (x) = cos (−x) cos (x) = 1 −
(−x)2 (−x)4 (−x)6 + − + ... 2! 4! 6! x2 x4 x6 =1− + − + ... 2! 4! 6! = cos (x) =1−
The series expansion of cos (x) contains only even terms, so Tx = −x changes nothing.
1.3 According to the group axioms (AB)(AB)−1 = E Multiply by A−1 from the left.
A−1 AB(AB)−1 = A−1 E B(AB)−1 = A−1 Multiply by B−1 from the left.
B−1 B(AB)−1 = B−1 A−1 (AB)−1 = B−1 A−1
FUNDAMENTAL CONCEPTS
3
1.4 Check whether the axioms that define a group (Sec. (1.3) of the text) are satisfied for positive and negative real integers under addition. The sum of two integers is also an integer, a member of the set, satisfying axiom (ii) that the group operation does not result in an element outside the set. The sum of an integer N and 0 is equal to N, so 0 acts as the identity element under addition, satisfying axiom (iii) that the group must have an identity element. The sum of an integer N and its negative −N is N − N, equal to the identity element 0. Every integer N has an inverse element −N under addition, so axiom (iv) is satisfied that the group must contain an inverse under the group operation for every element. Trivially, the addition of integers is associative: N1 + (N2 + N3 ) = (N1 + N2 ) + N3 , satisfying axiom(v).
1.5 Check whether the axioms that define a group (Sec. (1.3) of the text) are satisfied by the set of real integers under multiplication. The product of two integers is also an integer, a member of the set, satisfying axiom (ii) that the group operation does not result in an element outside the set. The product of an integer N and 1 is equal to N, so 1 acts as the identity element under multiplication, satisfying axiom (iii) that the group must have an identity element. Axiom (iv) requires that N × N −1 = 1. It follows that N −1 = N1 . The inverse of a real integer under multiplication is not an integer except for 1, so the real integers are not a group because the set of real integers does not include the inverses under multiplication.
4
FUNDAMENTAL CONCEPTS
1.6 The product table for the group {E, A, B} is E E A B
E A B
A A B E
B B E A
The product table shows that B2 = BB = A. Hence the cyclic set {E, B, B2 } has the same elements as the set {E, B, A}.
1.7 The group must have an identity element E. Let the group’s two elements be E and A, where A is distinct from E. A2 can only be A or E. If A2 = A, then A2 = A AA = A Multiplying by A−1 from the left gives
A=E so for this case A = E and A is not distinct. Hence the only possibility is AA = E so the product table must be E A
E E A
A A E
FUNDAMENTAL CONCEPTS
5
1.8 Products of the operations can be found by adding their rotations. For example, CC = 270◦ + 270◦ = 540◦ = 180◦ + 360◦ = BE = B. EE = 0◦ + 0◦ = E
EA = 0◦ + 90◦ = A
EB = 0◦ + 180◦ = B
EC = 0◦ + 270◦ = C
AE = 90◦ + 0◦ = A
AA = 90◦ + 90◦ = B
AB = 90◦ + 180◦ = C
AC = 90◦ + 270◦ = E
BE = 180◦ + 0◦ = B
BA = 180◦ + 90◦ = C BB = 180◦ + 180◦ = E
BC = 180◦ + 270◦ = A
CE = 270◦ + 0◦ = C CA = 270◦ + 90◦ = E CB = 270◦ + 180◦ = A CC = 270◦ + 270◦ = B
The product table is therefore E E A B C
E A B C
A A B C E
B B C E A
C C E A B
The group is cyclic – for example AE = A AA = B AB = AAA = C AC = AAAA = E and similarly for the other group elements.
1.9 Consider the hypothetical product table for the group of distinct operations {E, P, Q, R} where the Q column contains the operation P twice. E P Q R
E E P Q R
P P Q R E
Q Q R P P
R R E P Q
According to the product table QQ = P RQ = P QQ = RQ Multiplying by Q−1 from the right gives Q = R. This is a contradiction because the operations are assumed to be distinct.
