Introduction to Quantum Computing (Instructor Solution Manual, Solutions) [1 ed.] 3030693171, 9783030693176

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Solutions Manual Introduction to Quantum Computing R.R. 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√I1 I2 cos | E1 + E2 |2 = (E1 + E2)( E1 + E2)*  I12

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

 I2

 2√I1 I2 cos

 I12 = I1 + I2 + 2√I1 I2 cos ______________________________________________________________________________

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

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

4

π

π

π

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

=0

+∞

∫ −∞

1∗ 1 dx

=

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

=

2 L 2 π ∫ sin ( x) dx L 0 L 1

L

2π

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

1 L 1 2π L x| − sin ( x)| 0 L 0 2π L

=1

Similarly, +∞

∗2 2 dx = 1

∫ −∞

+∞

 ∫−∞ ∗i j dx = δij where i, j = {0, 1}  1(x) and 2(x) are orthonormal. ______________________________________________________________________________ Exercise 2.2. Prove Eq. (2.49): = |c1|2 a1 + |c2|2 a2 + … + |cn|2 an  = c11 + c22 + … cnn +∞

=

̂  dx ∫−∞ ∗ A +∞

∫−∞ ∗  dx

The numerator is: +∞

̂  dx ∗ A

∫ −∞

+∞

̂ (c1 1 + c2 2 +. . . cn n ) dx (c1∗ 1∗ + c2∗ 2∗ + cn∗ ∗n ) A

=∫ −∞

+∞

(c1∗ 1∗ + c2∗ 2∗ + cn∗ ∗n ) (a1 c1 1 + a2 c2 2 +. . . an cn n ) dx

=∫ −∞

= c1∗ c1 a1 + c2∗ c2 a2 +. . . cn∗ cn an = |c1 |2 a1 + |c2 |2 a2 +. . . |cn |2 an +∞

All other terms in the numerator, like a2 c1∗ c2 ∫−∞ 1∗ 2 dx, are zero since 1 and 2 are orthonormal.

Similarly, the denominator is: +∞

∫ −∞

∗  dx

+∞

=∫

(c1∗ 1∗ + c2∗ 2∗ + cn∗ ∗n ) (c1 1 + c2 2 +. . . cn n ) dx

−∞ +∞

1∗ 1 dx + c2∗ c2 ∫

= c1∗ c1 ∫ −∞

+∞

∗2 2 dx+. . . cn∗ cn ∫

−∞

+∞

∗n n dx

−∞

= |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 c1∗ c2 ∫−∞ 1∗ 2 dx, are zero.  =

+∞

̂  dx ∫−∞ ∗ A +∞

∫−∞ ∗  dx

= |c1 |2 a1 + |c2 |2 a2 +. . . |cn |2 an

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 ⟩ 1 1 2 1 0 √2 2 √2 2 + ħ Ŝx x− ⟩ = − 2 x− ⟩ ħ 0 1 1 ħ 1 −1 ħ 1 ( ) ( ) = ( ) = − x ⟩ −1 1 2 1 0 √2 2 √2 2 − ħ Ŝy y+ ⟩ = + 2 y+ ⟩ ħ 0 ħ 1 ħ −i 1 1 ( ) ( i ) = 2 (1i) = + 2 y+ ⟩ 2 i √2 √2 0 ħ Ŝy y− ⟩ = − 2 y− ⟩ ħ 0 ħ 1 ħ −i 1 1 ( ) ( −i ) = 2 (−1 ) = − 2 y− ⟩ i 2 i √2 √2 0 ħ Ŝz z+ ⟩ = + z+ ⟩ 2 ħ 1 ħ ħ 0 1 ( ) (0) = 2 (10) = + 2 z+ ⟩ 2 0 −1 ħ Ŝz z− ⟩ = − 2 z− ⟩ ħ 1 ħ ħ 0 0 ( ) (1) = (−10) = − z− ⟩ 2 0 −1 2 2

______________________________________________________________________________

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

I + σ ̂x + σ ̂y + σ ̂z = ( =(

1 0 0 ) + ( 0 1 1

α+γ β + δi

1 0 ) + ( 0 i

−i 1 ) + ( 0 0

0 ) −1

β − δi ) α−γ

Equate this to a general 2x2 complex matrix: α+γ ( β + δi

β − δi a )=( b α−γ

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 0 1 0 X2 = ( )( 1 0 1

Y2 = (

0 i

−i 0 )( 0 i

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

1 0 1 0 1 )( )=( 0 −1 0 −1 0

0 )=I 1

Z2 = (

− i XYZ = − i (

0 1 0 )( 1 0 i

0 1 0 )( 1 0 i

− i XY = − i (

1 0 0 ZY = ( )( 0 −1 i

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

−i 0 )=( 0 −i

1 0 i i )( )=−i( 0 i 0 0

0 1 )=( −i 0

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

0 )=Z −1

−i 0 1 )=−i( )=−iX 0 1 0

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

−i 0 )( 0 1

1 −i 0 1 0 )=( ) =−i( )=−iZ 0 0 i 0 −1

______________________________________________________________________________ ħ Exercise 3.5. Prove Eq. (3.50) from Eq. (3.49). Show that the Ŝn operator has eigenvalues ± 2 and

corresponding eigenvectors (

θ 2

cos

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

0 1

= 2 cossinθ (

ħ 1 0 ) + 2 sinsinθ ( 0 i

ħ −i 1 0 ) + 2 cosθ ( ) 0 0 −1

ħ

cosθ cossinθ − isinsinθ ) cossinθ + isinsinθ −cosθ

ħ

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

= 2( = 2(

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

Ŝn (

θ 2

θ

ħ cosθ e−i sinθ) ( cos2 ) = θ) = 2 ( i θ i e sin ei sin e sinθ −cosθ 2 2 cos

θ

θ

cosθcos + sinθsin ħ 2 2 ( ) 2 ei (sinθcosθ − cosθsinθ) 2

2

cos

ħ

= 2(

θ 2

)

θ 2

ei sin

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)

Ŝn (

θ 2

sin

)=

θ 2

− ei cos θ

sin ħ − ( i 2 θ) 2 −e cos 2

ħ ( cosθ 2 ei sinθ

θ

e−i sinθ) ( sin2 ) = θ − ei cos −cosθ 2

θ

2

θ

θ

cosθsin − sinθcos ħ 2 2 ( ) 2 ei (sinθsinθ + cosθcosθ) 2

=

ħ sin(−2) ( ) 2 ei cosθ 2

=

Chapter 4 ______________________________________________________________________________ 1

1

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

1 √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⟩ = ( |+⟩ + |−⟩ ) |1⟩ =

√2 1 √2

( |+⟩ − |−⟩ )

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

√2 1 √2

(⟨0| − ⟨1|)

The inner products are: ⟨+|−⟩ = 1

1 √2

(⟨0| + ⟨1|)

1 √2

( |0⟩ − |1⟩ )

= 2 (⟨0|0⟩ − ⟨0|1⟩ + ⟨1|0⟩ − ⟨1|1⟩) 1

= 2 (1 − 0 + 0 − 1) =0 ⟨+|+⟩ = 1

1 √2

(⟨0| + ⟨1|)

1 √2

( |0⟩ + |1⟩ )

= 2 (⟨0|0⟩ + ⟨0|1⟩ + ⟨1|0⟩ + ⟨1|1⟩) 1

= 2 (1 + 0 + 0 + 1) =1 ⟨−|−⟩ =

1 √2

(⟨0| − ⟨1|)

1 √2

( |0⟩ − |1⟩ )

1

= 2 (⟨0|0⟩ − ⟨0|1⟩ − ⟨1|0⟩ + ⟨1|1⟩) 1

= 2 (1 − 0 − 0 + 1) =1

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

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−> =

1 √2 1

(11), , |10> and |11>? Show that these four states are orthonormal.

