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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 =

∫ 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

−∞

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

Similarly, +∞

2∗ 2 dx = 1

∫ −∞

∫

+∞

∗  dx = δ where i, j = {0, 1} i

−∞

 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

+ c  +. . . c  ) dx 2

−∞

=∫

+∞

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

1 1

2 2

n n

−∞

= c∗c1a1 + c∗c2a2+. . . c∗ c a 1

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

=∫

+∞

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

∗  dx+. . . c∗ c ∫

+∞

−∞

= c∗c1 ∫

+∞

∗ dx + c∗c ∫

+∞

2 2

−∞

∗  dx

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

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

−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

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

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

− eicos

)

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

= ( 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

2 )=

2 ( i

e sin2

cosθ

e sinθ

e−isinθ ) ( cosθ i

−cosθ

e sin2

cosθcos θ+ sinθsin θ

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

2 ) = 2 ( i

2θ

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

Ŝ (

sin i2

−i

− e cos2 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)

sin2