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Introduction to Quantum Computing Solution Manual
Introduction to Quantum Computing Solution Manual
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, The Solutions Manual for Introduction to Quantum Computingby 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+)
E1E2* = (E01ei(kx−t))(E02ei(kx−t+)*
= E01E02e−i , assuming E01 is parallel to E02 (same polarization) so E01E02 = E01E02
E2E1* = (E02ei(kx−t+)(E01ei(kx−t))*
= E01E02e+i, assuming E01 is parallel to E02 (same polarization)
E1E2* + E2E1* = E01E02 (e+i + e−i)
= 2E01E02cos
2√I1I2cos
| E1 + E2 |2 = (E1 + E2)( E1 + E2)*
I12 = E1E1* + E2E2* + E1E2* + E2E1*
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
= c11 + c22 + … cnn
+∞
∫ ∗ 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
−∞
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