In general, for two qubits: > =  00> +  01> +  10> +  11> with matrix representation > = 







α00 α (α01 ) 10 α11

Thus, |00> =

1 (00) 0

, |01> =

0 (10) 0

, |10> =

0 (01) 0

, |11> =

0 (00) 1

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

Similarly, = = = = = = = = = = = 0

= (1 0 0 0)

1 (00) 0

=1

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

Exercise 4.8. Show that the following state is normalized: 1

i

’> = [(2 + 2) 00> + =

1 i 2 2 2

1 2

1 2 2

|( + )| +|( )| (√¾)

2

|01>] / √¾ =

1 1 1 + + 4 4 4

¾

=1

______________________________________________________________________________

Chapter 5 ______________________________________________________________________________ Exercise 5.1. Show that P(z−z+) = ½ P(z−z+) 1 =  − z−z+>)] 2 √2

= ½  − 2 = ½  − 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−) 1 =  − z−z+>)] 2 √2

= ½  − 2 = ½  − 2 = ½  (1) − (0) 2 = ½  2 1 Substituting = (|0> = |0> − |1> = |0> (1) − |1> (0) = |0> For |1> input: Z|1> = (|0> = |0> − |1> = |0> (0) − |1> (1) = −|1> Matrix representation: 1 0 1 Z|0> = ( ) (0) = (10) = |0> 0 −1 1 0 0 Z|1> = ( ) (1) = −(01) = −|1> 0 −1

Z=(

1 0 1 0 ) , Z† = ( ) 0 −1 0 −1

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

0 )=I 1

Z Z† = (

______________________________________________________________________________ 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. H=

1 √2

+> = −> =

(

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

( + |1> = For |1> input: H|1> = (|0> = |0> + |1> =

1

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

√2 1

(|0> − |1>) = |−>

√2

Matrix representation: 1 1 1 1 1 1 1 1 H|0> = ( ) (0) = (11) = (01) + (10) = (|0> + |1>) = |+> √2 1 −1 √2 √2 √2 √2 1 1 1 1 1 1 0 1 1 0 1 H|1> = ( ) (1) = (−1) = (0) − (1) = (|0> − |1>) = |−> √2 1 −1 √2 √2 √2 √2

H=

1 √2

(

1 1 1 1 ) , H† = ( √2 1 1 −1

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

1

√2

(

1 ) −1

1 1 1 2 0 ) = 2( )=( 0 2 0 −1

0 )=I 1

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

H|+> = H [ 1

1 √2

(0> + 1>) ] = 1

1

1 √2

H|0> + 1

1 √2

1 √2

1 √2

(0> + 1>), then we obtain |0>. Show

(0> − 1>), then we obtain |1>.

H|1> =

1

|+> +

√2

1

|−> =

√2

1 √2

[

1 √2

(0> + 1>) ] +

1 √2

[

1 √2

(0>

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

1 √2

(0> − 1>) ] = 1

1

1 √2

H|0> − 1

1 √2

H|1> =

1

|+> −

√2

1

|−> =

√2

1 √2

[

1 √2

(0> + 1>) ] −

1 √2

[

1 √2

(0>

− 1>) ] = 2|0> + 2|1> − 2|0> + 2|1> = |1> ______________________________________________________________________________ Exercise 7.7. Show that the output of the Hadamard gate can be represented as:

H|x> =

|0> + (−1)x |1> √2

|0> + (−1)x |1>

If x=0, the output is

√2 |0> + (−1)x |1>

If x=1, the output is

√2

= =

|0> + (−1)0 |1> √2 |0> + (−1)1 |0> √2

= =

|0> + |1> √2 |0> − |1> √2

= |+> = |−>

______________________________________________________________________________ Exercise 7.8. Show that H =

1

(X+Z) =

√2

0 [( √2 1 1

1 1 ) + ( 0 0

1

(X+Z).

√2

0 )] = −1

1 √2

1 1

(

1 )=H −1

______________________________________________________________________________ Exercise 7.9. Show that the identity matrix can also be written as |+> = 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 √2

(11) , |−> =

( 1) √2 −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, (01)(αβ) = (α0) β

1 (0 0 0

0 1 0 0

0 0 0 1

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

______________________________________________________________________________ Exercise 7.13. Show the following: H

H

H

H

=

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

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

1 = (0 0 0

0 0 0 1

0 0 1 0

0 0 0 1

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

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

= Z

By explicit calculation: 1 Left-hand side: CPHASE = |0> = ( 0 0 0

0 1 0 0

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

1 CPHASE|01> = ( 0 0 0

0 1 0 0

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

0 1 0 0 0 1 0 0

0 0 0 0) 1 0 0 −1 0 0 0 0) 1 0 0 −1

1 CPHASE|10> = ( 0 0 0

0 1 0 0

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

1 CPHASE|11> = ( 0 0 0

0 1 0 0

0 0 0 0 0 0 ) (0) = − (0) = −|11> 1 0 0 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 IH = ( 0

1 1 0 ) ( √2 1 1

1 0 CPHASE = ( 0 1 0 0 0 0

1 1( 1 1 )= ( 1 −1 √2 0 (1 1

1 ) −1 1 ) −1

0 1 0 0

0 0 0 1

H

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

0 0 0 0) 1 0 0 −1

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

Z

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

0 0) = CNOT 1 0

______________________________________________________________________________

Exercise 7.16. Check that the outer product notation gives the SWAP matrix. 1

0

0

0

0

0

0

1

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

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

= 1 CNOT = (0 0 0

0 1 0 0

0 0 0 1

0 0) 1 0

1 From Exercise 7.13, matrix for CNOT with 2nd qubit as control is (0 0 0 1 (0 0 0

0 0 0 1

0 0 1 0

0 1 1) (0 0 0 0 0

0 1 0 0

0 0 0 1

0 1 0) (0 1 0 0 0

0 0 0 1

0 0 1 0

0 1 1) = (0 0 0 0 0

0 0 1 0

0 1 0 0

0 0 0 1

0 0 1 0

0 1) 0 0

0 0) = SWAP 0 1

______________________________________________________________________________ Exercise 7.18. Using only Dirac notation, show that |0> = (|0> = |0>  I|0>  I|0> + |1>  SWAP|00> = |0> (1)  |0>  |0> + |1> (0)  SWAP|00> = |000> CSWAP|001> = (|0> = |0>  I|0>  I|1> + |1>  SWAP|01> = |0> (1)  |0>  |1> + |1> (0)  SWAP|01> = |001> CSWAP|010> = (|0>

