SOLUTIONS MANUAL for Introduction to Quantum Computing by Ray LaPierre. ISBN-13 978-3030693206.

The Solutions Manual for Introduction to Quantum Computing

by Ray LaPierre Chapter 1 Exercise 1.1. Derive Eq. (1.11): I12  | E1 + E2 |2 = I1 + I2 + 2√I1I2 cos  E1 = E01 ei(kx−t) E2 = E02 ei(kx−t+) E1E2* = (E01ei(kx−t))(E02ei(kx−t+)* = E01E02e−i , assuming E01 is parallel to E02 (same polarization) so E01E02 = E01E02 E2E1* = (E02ei(kx−t+)(E01ei(kx−t))* = E01E02e+i, assuming E01 is parallel to E02 (same polarization) E1E2* + E2E1* = E01E02 (e+i + e−i) = 2E01E02cos  2√I1I2cos | E1 + E2 |2 = (E1 + E2)( E1 + E2)*  I12

= E1E1* + E2E2* + E1E2* + E2E1*  I1

 I2

 2√I1I2cos

 I12 = I1 + I2 + 2√I1I2cos

Chapter 2 Exercise 2.1. Show that the first two states (two lowest energy levels) of the infinite quantum well are orthonormal.

2 π 1(x) = √ sin ( x) L L 2 2π 2 (x) = √ sin ( x) L L +∞

∫ −∞

1∗ 2 dx

π 2π = ∫ sin ( x) sin ( x) dx L L 0 L =

2

L

4

L

π

π

π

∫ sin (L x) sin (L x) cos (L x) dx, using sin(2x)=2sinxcosx L 0 4 L π π = ∫ sin2 ( x) cos ( x) dx L L 0 L 4 sin3 π L ( x)| = 3π 0 L 4 = [sin3(π) − sin3(0)] 3π =0

∫

+∞

∗  dx

−∞

2

1

1

π π = ∫ sin ( x) sin ( x) dx L L 0 L L 2 π = ∫ sin2 ( x) dx L 0 L 1 L 2π 2 = [1 − cos ( x)] dx, using 2sin (x) = 1−cos(2x) ∫ L 0 L 1 L 1 2π L | = x − sin ( x)| 0 L 0 2π L L

=1

Similarly, +∞

2∗ 2 dx = 1

∫ −∞

∫

+∞

∗  dx = δ where i, j = {0, 1} i

−∞

ij

j

 1(x) and 2(x) are orthonormal.

Exercise 2.2. Prove Eq. (2.49): <A> = |c1|2 a1 + |c2|2 a2 + … + |cn|2 an  = c11 + c22 + … cnn ∫

+∞

̂  dX ∗ A

<A> = −∞+∞

∫−∞ ∗ dX

The numerator is: +∞

̂  dx ∗ A

∫ −∞

=∫

+∞

̂ (c  (c ∗ ∗ + c ∗ ∗ + c ∗ ∗ ) A 1

1

2

2

n

n

1

1

+ c  +. . . c  ) dx 2

2

n

n

−∞

=∫

+∞

(c∗∗ + c∗∗ + c∗ ∗ ) (a c  + a c  +. . . a c  ) dx 1

1

2

2

n

n

1 1

1

2 2

2

n n

n

−∞

= c∗c1a1 + c∗c2a2+. . . c∗ c a 1

2

n n n

= |c1|2a1 + |c2|2a2+. . . |cn|2an All other terms in the numerator, like a c∗c ∫

+∞

2 1 2 −∞

orthonormal.

Similarly, the denominator is: +∞

∫ −∞

∗  dx

∗  dx, are zero since  and  are 1

2

1

2

=∫

+∞

(c∗∗ + c∗∗ + c∗ ∗ ) (c  + c  +. . . c  ) dx 1

1

2

2

n

n

1

1

2

2

n

n

∗  dx+. . . c∗ c ∫

+∞

−∞

= c∗c1 ∫

+∞

∗ dx + c∗c ∫

1

1

1

+∞

2 2

−∞

2

2

∗  dx

n n

−∞

n

n

−∞

= |c1|2 + |c2|2+. . . |cn|2 , assuming each i is normalized = 1 , since all probabilities must sum to 1 All other terms in the denominator, like c∗c ∫

+∞

1 2 −∞

∫

+∞

 <A> = −∞

∗̂

 A  dX

+∞ ∫−∞ ∗ dX

∗  dx, are zero.

1

= |c |2a + |c |2a +. . . |c |2a 1

1

2

2

n

n

2

Chapter 3

Exercise 3.1. How might you use the SG experiment to build a random number generator?

Pass atoms from an oven with random spin through a Stern-Gerlach apparatus. If the S-G apparatus passes spin up, then assign bit 0 to the atom. If the S-G apparatus passes spin down, then assign bit 1 to the atom. The result is a random sequence of 0’s and 1’s. A string of n bits can represent any number from 0 to 2n-1. Therefore, the random string of 0’s and 1’s can be converted to a random number.

Exercise 3.2. Show that Eq. (3.38) and (3.39) satisfy Eq. (3.33) to (3.36). ħ

ŜX x+ ⟩ = + 2 x+ ⟩ ħ 0 1 1 1 ( ) ( ) = ħ 1 (1) = + ħ x ⟩ 2 1 0 √2 1 2 √2 1 2 + ŜX x− ⟩ = − 2ħ x − ⟩ ħ 0 1 1 ( ) ( 1 ) = ħ 1 (−1) = − ħ x ⟩ 2 1 0 √2 −1 1 2 √2 2 − ħ Ŝ y ⟩ = + y ⟩ y

(0 2 i

+

ħ

2

+

−i ) 1 (1) = ħ 1 (1) = + ħ y  ⟩ 2 √2 i 2 + 0 √2 i

Ŝ y ⟩ = − ħ y ⟩ y − − ħ 0 −i ) 21 ( 1) = ħ 1 (−1) = − ħ y (  ⟩ 2 i 2 √2 i 2 − 0 √2 −i ħ Ŝz ⟩ = + z ⟩ z +

+

0 ) (21) = ħ ( 1) = + ħ z  ⟩ 0 2 0 −1 2 0 2 + ħ

(1

Ŝz ⟩ = − ħ z ⟩ z − − ħ 1 0 )2( 0) = ħ ( 0 ) = − ħ z (  ⟩ 1 2 0 −1 2 −1 2 −

Exercise 3.3. Prove that any complex 2x2 matrix can be written as I + ̂σX + σ ̂y + ̂σz where , ,  and  are complex numbers.

I + σ ̂X + σ ̂y + ̂σz = (

=(

1 0 0 1 + 0 ) ( ) + ( 1 0 0 1 i

α+γ β + δi

−i 0) ) + (1 0 −1 0

β − δi ) α−γ

Equate this to a general 2x2 complex matrix: (

α+γ

β − δi

a )=( b α−γ

β + δi

c ) where a, b, c, d are complex numbers d

a=α+γ b = β + δi c = β − δi d=α−γ To generate the complex 2x2 matrix, set the following parameters:  = (a+d)/2  = (b+c)/2  = (b−c)/2i  = (a−d)/2

Exercise 3.4. Prove the following relations: X2 = Y2 = Z2 = − i XYZ = I Z = − i XY ZY = − i X ZYX = − i I YX = − i Z X2 = (0 1) ( 0 1 0 1

Y2 = (

0 i

−i 0 )( 0 i

1 1 0 )=I )=( 0 0 1 −i ) = (1 0) = I 0 0 1

0 1 0 1 Z 2 = (1 )( )=( 0 −1 0 −1 0

0 )=I 1

0 1 0 − i XYZ = − i ( )( 1 0 i

0) = − i 0 −i 1 )( ( 0 0 −1 1

− i XY = − i 0 1) (0 ( 1 0 i

−i) = − i ( i 0 0

0 0 ZY = (1 )( 0 −1 i

−i ) = ( 0 0 −i

1 0 i) = − i (i )( 0 i 0 0

0 1 0 )=( )=I i 0 1

1 0 0 = )=Z ) ( −i 0 −1

−i ) = − i (0 1) = − i X 0 1 0

ZYX = (ZY)X = (− i X)X = − i (XX) = − i I YX = (0 i

−i 0 )( 0 1

1 −i )=( 0 0

0 1 0 ) =−i( )=−iZ i 0 −1

ħ

Exercise 3.5. Prove Eq. (3.50) from Eq. (3.49). Show that the Ŝ n operator has eigenvalues ± and 2 corresponding eigenvectors (

cos θ 2

eisin

θ) and ( 2

sin

θ

2

− eicos

θ

)

2

nx = cossinθ ny = sinsinθ nz = cosθ Ŝn = nX ŜX + ny Ŝy + nz Ŝz ħ cossinθ

=2

ħ

= ( 2

0 ħ 1) + sinsinθ (0 ( 2 1 0 i

cosθ cossinθ + isinsinθ

cosθ 2 (cos + isin)sinθ ħ

= (

−i ) + ħ cosθ ( 1 0) 2 0 0 −1

cossinθ − isinsinθ ) −cosθ (cos − isin)sinθ ) −cosθ

−i ħ = ( cosθ e sinθ) 2 eisinθ −cosθ

cos

Ŝn ( i

θ

ħ

2 )=

2 ( i

θ

e sin2

cosθ

e sinθ

e−isinθ ) ( cosθ i

−cosθ

ħ

θ

e sin2

θ

cosθcos θ+ sinθsin θ

cos ħ 2 ) ) = ( i 2 θ θ  e (sinθcos2 − cosθsin2) e sin2

2 ) = 2 ( i

2θ

2

where we have used the identities cox(x−y) = cos(x)cos(y) +sin(x)sin(y), and sin(x−y) = sin(x)cos(y) – cos(x)sin(y) θ

n

Ŝ (

sin i2

−i

θ

− e cos2 θ

2

θ

θ

θ

cosθsin2− e sinθ) ( i θ θ sinθcos2 θ ) ħ ħ sin(− i 2θ ) = 2 ( i )= 2( )= e sinθ −cosθ − e cos2 e (sinθsin2 + cosθcos2) e cos2

ħ i ) = 2 ( cosθ

sin2 ħ − 2 (−eicos θ)

θ

sin2

Chapter 4

1

Exercise 4.1. Show that the following state is normalized: ( + 1 i)  − 1  2

2

√2

1 1 2 1 2 1 1 | + i| + | | = + = 1 2 2 2 2 √2

Exercise 4.2. Rewrite 0> and 1> in terms of |+⟩ and |−⟩. Show that |+⟩ and |−⟩ are orthonormal. |+⟩ = |−⟩ =

1 √2 1 √2

( |0⟩ + |1⟩ ) ( |0⟩ − |1⟩ )

Rearranging gives: 1 |0⟩ = ( |+⟩ + |−⟩ ) √2 |1⟩ =

1

√2

( |+⟩ − |−⟩ )

The bras are: 1 ⟨+| = (⟨0| + ⟨1|) ⟨−| =

√2 1

(⟨0| − ⟨1|)

√2

The inner products are: 1

1

1

1

⟨+|−⟩ = (⟨0| + ⟨1|) ( |0⟩ − |1⟩ ) 1 ⟨ | ⟩√2 ⟨ | ⟩ ⟨ |√2⟩ ⟨ | ⟩ = 2( 0 0 − 0 1 + 1 0 − 1 1 ) 1 = (1 − 0 + 0 − 1) 2

=0 ⟨+|+⟩ = (⟨0| + ⟨1|) ( |0⟩ + |1⟩ ) 1 ⟨ | ⟩√2 ⟨ | ⟩ ⟨ |√2⟩ ⟨ | ⟩ = 2( 0 0 + 0 1 + 1 0 + 1 1 ) 1 = (1 + 0 + 0 + 1) 2

=1 ⟨−|−⟩ =

1

(⟨0| − ⟨1|)

√2

1

( |0⟩ − |1⟩ )

√2

1 ⟨ | ⟩ ⟨ | ⟩ ⟨ | ⟩ ⟨ | ⟩ = 2( 0 0 − 0 1 − 1 0 + 1 1 )

1

= (1 − 0 − 0 + 1) 2

=1

Exercise 4.3. Prove the following: (a) <12>* = <21> (b) <12> can be a complex number, but <11> is real and positive (c) <c12> = c*<12> (d) <1c2> = c<12> α δ Suppose |1> = (β) and |2> = (γ) Then <1| = (α∗β∗) and <2| = (δ∗γ∗)

(a)

δ ∗ <12>* = [(α∗β∗) (γ)] = [α∗δ + β∗γ]∗ = αδ∗ + βγ∗ α ∗ ∗ ∗ ∗ <  > ( ∗ ∗) 2 1 =[ δ γ (β)] = δ α + γ β = αδ + βγ  <12>* = <21> (b) δ <12> = (α∗β∗) ( ) = α∗δ + β∗γ , which in general is a complex number γ 2 ∗ ∗ <  > ( ∗ ∗) α | |2 | | , which is real and positive 1 1 = α β (β) = α α + β β = α + β (c) <c1| = (|c1>)† = (c|1>)† = c*<1|  <c12> = c*<12> (d) c2> = c2>  <1c2> = <1c2> = c<12>

Exercise 4.4. Prove the orthonormal relations listed above for the spin states.

Remember that the bras are the conjugate transpose of the kets. Therefore, x > = +

x−> = y > = +

y−> =

1

1

(1), <x  = +

√2 1 11

(1 1)

√2 1

( ) , <x− =

√2 1 1−1

1√2

( ) , <y  = +

√2 1 i1

(1

(1

√21

( ) , <y− =

√2 −i

√2

−1) −i)

(1

i)

z+> = (10) , <z+ = (1 0) z−> = (01), <z− = (0 1) <x x−> =

1

(1

1)

<x x > = √2 (1

1)

+

1

+ +

1

+

1

+ +

1

<x−x−> = √2 (1

1

1

( 1) = (0) = 0

√2 1 1−1

1 2

√2 1 1

2

( ) = (2) = 1 1

−1) ( 1) = (2) = 1 √2 √2 −1 1 1 12 <y y−> = (1 −i) ( 1) = (0) = 0 <y y > = √2 (1

−i)

<y−y−> = √2 (1 i) √2

√12 −i 1

1 2

( ) = (2) = 1

1 √2 1

i

12

( ) = (2) = 1

√2 −i

2

<z+z−> = (1 1 <z+z+> = (1 0)(10) = 1 <z−z−> = (0 1)(01) = 1 0)(0) = 0

Exercise 4.5. Show that each Pauli operator can be expressed in terms of an outer product as follows: 0 1 ̂ =[ σ ] = |1> <0| + |0> <1| X 1 0 0 −i ̂ =[ σ ] = i|1> <0| − i|0> <1| y i 0 1 0 ̂σ = [ ] = |0> <0| + |1> <1| z 0 −1 1 ) = ̂σX 1 0 1 (1 0) 0 (0 1) 1 0 0 −i i|1> <0| − i|0> <1| = i(0) (1 0) − i(1) (0 1) = i(0 (1 0)) − i(1 (0 1)) = ( )=σ ̂y 1 0 1 (1 0) 0 (0 1) i 0 |1> <0| + |0> <1| = (0) (1 0) + (1) (0 1) = (0 (1 0)) + (1 (0 1)) = (

0

|0> <0| − |1> <1| = (1) (1 0) − (0) (0 1) = (1 (1 0)) − (0 (0 1)) = ( 0

1

0 (1 0)

1 (0 1)

1

0

)=σ ̂z 0 −1

Exercise 4.6. Show that 0> <0 + 1> <1 = I, the identity matrix. This is called the “resolution of the identity” or the “completeness relation”.

