Introduction to Quantum Computing (Instructor Solution Manual, Solutions)

87 Pages·2021·0.603 MB·English

by  Ray LaPierre

Solutions Manual
Introduction to Quantum Computing
R.R. LaPierre

Chapter 1
______________________________________________________________________________

Exercise 1.1. Derive Eq. (1.11): I  | E + E |2 = I + I + 2√I I cos 
12 1 2 1 2 1 2

E = E ei(kx−t)
1 01
E = E ei(kx−t+)
2 02
E E * = (E ei(kx−t))(E ei(kx−t+)*
1 2 01 02
= E E e−i , assuming E is parallel to E (same polarization) so E E = E E
01 02 01 02 01 02 01 02
E E * = (E ei(kx−t+)(E ei(kx−t))*
2 1 02 01
= E E e+i, assuming E is parallel to E (same polarization)
01 02 01 02
E E * + E E * = E E (e+i + e−i)
1 2 2 1 01 02
= 2E E cos
01 02
 2√I I cos
1 2
| E + E |2 = (E + E )( E + E )*
1 2 1 2 1 2
= E E * + E E * + E E * + E E *
 I 1 1 2 2 1 2 2 1
12

 I1  I2  2√I1I2cos

 I = I +I +2√I I cos
12 1 2 1 2
______________________________________________________________________________
Chapter 2
______________________________________________________________________________
Exercise 2.1. Show that the first two states (two lowest energy levels) of the infinite quantum
well are orthonormal.
2 π
 (x) = √ sin( x)
1
L L
2 2π
 (x) = √ sin( x)
2
L L
+∞
∫ ∗ dx
1 2
−∞
2 L π 2π
= ∫ sin( x)sin( x) dx
L L L
0
4 L π π π
= ∫ sin( x)sin( x)cos( x) dx, using sin(2x)=2sinxcosx
L 0 L L L
4 L π π
= ∫ sin2( x)cos( x) dx
L L L
0
4 π L
= sin3( x)|
3π L 0
4
= [sin3(π)−sin3(0)]
3π
= 0

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

Similarly,
+∞
∫ ∗  dx = 1
2 2
−∞

 ∫+∞∗ dx = δ where i, j = {0, 1}
−∞ i j ij
  (x) and  (x) are orthonormal.
1 2
______________________________________________________________________________
Exercise 2.2. Prove Eq. (2.49): <A> = |c |2 a + |c |2 a + … + |c |2 a
1 1 2 2 n n

 = c  + c  + … c 
1 1 2 2 n n
∫+∞∗Â dx
<A> = −∞
∫+∞∗ dx
−∞
The numerator is:
+∞
∫ ∗Â dx
−∞
+∞
= ∫ (c∗∗ +c∗∗ +c∗∗ )Â(c  +c  +...c  ) dx
1 1 2 2 n n 1 1 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∗c a +c∗c a +...c∗c a
1 1 1 2 2 2 n n n
= |c |2a +|c |2a +...|c |2a
1 1 2 2 n n

All other terms in the numerator, like a c∗c ∫+∞∗  dx, are zero since  and  are
2 1 2 −∞ 1 2 1 2
orthonormal.

Similarly, the denominator is:
+∞
∫ ∗ dx
−∞
+∞
= ∫ (c∗∗ +c∗∗ +c∗∗ )(c  +c  +...c  ) dx
1 1 2 2 n n 1 1 2 2 n n
−∞
+∞ +∞ +∞
= c∗c ∫ ∗ dx+c∗c ∫ ∗ dx+...c∗c ∫ ∗ dx
1 1 1 1 2 2 2 2 n n n n
−∞ −∞ −∞
= |c |2 +|c |2+...|c |2 , assuming each  is normalized
1 2 n i
= 1 , since all probabilities must sum to 1
All other terms in the denominator, like c∗c ∫+∞∗  dx, are zero.
1 2 −∞ 1 2
∫+∞∗Â dx
 <A> = −∞ = |c |2a +|c |2a +...|c |2a
∫+∞∗ dx 1 1 2 2 n n
−∞
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).
Ŝ x ⟩ = +ħx ⟩
x + +
2
ħ(0 1) 1 (1) = ħ 1 (1) = +ħx ⟩
2 1 0 √2 1 2√2 1 2 +