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

0>

H

50% probability of measuring 0> 50% probability of measuring 1>

> = 0> + 1>

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. 0 1 )=X 1 0  X is Hermitian X† = (

0 −i )=Y i 0  Y is Hermitian Y† = (

1 0 )=Z 0 −1  Z is Hermitian Z† = (

1 1 )=H 1 −1  H is Hermitian H† =

1

√2

(

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

0 0) = CNOT 1 0

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

______________________________________________________________________________

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

H

|01> input: (HI)|01> = H|0> I|1> = 1

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

1

√2 1

√2

|10> input: (HI)|10> = H|1> I|0> = 1

1

1

(|00> − |10>)

√2

(|00> − |11>)

√2

1

(|0> − |1>)|1> =

√2 1

CNOT [ (|01> − |11>)] = √2

(|01> + |11>)

√2

(|01> + |10>)

(|0> − |1>)|0> =

√2 1

√2

1

1

√2

CNOT [ (|00> − |10>)] = |11> input: (HI)|11> = H|1> I|1> =

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

1

(|01> − |11>)

√2

(|01> − |10>)

√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> =

√2

1

(|000> + |100>)

√2

1

1

1

√2

√2

√2

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

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

1

(|000> +

√2

|110>) 1

1

√2

√2

(ICNOT) [ (|000> + |110>)] =

I|0> CNOT|00> +

1 √2

I|1>CNOT|10> =

1

(|000> + |111>)

√2

______________________________________________________________________________

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.

0> + 1>

Alice 1

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

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

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

0 ) 1 0 ) 1

1 0 1 1( 0

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

0 0( −i )=( i 0 0 1( i

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

0(

1 0 1 −1 ( 0 0(

0 0( 1 0 1( 1 −i ) 0 −i ) 0

0 i 0 0( i

0 0 1 0

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

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

1(

0 1 0 0

1 0 0 0

−i 0 ) 0 ) = (0 −i 0 ) 0 i

0 0) 0 1

0 0 0 1 0 0) 0 −1 0 0 0 −1 0 0 0 1

0 0) 1 0 0 0 −i 0

0 i 0 0

−i 0) 0 0

______________________________________________________________________________

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

Z

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

1 ) −1 1 ) −1

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

0 0) −1 1

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

0>

H

0> HI=

1 √2

(

1 CNOT = (0 0 0

1 1( 0 )= ( 0 √2 1 1 1( 0

1 1 1 )( 0 1 −1 0 1 0 0

0 0 0 1

0 ) 1 0 ) 1

1 1( 0 1 −1 ( 0

0 ) 1 )= 0 ) 1

1 (0 √2 1 0 1

0 1 0 1 0 1) 0 −1 0 1 0 −1

0 0) 1 0

1 (CNOT)(H  I) = (0 √2 0 0 1

1

0 1 0 0 1

|0>  |0> = (10)  (10) = (00) 0

0 0 0 1

0 1 0) (0 1 1 0 0

0 1 0 1 0 1) = 0 −1 0 1 0 −1

1 0 (0 1 √2 0 1 1 0 1

1 0 0 1) 0 −1 −1 0

1 (CNOT)(H  I)|00> = ( 0 √2 0 1 1

0 1 1 0

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

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

0>n1> =

1 (0) ⋮ 0

 (01) =

1(01) 0(01) ⋮ 0(01)

=

0 1 0 0 ⋮ 0 (0)

= |1>n+1

( ) ______________________________________________________________________________ Exercise 9.6. Show that CNOT = |0> = (10) , where A ̂ and B ̂ are the electron spin. Hint: is the expectation value given by = |10>) is the entangled state.

Measurement in the x-z plane can be decomposed as: cosθ1 0 ̂ = sin1X + cos1Z = sin1(0 1) + cos1(1 A )=( sinθ1 1 0 0 −1 ̂=( Similarly, B

cosθ2 sinθ2

sinθ1 ) −cosθ1

sinθ2 ) −cosθ2

cosθ2 sinθ2 cosθ2 sinθ2 ) sinθ1 ( ) cosθ1 ( sinθ −cosθ sinθ −cosθ 2 2 2 2 ̂ B ̂=( ) A cosθ2 sinθ2 cosθ2 sinθ2 ) −cosθ1 ( ) sinθ1 ( sinθ2 −cosθ2 sinθ2 −cosθ2

1

(|01> −

√2

cosθ1 cosθ2 cosθ1 sinθ2 =( sinθ1 cosθ2 sinθ1 sinθ2 |> =

1

cosθ1 sinθ2 −cosθ1 cosθ2 sinθ1 sinθ2 −sinθ1 cosθ2

(|01> − |10>) =

√2

1 √2

0

(10) − 0

cosθ1 cosθ2 ̂ B ̂ |> = 1 ( cosθ1 sinθ2 A √2 sinθ1 cosθ2 sinθ1 sinθ2 =

sinθ1 cosθ2 sinθ1 sinθ2 sinθ1 sinθ2 −sinθ1 cosθ2 ) −cosθ1 cosθ2 −cosθ1 sinθ2 −cosθ1 sinθ2 cosθ1 cosθ2 1 √2

0

(01) = 0

1 √2

cosθ1 sinθ2 −cosθ1 cosθ2 sinθ1 sinθ2 −sinθ1 cosθ2

0

(−11) 0

sinθ1 cosθ2 sinθ1 sinθ2 0 sinθ1 sinθ2 −sinθ1 cosθ2 ) (−11) −cosθ1 cosθ2 −cosθ1 sinθ2 0 −cosθ1 sinθ2 cosθ1 cosθ2

cosθ1 sinθ2 −sinθ1 cosθ2 −cosθ1 cosθ2 −sinθ1 sinθ2 ( ) √2 sinθ1 sinθ2 +cosθ1 cosθ2 −sinθ1 cosθ2 +cosθ1 sinθ2 1

̂ B ̂|> = 3? 1 1 1 1 1 H30>3 = √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 −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 0 −1 0 −1 0 1 = −1 0 1 0 1 0 −1 ) (0)

1 1 1 1 1 √8 1 1 1 (1)

The output is an equally weighted superposition. ______________________________________________________________________________

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 Uf0 = (0 1 0 0 0 1 0 0 0 Uf0 Uf†0

0 1 0 0 0) , U † = (0 1 0 f0 0 0 0 1 1 0 0 0

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

0 0) 0 1

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

0 0) = I 0 1

 Uf0 is unitary. ______________________________________________________________________________

Exercise 11.3. Check that Uf0 gives the expected output for each possible input. 1 0 0 Uf0 |00> = (0 1 0 0 0 1 0 0 0

0 1 1 0) (0) = (0) 0 0 00 0 1

1 0 0 Uf0 |01> = (0 1 0 0 0 1 0 0 0

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

1 0 0 Uf0 |10> = (0 1 0 0 0 1 0 0 0

0 0 0 0) (0) = (0) 1 0 10 0 1

1 0 0 Uf0 |11> = (0 1 0 0 0 1 0 0 0

0 0 0 0) (0) = (0) 0 0 01 1 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 Uf1 = (1 0 0 0 0 0 0 0 1