1 0> <0 + 1> <1 = (1) (1 0) + (0) (0 1) = (1 (1 0)) + (0 (0 1)) = ( 0

1

0 (1 0)

1 (0 1)

0 )= I 0 1

Exercise 4.7. How would you write the vectors for |00>, |01>, |10> and |11>? Show that these four states are orthonormal.

In general, for two qubits:

α00 > =  00> +  01> +  10> +  11> with matrix representation > = (αα1001)     α11

Thus, |00> =

1

0

0

0

( 0)

( 1)

( 0)

( 0)

0 0

, |01> =

0 0

, |10> =

1 0

, |11> =

0 1

0

<00|01> = (1 0 0 0)

( 1) = 0 0 0

Similarly, <00|10> = <00|11> = <01|00> = <01|10> = <01|11> = <10|00> = <10|01> = <10|11> = <11|00> = <11|01> = <11|10> = 0

1 0 <00|00> = (1 0 0 0) ( 0) = 1

0

Similarly, <01|01> = <10|10> = <11|11> = 1 |00>, |01>, |10> and |11> are orthonormal

Exercise 4.8. Show that the following state is normalized: 1 i 1 ’> = [( + ) 00> + |01>] / √¾ 2

2 1

2 i

2

1 2

1 1 1

2

4 4 4

|( + )| +|( )|

<’’> =

2 2

2

(√¾)

+ +

=

¾

=1

Chapter 5

Exercise 5.1. Show that P(z−z+) = ½ P(z−z+) =  <z−z+ [1 (z +z−> − z−z +>)] 2 √2

= ½  <z−z+z+z−> − <z−z+z−z+> 2 = ½  <z−z+> <z+|z−> − <z−|z−> <z+ z+> 2 = ½  (0)(0) – (1)(1) 2 =½

Exercise 5.2. Prove P(z+x−) = ¼, P(z−x+) = ¼, and P(z−x−) = ¼.

P(z+x−) =  <z+x− [1 (z +z−> − z−z +>)] 2 √2

= ½  <z+x−z+z−> − <z+x−z−z+> 2 = ½  <z+z+> <x−z−> − <z+z−> <x−z+> 2 = ½  (1) <x−z−> − (0) <x−z+> 2 = ½  <x−z−> 2 1 Substituting <x−|= (<z+ − <z−), we get: √2

P(z+x−) = ½  1 (<z+ − <z−) z−> 2 √2

= ½  1 (<z+ z−> − <z−z−>) 2 √2

= ½  1 (0 − 1) 2 √2

=¼ P(z−x+) =  <z−x+ [1 (z +z−> − z−z +>)] 2 √2

= ½  <z−x+z+z−> − <z−x+z−z+> 2 = ½  <z−z+> <x+z−> − <z−z−> <x+z+> 2 = ½  (0) <x+z−> − (1) <x+z+> 2 = ½  0 − <x+z+> 2 = ½  <x+z+> 2 1 Substituting <x+|= (<z+ + <z−), we get: √2

P(z−x+)

= ½  1 (<z  + <z−) z > 2 +

+

2 = ½  √2 1 (<z  z > + <z−z >)  +

√2

+

+

= ½  1 (1 + 0) 2 √2

=¼ P(z−x−) =  <z−x− [1 (z +z−> − z−z +>)] 2 √2

= ½  <z−x−z+z−> − <z−x−z−z+> 2 = ½  <z−z+> <x−z−> − <z−z−> <x−z+> 2 = ½  (0) <x−z−> − (1) <x−z+> 2 = ½  0 − <x−z+> 2 = ½  <x−z+> 2 1 Substituting <x−|= (<z+ − <z−), we get: √2

P(z−x−) = ½  1 (<z  − <z−) z > 2 +

+

2 = ½  √2 1 (<z  z > − <z−z >)  +

√2

+

= ½  1 (1 − 0) 2

+

√2

=¼

Exercise 5.3. Show that the other terms in Bell’s are: 1 inequality π P(b c ) = sin2 ~ 0.073 + +

P(a+c+) =

12

8

4

P(b+c+) =  <b c  1 ( z z−> − z−z > ) 2 1

+ +

+

+

=  <b c √2 z z−> − <b+c+z−z+> 2 1

+ + +

2

+ +

= 2  <b z > <c z−> − <b z−> <c z > 2 +

+

+ +

π

π

1

By definition, <b+| = cos 8 <z+| + sin 8 <z−| , <c+| = <x+| = √2 (<z+| + <z−|) π

π

π

<b+z+> = cos (<z+| + sin <z−|) |z+> = cos 8 8 8 <c+z−> =

1

√2

(<z+| + <z−|) |z−> = π

π

1

√2

π

<b+z−> = cos (<z+| + sin <z−|) |z−> = sin 8 8 8 1 1 <c+z+> = (<z+| + <z−|) |z+> = √2

√2

1P(b+c+) =  <b z > <c z−> − <b z−> <c z > 2 2 1

+ +

1

+

π

+

+ +

π 2

1

 √2 cos 8 − √2 sin 8  π π =  cos − sin 2 =

2 1

4

8

8

Use the following trigonometric identity: sin(x-y) = sinxcosy – cosxsiny Substitute x = /4, y = /8 π π π π π 1 π π sin(− ) = sin cos – cos sin = (cos – sin ) π4

18

8π √2

8 π

4 π

8 π

8 π

8

8

8

1

π

π

1

4

8

8

4

sin(− ) = sin =

√2

π

√2sin = cos – sin

8

8

8

 P(b c ) =  cos − sin 2 = + +

8

(cos – sin )

 √2sin 2 = sin2

π

1

π

8

2

8

P(a+c+) =  <a c  1 ( z z−> − z−z > ) 2 1

+ +

+

+

=  <a c √2 z z−> − <a +c+z−z +> 2 1

+ + +

2

+ +

= 2  <a z > <c z−> − <a +z−> <c +z +> 2 +

By definition, <a+ = <z+ , <c+| = <x+| = <a+z+> = <z+z+> = 1 <a+z−> = <z+z−> = 0 <c+z−> = <c+z+> =

1

1

1

1

(<z+| + <z−|) |z−> = √2 √2 (<z+| + <z−|) |z+> = √2 √2

1P(a+c+) =  <a z > <c z−> − <a z−> <c z > 2 = =

2 1 2 1 4

1

 √2

+ +

− 0 2

+

+

+ +

1 √2

(<z−| + <z −|)

Exercise 5.4. Show the above Bell inequality is satisfied if the initial state is not entangled.

Choose a separable two-qubit state, like |> = |z+z+> By definition, <a+ = <z+ , <a− = <z− , <b+| = cos π <z+| + sin 8

<a+|z+> = <z+|z +> = 1 π π π <b+z+> = cos (<z+| + sin <z−|) |z+> = cos 8

1

8

1

<c+|z+> = √2 (<z+| + <z−|) |z+> = √2  P(a+b+) =  <a+b+|z+z+> 2 =  <a+|z+> <b+|z+> 2 2 =  (1) (cos π)  8

π = cos2 8

P(b+c+) =  <b+c+z+z+> 2 =  <b+|z+> <c+|z+> 2 =  (cos π) ( 1 ) 2 1

8 π

2

8

= cos2

√2

P(a+c+) = 1<a+c+z+z+> 2 =  <a |z > <c |z > 2 +

+

2 1 = 1  (1) ( ) 2 2 √2 1

=

+

+

4

P(a+b+) + P(b+c+)  P(a+c+) π 1 π 1 cos2 + cos2  8

2

3 cos2 π  1

2

8

4

8

4

1.28  0.25  the Bell inequality is satisfied

8

π

1

<z−| , <c+| = √2 (<z+| + <z −|) 8

Exercise 5.5. For the entangled state of Figure 5.3, show in general that P = ½ sin2(/2) where  is the angle between Bob’s and Alice’s spin measurement direction. θ

θ

2

2

A general single-qubit state is: ⟩ = cos ( ) z+⟩ + ei sin ( ) z−⟩ A general two-qubit state is: θ1 12⟩ = [cos (

θ1

i

θ2

i

) z+⟩ + e 2 sin ( ) z−⟩] 2 2 θ1 θ2 θ1 θ2 2 1 i i ⟩ ⟩ z + e z z + e z z ⟩ + = cos (θ1 ) cos ( θ2) z cos ( ) sin ( ) sin ( ) cos ( ) + + + − − + 2 i( + ) 1 2 e

2

) z+⟩ + e 1 sin (

θ2

2

θ1

θ2

sin ( ) sin ( ) 2

2

z z ⟩

2

) z−⟩] [cos (

2

2

2

2

− −

P(12) =  <12  [1 (z +z−> − z−z +>)] 2 √2

θ

θ

1 2 θ1 ⟨ |+ θ2 i1 = | [cos (θ1 ) cos (θ2) z z | + ei2 cos ( ) sin ( ) ⟨ z+z− | + e sin ( ) cos ( ) z−z+ ⟨ + + 2 2 2 2 2 2

θ1

i( + )

e =

1

1 2

2

θ2

2

1

sin ( 2 ) sin ( 2 ) ⟨z−z−|] | √2 (|z+z−⟩ − |z−z+⟩) |

|ei 2 cos (

θ1

θ2

2

2

 ) sin (

) − e i 1 sin (

θ1

θ  ) cos ( 2) |

2

2

2

Choose x-z plane as the measurement plane, so 1 = 2 = 0: P(  ) = 1 1

=

1 2

2

2

θ1

θ2

θ1

θ2

|cos ( ) sin ( ) − sin ( ) cos ( ) | 2 2 2 2

θ −θ1

|sin ( 2

)|

2

2

1 θ = sin2 (2) where θ = θ2 − θ1 2

2

Chapter 6

Exercise 6.1. Why do we need single photons for QKD to be secure? Why not use “classical light” composed of many photons to represent 0 and 1?

With classical light, a fraction of the many photons (down to a single photon) comprising the state could be split off and measured without detection. A single photon, representing a single qubit, cannot be split.

Chapter 7 Exercise 7.1. Show that X is a unitary transformation.

0 1 ) , X† 0 1 X=( ) =( 1 0 1 0 0 1 0 1 1 0 XX† = ( )= I )= ( )( 1 0 1 0 0 1  X is unitary Exercise 7.2: Suppose > = 0> + 1>. Using the matrix and Dirac notation, determine NOT>.

Dirac notation: NOT> = NOT (0> + 1>) = NOT0> + NOT1> = 1> + 0>

Matrix representation: 0 1 α ) ( ) = (β) = β(1) + α(0) = 1> + 0> NOT> = ( α 0 1 1 0 β Exercise 7.3. Show that the outer product notation for Y gives the correct output. Show that the matrix for Y gives the correct output. Show that Y is unitary.

Y=(

0 i

−i

) = i|1> <0| − i|0> <1| 0

Dirac notation: For |0> input: Y|0> = (i|1> <0|−i|0> <1|)|0> = i|1> <0|0>−i|0> <1|0> = i|1> (1) − i|0> (0) = i|1> For |1> input: Y|0> = (i|1> <0|−i|0> <1|)|1> = i|1> <0|1>−i|0> <1|1> = i|1> (0) − i|0> (1) = −i|0> Matrix representation: 0 0 Y|0> = 0 −i 1 ( ) (0) = ( i ) = i(1) = i|1> 0 0 −i 1 Y|1> = 0i −i ( ) ( ) = ( ) = −i( ) = −i|0> 0 0 i 0 1

Y=(

0 i

−i

0 ) , Y† = ( i 0

−i) 0

Y Y† = ( 0i

−i 0 )( 0 i  Y is unitary

−i ) = (1 0) = I 0 0 1

Exercise 7.4. Show that the outer product notation for Z gives the correct output. Show that the matrix for Z gives the correct output. Show that Z is unitary.

1 Z=( 0

0 ) = |0> <0| − |1> <1| −1

Dirac notation: For |0> input: Z|0> = (|0> <0| − |1> <1|)|0> = |0> <0|0> − |1> <1|0> = |0> (1) − |1> (0) = |0> For |1> input: Z|1> = (|0> <0| − |1> <1|)|1> = |0> <0|1> − |1> <1|1> = |0> (0) − |1> (1) = −|1> Matrix representation: 1 0 1 ) ( ) = (1) = |0> Z|0> = ( 0 0 −1 1 0 00 ) ( ) = −(0) = −|1> Z|1> = ( 1 0 −1 1 1 Z=( 0

0 1 0 ) , Z† = ( ) −1 0 −1

0 1 0 1 Z Z† = (1 )( 0 −1 0 −1) = ( 0  Z is unitary

0 )=I 1

Exercise 7.5. Show that the outer product notation for H gives the correct output. Show that the matrix for H gives the correct output. Show that H is unitary.

1 ) = |0> <+| + |1> <−| √2 1 −1 1

(

+> =

1

H=

−> =

1

√2 1 √2

<+|0> =

( 0> + 1> ) ( 0> − 1> ) 1 √2

( <0| + <1| )|0> =

1 √2

( <0|0> + <1|0> ) =

1 √2

1 ( <0| − <1| )|0> = 1 ( <0|0> − <1|0> ) = 1 √2 √2

<−|0> = √2

<+|1> =

1 √2

( <0| + <1| )|1> =

1 √2

( <0|1> + <1|1> ) =

1 √2

1 ( <0| − <1| )|1> = 1 ( <0|0> − <1|1> ) = 1 − √2 √2

<−|1> = √2

Dirac notation: 1 For |0> input: H|0> = (|0> <+| + |1> <−|)|0> = |0> <+|0> + |1> <−|0> = (|0> + |1>) = |+> For |1> input: H|1> = (|0> <+| + |1> <−|)|1> = |0> <+|1> + |1> <−|1> =

√2 1 √2

(|0> − |1>) = |−>

Matrix representation: 1 1 1 0 1 1 1 1 1 H|0> = 1 (1 ) ( ) = ( ) = ( ) + ( ) = (|0> + |1>) = |+> √2 1 −1 0 1 1 1 √21 1 1 √21 0 0 √21 √2 1 1 1 0 H|1> = ( ) ( ) = ( ) = ( ) − ( ) = (|0> − |1>) = |−> 1 √2 0 √2 1 √2 √2 −1 √2 1 −1 H=

1 1 1 1 ) , H† = ( ) √2 1 −1 √2 1 −1 1

(

1

1 1 1 1 ( ) ( √2 1 −1 √2 1  H is unitary H H† =

1

1 1 1 2 0 −1) = 2 (0 2) = ( 0

0 1) = I

Exercise 7.6. Show that if we apply H to the superposition 1

that if we apply H to the superposition

1

1

√2

1 √2

(0> + 1>), then we obtain |0>. Show

(0> − 1>), then we obtain |1>.

