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