0 0) 1 0

0 1 0 Uf2 = (1 0 0 0 0 1 0 0 0

0 0) 0 1

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

0 0) 1 0

0 1 0 Uf1 |00> = (1 0 0 0 0 0 0 0 1

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

0 1 0 Uf1 |01> = (1 0 0 0 0 0 0 0 1

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

0 1 0 Uf1 |10> = (1 0 0 0 0 0 0 0 1

0 0 0 0) (0) = (0) = |11> 0 1 10 1 0

0 1 0 Uf1 |11> = (1 0 0 0 0 0 0 0 1

0 0 0 0) (0) = (0) = |10> 1 1 01 0 0

0 1 0 Uf2 |00> = (1 0 0 0 0 1 0 0 0

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

0 1 0 Uf2 |01> = (1 0 0 0 0 1 0 0 0

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

0 1 0 Uf2 |10> = (1 0 0 0 0 1 0 0 0

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

0 1 0 Uf2 |11> = (1 0 0 0 0 1 0 0 0

0 0 0 0) (0) = (0) = |11> 0 0 01 1 1

1 0 0 Uf3 |00> = (0 1 0 0 0 0 0 0 1

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

1 0 0 Uf3 |01> = (0 1 0 0 0 0 0 0 1

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

1 0 0 Uf3 |10> = (0 1 0 0 0 0 0 0 1

0 0 0 0) (0) = (0) = |11> 0 1 10 1 0

1 0 0 Uf2 |11> = (0 1 0 0 0 0 0 0 1

0 0 0 0) (0) = (0) = |10> 1 1 01 0 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 Uf1 = (1 0 0 0) , Uf†1 = (1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 1 Uf1 Uf†1 =

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

0 0) 1 0

0 0) = I 0 1

 Uf1 is unitary 0 1 0 Uf2 = (1 0 0 0 0 1 0 0 0

0 0 1 0 0 0) , U † = (1 0 0 0) f2 0 0 0 1 0 1 0 0 0 1

1 0 0 Uf2 Uf†2 = (0 1 0 0 0 1 0 0 0

0 0) = I 0 1

 Uf2 is unitary 1 0 0 Uf3 = (0 1 0 0 0 0 0 0 1

0 1 0 0 0 0) , U † = (0 1 0 0) f3 1 0 0 0 1 0 0 0 1 0

Uf3 Uf†3 =

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

0 0) = I 0 1

 Uf3 is unitary _____________________________________________________________________________ Exercise 11.5. As in the previous examples, prove that the output of the Deutsch circuit for f 3 is 1, indicating it is balanced.

Starting with Eq. (11.6), Uf 1

H|0>  H|1> → = = = =

1 2 1 2 1 2 1

2

(  0f3(0)> −  1f3(0)> + 1> 0f3(1)> −  1f3(1)> )

(  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 √2

1

1 1 (−1)f(1) |1⟩ = (−1)f(0) ( |⟩ + (−1)f(0)  f(1) |⟩) √2 √2 √2

(−1)f(0) |0⟩ +

If f(0) = 0 and f(1) = 0: 1 1 1 (−1)f(0) |0⟩ + (−1)f(1) |1⟩ = (−1)f(0) ( |⟩ + √2 1 √2

=

(−1)0 |0⟩ + 1

√2

|0⟩ +

1 √2

1

√2

√2

(−1)0 |1⟩ = (−1)0 ( 1

|1⟩ =

√2

1

|⟩ +

√2

1

|⟩ +

√2

1 (−1)f(0)  f(1) |⟩) √2 1 (−1)0  0 |⟩) √2

√2

|⟩

Left-hand side equals right-hand side If f(0) = 0 and f(1) = 1: 1 1 1 (−1)f(0) |0⟩ + (−1)f(1) |1⟩ = (−1)f(0) ( |⟩ + √2 1 √2 1 √2

(−1)0 |0⟩ + |0⟩ −

1 √2

1

√2

√2

|1⟩ = (

1 √2

|⟩ −

|⟩ +

√2

1 √2

(−1)f(0)  f(1) |⟩) √2 1 (−1)0  1 |⟩) √2

√2

1

(−1)1 |1⟩ = (−1)0 (

1

|⟩)

Left-hand side equals right-hand side If f(0) = 1 and f(1) = 0: 1 1 1 (−1)f(0) |0⟩ + (−1)f(1) |1⟩ = (−1)f(0) ( |⟩ + √2 1 √2

(−1)1 |0⟩ +

=−

1 √2

|0⟩ +

1

1

√2

√2

(−1)0 |1⟩ = (−1)1 (

|1⟩ = − (

√2

1 √2

|⟩ −

1 √2

1 (−1)f(0)  f(1) |⟩) √2 1 (−1)1  0 |⟩) √2

√2

1

|⟩ +

√2

|⟩)

Left-hand side equals right-hand side If f(0) = 1 and f(1) = 1: 1 1 1 (−1)f(0) |0⟩ + (−1)f(1) |1⟩ = (−1)f(0) ( |⟩ + √2 1 √2

(−1)1 |0⟩ +

=−

1 √2

|0⟩ −

1

1

√2

√2

√2

(−1)1 |1⟩ = (−1)1 ( 1

|1⟩ = − (

√2

|⟩ +

1 √2

1 (−1)f(0)  f(1) |⟩) √2 1 (−1)1  1 |⟩) √2

√2

1 √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  (−1)f(0) |0⟩ + (−1)f(1) |1⟩ = (−1)f(0) ( |⟩ + (−1)f(0)  f(1) |⟩) √2

√2

√2

√2

______________________________________________________________________________ 1

Exercise 11.8. Show that H (

√2

transformation.

|⟩ +

1 √2

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

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

1 √2 1 √2 1 √2

(−1)a |1⟩) = |a⟩ (−1)0 |1⟩) = |0⟩ |1⟩) = |0⟩

If a = 1: 1 1 (−1)a |1⟩) = |a⟩ H ( |⟩ + √2 √2 1 1 (−1)1 |1⟩) = |1⟩ H ( |⟩ + √2 √2 1 1 H ( |⟩ − |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 − 1 2/N D† = ( ⋮ 2/N 2/N − 1 2/N D D† = ( ⋮ 2/N

2/N 2/N ) ⋮ 2/N − 1

⋯ ⋯ ⋱ ⋯

2/N 2/N − 1 ⋮ 2/N

2/N 2/N )=D ⋮ 2/N − 1

⋯ ⋯ ⋱ ⋯

2/N 2/N − 1 ⋮ 2/N

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

⋯ ⋯ ⋱ ⋯

2/N 2/N − 1 ⋮ 2/N

⋯ ⋯ ⋱ ⋯

2/N 2/N ) ⋮ 2/N − 1

Each diagonal entry is: 2

2

2 2

4

4

4

4

(N − 1) + (N − 1) (N) = N2 − N + 1 + N − N2 = 1 Each non-diagonal entry is: 2

2

2 2

2 (N − 1) (N) + (N − 2) (N) =

8 N2

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). 1 1 Hn = ( √N ⋮ 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 0 0 Hn( ⋮ ⋮ 0 0