1

1

1

1

1

1

1

H|+> = H [ (0> + 1>) ] = H|0> + H|1> = |+> + |−> = [ (0> + 1>) ] + [ (0> √2 1 √2 √2 √2 √2 √2 √2 √2 1 √2 1 1 − |0> + |1> + |0> − |1> = |0> 1>) ] = 2 2 2 2 H|−> = H [

1

1 √2

(0> − 1>) ] = 1

1

1

H|0> − 1 H|1> = 1 |+> − 1 |−> =

√2 1

√2

− 1>) ] = 2|0> + 2|1> − 2|0> + 2|1> = |1>

√2

√2

1 √2

[

1

(0> + 1>) ] − 1 [

1

√2

√2

√2

Exercise 7.7. Show that the output of the Hadamard gate can be represented as:

H|x> =

|0> + (−1)x|1> √2

(0>

|0> + (−1)x|1>

If x=0, the output is

√2 |0> + (−1)x|1>

If x=1, the output is

√2

Exercise 7.8. Show that H =

1

1

(X+Z) =

√2

√2

[(

0 1

1 √2

= =

|0> + (−1)0|1> √2 |0> + (−1)1|0> √2

= =

|0> + |1> √2 |0> − |1> √2

= |+> = |−>

(X+Z).

1 1 1 1 0 ) + ( ( )] = √2 1 0 −1 0

1 )=H −1

Exercise 7.9. Show that the identity matrix can also be written as |+> <+| + |−> <−|, where +> = 1 1 (0> + 1>) and |−> = (0> − 1>). This is another form of the “resolution of the identity” that √2

√2

was mentioned in Exercise 4.6.

|+> =

1

(1) , |−> =

√2 1

1

( 1)

√2 −1

|+> <+| + |−> <−| =

1

(1)

1

1

2 1 (1 1)

2 −1 (1 −1)

1 1 = ( 1 1) + ( 1 −1) 2 1 1 2 −1 1

= (1 0) 0 1 =I

(1 1) +

√2 1 √2

= (1 (1 1)) + ( 1 (1 −1) )

1 2 0 = ( ) 2 0 2

1

1

( 1)

1

(1 −1)

√2 −1 √2

Exercise 7.10. Show the following:

Remember that the corresponding unitary operators (matrices) must be multiplied together from right to left, so that the earliest gates are applied to the state first.

HXH =

1 1 0 1 1 1 0 1 0 1 1 1 1 1 −1 1 2 ( )( ( ) = ( )=Z ( ) ( ) = ) ( ) = 2 √2 1 −1 0 −2 0 −1 1 1 0 √2 1 −1 2 1 −1 1

HZH =

1 1 1 0 1 1 ( )( 2 √ 1 −1 0 −1) √2 (1

1

1

1 1 1 ( −1) = 2 1

1 1 )( −1 −1

1 1 0 2 0 )= 2( )=( 1 2 0 1

XX = (0 1) ( 0 1) = ( 1 0) = I 1 0 1 0 0 1 0 −i 0 −i 1 0 YY = ( )( )= )=I ( i 0 i 0 0 1 0 1 0 1 0 ZZ = (1 )=I )( )=( 0 −1 0 −1 0 1 0 1 1 XYZ = ( 0 −i 1 0) = 0 1 0 i ) = ( i 0 = i ( ) ( ( ) ( )( ) 1 0 i 0 0 −1 0 1 0 i 0 0 i

Exercise 7.11. Show that CNOT is unitary.

1 CNOT = (0 0 0

0 1 0 0

0 0 0 1

0 1 0) , CNOT† = (0 1 0 0 0

0 1 0 0

0 0 0 1

0 0) 1 0

0 ) 1

= iI

1 )=X 0

1 (CNOT)( CNOT†) = (0 0 0

0 1 0 0

0 0 0 1

0 1 0 0 0) (0 1 0 1 0 0 0 0 0 0 1

0 1 0 0) = (0 1 1 0 0 0 0 0

0 0 1 0

0 0) = I 0 1

 CNOT is unitary

Exercise 7.12. What is the CNOT output for a control qubit of 1> and a target qubit of t> = 0> + 1>?

A control qubit of |1> will apply the NOT gate to the target: NOT(0> + 1>) = 1> + 0> Therefore, the output is |1>( 1> + 0>) = 11> + 10> 0

Alternatively, (0)(α) = (0) 1

1 0 0 1 ( 0 0 0 0

α β

β

0 0 0 0 0) (0) = (0)0= (0)(β) = |1>( 1> + 0>) = 11> + 10> β 1 α 0 1 αβ α 1 0

Exercise 7.13. Show the following: H

H

H

H

=

1 1 (1 ) 1 1 1 1 1 1 ) = ( 1 −1 HH = 1 ( ) ( 2 1 1 √2 1 −1 √2 1 −1 1( ) 1 −1 1 CNOT = (0 0 0

0 1 0 0

0 0 0 1

0 0) 1 0

1) 1 1 (1 1 −1 ) = 1 (1 1 1 2 1 −1 ( ) 1 −1 1

1 1 1 −1 1 −1) 1 −1 −1 −1 −1 1

1 0 1 1 1 1 1 −1 1 −1 0 1 (HH)(CNOT) (HH) = ( )( 2 1 1 −1 −1 0 0 1 −1 −1 1 0 0 1

1 0 = (0 0 0 0 0 1

0 0 0 1

0 1 1 1 1 0) 1 (1 −1 1 −1) 1 2 1 1 −1 −1 0 1 −1 −1 1

0 0 0 1) , which is equal to the CNOT gate with second qubit as control.

1 0 0 0

Exercise 7.14. Show that the CPHASE gate is symmetric; i.e., the transformation matrix is identical whether the control gate is at the top or bottom:

= Z

By explicit calculation: 1 0 Left-hand side: CPHASE = |0><0|  I + |1><1|  Z = ( 0 0

0 0 0 1 0 0) 0 0 1 0 0 −1 0 1 0 0 0 1 0 0) Right-hand side: CPHASE = I  |0><0| + Z  |1><1| = ( 0 0 0 1 0 0 0 −1 1 CPHASE|00> = (0 0 0

0 1 0 0

1 1 0 0 0 0) (0 ) = (0) = |00> 0 1 0 0 0 −1 0 0

1 0 CPHASE|01> = ( 0 0

0 1 0 0

0 0 0 0 0 0) (1 ) = (1) = |01> 0 1 0 0 0 −1 0 0

1 CPHASE|10> = (0 0 0

0 1 0 0

0 0 0 0 0 0) (0 = 0) = |10> ) ( 0 1 1 1 0 −1 0 0

1 0 CPHASE|11> = ( 0 0

0 1 0 0

0 0 0 0 0 0 0) ( ) = − (0) = −|11> 0 0 1 0 0 −1 1 1

The CPHASE matrix gives the correct output regardless of whether the control is on the first or second qubit.

Exercise 7.15. Show the following:

= H 1) 1 (1 1) = 1 ( 1 −1 IH = (1 0)  1 (1 √2 1 1 √2 1 −1 0 1 0( ) 1 −1 1 0 CPHASE = (0 1 0 0 0 0

Z

H

1) 1 1 0 0 0 (1 1 0) 1 −1 ) = (1 −1 0 1 1 √2 0 0 1 1 1( ) 1 −1 0 0 1 −1

0 0 0 0) 1 0 0 −1 1 0 0 0 1 1 0 0 1 1 0 0 (1 −1 0 0) (0 1 0 0) 1 (1 −1 0 0) √2 0 0 1 1 0 0 1 0 √2 0 0 1 1 0 0 1 −1 0 0 0 −1 0 0 1 −1

(IH)(CPHASE)(IH) =

1 0 = (0 1 0 0 0 0

1

0 0 0 0) = CNOT 0 1 1 0

Exercise 7.16. Check that the outer product notation gives the SWAP matrix.

1

0

0

0

0 0

0 0

1 0

0 1

|00> = (0), |01> = (1) , |10> = (0) , |11> = (0)

SWAP = |00> <00| + |10> <01| + |11> <11| + |01> <10| 1

0

0

0

0 0

1 0

0 1

0 0

= (0)  (1 0 0 0) + (0)  (0 1 0 0) + (0)  (0 0 0 1) + (1)  (0 0 1 0) 1 0 = (0 0 0 0 0 0

0 0 0 0

0 0 0) + (0 0 0 0 0

0 0 1 0

1 0 = (0 0 0 1 0 0

0 0 1 0) = SWAP 0 0 0 1

0 0 0 0

0 0 0 0) + (0 0 0 0 0 0 0 0

0 0 0 0

0 0 0) + (0 0 0 1 0

0 0 0 0

Alternatively, SWAP|00> = (|00> <00| + |10> <01| + |11> <11| + |01> <10|)|00> = |00> <00|00> + |10> <01|00> + |11> <11|00> + |01> <10|00> = |00> SWAP|01> = (|00> <00| + |10> <01| + |11> <11| + |01> <10|)|01> = |00> <00|01> + |10> <01|01> + |11> <11|01> + |01> <10|01> = |10> SWAP|10> = (|00> <00| + |10> <01| + |11> <11| + |01> <10|)|10> = |00> <00|10> + |10> <01|10> + |11> <11|10> + |01> <10|10> = |01> SWAP|11> = (|00> <00| + |10> <01| + |11> <11| + |01> <10|)|11> = |00> <00|11> + |10> <01|11> + |11> <11|11> + |01> <10|11> = |11> The output matches that obtained from the matrix.

0 1 0 0

0 0) 0 0

Exercise 7.17. Show that a 2-qubit SWAP gate can be made from 3 CNOT gates as follows:

= 1 0 CNOT = ( 0 0

0 1 0 0

0 0 0 1

0 0) 1 0 1 0 0 0 0 0 1

From Exercise 7.13, matrix for CNOT with 2nd qubit as control is (0

1 0 ( 0 0

0 0 0 1

0 0 1 0

0 1 0 1) (0 1 0 0 0 0 0 0

0 0 0 1

0 1 0) (0 1 0 0 0

0 0 0 1

0 0 1 0

0 1 0 1) = (0 0 0 0 1 0 0 0

0 1 0 0

0 0 1 0

0 1) 0 0

0 0) = SWAP 0 1

Exercise 7.18. Using only Dirac notation, show that |0> <0|  I  I + |1> <1|  SWAP gives the correct output for the CSWAP gate.

Determine CSWAP output for each of the basis state inputs: CSWAP|000> = (|0> <0|  I  I + |1> <1|  SWAP)|000> = |0> <0|0>  I|0>  I|0> + |1> <1|0>  SWAP|00> = |0> (1)  |0>  |0> + |1> (0)  SWAP|00> = |000> CSWAP|001> = (|0> <0|  I  I + |1> <1|  SWAP)|001> = |0> <0|0>  I|0>  I|1> + |1> <1|0>  SWAP|01> = |0> (1)  |0>  |1> + |1> (0)  SWAP|01> = |001> CSWAP|010> = (|0> <0|  I  I + |1> <1|  SWAP)|010>

= |0> <0|0>  I|1>  I|0> + |1> <1|0>  SWAP|10> = |0> (1)  |1>  |0> + |1> (0)  SWAP|10> = |010> CSWAP|011> = (|0> <0|  I  I + |1> <1|  SWAP)|011> = |0> <0|0>  I|1>  I|1> + |1> <1|0>  SWAP|11> = |0> (1)  |1>  |1> + |1> (0)  SWAP|11> = |011> CSWAP|100> = (|0> <0|  I  I + |1> <1|  SWAP)|100> = |0> <0|1>  I|0>  I|0> + |1> <1|1>  SWAP|00> = |0> (0)  |0>  |0> + |1> (1)  SWAP|00> = 0 + |1>SWAP|00> = |1>|00> = |100> CSWAP|101> = (|0> <0|  I  I + |1> <1|  SWAP)|101> = |0> <0|1>  I|0>  I|1> + |1> <1|1>  SWAP|01> = |0> (0)  |0>  |1> + |1> (1)  SWAP|01> = 0 + |1>SWAP|01> = |1>|10> = |110> CSWAP|110> = (|0> <0|  I  I + |1> <1|  SWAP)|110> = |0> <0|1>  I|1>  I|0> + |1> <1|1>  SWAP|10> = |0> (0)  |1>  |0> + |1> (1)  SWAP|10> = 0 + |1>SWAP|10> = |1>|01> = |101> CSWAP|111> = (|0> <0|  I  I + |1> <1|  SWAP)|111> = |0> <0|1>  I|1>  I|1> + |1> <1|1>  SWAP|11> = |0> (0)  |1>  |1> + |1> (1)  SWAP|11> = 0 + |1>SWAP|11> = |1>|11> = |111>

Exercise 7.19. Draw a quantum circuit that implements a random number generator.

Repeat the measurement n times for successive inputs of |0>. The result is a random sequence of 0’s and 1’s that can represent a number from 0 to 2n−1.

Exercise 7.20. Show that the gates X, Y, Z, H, and CNOT are Hermitian. Show that the phase gate, R, is not Hermitian.

X† = (0 1) = X 1 0  X is Hermitian Y† = ( 0i

−i ) = Y 0  Y is Hermitian 0 )=Z 0 −1  Z is Hermitian Z† = (

1

1 1 )=H ( √2 1 −1  H is Hermitian H† =

1

1 0 0 0 1 0 0) = CNOT 0 0 0 1 0 0 1 0  CNOT is Hermitian CNOT† = (0

1 0 1 0 R = [ ] , R† = [ ] i  −i 0 e 0 e R  R†  R is not Hermitian

Exercise 7.21. For the Bell state circuit, what is the output if the input qubits are 01>, 10>, or 11>?

H

|01> input: 1 1 (HI)|01> = H|0> I|1> = (|0> + |1>)|1> = (|01> + |11>) 1

√2 1

√2

√2

CNOT [ (|01> + |11>)] =

√2

(|01> + |10>)

|10> input: (HI)|10> = H|1> I|0> = 1 (|0> − |1>)|0> = 1 (|00> − |10>) √2

1

CNOT [ (|00> − |10>)] = 1 (|00> − |11>) √2

√2

√2

|11> input: (HI)|11> = H|1> I|1> = 1 (|0> − |1>)|1> = 1 (|01> − |11>) √2

1

CNOT [ (|01> − |11>)] = 1 (|01> − |10>) √2

√2

√2

Exercise 7.22. Show that the circuit in Figure 7.15 produces the indicated output. 0>

H

0> 0>

(HII)|000> = H|0>  I|0>  I|0> =

1

(|0> + |1>)|00> =

1

√2 1

√2

√2

1

(|000> + |100>)

√2

(CNOTI) [ (|000> + |100>)] = CNOT [ (|00> + |10>)]|0> =

1

(|00> + |11>)|0> =

√2

1 √2

(|000> +

|110>) 1

(ICNOT) [ (|000> + |110>)] = 1 I|0> CNOT|00> + √2

√2

1 √2

I|1>CNOT|10> =

1 √2

(|000> + |111>)

Chapter 8

Exercise 8.1. How might we use teleportation to design a quantum optical repeater? Sketch the overall design of your repeater.

