SOLUTIONS MANUAL for Introduction to Quantum Computing by Ray LaPierre | Chapters 1-26

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


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

, =1


Similarly,
+∞
∫ 2∗ 2 dx = 1
−∞


+∞
∫ ∗  dx = δ where i, j = {0, 1}
−∞ i j ij

 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
+∞
∫ ∗ A
̂  dX
<A> = −∞
+∞
∫−∞ ∗ dX

The numerator is:
+∞
∫ ∗ A
̂  dx
−∞
+∞
=∫ (c ∗ ∗ + c ∗ ∗ + c ∗ ∗ ) A
̂ (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∗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 ∫ ∗  dx, are zero since  and  are
2 1 2 −∞ 1 2 1 2

orthonormal.


Similarly, the denominator is:
+∞
∫ ∗  dx
−∞