6
FUNDAMENTAL CONCEPTS
1.10 The product table for S3 is E P2 P3 P4 P5 P6
E E P2 P3 P4 P5 P6
P2 P2 E P5 P6 P3 P4
P3 P3 P6 E P5 P4 P2
P4 P4 P5 P6 E P2 P3
P5 P5 P4 P2 P3 P6 E
P6 P6 P3 P4 P3 E P5
According to the product table for S3 the set {E, P5 , P6 } has the product table E E P5 P6
E P5 P6
P5 P5 P6 E
P6 P6 E P5
The set {E, P5 , P6 } satisfies the group axioms. Cayley’s theorem says that S 3 has every subgroup of order 3. {E, P5 , P6 } is the only subgroup of order 3 found in S3 and there are no others.
1.11
E K L M
E E K L M
K K E M L
L L M E K
M M L K E
Check that the group axioms are satisfied: (ii) The product of any two operations is a member of the set. (iii) There is an identity element E such that KE = K, etc. (iv) Every element of the set has an inverse. KK = E so that K−1 = K etc. The group is Abelian; all the elements commute. KL = LK = M etc. The group is not cyclic. K2 = E etc. The group elements cannot be obtained by taking powers of one element.
FUNDAMENTAL CONCEPTS
7
1.12 The product table for a group {E, A} of order 2 is E A
E E A
A A E
For distinct elements it follows that AA = E. Inspection of the product table for S3 (see Problem 1.10) shows that Pn Pn = E for n = 2, 3, 4. The three subgroups of order 2 are therefore {E, P2 }, {E, P3 } and {E, P4 }.
1.13 (a) 1 2 3 2 3 1 For the interchange 3 ↔ 1 the bottom row becomes (213). A second interchange 2 ↔ 1 converts the bottom row to standard form (123). Because two interchanges are required, this permutation is even. (b) 1 2 3 4 2 3 1 4 For the interchange 3 ↔ 1 the bottom row becomes (2134). A second interchange 2 ↔ 1 converts the bottom row to standard form (1234). Because two interchanges are required, this permutation is even. (c) 1 2 3 4 4 3 2 1 Converting to standard form requires 2 interchanges: 4 ↔ 1. The bottom row becomes (1324). 3 ↔ 2. The bottom row becomes (1234). This permutation is even.
8
FUNDAMENTAL CONCEPTS
1.14 In cycle notation for permutations an entry maps the entry on its right. Hence for the cycle (536142), 6 7→ 1.
1.15 In cycle notation the entry at the end of the cycle maps around to the entry at the beginning. Hence for the cycle (326415), 5 7→ 3.
1.16
Let c be the length of the side of the equilateral triangle. The “flip” in this problem is a rotation about aa by π. Before the flip the coordinates are √ 3 xt = 0, yt = 2 c (top apex) xr = 2c , yr = 0 (bottom right apex) x` = − 2c , y` = 0 (bottom left apex) After the flip√the coordinates are xt0 = 0, y0t = 23 c xr0 = − 2c , y0r = 0 x`0 = + 2c , y0` = 0 continued next page =⇒
FUNDAMENTAL CONCEPTS
9
The representation matrix has the general form ! ! ! A11 A12 xi xi0 = A21 A22 yi y0i
Written out, the matrix form is xi0 = A11 xi + A12 yi y0i = A21 xi + A22 yi To start evaluating the terms Ai j , note that y0i is independent of xi . Therefore A21 = 0. Because y0i = yi , A22 = 1. Similarly A12 = 0 because xi0 is independent of yi . A11 = −1 because xr and x` change sign from the flip. The result for the representation matrix is −1 0 0 1
1.17 The product table for the group Γ = {E, A, B} is
E A B
E E A B
A A B E
B B E A
A correct homomorphism must give product results that agree with the product table. For example, according to the product table AB = E. The corresponding homomorphism product (-1) (-1) = (1) agrees. However, AA = B. The homomorphism (-1) (-1) = (1) does not agree. The given matrices are not a homomorphism for Γ.