A repeated series of teleportation circuits can transport a state over a long distance. Alice 1 and Bob 1 share an entangled state, Alice 2 and Bob 2 share an entangled state, etc. Each pair of Alice and Bob constitute a single teleportation circuit. The double lines indicate the transmission of the classical bits in each teleportation circuit. Thus, one can employ a process of entanglement swapping: Rev. Mod. Phys. 83 (2011) 33.

Alice 1

Bob 1

0> + 1>

Alice 2 0> + 1>

Bob 2

…

Alice k

Bob k

0> + 1> 0> + 1>

Chapter 9

Exercise 9.1. What is the matrix for each of the following?

I I Z I IX XY I is the 2x2 identity matrix and X, Y and Z are the Pauli matrices.

1 1( 1 0 0 I  I = (0 1)  (0 1) = ( 1 0( 0 1 0

0 ) 1 0 ) 1

0 1 ) 0 1 ) = (0 1 0 0 1( ) 0 0 1 0(

1

1 1 ( 0 0)  (1 0) = ( Z  I = (1 0 −1 0 1 0 (1 0

0 1) 0) 1

0 1 ( I  X = ( 1 0)  ( 0 1) = ( 1 0 1 1 0 0 (0 1

1 0 1 0 1 ) 0 ( ) = ( 0 1 0 ) 1) 1 (0 1 0 ) 0 1 0 0

X  Y = (0 1)  (0 1 0 i

0 0 ( −i) = ( i 0 1 (0 i

−i

1 0 (0 −1 ( 1 0

) 0 −i ) 0

0 1( i 0 (0 i

0 0 0 1 0 0) 0 1 0 0

0 1 0 1) ) = ( 0) 0 1 0

−i

0 1 0 0 0 1 0 0) 0 −1 0 0 0 −1

1 0 0 0

0 ) 0 ) = (0 −i ) 0 i

0 0 0 1

0 0) 1 0 0 0 0 i −i 0 0 0

−i 0 ) 0 0 _

Exercise 9.2. What is the matrix representation for the following circuit?

Z H 1 1) 1 1 0 1 (1 0 (1 ) 1 1 1 1 1 0 1 0 ) = ( 1 −1 1 −1 ) = (1 −1 ZH= ( ) ( √2 1 1 1 1 √2 1 −1 √2 0 0 −1 0 −1 0( ) −1 ( ) 1 −1 1 −1 0 0 −1

Exercise 9.3. Calculate the output of the following Bell state circuit, employing the tensor product.

0>

H

0> 1(

0 1 0 0

0 0 0 1

1 ) 0 1 ) = 1 (0 1 0 √2 1 −1 ( ) 0 1 0 1(

1

0

0

1

0

1 0 0 −1

1) 0

1

−1

0

0 0) 1 0

1 0 (CNOT)(H  I) = ( √2 0 0 1

0 1 0 0 1

|0>  |0> = (1)  (1) = (0) 0

0

) 0 = 1( 0 1 1 0 1) √2 1( ) 0 1

1 1 H  I = √12 (1 1 −1)  (0

1 CNOT = (0 0 0

1

0

0 0

0 0 0 1

0 1 0) (0 1 1 0 0

0 1 0 1 0 1 0 1 1 0 1) = 0 1) (0 1 0 −1 0 √2 0 1 0 −1 1 0 −1 1 0 −1 0

0 0) −1 1

1 0 (CNOT)(H  I)|00> = ( √2 0 1

0 1 1 0

1

1 0 1 1 0 1 0 ) ( 0) = 1 (00 ) 0 −1 0 √2 1 −1 0

Exercise 9.4. Show that 26>5 = 3>2  2>3 where the superscript denotes the number of qubits. 3>2 = |11> 2>3 = |010> 3>2  2>3 = |11>  |010> = |11010> = |26>5 11010 is the binary representation of 26.

Exercise 9.5. Show that 0>n1> = 1>n+1 where the superscript denotes the number of qubits. 0>n1> =|0…0> |1> = |0…01> = |1>n+1 Alternatively, n

1

n

0> = (0) , i.e., 1 followed by 2 −1 zeros ⋮ 0

1

1(0)

0 1

1

0 ⋮ 0 𝗁0)

0>n1> = (0)  ( 0) = I0(01)I = I0I = |1> ⋮ 0

1

𝗁

⋮ 0(10)

)

n+1

Exercise 9.6. Show that CNOT = |0> <0|  I + |1> <1|  X |0> = (10) , <0| = (1 0) 1 0 |0> <0| = (1)  (1 0) = ( ) 0 0 0 0 0 |1> <1| = (0)  (0 1) = ( ) 1 0 1

0 |1> <1|  X = ( 0

0 0 (1 1 )=( 0 0 0( 1

0 0 )( 1 1

1 0) 1 ) 0

0 0 ) 1 0 ) = (0 0 0 1 0 0 1( ) 1 0 0 0

0(

1 1 ( |0> <0|  I = ( 1 0) ( 1 0) = ( 10 0 0 0 1 0( 0

0 1) 0) 1

1 0 |0> <0|  I + |1> <1|  X = ( 0 0

0 0 0) + (0 0 0 0 0

0 1 0 0

0 0 0 0

1 0 (0 0 (1 0

0 1

0 1 ) 1 ) = (0 0 0 ) 1 0 0 0 0 0

0 0 0 1

0 1 0 0

0 0 0 0 0 1) 1 0

0 0 0 0

0 0) 0 0

0 1 0 0 0 0) = (0 1 0 0) = CNOT 1 0 0 0 1 0 0 0 1 0

Exercise 9.7. Show that CNOT = (I  H) (CPHASE) (I  H) See Exercise 7.15.

Exercise 9.8. Referring to the CHSH inequality in Chapter 5, show that quantum mechanics predicts <ab> = −cos() where  is the angle between any two measurement directions of ̂B ̂ |> where A ̂ and B ̂ are the electron spin. Hint: <ab> is the expectation value given by <|A 1

spin operators along an arbitrary direction in the x-z measurement plane, and |> = |10>) is the entangled state.

Measurement in the x-z plane can be decomposed as: ̂ = sin X + cos Z = sin A 1

1

̂=( Similarly, B

̂B ̂=( A

1(

0 1 cosθ1 ) + cos 1 0 )=( 1( sinθ1 1 0 0 −1

cosθ2

sinθ2

sinθ2

−cosθ2

sinθ1 −cosθ1

)

cosθ1 (cosθ2 sinθ2

sinθ2 ) −cosθ2

sinθ1 ( cosθ2 sinθ2

sinθ2 ) −cosθ2

sinθ1 (cosθ2 sinθ2

sinθ2 ) −cosθ2 ) sinθ2 cosθ2 ) −cosθ1 ( sinθ2 −cosθ2

)

√2

(|01> −

cosθ1cosθ2 cosθ1sinθ2 cosθ1sinθ2 −cosθ1cosθ2 = ( sinθ1cosθ2 sinθ1sinθ2 sinθ1sinθ2 −sinθ1cosθ2

sinθ1cosθ2 sinθ1sinθ2 sinθ1sinθ2 −sinθ1cosθ2 ) −cosθ1cosθ2 −cosθ1sinθ2 −cosθ1sinθ2 cosθ1cosθ2

0

0

1

1 0

( 1)

√2

−1 0

|> = 1 (|01> − |10>) = 1 (1) − 1 (0) = √2

√2

0 0

√2

cosθ1cosθ2 cosθ1sinθ2 cosθ1sinθ2 −cosθ1cosθ2 ̂B ̂ |> = 1 A ( sinθ1cosθ2 sinθ1sinθ2 √2 sinθ1sinθ2 −sinθ1cosθ2

0

sinθ1cosθ2 sinθ1sinθ2 0 sinθ1sinθ2 −sinθ1cosθ2 ) ( −11) −cosθ1cosθ2 −cosθ1sinθ2 0 −cosθ1sinθ2 cosθ1cosθ2

cosθ1sinθ2−sinθ1cosθ2 1 −cosθ1cosθ2−sinθ1sinθ2 = 2 ( sinθ1sinθ2+cosθ1cosθ2 ) √ −sinθ1cosθ2+cosθ1sinθ2

̂ B ̂|> = 1 <|A

cosθ1sinθ2−sinθ1cosθ2 −cosθ1cosθ2−sinθ1sinθ2 (0 1 −1 0) (sinθ sinθ +cosθ cosθ ) 1 2 1 2 √2 √2 −sinθ1cosθ2+cosθ1sinθ2

1

1

= 2(−cosθ1cosθ2 − sinθ1sinθ2 − sinθ1sinθ2 − cosθ1cosθ2) = −cosθ1cosθ2 − sinθ1sinθ2 = −cos (θ1 − θ2) = −cos (θ) where θ = θ1 − θ2

Chapter 10 Exercise 10.1. What is the matrix for H2?

1 1 (1 ) 1 1 1 1 1 1 1 ) = ( 1 −1 ) ( H2 = ( 2 1 1 √2 1 −1 √2 1 −1 1( ) 1 −1

1) 1 1 1 1 1 (1 1 −1) 1 −1 ) = 1 (1 −1 1 1 2 1 1 −1 −1 −1 ( ) 1 −1 −1 1 1 −1

Exercise 10.2. What is the matrix for H3?

1 1 1 1 1 1 ) 1 (1 1) 1 (1 1) = 1 1 1 1 −1 1 −1 )   ( ) ( √2 1 −1 √2 1 −1 √2 1 −1 √8 1 −1 1 1 −1 −1 1 −1 −1 1 1 1 1 1 1 1 1 1 1 −1 1 −1 1 −1 1 −1 l1 1 −1 −1 1 1 −1 −1 1 −1 1 1 −1 −1 1 1 −1 = √8 1 1 1 1 −1 −1 −1 −1 I1 −1 1I 1 −1 −1 1 −1 1 1 −1 −1 −1 −1 1 1 𝗁1 −1 −1 1 −1) 1 −1 1 1 H3 = (

Exercise 10.3. What is the output vector for H30>3?

1 1 l1 H30>3 = 1 1 √8 1 I1 1 𝗁1

1 −1 1 −1 1 −1 1 −1

1 1 −1 −1 1 1 −1 −1

1 −1 −1 1 1 −1 −1 1

1 1 1 1 −1 −1 −1 −1

1 −1 1 −1 −1 1 −1 1

1 1 −1 −1 −1 −1 1 1

The output is an equally weighted superposition.

1 1 1 0 1 −1 l l 0 1 −1 1 1 1 I0I = −1 0 √8 1 I1I 1 I I0I 1 0 1 −1) 𝗁0) 𝗁1)

Chapter 11 Exercise 11.1. What is the output for each possible input of 00>, 01>, 10> and 11> in Figure 11.3 for each function f0, f1, f2 and f3?

The input is |x>|y> and the output is |x>|yf(x)> Input

f0

f1

|00>

|0>|0+f0(0)>

|0>|0+f1(0)>

|0>|0+f2(0)> |0>|0+f3(0)>

= |00>

= |01>

= |01>

|0>|1+f0(0)>

|0>|1+f1(0)>

|0>|1+f2(0)> |0>|1+f3(0)>

= |01>

= |00>

= |00>

|1>|0+f0(1)>

|1>|0+f1(1)>

|1>|0+f2(1)> |1>|0+f3(1)>

= |10>

= |11>

= |10>

|1>|1+f0(1)>

|1>|1+f1(1)>

|1>|1+f2(1)> |1>|1+f3(1)>

= |11>

= |10>

= |11>

|01>

|10>

|11>

f2

f3

= |00>

= |01>

= |11>

= |10>

Note that the output is unique for each input. Therefore, |x>|yf(x)> is reversible.

Exercise 11.2. Check that the Uf0 matrix is unitary and can therefore be implemented in a quantum circuit.

1 0 0 0 1 0 0 1 0 0 † 0 1 Uf0 = ( ) , Uf0 = ( 0 0 1 0 0 0 0 0 0 1 0 0

0 0 0 0) 1 0 0 1

1 0 0 0 1 0 0 0 1 0 Uf U† = (0 1 0 0) (0 1 0 0) = (0 1 0 f0 0 0 1 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 1  Uf0 is unitary.

0 0 0 0) = I 1 0 0 1

Exercise 11.3. Check that Uf0 gives the expected output for each possible input.

1 0 0 0 1 1 0 1 0 0 0 0 Uf |00> = ( )( ) = ( ) 0 0 0 0 1 0 00 0 0 0 0 1 1 0 0 0 0 0 0 1 0 0 1 Uf |01> = ( ) ( ) = (1) 0 0 0 0 1 0 00 0 0 0 0 1 1 0 0 0 0 0 0 1 0 0 0 0 Uf |10> = ( )( ) = ( ) 0 1 0 0 1 0 01 0 0 0 0 1 1 0 0 1 Uf |11> = ( 0 0 0 0 0

0 0 0 0

0

0

) (0) = (0)

1 0 0 1

0

1

0

1

Note that we could have saved ourselves a lot of work by realizing that Uf0 is the identity matrix.

Therefore, Uf0 |xy> = |xy>

Exercise 11.4. Show that the following matrices implement f1, f2 and f3, and show they are unitary.

0 1 0 0 1 0 0 0) Uf1 = ( 0 0 0 1 0 0 1 0 0 1 1 0 Uf2 = ( 0 0 0 0

0 0 0 0) 1 0 0 1

1 0 0 0 0 1 0 0) Uf3 = ( 0 0 0 1 0 0 1 0

0 0 1 0 0 1 1 0 0 0 0 1 Uf |00> = ( ) ( ) = ( ) = |01> 1 0 0 0 0 1 00 0 0 0 1 0

0 1 0 0 0 1 1 0 0 0 1 0 Uf |01> = ( ) ( ) = ( ) = |00> 1 0 0 0 0 1 00 0 0 0 1 0 0 1 0 0 0 0 1 0 0 0 0 0 Uf |10> = ( ) ( ) = ( ) = |11> 1 0 0 0 0 1 01 1 0 0 1 0 0 1 0 0 0 0 1 0 0 0 0 0 Uf |11> = ( ) ( ) = ( ) = |10> 1 1 0 0 0 1 10 0 0 0 1 0 0 1 0 0 1 0 1 0 0 0 0 1 Uf |00> = ( ) ( ) = ( ) = |01> 2 0 0 0 1 0 00 0 0 0 0 1 0 1 0 0 0 1 1 0 0 0 1 Uf |01> = ( ) ( ) = (0) = |00> 2 0 0 0 1 0 00 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 0 Uf |10> = ( ) ( ) = (0) = |10> 2 1 0 0 1 0 01 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 0 0 Uf |11> = ( ) ( ) = ( ) = |11> 2 0 0 0 1 0 10 1 0 0 0 1 1 0 0 1 Uf |00> = ( 3 0 0 0 0

0 0 0 0

1

1

) (0) = (0) = |00>

0 1 1 0

0

0

0

0

0 1 0 0 0 0 0 1 0 0 1 1 Uf |01> = ( ) ( ) = ( ) = |01> 3 0 0 0 0 1 00 0 0 0 1 0

1 0 0 0 0 0 0 1 0 0 0 0 Uf |10> = ( ) ( ) = ( ) = |11> 3 0 0 0 0 1 01 1 0 0 1 0 1 0 0 1 Uf |11> = ( 2 0 0 0 0

0 0 0 0

0

0

) (0) = (0) = |10>

0 1 1 0

0

1

1

0

Comparing the above outputs with the results of the table in Exercise 11.1, we see that each matrix produces the correct output.