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

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

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

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

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

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

I + σ̂ + σ̂ + σ̂
x y z
1 0 0 1 0 −i 1 0
= ( ) + ( ) + ( ) + ( )
0 1 1 0 i 0 0 −1
α+γ β−δi
= ( )
β+δi α−γ
Equate this to a general 2x2 complex matrix:
α+γ β−δi a c
( ) = ( ) where a, b, c, d are complex numbers
β+δi α−γ b 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 1 1 0
X2 = ( )( ) = ( ) = I
1 0 1 0 0 1

0 −i 0 −i 1 0
Y2 = ( )( ) = ( ) = I
i 0 i 0 0 1
1 0 1 0 1 0
Z2 = ( )( ) = ( ) = I
0 −1 0 −1 0 1

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

0 1 0 −i i 0 1 0
− i XY = − i ( )( ) = − i ( ) = ( ) = Z
1 0 i 0 0 −i 0 −1

1 0 0 −i 0 −i 0 1
ZY = ( )( ) = ( ) = − i ( ) = − i X
0 −1 i 0 −i 0 1 0

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

0 −i 0 1 −i 0 1 0
YX = ( )( ) = ( ) = − i ( ) = − i Z
i 0 1 0 0 i 0 −1

______________________________________________________________________________

Exercise 3.5. Prove Eq. (3.50) from Eq. (3.49). Show that the Ŝ operator has eigenvalues ± ħ and
n
2
θ θ
cos sin
corresponding eigenvectors ( 2 ) and ( 2 )
eisinθ − eicosθ
2 2

n = cossinθ
x
n = sinsinθ
y
n = cosθ
z

Ŝ = n Ŝ + n Ŝ + n Ŝ
n x x y y z z

ħ 0 1 ħ 0 −i ħ 1 0
= cossinθ ( )+ sinsinθ( )+ cosθ( )
2 1 0 2 i 0 2 0 −1

cosθ cossinθ−isinsinθ
ħ
= ( )
2 cossinθ+isinsinθ −cosθ

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

Ŝ ( cosθ2 ) = ħ( cosθ e−isinθ)( cosθ2 ) = ħ( cosθcosθ2 + sinθsinθ2 ) = ħ( cosθ2 )
n eisinθ 2 eisinθ −cosθ eisinθ 2 ei(sinθcosθ − cosθsinθ) 2 eisinθ
2 2 2 2 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)

Ŝ ( sinθ2 ) = ħ( cosθ e−isinθ)( sinθ2 ) = ħ( cosθsinθ2− sinθcosθ2 ) = ħ(sin(−θ2)) =
n − eicosθ 2 eisinθ −cosθ − eicosθ 2 ei(sinθsinθ + cosθcosθ) 2 eicosθ
2 2 2 2 2
θ
ħ sin
− ( 2 )
2 −eicosθ
2
Chapter 4
______________________________________________________________________________
1 1 1
Exercise 4.1. Show that the following state is normalized: ( + i)− 
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
|+⟩ = ( |0⟩ + |1⟩ )
√2
1
|−⟩ = ( |0⟩ − |1⟩ )
√2

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

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

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

______________________________________________________________________________
Exercise 4.3. Prove the following:
(a) <  >* = <  >
1 2 2 1
(b) <  > can be a complex number, but <  > is real and positive
1 2 1 1
(c) <c  > = c*<  >
1 2 1 2
(d) < c > = c<  >
1 2 1 2

α δ
Suppose | > = ( ) and | > = ( )
1 β 2 γ
Then < | = (α∗β∗) and < | = (δ∗γ∗)
1 2

(a)
∗
δ
<  >* = [(α∗β∗)( )] = [α∗δ+β∗γ]∗ = αδ∗ + βγ∗
1 2 γ
α
<  > = [(δ∗γ∗)( )] = δ∗α+γ∗β = αδ∗ + βγ∗
2 1 β
 <  >* = <  >
1 2 2 1

(b)
δ
<  > = (α∗β∗)( ) = α∗δ+β∗γ , which in general is a complex number
1 2 γ
α
<  > = (α∗β∗)( ) = α∗α+β∗β = |α|2 +|β|2 , which is real and positive
1 1 β

(c)
<c | = (|c >)† = (c| >)† = c*< |
1 1 1 1
 <c  > = c*<  >
1 2 1 2

(d)
c > = c >
2 2
 < c > = < c > = c<  >
1 2 1 2 1 2

______________________________________________________________________________