1 ⋯ 0 ⋯ 0 1 1 ⋱ ⋮ ) = √N ( ⋮ ⋯ 0 1

1

2 0 0 0 Hn( ⋮ ⋮ 0 0

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

⋮

⋯ ⋯ ⋱ ⋯

1

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

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

0 0 ) ⋮ 0 ⋯ 2 ⋯ 2 ⋱ ⋮) ⋯ 2

______________________________________________________________________________

Exercise 12.4. Using the tensor product for the vectors |0> and = ( ) , 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

< |> = 1 ; |> and |x*> are orthogonal so = 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 β0 = 2 ∑3j=0 αj = 2 (α0 + α1 + α2 + α3 ) 1

1

β1 = 2 ∑3j=0 αj eπij/2 = 2 (α0 + α1 eiπ/2 + α2 eiπ + α3 e3iπ/2 ) 1

1

β2 = 2 ∑3j=0 αj eπij = 2 (α0 + α1 eiπ + α2 e2iπ + α3 e3iπ ) 1

1

β3 = 2 ∑3j=0 αj e3πij/2 = 2 (α0 + α1 e3iπ/2 + α2 e3iπ + α3 e9iπ/2 ) 1

1

1

1

1

1

1

1

1

1

1

1

For basis state |0>: 0 = 2,  1 = 2,  2 = 2,  3 = 2 1

1

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

1

1

1

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

1

Thus, QFT4|1> = 2 (−1i ) −i

1

1

For basis state |2>:  0 = 2,  1 = 2 eiπ = − 2,  2 = 2,  3 = − 2 1

1

Thus, QFT4|2> = 2 (−11) −1

1

1

1

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

1 −i ( ) 2 −1 i 1

Now show that the 4 states,

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

and

1 −i ( ), 2 −1 i 1

are orthonormal.

First, show these vectors are normalized: 1

1

1

1

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

1

1

1

1

1

(1) is normalized 2 1

1

1

1

1

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

1

−i

1

1

1

1

1

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

1 2

(1 i −1 −i)

1 −i ( ) 2 −1 i 1

−1

1

= 4(1+1+1+1) = 1, 

Next, check orthogonality: 1

1

1

(1 1 1 1) 2 (−1i ) = 0 2

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

1

1 −i ( ) 2 −1 i 1

is normalized

______________________________________________________________________________ Exercise 13.2: Show QFT4 is a unitary operator, and therefore can be implemented. 1 1 1 1 1 i −1 −i ) QFT4 = 2 ( 1 −1 1 −1 1 −i −1 i 1

1 1 1 1 i −1 −i ) QFT4† = 2 (1 1 −1 1 −1 1 −i −1 i 1

1 1 1 1 1 1 1 1 i −1 −i ) (1 i −1 QFT4 QFT4† = 2 (1 1 −1 1 −1 2 1 −1 1 1 −i −1 i 1 −i −1 1 0 0 0 (0 1 0 0) = I 0 0 1 0 0 0 0 1 1

4 1 1 −i ) = (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 QFT4 = ( 2 1 1

1 1 1 i −1 −i ) −1 1 −1 −i −1 i

1 1 1 1 1 1 i −1 −i ) (00) = QFT4|0> = 2 ( 1 −1 1 −1 0 1 −i −1 i 1

1 1 1 1 0 i −1 −i ) (1) = QFT4|1> = 2 (1 1 −1 1 −1 00 1 −i −1 i 1

1 1 1 1 0 i −1 −i ) (0) = QFT4|2> = 2 (1 1 −1 1 −1 10 1 −i −1 i 1

1

1

(1) 2 1 1

1

1

( i) 2 −1 −i

1

1

(−11) 2 −1

1 1 1 1 0 1 i −1 −i ) (00) = QFT4|3> = 2 ( 1 −1 1 −1 1 1 −i −1 i 1

1

1

( −i ) 2 −1 i

The result is identical to Exercise 13.1. ______________________________________________________________________________ Exercise 13.4. Show that QFTN is unitary. 1 1 1 QFTN = 1 √N ⋮ 1 (

1 1 2 ω ω ω4 ω2 ⋮ ⋮ ωN−1 ω2(N−1)

1 1 1 QFTN† = 1 √N ⋮ 1 (

1 1 ∗ ω ω∗ 2 2 ω∗ ω∗ 4 ⋮ ⋮ ω∗ N−1 ω∗ 2(N−1)

Ignoring the pre-factor of

1

1 ω3 ω6 ⋮

… … …

1 ωN−1 ω2(N−1) ⋮

where ω = e2πi/N

ω3(N−1) … ω(N−1)(N−1) ) 1 … ∗ N−1 … ω … ω∗ 2(N−1) where ω∗ = e−2πi/N ⋱ ⋮ ∗ 3(N−1) … ∗ (N−1)(N−1) ω ω ) 1 ω∗ 3 ω∗ 6 ⋮

, the entry in the jth row and kth column of QFTN 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 (QFTN QFTN† ) is: (QFTN QFTN† )jk 1

jm ∗ mk = N ∑N−1 m=0(ω) (ω ) 1

2πi/N jm −2πi/N mk = N ∑N−1 ) (e ) m=0(e 1

2πim(j−k)/N = N ∑N−1 m=0 e

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

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.

j2

2 QFT32 k

j r=8

k N/r = 32/8 = 4

The input column vector is: 1 √3

(|01000⟩ + |10000⟩ + |11000⟩) =

1 √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

′

cosθ2 −isin

−isinθ2

θ′ 2

cos

θ′ 2

′

̂ y (θ′) = ( ),R

′

cosθ2

−sinθ2

θ′ 2

θ′ 2

sin

cos

̂ z (θ′) = (e ),R

−iθ′ /2

0

e

0 )

iθ′ /2

 

′

̂ x > = ( R

θ

′

cosθ2

−isinθ2 ′

−isinθ2

′

cosθ2

)(

cos 2

θ

eisin 2

)=(

cos

θ′ 2

θ

θ

′

cos 2 − ieisinθ2 sin 2 θ

θ′ 2

−isin cos 2

θ +eicos

′

2

θ

)

sin 2

Suppose  = /2: ̂ x > = ( R

cos

θ′ 2

θ

θ′

θ

−isin 2 cos 2 +eiπ/2 cos

(θ−θ′ )

′

θ

cos 2 − ieiπ/2 sinθ2 sin 2 θ′ 2

cos

θ′

θ

′

θ

cos 2 + sinθ2 sin 2

cos

(θ−θ′ )

2 2 )=( )=( )= θ θ′ θ (θ−θ′ ) θ′ −i (sin 2 cos 2 − cos 2 sin 2) sin 2 i (sin ) θ

2

cos 2 ( ) (θ−θ′ ) eiπ/2 (sin 2 ) Thus, θ becomes θ – θ′ . This is exactly what you would expect for counter-clockwise rotation of a vector about the x-axis.