0 1 0 0 0 1 0 0 Uf1 = (1 0 0 0) , Uf†1 = (1 0 0 0) 0 0 0 1 0 0 0 1 0 0 1 0 0 0 1 0 1

0 1 0 0 0 0

Uf U† = (0 1

f1

0 0 0 0 )=I 1 0 0 1

 Uf1 is unitary 0 1 0 0 0 1 0 0 Uf2 = (1 0 0 0) , Uf†2 = (1 0 0 0) 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 1 1 0 0 0 Uf U† = (0 1 0 0) = I 2 f2 0 0 1 0 0 0 0 1  Uf2 is unitary 1 0 0 0 1 0 0 0 0 1 0 0 † 0 1 0 0) Uf3 = ( ) , Uf3 = ( 0 0 0 1 0 0 0 1 0 0 1 0 0 0 1 0

1 0 0 0 Uf U† = (0 1 0 0) = I 3 f3 0 0 1 0 0 0 0 1  Uf3 is unitary

Exercise 11.5. As in the previous examples, prove that the output of the Deutsch circuit for f3 is 1, indicating it is balanced.

Starting with Eq. (11.6), U𝐹 1 H|0>  H|1> → (  0f3(0)> −  1f3(0)> + 1> 0f3(1)> −  1f3(1)> ) = = = =

1 2 1 2 1

2 1

2

(  00> −  10> + 1> 01> −  11> ) ( 0> −  1> + 1>1> − 0> ) ( 0> − 1> + 11> − 0> )

√2

( 0 > − 1> )

1 √2

( 0 > − 1> )

We trash the second qubit, and apply the Hadamard gate to the first qubit: H[

1 √2

( 0 > − 1> ) ] = 1

The result is 1, indicating f3 is balanced.

Exercise 11.6. Show that 0f(x)> − 1f(x)> = (−1)f(x)( 0> − 1> ) Hint: You can check that this works for each of the possible values of f(x) = {0, 1} and using the rules for binary addition.

If f(x) = 0: The left-hand side gives 0f(x)> − 1f(x)> = 00> − 10> = 0> − 1> The right-hand side gives (−1)f(x)( 0> − 1> ) = (−1)0( 0> − 1> ) = 0> − 1> = LHS If f(x) = 1: The left-hand side gives 0f(x)> − 1f(x)> = 01> − 11> = 1> − 0> The right-hand side gives (−1)f(x)( 0> − 1> ) = (−1)1( 0> − 1> ) = 1> − 0> = LHS

Exercise 11.7. Prove the following: 1

1

(−1)f(0)|0⟩ +

√2

(−1)f(1)|1⟩ = (−1)f(0) (

√2

√2

If f(0) = 0 and f(1) = 0: 1 (−1)f(0)|0⟩ + 1 (−1)f(1)|1⟩ = (−1)f(0) ( 1 |⟩ +

1

√2

√2

√2

√2

1 (−1)0|0⟩ + 1 (−1)0|1⟩ = (−1)0 ( 1 |⟩ +

√2

=

1

1

|0⟩ +

√2

√2

1

|1⟩ =

√2

√2

1

|⟩ +

√2

1

1

√2

|⟩ +

1 √2

(−1)f(0)  f(1)|⟩)

(−1)f(0)  f(1)|⟩)

(−1)0  0|⟩)

|⟩

√2

Left-hand side equals right-hand side If f(0) = 0 and f(1) = 1: 1 (−1)f(0)|0⟩ + 1 (−1)f(1)|1⟩ = (−1)f(0) ( 1 |⟩ +

1

√2

√2

√2

√2

1 (−1)0|0⟩ + 1 (−1)1|1⟩ = (−1)0 ( 1 |⟩ +

1

√2

√2

√2

√2

1 |0⟩ − 1 |1⟩ = ( 1 |⟩ − 1 |⟩)

√2

√2

√2

(−1)f(0)  f(1)|⟩)

(−1)0  1|⟩)

√2

Left-hand side equals right-hand side If f(0) = 1 and f(1) = 0: 1 (−1)f(0)|0⟩ + 1 (−1)f(1)|1⟩ = (−1)f(0) ( 1 |⟩ +

1

√2

√2

√2

√2 1

1 (−1)1|0⟩ + 1 (−1)0|1⟩ = (−1)1 ( 1 |⟩ +

√2

=−

1

|0⟩ +

√2

1

√2

|1⟩ = − (

√2

1

|⟩ −

√2

1

√2

(−1)f(0)  f(1)|⟩)

(−1)1  0|⟩)

√2

|⟩)

√2

Left-hand side equals right-hand side If f(0) = 1 and f(1) = 1: 1 (−1)f(0)|0⟩ + 1 (−1)f(1)|1⟩ = (−1)f(0) ( 1 |⟩ +

1

√2

√2

√2

√2 1

1 (−1)1|0⟩ + 1 (−1)1|1⟩ = (−1)1 ( 1 |⟩ +

√2

=−

1 √2

|0⟩ −

1

√2

|1⟩ = − (

√2

1

|⟩ +

√2

1

√2

(−1)f(0)  f(1)|⟩)

(−1)1  1|⟩)

√2

|⟩)

√2

Left-hand side equals right-hand side The left-hand side equals right-hand side for all possible values of f(0) and f(1) 1 1 1 (−1)f(1)|1⟩ = (−1)f(0) ( |⟩ + (−1)f(0)  f(1)|⟩)  1 (−1)f(0)|0⟩ + √2

√2

√2

1

1

√2

√2

Exercise 11.8. Show that H ( |⟩ + transformation.

√2

(−1)a|1⟩) = |a⟩ for a = 0 or 1 where H is the Hadamard

If a = 0: H ( 1 |⟩ + 1 (−1)a|1⟩) = |a⟩ √2 √2 1 H ( |⟩ + 1 (−1)0|1⟩) = |0⟩ √2 √2 1 H ( |⟩ + 1 |1⟩) = |0⟩ √2 √2 |0⟩ = |0⟩ If a = 1: H ( 1 |⟩ + 1 (−1)a|1⟩) = |a⟩ √2 √2 1 H ( |⟩ + 1 (−1)1|1⟩) = |1⟩ √2 √2 H ( 1 |⟩ − 1 |1⟩) = |1⟩ √2 √2 |1⟩ = |1⟩

Chapter 12 Exercise 12.1. Show that D is unitary, and therefore can be implemented as a quantum circuit.

2/N − 1 2/N D=( ⋮ 2/N

2/N 2/N − 1 ⋮ 2/N

⋯ ⋯ ⋱ ⋯

2/N 2/N ⋮ 2/N − 1

2/N − 1 2/N D† = ( ⋮ 2/N

2/N 2/N − 1 ⋮ 2/N

⋯ ⋯ ⋱ ⋯

2/N 2/N )=D ⋮ 2/N − 1

2/N − 1 2/N D D† = ( ⋮ 2/N

2/N 2/N − 1 ⋮ 2/N

⋯ ⋯ ⋱ ⋯

2/N 2/N − 1 2/N 2/N )( ⋮ ⋮ 2/N − 1 2/N

)

2/N 2/N − 1 ⋮ 2/N

⋯ ⋯ ⋱ ⋯

2/N 2/N ) ⋮ 2/N − 1

Each diagonal entry is: 2 2 2 4 4 4 4 2 ( − 1) + (N − 1) ( ) = − + 1 + − 2= 1 N

N2

N

Each non-diagonal entry is: 2 2 2 2 2 ( − 1) ( ) + (N − 2 )( ) = N

N

N

N

8 N2

N

N

4

4

8

N

N

N2

− + −

=0

 DD† = I and D is unitary

Exercise 12.2. Show that the circuit of Figure 12.8 implements an output of −|x> for an input of |x> when x  0.

Suppose the input is: |0> = |1>|0> … |0>|0> i.e, the first qubit is |1>. The NOT (X) gates transform the state to: |1> = |0>|1> … |1>|1> The iI and H gate on the first and last qubit, respectively, transforms the state to:

|2> = i|0> |1> … |1> |−> The multi-control CNOT gate transforms the state to: |3> = i|0>|1> … |1>|−> i.e., nothing happens. A second application of the iI and H gate on the first and last qubit, respectively, transforms the state to: |4> = i  i |0>|1> … |1>|1> = −|0>|1> … |1>|1> Finally, the NOT gates give: |5> = −|1>|0> … |0>|0> which is −|0>.

Next, suppose the input is: |0> = |0>|0> … |0>|1> i.e., the last qubit is |1>. The NOT (X) gates transform the state to: |1> = |1>|1> … |1>|0> The iI and H gate on the first and last qubit, respectively, transforms the state to: |2> = i|1>|1> … |1>|+> The multi-control CNOT gate transforms the state to: |3> = i|1>|1> … |1>|+> i.e., nothing happens because CNOT|+> = |+>. A second application of the iI and H gate on the first and last qubit, respectively, transforms the state to: |4> = i  i |1>|1> … |1>|0> = −|1>|1> … |1>|0> Finally, the NOT gates give: |5> = − |0>|0> … |0>|1> which is −|0>.

Finally, suppose the input is: |0> = |0>|0> … |1> … |0>|0> i.e., one of the middle qubits is |1>. The NOT (X) gates transform the state to:

|1> = |1>|1> … |0> … |1>|1> The iI and H gate on the first and last qubit, respectively, transforms the state to: |2> = i|1>|1> … |0> … |1>|−> The multi-control CNOT gate transforms the state to: |3> = i|1>|1> … |0> … |1>|−> i.e., nothing happens because one of the control qubits is |0>. A second application of the iI and H gate on the first and last qubit, respectively, transforms the state to: |4> = i  i |1>|1> … |0> … |1>|1> = −|1>|1> … |0> … |1>|1> Finally, the NOT gates give: |5> = − |0>|0> … |1> … |0>|0> which is −|0>.

Thus, if any of the n qubits are |1>; |x> is transformed to −|x>.

Exercise 12.3. Prove Eq. (12.17) and (12.18). n

H =

1 √N

(

1 1

1

⋯ ⋯

1

⋮ 1

⋮

⋱ ⋯

⋮

)

Note that all elements of the first row and column of Hn are 1. Thus, explicit calculation gives: 2 0 H ( ⋮ 0 2 n 0 H ( ⋮ 0 n

0 0 ⋮ 0 0 0 ⋮ 0

1 1 ⋯ 1 ⋯ 0 2 0 ⋯ 0 ⋯ 1 1 0 0 )= ( ⋱ ⋮ ) (⋮ ⋮ ⋱ ⋮ √N ⋮ ⋮ ⋯ 0 ⋯ 1 0 0 1 ⋯ 0 2 0 ⋯ 0 ⋯ 0 n 1 2 0 ⋯ 0 1 1 ) ( = ( √N ⋮ ⋮ ⋱ ⋮ √N ⋮ ⋱ ⋮ )H ⋯ 0 2 0 ⋯ 0 1

⋯ 0 2 0 ⋯ 0 ⋯ 0 2 0 ⋯ 0 1 )= ( ) ⋱ ⋮ √N ⋮ ⋮ ⋱ ⋮ ⋯ 0 2 0 ⋯ 0 1 ⋯ 1 2 2 ⋯ 2 ⋯ ⋯ 2 1 2 2 ⋮ ⋱ ⋮ )= N(⋮ ⋮ ⋱ ⋮) ⋯ 2 2 ⋯ 2

Exercise 12.4. Using the tensor product for the vectors |0> and <0|, show the conditional phase shift is 20> <0 − I.

1 0 |0> = ( ) , <0| = (1 0 … 0) ⋮ 0 1 1 0 ⋯ 0 0 0 ⋯ 0 0 0> <0 = ( )  (1 0 … 0) = ( ⋮ ⋮ ⋱ ⋮ ) ⋮ 0 0 ⋯ 0 0 1 0 ⋯ 0 1 0 0 0 ⋯ 0 0 −1 20> <0 − I = 2( ⋮ ⋮ ⋱ ⋮ ) − I = ( ⋮ ⋮ ⋯ 0 0 0 0 0

⋯ ⋯ ⋱ ⋯

0 0 ⋮) −1

θ N−1 Exercise 12.5. Show that the angle in Figure 12.1 0 is given by cos = √ (Hint: take the scalar 2

N

product between  > and α >). |> = √

N−1 N

|> +

1

|x*>

√N

|> and |> are vectors in a two-dimensional vector space, so we can calculate the scalar product: <|> = √

N−1 N

<|> +

1 √N

<|x*>

< |> = 1 ; |> and |x*> are orthogonal so <|x*> = 0  <|> = √

N−1 N

In a two-dimensional vector space, the scalar product is: <|> = |||| cos(/2) where /2 is the angle between the vectors |> and |> || = <|> = 1, || = <|> = 1, since |> and |> are normalized.  <|> =√

N−1 N

= cos(/2)

Chapter 13 Exercise 13.1. Derive the QFT4 output vector for each of the two-qubit basis states. Show that the resulting states are orthonormal.

The general two-qubit state is: ⟩ = α00⟩ + α11⟩ + α22⟩ + α33⟩ For basis state |0>: 0 = 1, 1 = 0, 2 = 0, 3 = 0 For basis state |1>: 0 = 0, 1 = 1, 2 = 0, 3 = 0 For basis state |2>: 0 = 0, 1 = 0, 2 = 1, 3 = 0 For basis state |3>: 0 = 0, 1 = 0, 2 = 0, 3 = 1 By definition: 1 1 β = ∑3 1 α3) j = 2 (α0 + 1 α1 + α2 +iπ/2 iπ β0 = 2 ∑3j=0 α α eπij/2 + α e + α e3iπ/2) 1 1 j = (α + α1e 2 3 β = 2 ∑3j=0 α eπij 1 2 0 iπ 2iπ + α e3iπ ) + α 2e 3 1 2 = (α + α e j=0 j 0 1 3 2 β = ∑ α e3πij/2 2 1 3iπ/2 + α e3iπ + α e9iπ/2) 3 = 2 (α0 + α1e 2 3 2 j=0 j For basis state |0>: 0 = ,  1 = ,  2 = ,  3 = 1

1

1

1

1

2

2

2

2

Thus, QFT4|0> = 1 (11) 2

1 1

1

1

1

1

2

2

2

2

2

1

1

2

2

For basis state |1>:  0 = 1,  1 = eiπ/2 = i,  2 = eiπ= − ,  3 = − i 1 1 Thus, QFT4|1> = (−1i ) 2 −i

2

For basis state |2>:  0 = 1,  1 = eiπ= − ,  2 = 1,  3 = − 1 1 1 Thus, QFT4|2> = 2 (−11 ) −1

2

2

2

1

1

2

2

For basis state |3>:  0 = 1,  1 = 1 e3iπ/2= − i,  2 = − ,  3 = 1 i 1 −i Thus, QFT4|3> = 1 (−1 ) 2 i

2

2

1

1

1

Now show that the 4 states, (1) , ( 2

1 1

2

1 i ),

−1 −i

1

1

(−1) and

2

1 −1

2

1

1

(−i), are orthonormal.