10
FUNDAMENTAL CONCEPTS
1.18 Consider C = AB where A and B are n × n diagonal matrices. The elements of A and B satisfy Ai j = 0
i, j
Bk` = 0
k,`
The elements of C are Ci` =
n X
Ai j B j`
j
Ai j = 0 if i , j.
Ci` = Aii Bi` Bi` = 0 if i , ` so C has only diagonal elements Cii = Aii Bii .
1.19 The group operations are E: rotate by 0◦ A: rotate by 90◦ B: rotate by 180◦ C: rotate by 270◦ The required matrix was derived in Sec. (1.5.3). cos θ sin θ − sin θ cos θ 1 0 0 1
0 1 −1 0
−1 0 0 −1
0 −1 1 0
D(E)
D(A)
D(B)
D(C)
FUNDAMENTAL CONCEPTS
11
1.20 A matrix has an inverse only if its determinant is not 0. Otherwise, it would be possible to use Cramer’s rule to calculate an inverse. (a) det
1 0 = (1)(−1) = −1 0 −1
This matrix has an inverse. (b) det
1 −1 = (1)(−1) − (1)(−1) = 0 1 −1
This matrix does not have an inverse. (c) 1 3 2 det 3 0 −1 = (1)(−2) − (3)(3 + 2) + (2)(−6) = −29 2 −2 1 This matrix has an inverse.
12
FUNDAMENTAL CONCEPTS
1.21 Let C be the resultant matrix in each case. C = A+B Ci j = Ai j + Bi j 3 2 −3 5 2 −3 C = −2 3 2 + 1 3 1 1 0 4 −2 0 5 8 4 −6 = −1 6 3 −1 0 9 C = A−B Ci j = Ai j − Bi j −2 0 0 C = −3 0 1 3 0 −1 C = AB X Ci j = Aik Bk j k
(3)(5) + (2)(1) + (−3)(−2) (3)(2) + (2)(3) + 0 (3)(−3) + (2)(1) + (−3)(5) C = (−2)(5) + (3)(1) + (2)(−2) (−2)(2) + (3)(3) + 0 (−2)(−3) + (3)(1) + (2)(5) (1)(5) + 0 + (4)(−2) (1)(2) + 0 + 0 (1)(−3) + 0 + (4)(5) 23 12 −22 = −11 5 19 −3 2 17 C = BA X Ci j = Bik Ak j k
(5)(3) + (2)(−2) + (−3)(1) (5)(2) + (2)(3) + 0 (5)(−3) + (2)(2) + (−3)(4) C = (1)(3) + (3)(−2) + (1)(1) (1)(2) + (3)(3) + 0 (1)(−3) + (3)(2) + (1)(4) (−2)(3) + 0 + (5)(1) (−2)(2) + 0 + 0 (−2)(−3) + 0 + (5)(4) 8 16 −23 = −2 11 7 −1 −4 26
FUNDAMENTAL CONCEPTS
13
1.22
Let C be the resultant matrix in each case.