′

̂ y > = ( R

θ′

θ

′

θ

′

θ

cos 2 cos 2 − eisinθ2 sin 2 )=( ) ′ )( θ θ θ′ θ θ′ i cosθ2 eisin 2 sin 2 cos 2 +e cos 2 sin 2

cosθ2

cos 2

−sinθ2

′

sinθ2

Suppose  =  cos

θ′

θ

(θ+θ′ )

θ

′

cos 2 − sinθ2 sin 2

cos 2 2 ̂ y > = ( ) ( ) R = ′ ′ θ θ θ (θ+θ′ ) sinθ2 cos 2 +cos 2 sin 2 sin 2 Thus, θ becomes θ + θ′ . This is exactly what you would expect for counter-clockwise rotation of a vector about the y-axis. ̂ z > = (e R

−iθ′ /2

θ

θ

′

e−iθ /2 cos 2

cos 2

0 )( )=( ′ )=e θ θ e eisin 2 eiθ /2 eisin 2 iθ′ /2

0

θ

−iθ′ /2

cos 2

( ) θ ei(+θ') sin 2

′

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. cosθ2 −isinθ2 cosθ2 ̂ x (θ) = ( ̂ R ) , R (θ) = ( y −isinθ2 cosθ2 sinθ2 cosθ2 † ̂ R x (θ) = ( isinθ2 ̂ x (θ)R ̂ †x (θ) = ( R

isinθ2

̂† θ ) , R y (θ) = (

cos2 cosθ2 −isin

̂ y (θ)R ̂ †y (θ) = ( R

cosθ2 sin

θ 2

−isinθ2 θ 2

cosθ2 −sin

cosθ2

)( cos isinθ2 θ 2

−sinθ2

cosθ2

)( cos −sinθ2 θ 2

−sinθ2

−iθ/2 ̂ z (θ) = (e ),R cos 0 θ 2

sinθ2

iθ/2 ̂ †z (θ) = (e ),R cos 0

θ 2

θ 2

isinθ2 cos sinθ2 cos

θ 2

θ 2

e

0 ) eiθ/2

0 )

−iθ/2

1 0 ̂ x (θ) is unitary )R 0 1

)=(

1 0 ̂ y (θ) is unitary )R 0 1

)=(

−iθ/2 0 ) (eiθ/2 0 ) = (1 0 )  R ̂ z (θ)R ̂ †z (θ) = (e ̂ z (θ) is unitary R iθ/2 −iθ/2 0 1 0 e 0 e ______________________________________________________________________________

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.

Z

Ry(/2)

̂ y (θ) = ( R

cos2

θ

−sin2

θ

cos2

sin2

cosπ ̂ y (π/2 ) = ( π4 R sin 4

0 1 0 0

Ry(−/2)

θ θ

)

−sinπ4 1 1 )= ( √2 1 cosπ4

cos(−π4) ̂ y (−π/2 ) = ( R sin(−π4)

−sin(−π4) π 4

cos(− )

−1 ) 1 )=

1 √2

(

1 −1

1 ) 1

I  R y (π/2 ) 1 =( 0

1 1 0 ) ( √2 1 1

1 1( 1 −1 )= ( 1 1 √2 0 (1 1

−1 ) 1 −1 ) 1

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

I  R y (−π/2 ) 1 =( 0

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

1 0 CPHASE = ( 0 1 0 0 0 0

1 ) 1 1 ) 1

0 0 0 0) 1 0 0 −1

I  R y (−π/2 ) (CPHASE) (I  R y (π/2 ))

1 −1 1 1( −1 0(

1 1 ) 1 ) = 1 (−1 √2 1 0 ) 1 0

0 1 1 0 0 1 0 −1

0 0) 1 1

0 0 0 0 ); 0 −1 −1 0

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

0 1 0 0 −1 1 0 0

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

0 0 0 0) 1 −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

̂

U†(t) = e ħ Ht i

̂

U(−t) = e ħ Ht i

̂

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

π

Exercise 14.5. Show that ei 2 σ̂z ei 4 σ̂y = global phase factor of i. e

π 2

̂z i σ

π

i ̂ z (−π) = (e 2 =R 0

0 ) = (i π 0 e−i 2

i √2

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

0 1 0 )=i( ) −i 0 −1

e

π 4

̂y i σ

π

cos(−π4) −sin(−π4) 1 1 1 ̂ = R y (−π/2) = ( )= ( ) π π √2 −1 1 sin(− 4 ) cos(− 4 ) π

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

ei 2 σ̂z ei 4 σ̂y = i (

̂ in the computational basis is H ̂= Exercise 14.6. Show that the dual vector representation for H μB Bo ( 0> = (10)  (1 0) = (

0 0 ) 0 1

|1> = (01)  (0 1) = ( 1 0

0> state, describe how you would achieve the (|0⟩ + |1⟩) state. What √2

Larmor frequency and ESR pulse duration would be required? E = 2μB Bo = 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

=

μB B1 ħ

=

(9.27 x 10−24 J/T)(10−3 T) 6.626 x 10−34 Js/2π

= 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 B1 cos (ωo t)ŷ for a duration of 18 ns, causing

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 ̂ ′ = (W11 H W21

W12 ) W22

̂ ′|0⟩ = (1 0) (W11 ⟨0|H W21

W12 1 11 ) ( ) = (1 0) (W ) = W11 W21 W22 0

̂ ′|1⟩ = (1 0) (W11 ⟨0|H W21

W12 0 12 ) ( ) = (1 0) (W ) = W12 W22 W22 1

̂ ′|0⟩ = (0 1) (W11 ⟨1|H W21

W12 1 11 ) ( ) = (0 1) (W ) = W21 W21 W22 0

̂ ′|1⟩ = (0 1) (W11 ⟨1|H W21

W12 0 12 ) ( ) = (0 1) (W ) = W22 W22 W22 1

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

μB B1 ħ

Eo = μB B0 E1 = −μB B0 π

t = 2 t 2 E0 ħ E1 ħ

π

=4 t=

μB B0 π ħ

2

=

μB B0 π ħ

ħ

2 μB B1

πB

= 2B 0 1

πB

t = − 2B 0 1

⟩ = (αβ) = (

t 2

πB0

E −i 0 t ħ

cos( ) e

E t −i 1 t −i sin( ) e ħ 2

−i π cos( ) e 2B1

)=(

4

π 4

i

−i sin( ) e

πB0 2B1

)=

1 √2

e

πB 1

−i 2B 0

|0> −

1 √2

ie

πB 1

+i 2B 0

|1>

______________________________________________________________________________

Exercise 16.3. Derive Eq. (16.32). Solving the two coupled differential equations,

∂α(t) ∂t

=

μB B1 2iħ

β(t) and

∂β(t) ∂t

=

μB B1 2iħ

α(t), with initial

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)

ω t −i 0 t 2 sin ( ) 2 ω0 t +i t 2 e cos( ) 2

−ie

=(

)

______________________________________________________________________________ Exercise 16.4. Derive Eq. (16.51) to (16.54). H + H′ = ( E + W11 ( 0 W21