2

−1 i

First, show these vectors are normalized: 1 (1 1 1 1) 1

2

1

1

1 (1) = (1+1+1+1) = 1,  1 (11) is normalized 2 1 4

1

2

1

1 1 1 i ) is normalized i (1+1+1+1) = 1,  1 ( −1 ( ) = 2 2 −1 4 −i −i

1 (1 −i −1 i) 1

2

1

1

1 (1 −1 1 −1) 1 (−1) = 1(1+1+1+1) = 1,  1 (−1) is normalized

1 −1

2

2

1 −1

2

4

1

1

1 (1 i −1 −i) 1 ( −i ) = 1(1+1+1+1) = 1,  1 ( −i ) is normalized

2

−1 i

2

4

Next, check orthogonality: 1 (1 1 1 1) 1 (

2 1 2

2 1

1 i) = 0 −1 −i 1

(1 1 1 1) (−11) = 0 2

−1 1

1 (1 1 1 1) 1 ( −i ) = 0

2

2

−1 i 1

1 (1 −i −1 i) 1 (1) = 0

2

2

1 1

1

1 (1 −i −1 i) 1 (−1) = 0

2

2

1 −1 1

1 (1 −i −1 i) 1 ( −i ) = 0

2

2

−1 i

1

1 (1 −1 1 −1) 1 (1) = 0

2

1 1

2

1 1 (1 −1 1 −1) 1 ( i ) = 0 2 −1 2 −i 1

1 (1 −1 1 −1) 1 ( −i ) = 0

−1 i

2

2

1

1 (1 i −1 −i) 1 (1) = 0

2

2

1 1

1

1 (1 i −1 −i) 1 ( i ) = 0

2

2

−1 −i 1

1 (1 i −1 −i) 1 (−1) = 0

2

2

1 −1

2

−1 i

Exercise 13.2: Show QFT4 is a unitary operator, and therefore can be implemented.

1 1 1 1 1 i −1 −i QFT = (1 ) 4 2 1 −1 1 −1 1 −i −1 i 1

1 1 1 i −1 −i ) −1 1 −1 1 −i −1 i

1 QFT4 = 2 (1 1 †

1 1 1 1 QFT QFT† = (1 i −1 4 4 2 1 −1 1 1 −i −1 1 0 0 0 0 1 0 0) = I ( 0 0 1 0 0 0 0 1

1 1 1 1 −i ) 1 (1 i −1 −1 2 1 −1 1 i 1 −i −1

1 4 −i ) = 1 (0 4 0 −1 i 0

0 4 0 0

0 0 4 0

0 0) = 0 4

 QFT4 is unitary Exercise 13.3: Using the matrix for QFT4, what is the output vector for an input state of 0>, 1>, 2> and 3>?

1 1 1 QFT = ( 4 2 1 1

1 1 1 i −1 −i ) −1 1 −1 −i −1 i

QFT |0> =1 (1

1 1 1

1 1 i −1 −1 1 −i −1

1 1 1 −i ) (0) = 1 (1) 2 1 −1 00 1 i

1 QFT4|1> =1 (1 2 1 1

1 1 i −1 −1 1 −i −1

1 0 1 1 −i ) (1) = ( i ) 2 −1 −1 00 −i i

1

1 1 i −1 −1 1 −i −1

1 1 0 −i ) (0) = 1 (−1) 1 2 −1 01 −1 i

4

2

QFT |2> =1 (1 4

2

1 1

1 QFT |3> =1 (1 4

2

1 1

1 1 i −1 −1 1 −i −1

1 0 1 −i ) (0) = 1 ( −i) 2 −1 −1 10 i i

The result is identical to Exercise 13.1.

Exercise 13.4. Show that QFTN is unitary.

1 1 1 l QFTN = √N I1 ⋮ 𝗁1

1 1 ω ω2 ω2 ω4 ⋮ ⋮ N−1 2(N−1) ω ω

1 1 ∗ l 1 ω 1 QFT † = 1 ∗2 ω N √N I ⋮ ⋮ N−1 𝗁1 ω∗ Ignoring the pre-factor of

1 ω∗2 ω∗4 ⋮ ω∗2(N−1) 1

…

1 … ωN−1 … ω2(N−1) where ω = e2πi/N I ⋮ …

3(N−1) (N−1)(N−1) ω ω ) 1 ω3 ω6 ⋮

1 ω∗3 ω∗6 ⋮

1 … ∗N−1 … … ωω∗2(N−1) where ω∗ = e−2πi/N I ⋱ ⋮ …

(N−1)(N−1) ∗ ) ω∗3(N−1) ω

, the entry in the jth row and kth column of QFT is jk, and the entry

√N

in the jth row and kth column of QFTN† is ()jk.  The entry of the jth row and kth column of (QFTNQFTN†) is: (QFTNQFTN†)jk 1

= ∑N−1 (ω)jm(ω∗)mk N

m=0

1 N−1 = ∑m=0 (e2πi/N)jm(e−2πi/N)mk N

1 N−1 2πim(j−k)/N = ∑m=0 e N

= jk (Kronecker delta)  QFTNQFTN† = I and QFTN is unitary.

N

Exercise 13.5. Consider the 5-qubit state: |⟩ =

1 √3

(|01000⟩ + |10000⟩ + |11000⟩) =

1 √3

(|8⟩ + |16⟩ + |24⟩)

Show that QFT32|⟩ has frequency N/r as illustrated below.

The input column vector is: 1 (|01000⟩ + |10000⟩ + |11000⟩) = 1

√3

√3

(0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0

0 0)T Multiply the QFT32 matrix by the above column vector. This is tedious but can be done. It is beneficial to use the fft function in MATLAB (or equivalent software) to calculate the QFT output.

The QFT output includes all states from |0> to |31>. However, the states |0>, |4>, |8>, |12>, |16>, |20>, |24> and |28> have the largest amplitude, and all other states have a lower amplitude, as illustrated qualitatively above. The QFT output has frequency N/r = 4.

Chapter 14 ̂ X (θ), R ̂ y (θ) and R ̂ z (θ) result in a rotation angle Exercise 14.1 Show that the rotation matrices R of  about the x, y and z axes, respectively, in the Bloch sphere.

̂ X (θ′) = ( R

𝘍

𝘍

−isinθ

𝘍

−sinθ2

𝘍

−isinθ2

cosθ2

̂ y (θ′) = ( cosθ2 ),R 𝘍 𝘍 cosθ2 sinθ2

𝘍

2

cosθ

𝘍

−iθ𝘍/2

̂ z (θ′) = (e ),R

0

2

0 )

𝘍 eiθ /2

 

̂ > = ( R X

θ𝘍

cos 2 𝘍 −isinθ

θ𝘍

2

Suppose  = /2:

θ

cos 2

2

2

cos iπ/2

e

(θ−θ𝘍)

(sin

θ

𝘍

cos − ieisinθ sin

− ieiπ/2sinθ

𝘍

2

θ

̂ > = ( cos 2 𝘍 R θ θ𝘍 θ x −isinθ cos +eiπ/2cos 2 sin sin 2 (

θ𝘍

θ

−isin 2 ) ( 2 2 2 2 2 𝘍 𝘍 θ) = ( θ θ𝘍 θ) cosθ eisin −isinθ cos +eicos sin

2

θ𝘍

cos

cos θ

2

2

2

cos )=(

2 θ𝘍

2

2

θ

𝘍

2𝘍

2θ

2θ𝘍

2

2

2

cos + sinθ sin

−i (sinθ cos − cos

θ 2

cos (θ−θ𝘍) θ ) = (

sin ) 2

2

i(

sin

(θ−θ𝘍))

)=

2

2 (θ−θ𝘍) ) 2

)

Thus, θ becomes θ – θ′. This is exactly what you would expect for counter-clockwise rotation of a vector about the x-axis.

θ𝘍

̂ > = ( cos 2 R 𝘍 y sinθ

cos

cos θ

θ𝘍

θ𝘍

θ

𝘍

cos − eisinθ sin

θ

−sin 2 ) ( 2 2 2 2 2 𝘍 𝘍 θ) = ( θ θ𝘍 θ) cosθ eisin sinθ cos +eicos sin

2

2

2

2

2

2

2

Suppose  =  𝘍 𝘍 θ θ θ (θ+θ𝘍) cos cos − sinθ2 sin 2 cos 2𝘍 2 2 ) ̂ > = ( R θ θ𝘍 θ) = ( y sinθ cos +cos sin (θ+θ𝘍) sin 2 2 2 2 2

Thus, θ becomes θ + θ′. This is exactly what you would expect for counter-clockwise rotation of a vector about the y-axis. θ

θ

θ 𝘍 cos e−iθ /2cos 2 cos 2 𝘍 ̂R> = ( e −iθ /2 2 )=( 0 )( ) = e−iθ /2 ( ) 𝘍 θ 𝘍 θ z iθ /2 i θ 0 e e sin eiθ /2eisin ei(+θ') sin 𝘍

2

2

2

𝘍

Within a global phase factor of e−iθ /2, this is exactly what you would expect for rotation of a vector about the z-axis.

Exercise 14.2. Show that the rotation matrices are unitary. ̂ X (θ) = ( R

cosθ

−isinθ

cosθ

2) , R ̂ (θ) = (

2

−isinθ2

cosθ2 isinθ2

̂ X (θ)R ̂† (θ) = ( R

cosθ2

−isinθ

cosθ

2

2

−isinθ2

cosθ2

̂ y (θ)R ̂† (θ) = ( R y

̂ (θ)R ̂† (θ) = (e R z

z

̂ † (θ) = ( cosθ2 ),R y cos2θ −sinθ2 )(

isinθ2

cosθ2

−sinθ

cosθ

2

2

sinθ2

cos2

−sinθ2

−iθ/2

0

)( θ

0

2) , R ̂ (θ) = (

sin2θ

̂ † (θ) = ( cosθ2 R X isinθ2

x

−sinθ

2

y

eiθ/2 ) ( eiθ/2 0

cosθ2

z

θ

e−iθ/2 0

sin 2 ) , R ̂ † (θ) = (e iθ/2 z 0 cos2θ

eiθ/2

0 ) e−iθ/2

isinθ2

1 0)  R ̂ X (θ) is unitary )=( 0 1 cosθ2

sinθ2

̂ y (θ) is unitary ) = (1 0)  R 0 1 cosθ2 0 )= e−iθ/2

(

1 0 ̂ (θ) is unitary )R 0 1

z

0 )

1 Exercise 14.3. Show the transformation matrix for the following circuit is ( 0 0 0 i.e., a CNOT gate but with a phase shift of −1.

̂ y (θ) = ( R

cosθ

−sinθ

sinθ

cosθ

2

2

)

2

2

̂ y (π/2 ) = ( R

cosπ

−sinπ

sinπ4

4

4

̂ y (−π/2 ) = ( R

0 1 0 0

1

−1 ) 1

−sin(−π)

1

1

4) = ( √2 1 cosπ

cos(−π)

4 sin(−π) 4

1 1

4

I  Ry(π/2 )

1 −1 1( ) 1 1 −1 1 1 0 1 1 = (0 1)  √2 (1 1 ) = √2 ( 1 −1 0( ) 1 1 I  Ry(−π/2 )

1 ) 1 ( 1 1) = 1 ( 1 0  −1 1 = (0 1) √2 −1 1 √2 0 ( 1 1) −1 1 1(

1 0 0 1 CPHASE = ( 0 0 0 0

1

4 )= ( √2 −1 cos(−π)

1

0 0 0 0) 1 0 0 −1

I  Ry(−π/2 ) (CPHASE) (I  Ry(π/2 ))

)

1 −1 1 −1 0 0 ) 1 1 0 1 1 1 0 1 −1 ) = √2 ( 0 0 1 −1 ) 1( ) 0 0 1 1 1 1

0(

1 1 1 0 ) 0 −1 1 ) = 1 (−1 1 1 1 √2 0 0 1 1( ) 0 0 −1 −1 1 0(

1

0 0) 1 1

0 0 0 0); 0 −1 −1 0

1 1 1 0 0 −1 1 0 0 0 = ( )( √2 0 0 1 1 0 0 0 −1 1 0 1 1 0 0 1 1 0 0) (1 = (−1 1 2 0 0 1 1 0 0 0 −1 1 0 2 0 0 0 1 0 0) = (0 2 2 0 0 0 −2 0 0 −2 0 1 0 0 0 0 0) = (0 1 0 0 0 −1 0 0 −1 0

0 0 0 1 −1 0 0 1 0 0) 1 (1 1 0 0) 0 1 0 √2 0 0 1 −1 0 0 −1 0 0 1 1 −1 0 0 1 0 0) 0 1 −1 0 −1 −1

1

̂ (t))-1 = (U ̂ (t))† = U ̂ (−t) for a time-independent H ̂ ; i.e., the evolution Exercise 14.4. Show that (U operator is unitary and reversible. i

̂

U(t) = e− ħ H(t−t0) Choose t0 = 0 for simplicity i

̂

U(t) = e− ħ Ht †

i ̂

H t

U (t) = e

ħ i ̂

U(−t) = e ħ

H t

†

†

-1

†

i ̂

H t

Note that UU = U U = I,  U is unitary and U = U = e ħ From the above, U-1(t) = U†(t) = U(−t)

π

π

Exercise 14.5. Show that ei2σ̂z ei4σ̂y =

i √2

(

1 1

1

); i.e., it is equal to the Hadamard gate within a −1

global phase factor of i. π

e ̂z i 2σ

̂ ( ) ei2π = Rz −π = ( 0

0

π) = (

e−i2

i 0

0 1 0 )=i( ) −i 0 −1

π

̂ (−π/2) = ei 4 σ̂y = R y

(

cos(−π) −sin(−π) 1 1 4 4 )= ( √2 −1 sin(−π) cos(−π) 4

π

e

̂z i 2σ

π îy4σ

e

4

1 1 1 0 1 1 1 =i( ) ( )=i ( √2 1 0 −1 √2 −1 1

1 −1

1 ) 1

) = iH

̂ in the computational basis is H ̂= Exercise 14.6. Show that the dual vector representation for H μBBo ( 0> <0 − 1> <1 ) 1 0 |0> <0> = (1)  (1 0) = ( ) 0 0 0 0 0 |1> <1> = (0)  (0 1) = ( ) 1 0 1 1 0> <0 − 1> <1 = ( 0

0 ) −1

μBBo ( 0> <0 − 1> <1 ) = μBBo (

1 0 ) 0 −1

Exercise 14.7. Calculate the Zeeman splitting for an electron in a 10 T magnetic field.