C = A+B Ci j = Ai j + Bi j −1 5 4 2 −2 −1 C = −3 3 4 + 5 1 4 2 0 −3 −3 0 2 1 3 3 = 2 4 8 −1 0 −1 C = A−B Ci j = Ai j − Bi j −3 7 5 C = −8 2 0 5 0 −5 C = AB X Ci j = Aik Bk j k
(−1)(2) + (5)(5) + (4)(−3) (−1)(−2) + (5)(1) + 0 (−1)(−1) + (5)(4) + (4)(2) C = (−3)(2) + (3)(5) + (4)(−3) (−3)(−2) + (3)(1) + 0 (−3)(−1) + (3)(4) + (4)(2) (2)(2) + 0 + (−3)(−3) (2)(−2) + 0 + 0 (2)(−1) + 0 + (−3)(2) 11 7 29 = −3 9 23 13 −4 −8 C = BA X Ci j = Bik Ak j k
(2)(−1) + (−2)(−3) + (−1)(2) (2)(5) + (−2)(3) + 0 (2)(4) + (−2)(4) + (−1)(−3) C = (5)(−1) + (1)(−3) + (4)(2) (5)(5) + (1)(3) + 0 (5)(4) + (1)(4) + (4)(−3) (−3)(−1) + 0 + (2)(2) (−3)(5) + 0 + 0 (−3)(4) + 0 + (2)(−3) 2 4 3 = 0 28 12 7 −15 −18
14
FUNDAMENTAL CONCEPTS
1.23
2 A = 2 3 1 B = 0 2 6 AB = 8 1
−1 2 0 3 −1 −1 2 −2 3 −1 1 −2 3 −7 7 −10 2 −3
det(A) = (2)(3) − (−1)(−11) + (2)(−2) = −9 det(B) = (1)(−5) − (2)(2) + (−2)(−6) =3 det(A) det(B) = (−9)(3) = −27 det(AB) = (6)(−1) − (3)(−14) + (−7)(9) = −27
1.24 Ã is the transpose of matrix A so its elements are (Ã)i j = A ji . f i j = (AB) ji (AB) X = A jk Bki =
k X
Ãk j B̃ik
k
=
X k
= B̃Ã
B̃ik Ãk j
FUNDAMENTAL CONCEPTS
1.25
1 5 2 A = 3 0 −1 4 −2 1 (A∗ )i j = A∗i j 1 5 2 = 3 0 −1 4 −2 1 (Ã)i j = A ji 1 3 4 = 5 0 −2 2 −1 1 (A† )i j = (A∗ ) ji 1 3 4 = 5 0 −2 2 −1 1
15
16
FUNDAMENTAL CONCEPTS
1.26
1 3+i 2i A = 3 0 −1 + 2i 4 − 3i −2 1−i (A∗ )i j = A∗i j 1 3−i −2i = 3 0 −1 − 2i 4 + 3i −2 1+i (Ã)i j = A ji 1 3 4 − 3i = 3 + i 0 −2 2i −1 + 2i 1 − i (A† )i j = (A∗ ) ji 1 3 4 + 3i = 3 − i 0 −2 −2i −1 − 2i 1 + i
2.1 In the problem the reference to Sec. (1.4) should instead be to Sec. (1.5.3). Let T be rotation by θ and let φ1 = x, φ2 = y.If x and y are truly basis functions for this operation they must satisfy Tφ j = Tx = D11 x + D12 y Tφ2 = Ty = D21 x + D22 y In Sec.(1.5.3) Eqs. (1.1) and (1.2), the expression for rotation of the x − y axes is Tx = x cos θ + y sin θ Ty = −x sin θ + y cos θ The operation of T on x and y give terms involving only x and y. Hence x and y qualify as basis functions. continued next page =⇒
18
MATRIX REPRESENTATIONS OF DISCRETE GROUPS
D11 = cos θ D12 = sin θ D21 = − sin θ D22 = cos θ cos θ sin θ D(T) = − sin θ cos θ This is the same as the matrix in Sec. (1.5.3) for rotated axes. The matrix D(T) for θ = 120◦ is therefore √ 3 − 1 2 2 √ D(T) = 3 1 − 2 −2
2.2 In the problem the reference to Sec. (1.4) should instead be to Sec. (1.5.3). In Sec.(1.5.3) Eqs. (1.1) and (1.2), the expression for rotation of the x − y axes is Tx = x cos θ + y sin θ Ty = −x sin θ + y cos θ Tx2 = (x cos θ + y sin θ)2 = x2 cos2 θ + y2 sin2 θ + 2xy cos θ sin θ Ty2 = (−x sin θ + y cos θ)2 = x2 sin2 θ + y2 cos2 θ − 2xy cos θ sin θ T(x2 − y2 ) = (x2 − y2 )(cos2 θ − sin2 θ) + 4xy cos θ sin θ T(xy) = (x cos θ + y sin θ)(−x sin θ + y cos θ) = −(x2 − y2 ) cos θ sin θ + xy(cos2 θ − sin2 θ) T(x2 −y2 ) and T(xy) contain only the functions (x2 −y2 ) and (xy) so these are indeed basis functions referring to the same subspace. continued next page =⇒