E0 + W11 W21

W12 ) E1 + W22

W12 ) (α) = E (αβ) E1 + W22 β

E + W11 − E W12 ( 0 ) (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: E + W11 − E det ( 0 W21

W12 )=0 E1 + W22 − E

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

(E0 + W11 − E)(E1 + W22 − E) − |W21 |2 = 0 E 2 − (E0 + E1 + W11 + W22 )E + E0 E1 + E0 W22 + W11 E1 + W11 W22 − |W21 |2 = 0 Using the quadratic equation: E=

E0 +E1 +W11 +W22 ±√(E0 +E1 +W11 +W22 )2 −4(E0 E1 +E0 W22 +W11 E1 +W11 W22 )−|W21 |2 ) 2 1

1

E± = 2 (E0 + W11 + E1 + W22 ) ± 2 √(E0 + W11 − E1 − W22 )2 + |2W21 |2

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

E + W11 − E+ ( 0 W21

This leads to the following equation: W21 α+ + (E1 + W22 − E+ ) β+ = 0 Rearranging: β+ =

W21 α (E+ −E1 −W22 ) +

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

1 |W21 |2 1+ (E+ −E1 −W22 )2

|W21 |2 |α |2 (E+ − E1 − W22 )2 +

= |W

(E+ −E1 −W22 )2

21 |

2 +(E −E −W )2 + 1 22

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

and β+ =

W21 α (E+ −E1 −W22 ) +

=

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

=

W21 √|W21 |2 +(E+ −E1 −W22 )2

Introduce W21 = |W21|eiφ β+ =

|W21 |eiφ 2 √|W21 | +(E+ −E1 −W22 )2

(αβ+) = +

1 √|W21 |2 +(E+ −E1 −W22 )2

(

E+ −E1 −W22 ) |W21 |eiφ

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

1 √|W21 |2 +(E− −E1 −W22 )2

Define E0 + W11 + E1 + W22 2

Ea = and ∆=

E0 + W11 − E1 − W22 2

Thus,

(

E− −E1 −W22 ) |W21 |eiφ

E± = Ea ± √∆2 + |W21 |2 and α+ 1 E+ − E1 − W22 ( )= ( ) |W21 | β+ √|W21 |2 + (E+ − E1 − W22 )2 2

1

=

√|W21 |2 +(∆+√∆2 +|W12 |2 )

2

|2

(∆+√∆|W+|W| 12 ) 21

Similarly, α− 1 E− − E1 − W22 ( )= ( ) |W21 |eiφ β− √|W21 |2 + (E− − E1 − W22 )2 e−iφ (−∆+√∆2 +|W21 |2 )

1

=

√|W21 |2 +(−∆+√∆2 +|W21 |2 )

( 2

|W21 |eiφ

)

To simplify the above, let us define tanθ =

|W21 | ∆

Then start with the trigonometric identity tanθ =

θ 2 θ 1−tan2 2

2tan

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

Solve this quadratic equation for tan 2: θ −1 ± √1 + tan2 θ tan = 2 tanθ θ

On the Bloch sphere, 0    , so tan 2 is positive. This means we take the positive square root θ −1 + √1 + tan2 θ tan = 2 tanθ Substituting tanθ =

|W21 | ∆

gives:

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

θ

Now find sin 2: θ sin = 2

θ tan 2 √1 +

θ tan2

1 (−∆ + √∆2 + |W21 |2 ) |W21 |

=

√1 +

2

2 1 (−∆ + √∆2 + |W21 |2 ) 2 |W21 |

−∆ + √∆2 + |W21 |2

=

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

1 √1 + tan2 θ 2

2

1

= √1 +

2 1 2 + |W |2 ) (−∆ + √∆ 21 |W21 |2

|W21 |

=

√|W21 |2 + (−∆ + √∆2 + |W21 |2 )

2

Thus, α− ( )= β−

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

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

α+ ( )= β+

|W21 | 2

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

2

√|W21 |2 + (∆ + √∆2 + |W21 |2 )

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

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

2

θ 2

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

= 4 sin2 θ (2 − e−i(E+ −E−)t/ħ + e+i(E+ −E−)t/ħ ) 1

E+ −E−

= 4 sin2 θ [2 − 2cos ( 1

= 2 sin2 θ [1 − cos ( = sin2 θ sin2 (

ħ

E+ −E− ħ

𝑡)]

𝑡)]

E+ − E− t) 2ħ

By definition tanθ =

|W21 | ∆

sin2 θ =

|W21 |2 tan2 θ = 1 + tan2 θ ∆2 + |W21 |2

P1 (t) =

|W21 |2 E+ − E− sin2 ( t) 2 2 ∆ + |W21 | 2ħ 1

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

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

ħ ω 2 d )=0 ħ ∆ω − E 2

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

ħ E± = ± √∆ω2 + ωd 2 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 − ∆ω − E+ ( 2 ħ ω 2 d

ħ ω 2 d ħ ∆ω − E+ 2

α+ )( ) = 0 β+

This leads to the following equation: ħ ħ ωd α+ + ( ∆ω − E+ ) β+ = 0 2 2 Rearranging: β+ =

ħ ω 2 d ħ (E+ − ∆ω) 2

α+

Also, |α+ |2 + |β+ |2 = 1 (normalization condition) 2 ħ ( ωd ) 2 |α+ |2 = 1 − |β+ |2 = 1 − |α+ |2 2 ħ (E+ − 2 ∆ω)

|α+

|2

α+ =

=

1 2 ħ ( ωd ) 1+ 2 2 ħ (E+ −− ∆ω) 2

ħ 2

(E+ − ∆ω)

=

2

ħ

2 2

ħ

( ωd ) +(E+ − ∆ω) 2 2

ħ 2

(E+ − ∆ω) 2

√(ħωd ) +(E+ −ħ∆ω) 2

2

2

and β+ =

(αβ+) +

ħ ω 2 d ħ (E+ − ∆ω) 2

ħ ω 2 d ħ (E+ − ∆ω) 2

α+ =

2

2

√(ħωd ) +(E+ −ħ∆ω) 2

2

2

ħ 2

(

ħ ω 2 d

2

2

√(ħωd ) +(E`+ −ħ∆ω)

E+ − ∆ω

1

=

ħ 2

(E+ − ∆ω)

=

2

ħ ω 2 d 2

√(ħωd ) +(E+ −ħ∆ω) 2

2

)

The eigenvector corresponding to E− is solved in an identical manner: (αβ−) −

=

ħ 2

E− − ∆ω

1 2

√(ħωd ) +(E− −ħ∆ω) 2

2

(

ħ ω 2 d

2

)

ħ

Substitute E± = ± 2 √∆ω2 + ωd 2 α+ ( )= β+

ħ E+ − 2 ∆ω ( ħ ) 2 2 ωd

1 2 √(ħ ωd ) + (E+ − ħ ∆ω) 2 2 1

=

((√∆ω 2

√ωd 2 +(√∆ω2 +ωd 2 −∆ω)

2 +ω 2 −∆ω) d

ωd

)