Bohr magneton, μB = 9.27 x10−24 J/T E = 2μBBo = 2(9.27 x10−24 J/T)(10 T) = 1.85 x 10−22 J = 1.16 meV Exercise 14.8. Calculate the Zeeman splitting for a proton in a 10 T magnetic field. q

From Eq. (3.19), γ = g 2m

p

g factor for proton is ~5.59 mass of proton, mp = 1.67 × 10−27 kg γ=g

q 2mp

=

(5.59)(1.602X10−19 C) 2(1.67 X 10 −27 kg)

= 26.8 x 107 C/kg = 26.8 x 107 /Ts 7

magnetic moment,  = γS = γ ħ/2 = gqħ = (26.8 X 10 /Ts)(6.626 X 10 4mp

2(2π)

E = 2Bo = 2(1.41 x 10-26 J/T)(10 T) = 2.82 x 10−25 J = 1.76 eV

−34 Js)

= 1.41 x 10-26 J/T

Exercise 14.9. Estimate the magnetic field that would be required for the electron Zeeman splitting to exceed the thermal energy at room temperature? At 4 K? E = 2μBBo = kT Bo = kT/2μB 300 K: Bo = kT/2μB = [(1.38 x 10−23 J/K)(300 K)] / [2(9.27 x10−24 J/T)] = 223 T 4 K: Bo = kT/2μB = [(1.38 x 10−23 J/K)(4 K)] / [2(9.27 x10−24 J/T)] = 2.98 T

Chapter 15 Exercise 15.1. What Larmor frequency is required for electron spin resonance at 1 T? Suppose B1 1 = 10-3B . Starting in the |0> state, describe how you would achieve the (|0⟩ + |1⟩) state. What o

√2

Larmor frequency and ESR pulse duration would be required? E = 2μBBo = 2(9.27 x10−24 J/T)(1 T) = 1.85 x 10−23 J o = E/ħ = 1.85 x 10−23 J / (6.626x10-34 Js/2) = 1.75 x 1011 s-1 B1 = 10-3Bo = 10-3 T =

eB1 2me

μ B

(9.27 X 10−24 J/T)(10−3 T)

ħ

6.626 X 10−34 Js/2π

= B 1=

= 8.79 x 107 s-1

t = (/2)/ = (/2)/ (8.79 x 107 s-1) = 18 ns Transform |0> to

1 √2

(|0⟩ + |1⟩) by applying field B cos (ω t) ŷfor a duration of 18 ns, causing 1

o

rotation of |0> state by /2 about the y-axis in the Bloch sphere. Note that this neglects the rotation about the z-axis due to the Larmor precession, which could be corrected by free precession.

Chapter 16 ̂ ′ are W11 = ⟨0|H ̂ ′|0⟩, W12 = ⟨0|H ̂ ′|1⟩, W21 = Exercise 16.1. Show that the matrix elements of H ̂ ′|0⟩, and W22 = ⟨1|H ̂ ′|1⟩. ⟨1|H W12 ̂ ′ = (W11 H ) W21 W22 W12 1 ̂ ′|0⟩ = (1 0) (W11 ⟨0|H ) ( ) = (1 0) (W11) = W 11 W21 W21 W22 0 W12 0 ̂ ′|1⟩ = (1 0) (W11 ⟨0|H ) ( ) = (1 0) (W12) = W 12 W22 W21 W22 1 W12 1 ̂ ′|0⟩ = (0 1) (W11 ⟨1|H ) ( ) = (0 1) (W11) = W 21 W21 W21 W22 0 W12 0 ̂ ′|1⟩ = (0 1) (W11 ⟨1|H ) ( ) = (0 1) (W12) = W 22 W22 W21 W22 1

Exercise 16.2. Write out the explicit form of the state |> at a time t = /2. μBB1

=

ħ

Eo = μBB0 E1 = −μBB0 π

t = 2 t 2 E0 ħ E1 ħ

=

π 4

t=

μBB0 π ħ

t=−

2

μ B π

= B 0 ħ

ħ

2 μBB1

=

πB0 2B1

πB0 2B1

⟩ = (α) = ( β

cos(

)e

E −i 0t ħ

t

−i

t

2

−i sin( 2 ) e

) E1 = ( t

ħ

π

πB −i 0 2B1 πB 0 i

−i sin(

) e 2B1

π) e cos(4

4

)=

1 √2

e

−i

πB0 2B1

|0> −

1 √2

+i

ie

πB0 2B1

|1>

Exercise 16.3. Derive Eq. (16.32).

Solving the two coupled differential equations,

∂α(t) ∂t

=

μBB1 2iħ

β(t) and

∂β(t)

=

μBB1

∂t

α(t), with initial

2iħ

condition of (0) = 0 and (0) = 1 (state starts in 1) gives: t

α(t) = −i sin ( ) 2

t

β(t) = cos ( ) 2

⟩

α(t)

= (β(t)) = (

t −iω0 t 2 sin ( ) 2 ω0 t e+i 2 tcos( ) 2

−ie

)

Exercise 16.4. Derive Eq. (16.51) to (16.54).

E + W11 W12 H + H′ = ( 0 W21 E1 + W22 ) E0 + W11 W12 α α ( ) (β) = E (β) W21 E1 + W22 E0 + W11 − E W12 α ( )( ) = 0 W21 E1 + W22 − E β The eigenvalues are found by diagonalizing the matrix, H + H′. This is done by setting the determinant of the matrix to zero: det (

E0 + W11 − E W12 )=0 W21 E1 + W22 − E

(E0 + W11 − E)(E1 + W22 − E) − W12W21 = 0 ∗ (perturbation matrix is Hermitian), we have W12W21 = |W21|2. Hence, Since W12 = W21

(E0 + W11 − E)(E1 + W22 − E) − |W21|2 = 0 E2 − (E0 + E1 + W11 + W22)E + E0E1 + E0W22 + W11E1 + W11W22 − |W21|2 = 0 Using the quadratic equation: E +E +W +W ±√(E +E +W +W )2−4(E E +E W +W E +W W )−|W |2)

0 1 11 22 0 1 11 22 0 1 0 22 11 1 11 22 21 E = 2 1 1 ( E = (E + W + E + W ) )2 + |2W |2 ± √ E + W − E − W 22 11 1 ± 0 0 11 1 22 21 2

2

First, let us solve the eigenvector corresponding to E+. The eigenvector is found by solving: α+ E0 + W11 − E+ W12 ( )( ) = 0 W21 E1 + W22 − E+ β +

This leads to the following equation: W21α+ + (E1 + W22 − E+) β+ = 0 Rearranging: β =

W21

+

(E+−E1−W22)

α +

Also, |α+|2 + |β+|2 = 1 (normalization condition) |W21|2 |α+|2 = 1 − |β+|2 = 1 − 1

|α+ | = 2

1+

(E+ − E1 − W22)2 (E+−E1−W22)2

=

|W21|2

|α+|2

|W21|2+(E+−E1−W22)2

(E+−E1 −W22)2

(E+−E1−W22)

α = +

√|W21|2+(E+−E1−W22)2

and W21

β = +

α = +

(E+−E1−W22)

W21

(E+−E1−W22)

(E+−E1−W22) √|W21|2+(E+−E1−W22)2

=

W21 √|W21|2+(E+−E1−W22)2

Introduce W21 = |W21|eiφ |W21|eiφ

β = +

√|W21|2+(E+−E1−W22)2 1

(α+) = β+

√|W21|2+(E+−E1−W22)2

22 (E+−E1−W ) iφ

|W21|e

The eigenvector corresponding to E− is solved in an identical manner: (α−) = β−

1 √|W21|2+(E−−E1−W22)2

Define E0 + W11 + E1 + W22 2

Ea = and ∆=

E0 + W11 − E1 − W22 2

Thus,

22 (E−−E1−W ) iφ

|W21|e

E± = Ea ± √∆2 + |W21|2 and α+ (

β+

1 )=

E+ − E1 − W22

√|W21|2 + (E+ − E1 − W22)2

|W21|

2

1

=

(

√|W21 |2+(∆+√∆2+|W12 |2)

2

)

2

12| (∆+√∆|W+|W ) | 21

Similarly, (

α−

)=

β−

1

E− − E1 − W22 ( |W21|eiφ ) √|W21|2 + (E− − E1 − W22)2 1

=

√|W21 |2+(−∆+√∆2+|W21 |2)

2

(

e−iφ(−∆+√∆2+|W21|2)

)

|W21|eiφ

To simplify the above, let us define tanθ =

|W21| ∆

Then start with the trigonometric identity tanθ =

2tan

θ

2

1−tan2

θ 2

Rearranging, θ θ tanθtan2 + 2tan − tanθ = 0 2 2 θ

Solve this quadratic equation for tan : 2

θ −1 ± √1 + tan2θ tan = 2 tanθ θ

On the Bloch sphere, 0    , so tan is positive. This means we take the positive square root 2

θ −1 + √1 + tan2θ tan = 2 tanθ Substituting tanθ =

|W21| ∆

gives:

∆ θ ∆ )2 + 1 = 1 √ tan = − (−∆ + √∆2 + |W21|2) |+ ( | | | | |W21 W21 2 W21

θ

Now find sin : 2

1 (−∆ + √∆2 + |W |2) θ 21 |W21| tan 2 θ sin 2 = = 2 √1 + tan2 θ 1 2 2 2 √1 + |W21 |2 (−∆ + √∆ + |W21| ) −∆ + √∆2 + |W21|2

=

√|W21|2 + (−∆ + √∆2 + |W21|2) θ cos 2 =

1 √1 + tan2 θ 2

2

1

=

2

1 √1 +

|W21 |2 (−∆ + √∆ + |W21| ) 2

2

|W21|

=

√|W21|2 + (−∆ + √∆2 + |W21|2)

2

Thus, α− (β ) = −

θ −sin −(−∆ + √∆2 + |W21|2) ) = ( iφ 2 ) 2( |W |eiφ θ 21 √|W21|2 + (−∆ + √∆2 + |W21 |2) e cos 2 1

Also, after some arduous arithmetic, you can show that: θ sin = 2 θ cos = 2

α+ (β ) = +

|W21| √|W21|2 + (∆ + √∆2 + |W21|2)

2

∆ + √∆2 + |W21|2 √|W21|2 + (∆ + √∆2 + |W21|2)

2

θ cos (∆ + √∆2 + |W21|2) ) = ( iφ 2 ) 2( |W |eiφ θ 21 √|W21|2 + (∆ + √∆2 + |W21 |2) e sin 2 1

Using the above coefficients (probability amplitudes), we can rewrite the wavefunctions as: θ

θ

|+⟩ = cos 2 | 0⟩ + eiφ sin 2 |1 ⟩

θ

θ

|−⟩ = −sin 2 | 0⟩ + eiφ cos 2 |1 ⟩ Or, if we multiply by the global phase factor of e−iφ2 θ

θ

|+ ⟩ = e−iφ2 cos 2 |0 ⟩ + e+iφ2 sin 2 |1 ⟩ θ

θ

|− ⟩ = −e−iφ2 sin 2 |0 ⟩ + e+iφ2 cos 2 |1 ⟩ where we previously defined tanθ =

|W21| ∆

and W = |W |eiφ 21 21

Exercise 16.5. Derive Eq. (16.67). θ θ −iE+t/ħ ⟨ |(t)⟩ = e+i − e−iE−t/ħ) sin cos (e 1 2 2 P1(t) = |⟨1|(t)⟩|

2

θ 2

= (sin θ2 cos 2) (e−iE+t/ħ − e−iE− t/ħ )(e+iE+t/ħ − e+iE− t/ħ) 1

= sin2 θ (2 − e −i(E+−E−)t/ħ + e+i(E+−E−)t/ħ) 4

1

E+−E−

2

= sin θ [2 − 2cos ( 4

ħ

𝑡)]

1

= sin2 θ [1 − cos ( E+−E− 𝑡)] ħ

2

= sin2θ sin2 (

E+ − E−

t)

2ħ By definition tanθ =

|W21| ∆

|W21|2 tan2θ = 1 + tan2θ ∆2 + |W21|2 |W21|2 sin2 (E+ − E− t) P1(t) = ∆2 + |W21|2 2ħ sin2θ =

1

Substitute  = ½ (E0 − E1) and E± = Ea ± √∆2 + |W21|2 2 P1(t) =

|2W21|2 (E0 − E1)2 + |2W21|

sin2 ( 2

√(E0 − E1)2 + |2W21|2 2ħ

t)

Exercise 16.6. Derive Eq. (16.78), (16.79) and (16.80). ħ −∆ω ωd H + H′ = ( ) ∆ω 2 ωd The eigenvalues are found by diagonalizing the above matrix:

det (

ħ − 2 ∆ω − E

ħ 2 ωd

ħ 2

ħ ωd

2

)=0 ∆ω − E

2 ħ (− ∆ω − E) ( ∆ω − E) − ( ωd) = 0 2 2 2

ħ

ħ

2 2 ħ ħ 2−( + E − ( ∆ω) ω d) = 0 2 2 ħ E± = ± √∆ω2 + ωd2 2 1 E± = ± √(E0 − E1 − ħω)2 + |2W21|2 2

First, let us solve the eigenvector corresponding to E+. The eigenvector is found by solving: ħ ħ − ∆ω − E+ ωd α+ 2 2 ( )( ) = 0 ħ ħ β+ ωd ∆ω − E+ 2 2 This leads to the following equation: ħ

ħ ωdα+ + ( ∆ω − E+) β+ = 0 2 2 Rearranging: β+ =

ħ ω 2 d ħ (E+−2∆ω)

α+

Also, |α+|2 + |β+|2 = 1 (normalization condition) |α+ |2 = 1 − |β+ | = 1 − 2

ħ ωd) (2

2

2 ħ (E+ − 2 ∆ω)

|α+|2

|α+

|2 =

2

ħ

1 2 ħ (2ωd) 1+ 2 ħ (E +−− ∆ω) 2

(E+−2∆ω)

= ħ

2

2

ħ

( ωd) +(E+− ∆ω) 2

2

ħ

α+ =

(E+−2∆ω) 2 ħ √ ħ (2ωd) +(E+−2∆ω)

2

and ħ

2ωd

β+ =

ħ

(E+− ∆ω) √

(E+−2∆ω)

