Exam (elaborations)
Introduction to Quantum Computing Solution Manual
Introduction to Quantum Computing Solution Manual
[Show more]
Connected book
Book Title:
Author(s):
- Edition:
- ISBN:
- Edition:
Content preview
, 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
−∞
The benefits of buying summaries with Stuvia:
Guaranteed quality through customer reviews
Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.
Quick and easy check-out
You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.
Focus on what matters
Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!
Frequently asked questions
What do I get when I buy this document?
You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.
Satisfaction guarantee: how does it work?
Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.
Who am I buying these notes from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller Succeed. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for $21.49. You're not tied to anything after your purchase.
Can Stuvia be trusted?
4.6 stars on Google & Trustpilot (+1000 reviews)
53022 documents were sold in the last 30 days
Founded in 2010, the go-to place to buy study notes for 14 years now