Similarly, (αβ−) = −

2

√(ħωd ) +(E− −ħ∆ω) 2

( 2

2

1

=

ħ 2

E− − ∆ω

1

√ωd 2 +(√∆ω2 +ωd 2 ∆ω)

ħ ω 2 d

)

(−(√∆ω2 +ωd 2 +∆ω)) ) ωd

( 2

To simplify the above, let us define tanθ = −

ωd ∆ω

From Exercise 16.4:

2

θ −1 + √1 + tan2 θ tan = 2 tanθ θ ∆ω ∆ω 2 1 tan = + √( ) + 1 = (∆ω + √∆ω2 + ωd 2 ) 2 ωd ωd ωd θ

Now find sin 2: θ sin = 2 θ cos = 2

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

1 √1 + tan2 θ 2

1

= √1 +

2 1 (∆ω + √∆ω2 + ωd 2 ) 2 ωd

=

ωd √ωd 2 + (∆ω + √∆ω2 + ωd 2 )

2

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

ωd √ωd 2 + (−∆ω + √∆ω2 + ωd 2 )

2

−∆ω + √∆ω2 + ωd 2 √ωd 2 + (−∆ω + √∆ω2 + ωd 2 )

2

θ cos 2 α+ ( )=( θ) β+ sin 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 |0 ⟩ = cos |+ ⟩ − sin |− ⟩ 2 2 The wavefunction at some later time t is: θ

θ

2

2

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

=e

−i

2

t

√∆ω2 +ωd 2

θ

cos 2 |+ ⟩ − e

+i

2

t

θ

sin 2 |− ⟩

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

θ

2

2

⟨1|(t)⟩ = e−iE+t/ħ cos ⟨1 |+ ⟩ − e−iE−t/ħ sin ⟨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)⟩| θ

2

θ 2

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

= 4 sin2 θ (2 − e−i(E+ −E−)t/ħ + e+i(E+ −E−)t/ħ ) 1

E+ −E−

= 4 sin2 θ [2 − 2cos ( 1

= 2 sin2 θ [1 − cos ( = sin2 θ sin2 (

ħ

E+ −E− ħ

E+ − E− t) 2ħ ω

By definition tanθ = − ∆ωd

𝑡)]

𝑡)]

tan2 θ ωd 2 sin θ = = 1 + tan2 θ ∆ω2 + ωd 2 2

ωd 2 E+ − E− 2 ( P1 (t) = sin t) ∆ω2 + ωd 2 2ħ ħ

Substitute E± = ± 2 √∆ω2 + ωd 2 P1 (t) =

ωd 2 √∆ω2 + ωd 2 2 ( sin t) ∆ω2 + ωd 2 2

______________________________________________________________________________

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 0 0 ħ 0 ω − ω − 2J/ħ 0 0 1 2 ̂= ( ) H −ω1 + ω2 − 2J/ħ 0 2 0 0 0 −ω1 − ω2 + 2J/ħ 0 0 ̂. The energies are equal to the diagonal entries of H E00 =

ħ (ω + ω2 + 2J/ħ) 2 1

E01 =

ħ (ω − ω2 − 2J/ħ) 2 1

E10 =

ħ (−ω1 + ω2 − 2J/ħ) 2

E11 =

ħ (−ω1 − ω2 + 2J/ħ) 2

E00−10 = E00 − E10 =

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

E01−11 = E01 − E11 =

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

2J

ħ

E10−11 = E10 − E11 = 2 (−ω1 + ω2 − ħ ) − 2 (−ω1 − ω2 + 2J/ħ) = ħ(ω2 − 2J/ħ) ______________________________________________________________________________

Exercise 17.2. Show the following: −i ̂ (t = π ħ) = √i ( 0 U 4J 0 0 1 0 1( ) 1 0 1 0 0 −1 ̂z1  σ σ ̂z2 = ( )( )=( 1 0 0 −1 0 −1 0( ) 0 −1 î

0 1 0 0

0 0 1 0

0 0) 0 −i

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

π

̂ (t) = e− ħHt = e− i 4 σ̂z1  σ̂z2 = cosπ Î − isinπ(σ From Eq. (17.13), U ̂z1  σ ̂z2 ) 4 4 1 = (0 √2 0 0 1

0 1 0 0

1−i = ( 0 √2 0 0 1

0 0 1 0

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

0 0 1+i 0 0 1+i 0 0 1−i

0 0 1 0 0 1 0 0

1+i

=

1 √2

(1 + i)

0 0 (0

The matrix element 1−i 1+i

1−i

1+i

2

1−i 1+i 1

0 0 ) 0 1−i 0 0 0 1−i 1+i)

can be written as:

i

= 1+i ∗ 1+i = 2i = i ∗ i = −i

The pre-factor can be written as: 1 √2

(1 + i) =

1 √2

−i  U = √i ( 0 0 0

1/2

(1 + i) = cos(/4) + isin(/4) = ei/4 = (eiπ/2 )

0 1 0 0

0 0 1 0

= √i

0 0) 0 −i

______________________________________________________________________________

Chapter 18 No Exercises

Chapter 19 ______________________________________________________________________________ Exercise 19.1. What is N/N for 1H at 10 T and T = 300 K? E = ħo = ħ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 En = p2/2m = ħ2k2/2m = ħ2(kx2+ky2+kz2)/2m = 3ħ2(n/L)2/2m =

3h2 2 n 8mL2

3h2

n = 1: E1 = 8mL2 3h2

n = 2: E2 = 4 8mL2 3h2

3h2

9h2

E = E2 – E1 = 4 8mL2 − 8mL2 = 8mL2 9h

 = E/h = 8mL2 9h

9(6.626 x 10−34 Js)

Rearranging, L = √8m = √8(9.11x10−31 kg)(10x109 Hz) = 286 nm ______________________________________________________________________________

Chapter 21 ______________________________________________________________________________ Exercise 21.1. Estimate a typical ion speed at 10 mK and calculate the corresponding Doppler frequency shift. 1

mv2 ~ kT

2

2kT

speed, v ~ √

m

Choose Ca ion, m = 6.64 x 10−26 kg 2kT

v~√

m

2(1.38 x 10−23 J/K)(10−2 K)

=√

6.64 x 10−26 kg

= 2.04 m/s

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 1+v/c

Note that v/c Io. The potential is U() = −EJ (cos + IIo ) dU d dU d

= 0 at a potential minima = EJ (sin − IIo )

If I > Io, then I/Io > 1 dU

 d < 0 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: M uMl = MlMu

−1 0 ) 0 1

Mu = ( Ml = (

1 0

0 ) −1 −1 0 1 )( 0 1 0

Mu Ml = (

0 −1 0 )=( ) −1 0 −1

1 0 −1 0 −1 0 )( )=( ) 0 −1 0 1 0 −1

Ml Mu = (

 Mu Ml = Ml Mu ______________________________________________________________________________

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

> 0>

’>

0> 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, Fy = 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 B

Combining the above equations gives the Hall coefficient, RH = Vy/Ix = entz For a two-dimensional electron gas (2DEG), nt becomes the charge per unit area. ______________________________________________________________________________

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