2

ħ

(E+−2∆ω) 2

ħ

ħ ħ

2

=

( ωd) +(E`+− ∆ω) 2

2

ħ

22

ωd

ħ

√( ωd) +(E+− ∆ω) 2

2

ħ

1

(α+ β+ ) =

ħ

ħ

2ωħd

α+ =

2

ħ

√( ωd) +(E+− ∆ω) 2

2

(

E+− ∆ω 2

ħω

2

)

d

2

The eigenvector corresponding to E− is solved in an identical manner: ħ

1

(α− β− ) =

ħ

2

ħ

√( ωd) +(E−− ∆ω) 2

Substitute

(

E−− ∆ω 2

ħω

)

d

2

2

ħ

2

2

E± = ± 2 √∆ω + ωd

α+

)= β+

( 2

ħ E+ − 2 ∆ω ( ħ ) 2 d ω 2

1 ħ 2 √ (ħ ωd) + (E+ − ∆ω) 2 2

2

1

= √ω 2+(√∆ω2+ω 2

2

d −∆ω)

d

2

d −∆ω) ((√∆ω +ω ) ωd

Similarly, (α−) = β−

ħ

1 ħ

2

ħ

√( ωd) +(E−− ∆ω) 2

=

2

(

2

ħω 2

2

)

d 2

1 √ωd 2+(√∆ω2+ωd 2∆ω)

E−− ∆ω

2

To simplify the above, let us define ωd tanθ = − ∆ω From Exercise 16.4:

2

d +∆ω)) ((−(√∆ω +ω ) ω d

2

θ −1 + √1 + tan2θ tan = 2 tanθ 1 θ ∆ω ∆ω 2 (∆ω + √∆ω2 + ωd2) + √( ) + 1 = tan = 2 ωd ωd ωd θ

Now find sin : 2

1

θ tan 2

(∆ω + √∆ω2 + ω 2)

2

2

d ∆ω + √∆ω + ωd ωd = = 1 2 2 2θ √ωd2 + (∆ω + √∆ω2 + ωd2) √1 + 2 (∆ω + √∆ω 2 + ωd 2) √1 + tan ωd 2 θ 1 1 ωd cos = = = 2 1 2 2 2θ 2 √ωd2 + (∆ω + √∆ω2 + ωd2) + ωd2) √1 + √1 + tan (∆ω + √∆ω 2 ωd 2

θ sin 2 =

Thus, θ −sin α − 2 ( )=( ) θ β− cos 2 Also, after some arduous arithmetic, you can show that: θ sin = 2 θ cos = 2

ωd √ωd2 + (−∆ω + √∆ω2 + ωd2)

2

−∆ω + √∆ω2 + ωd2 √ωd2 + (−∆ω + √∆ω2 + ωd2)

2

θ cos α + 2 ( )=( θ) sin β+ 2 Using the above coefficients (probability amplitudes), we can rewrite the wavefunctions as: θ

θ

|+⟩ = cos 2 | 0⟩ + sin 2 |1 ⟩ θ

θ

|−⟩ = −sin 2 | 0⟩ + cos 2 |1 ⟩

Exercise 16.7. Derive Eq. (16.81). θ

θ

|+⟩ = cos 2 | 0⟩ + sin 2 |1 ⟩ θ

θ

|−⟩ = −sin 2 | 0⟩ + cos 2 |1 ⟩ Solve for |0⟩: θ θ ⟩ = cos |0 | ⟩ − sin |−⟩ 2 2 + The wavefunction at some later time t is: θ

θ

|(t)⟩ = e−iE+t/ħ cos 2 |+⟩ − e−iE−t/ħ sin 2 |−⟩ √∆ω2+ω d2

= e−i

2

tcos θ

2

√∆ω2+ω d2

|+ ⟩ − e

+i

2

tsin θ

2

|−⟩

The probability amplitude of finding the system in the state |1> at some time t is: θ

θ

⟨ |(t)⟩ = e−iE+t/ħcos ⟨ | ⟩ − e−iE−t/ħsin ⟨ | ⟩ 1

2

1

+

2

1

−

θ θ ⟨1|(t)⟩ = sin cos (e−iE+t/ħ − e−iE−t/ħ) 2 2 Finally, the probability of finding the system in the state |1> at time t is: P1(t) = |⟨1|(t)⟩|

2

θ 2

= (sin θ2 cos 2) (e−iE+t/ħ − e−iE− t/ħ )(e+iE+t/ħ − e+iE− t/ħ) 1

= sin2 θ (2 − e −i(E+−E−)t/ħ + e+i(E+−E−)t/ħ) 4

1

E+−E−

2

= sin θ [2 − 2cos ( 4

ħ

1

𝑡)]

= sin2 θ [1 − cos ( E+−E− 𝑡)] ħ 2

= sin2θ sin2 (

E+ − E−

t)

2ħ By definition tanθ = −

ωd ∆ω

sin2θ = P1(t) =

tan2θ ωd2 = 1 + tan2θ ∆ω2 + ωd2 ωd2

sin2 (

E+ − E−

∆ω2 + ωd2 2ħ ħ 2 2 Substitute E± = ± 2 √∆ω + ωd

t)

√∆ω2 + ωd2 ωd2 2( t) sin P1(t) = 2 ∆ω2 + ωd2

Chapter 17

Exercise 17.1. Derive the following energy separations: E00−10 = ħ(ω1 + 2J/ħ) E01−11 = ħ(ω1 − 2J/ħ) E10−11 = ħ(ω2 − 2J/ħ)

ω1 + ω2 + 2J/ħ 0 ̂= ( H 0 2 0 ħ

0 ω1 − ω2 − 2J/ħ 0 0

0 0 −ω1 + ω2 − 2J/ħ 0

0 0 ) 0 −ω1 − ω2 + 2J/ħ

̂.

The energies are equal to the diagonal entries of H ħ E00 =

2

(ω1 + ω2 + 2J/ħ)

ħ E01 =

2

(ω1 − ω2 − 2J/ħ)

ħ E10 =

2

(−ω1 + ω2 − 2J/ħ)

ħ E11 =

2

(−ω1 − ω2 + 2J/ħ)

E00−10 = E00 − E10 =

ħ 2J ħ (ω1 + ω2 + ) − (−ω1 + ω2 − 2J/ħ) = ħ(ω1 + 2J/ħ) 2 ħ 2

ħ 2J ħ E01−11 = E01 − E11 = (ω1 − ω2 − ) − (−ω1 − ω2 + 2J/ħ) = ħ(ω1 − 2J/ħ) 2 ħ 2 E 10−11

=E 10

−E

ħ

2J

ħ (−ω

11 = 2 (−ω1 + ω2 − ħ ) − 2

− ω + 2J/ħ) = ħ(ω − 2J/ħ) 1

2

2

Exercise 17.2. Show the following: −i 0 0 0 0 1 ̂ (t = ) = √i ( 0 0) U 0 4J 0 0 1 0 0 0 −i πħ

0 (1 0) 1 0 0 1 0 0 (1 1 ) 0 0 −1 0 −1 0 −1 ̂z1  σ σ ̂z2 = ( )( )=( )=( 1 0 1 0 0 0 −1 0 −1 0 −1 0( ) −1 ( ) 0 −1 0 −1 0 0 0 i π ̂t − H − i σ ̂  σ ̂ π π z 1 z 2 ̂ ̂ ̂z2 ) From Eq. (17.13), U(t) = e ħ = e 4 = cos 4 I − isin 4 (̂σz1  σ 1 = (0 √2 0 0 1

0 1 0 0

1−i = ( 0 0 √2 0

0

0 0 1 0

0 1 0 0 0 1 0) − i (0 −1 0 0) 0 √2 0 0 −1 0 1 0 0 0 1

0 1+i 0 0

1

1−i l 1+i

I

√2

0 0 1+i 0

0 𝗁0

0 0 ) 0 1−i

0 0 0 0 1 0I 1−i 0 1+i)

0 1

(1 + i) 0

1

=

1

0 0

1−i

The matrix element 1+i can be written as: 1−i 1+i

= 1−i ∗ 1+i = 2 = 1 ∗ i = −i 1+i

1+i

2i

i

i

The pre-factor can be written as: 1 (1 + i) = 1 (1 + i) = cos(/4) + isin(/4) = ei/4 = (eiπ/2 )1/2=

√2

√2

−i 0  U = √i ( 0 0

0 1 0 0

0 0 0 0) 1 0 0 −i

√i

0 0) 0 1

Chapter 18 No Exercises

Chapter 19

Exercise 19.1. What is N/N for 1H at 10 T and T = 300 K? E = fio = fiBo = (6.626 x 10-34 Js/2)(26.8 x 107 rad/Ts)(10 T) = 2.82 x 10−25 J N / N = exp (−E / kBT) = exp [−2.82 x 10−25 J / [(1.38 x 10−23 J/K)(300 K)] N/N = 0.99993 Therefore, there are only slightly more spins up compared to down.

Chapter 20 Exercise 20.1. Estimate the dimensions of a quantum dot that gives a Larmor frequency of =10 GHz.

Approximate the dot as an infinite quantum well in three directions (a “quantum cube”): kx = n/L , ky = n/L , kz = n/L E = p2/2m = fi2k2/2m = fi2(k 2+k 2+k 2)/2m = 3fi2(n/L)2/2m = n

x

y

z

3h2

n2

8mL2

3h2

n = 1: E1 = 8mL2 n = 2: E2 = 4

3h2 8mL2

3h2 3h2 9h2  E = E2 – E1 = 4 8mL2 − 8mL2 = 8mL2

 = E/h =

9h 8mL2

Rearranging, L = √

9h 8m

=√

9(6.626 X 10−34Js) 8(9.11X10−31 kg)(10X10 9 Hz)

= 286 nm

Chapter 21 Exercise 21.1. Estimate a typical ion speed at 10 mK and calculate the corresponding Doppler frequency shift.

1mv2~ kT 2

speed, v ~ √

2kT m

Choose Ca ion, m = 6.64 x 10−26 kg v~√

2kT m

=

2(1.38 X 10−23 J/K)(10−2K) = 2.04 m/s

√

6.64 X 10−26 kg

Choose the  = 397 nm transition in Ca. frequency,  = c/ = (3x108 m/s) / (397 x 10−9 m) = 7.56 x 1014 Hz 1+v/c

ion = √1−v/c Note that v/c << 1, so  = 

–= ion

1+v/c

√

1−v/c

−1~

√

1−v/c

1

1+2v/c

~

1

1−2v/c

1+v/c –  =  ( 1−v/c

1+1 v/c 2

1

1+v/c

√

1−2v/c

1+v/c

√

1v

−1=

− 1)

1−v/c 1v

(1+ )− (1− ) 2c

1

2c

=

1−2v/c

 =  (v/c) = (7.56 x 1014 Hz)(

2.03 m/s

v/c

~ v/c

1

1−2v/c

) = 5.1 MHz

3 X 108m/s

Chapter 22

Exercise 22.1. Show that the potential minimum in Figure 22.20(a) disappears when I > I o.

The potential is U() = −EJ(cos + IIo) dU d

= 0 at a potential minima

dU = E (sin − II )

J

d

o

If I > Io, then I/Io > 1  dU < 0 d

There are no potential minima.

Chapter 23 No Exercises

Chapter 24 Exercise 24.1. Show that the order of matrix multiplication for the mirrors does not matter: MuMl = MlMu −1 0 Mu = ( 0 1) 1 Ml = ( 0

0 ) −1

−1 M uM l = ( 0 MlMu = (

0 1 )( 1 0

1

0 −1 )( 0 −1 0

 MuMl = MlMu

0 −1 )=( −1 0

0 ) −1

0 −1 0 )=( ) 1 0 −1

Chapter 25 Exercise 25.1. Check that the circuit works; i.e, 0> → 000>, 1> → 111>, and 0> + 1> → 000> + 111>

Input |000>: First CNOT does nothing. Second CNOT does nothing. Output is |000> Input |100>: First CNOT produces |110>. Second CNOT produces |111> Input 0> + 1>: Input is (0> + 1>)|00> = 000> + 100> According to the above, the output becomes 000> + 111>

Exercise 25.2. If the probability of a bit flip error occurring is p, show that the probability of two or more errors occurring in a 3-qubit encoding is 3p2 − 2p3.

Probability of a bit flip error occurring is p Probability of a bit flip error not occurring is 1−p Start with |000> Probability of |110> occurring = (p)(p)(1-p) = p2 – p3 Probability of |011> occurring = (1-p)(p)(p) = p2 – p3 Probability of |101> occurring = (p)(1-p)(p) = p2 – p3 Probability of |111> occurring = (p)(p)(p) = p3 Total probability = 3p2 − 2p3

Exercise 25.3. Verify the syndromes in Table 25.1.

’>

0>

Bit flip error

>

0> 3-qubit encoding circuit

Apply correction

’>

0> 0>

Determine the effect of the CNOT gates on the last two qubits: If the bit flip error results in |000> (no error): The CNOT gates have no effect Measurement results in |00> If the bit flip error results in |001>: After the 1st CNOT, we get |00> on the last two qubits After the 2nd CNOT, we get |00> After the 3rd CNOT, we get |00> After the 4th CNOT, we get |01> Measurement results in |01> If the bit flip error results in |010>: After the 1st CNOT, we get |00> on the last two qubits After the 2nd CNOT, we get |10> After the 3rd CNOT, we get |10> After the 4th CNOT, we get |10> Measurement results in |10> If the bit flip error results in |011>: After the 1st CNOT, we get |00> on the last two qubits After the 2nd CNOT, we get |10> After the 3rd CNOT, we get |10> After the 4th CNOT, we get |11> Measurement results in |11>

If the bit flip error results in |100>: After the 1st CNOT, we get |10> on the last two qubits After the 2nd CNOT, we get |10> After the 3rd CNOT, we get |11> After the 4th CNOT, we get |11> Measurement results in |11> If the bit flip error results in |101>: After the 1st CNOT, we get |10> on the last two qubits After the 2nd CNOT, we get |10> After the 3rd CNOT, we get |11> After the 4th CNOT, we get |10> Measurement results in |10> If the bit flip error results in |110>: After the 1st CNOT, we get |10> on the last two qubits After the 2nd CNOT, we get |00> After the 3rd CNOT, we get |01> After the 4th CNOT, we get |01> Measurement results in |01> If the bit flip error results in |111> (no error): After the 1st CNOT, we get |10> on the last two qubits After the 2nd CNOT, we get |00> After the 3rd CNOT, we get |01> After the 4th CNOT, we get |00> Measurement results in |00> The above results reproduce Table 25.1.

Chapter 26

Exercise 26.1. Referring to Figure 26.5, derive Eq. (26.4).

Bz

Bz t

Vy vx

W L Ix

A = tW Ix

The carriers experience a Lorentz force from the applied magnetic field, F y = evxBz. Carriers are deflected by Lorenz force producing a counteracting electrostatic force, eEy. At equilibrium, eEy = evxBz Ey = vxBz The carrier velocity, vx = Jx/ne Current density, Jx = Ix/A = Ix/tW Electric field, Ey = Vy / W Combining the above equations gives the Hall coefficient, R = V /I = Bz H

y x

ent

For a two-dimensional electron gas (2DEG), nt becomes the charge per unit area.