Feedback — Homework 2

Feedback — Homework 2 You submitted this homework on Thu 4 Apr 2013 9:35 PM EDT -0400. You got a score of 10.00 out of 10.00.

Question 1 Suppose we have a state

|ψ

otherwise unspecified, and

3 that is a linear combination of two other states |ψ3 =

3 and |23 are orthonormal (21|23 = 22|13 = 0, 21|13 = 22|23 = 1). What is 21|ψ3?

|1

Your Answer

Score ✔

a

b

3 + b|23, where a and b are non-zero, but

a|1

Explanation

1.00

∗

2

|a|

b

Total

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Question Explanation

21|ψ3 = 21|(a|13 + b|23)

2

3 + b21|23 =

= a 1|1

a

Question 2 The probability for particle to be at between points

⃗ r1

→ r2

2

at t2 if it was at r 1⃗ at t1 is given by ∣G(r 2⃗ , t2 ; r 1⃗ , t1 )∣ . Let us consider a propagator

and r 2⃗ at time t1 and t2 which has the form

⃗ , t2 ; r 1 ⃗ , t1 ) = a + b G(r 2

, where a and b are complex and

non-zero. This can, in some sense, be considered the case where there are two (and only two) possible ways for particle to travel from

⃗ , t1 ) (r 1

to (r 2⃗ , t2 ). What is the probability for the particle to go from

Recall that the complex conjugate of a complex number

z = x + iy

is z ∗

Your Answer 2

|a|

2

+ |b|

⃗ , t1 ) (r 1

to (r 2⃗ , t2 )?

= x − iy ,

where

x

Score ✔

∗

∗

+ a b + b a

and

y

are real numbers. Explanation

1.00

a + b

a

2

+ b

2

|a|

2

+ 2ab

2

+ |b|

∗

+ 2a b

∗

Total

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Question Explanation The probability is given by 2

⃗ , t2 ; r 1 ⃗ , t1 )∣ ∣G(r 2

2

= |a + b|

∗

= (a + b ) (a + b) = (a

∗

∗

∗

+ b )(a + b) = a a + b

∗

b + ab

∗

+ ba

∗

2

= |a |

2

+ |b |

+ a

∗

∗

b + b a

Question 3 Consider the time evolution operator what is

^ U (τ )|E

^ U (τ ) = exp(−

3, where τ is some time τ

i ℏ

^ H τ ).

−

i ℏ

= E|E

3, at t = 0,

> 0.

Your Answer E|E

^ Given an eigenstate of the Hamiltonian, H |E3

Score

Explanation

3 Eτ |E

exp(

i ℏ

Eτ )|E

exp(−

exp(−

3

i

Eτ )

ℏ

i ℏ

3

Eτ )|E

✔

3

Total

1.00 1.00 / 1.00

Question Explanation For

^ τ > 0 , U (τ ) = exp(−

Thus, we have: exp(−

i ℏ

i ℏ

^ H τ ).

Eτ )|E

Since

|E

^ ^ 3 is an eigenstate of H , every time a H hits the state we will get back an E .

3

Question 4 Recall that in lecture 4, parts II and III, we considered the effects of localization in weakly disordered systems. In this class of systems, why are closed loops the dominant contribution to localization? Your Answer

Score

Explanation

Closed loops have less energy than other paths. Closed loops can be traversed either way, giving two different paths with the same path length which add constructively.

✔

1.00

Loops are not the dominant contribution to localization in these systems. Closed loops have a smaller actions than normal paths. Total

1.00 / 1.00

Question 5 For the same class of weakly disordered systems as problem 4, in which of the following dimensions might we expect to find localization effects causing an increase in resistance at low temperatures? (Check all that apply) Your Answer 1

Score ✔

0.25

Explanation

7

✔

0.25

3

✔

0.25

4

✔

0.25

Total

1.00 / 1.00

Question 6 Which of the following satisfy the diffusion equation in one dimension (for

t > 0)

(including, if applicable, unbounded solutions which

however satisfy the equation at any finite x and t)? (Check all that apply) Your Answer 1 √2πDt

−

e

Score 2 x

4 Dt

exp(iD(px − Et))

A ⋅ (t +

−

e

1 2D

where

p

and

E

are non-zero real numbers.

2

x )

2 x 4 Dt

Total

✔

0.25

✔

0.25

✔

0.25

✔

0.25

Explanation

1.00 / 1.00

Question Explanation The diffusion equation is

∂ϕ ∂t

= D

∂

2

ϕ

∂ x2

Question 7 Which of the following is most suitable for a saddle point approximation (Laplace's method)? Your Answer

Score

✔

1.00

Explanation

Total

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Question 8 Why are interference effects for particles less important as we go to the classical regime? Your Answer

Score

Explanation

Interference is significant in the classical limit. As ℏ

→ 0

the phases of paths, given by eiS/ℏ , oscillate rapidly, so only paths that interfere

✔

1.00

constructively (classical paths) contribute. Paths are restricted to be within a certain distance of each other at all times, a distance proportional to ℏ . As ℏ → 0 , the paths cannot get far enough apart to interfere. The number of interfering paths is proportional to the classical paths

ℏ

and thus goes to 0 as ℏ

→ 0,

leaving only

Total

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Question 9 Suppose we have the Hamiltonian

H =

p

2

2m

+

1 2

2

2

mω x

. What is the corresponding action?

Your Answer S = ∫ dt[

S = ∫ dt[

Score p

− ωx]

m

1 2

2

mx ˙

−

1 2

2

✔

2

mω x ]

1.00

S = ∫ dt[mx ¨ − mωx]x

S = ∫ dt[ψ

∗

(iℏ

∂ ∂t

Total

Question Explanation

−

p

2

2m

−

1 2

2

2

m ω x )ψ]

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Explanation

The action of a system is the time integral of the Lagrangian. The Lagrangian may be obtained from the hamiltonian as L = px ˙ − H . In our case L = px ˙ − H = px ˙ − (

S = ∫ dtL = ∫ dt [

1 2

p

2

2

2m 2

mx ˙

+

−

1 2 1 2

2

˙) (mx

2

mω x ) = (mx ˙)x ˙ − 2

−

2m

1 2

2

2

mω x

=

1 2

2

mx ˙

−

1 2

2

2

mω x

. The action is then:

2

mω x ]

Question 10 Say we have the action

S = ∫ dt[

1 2

2

mx ˙

+ qEx].

What equation of motion does the principle of least action give?

Your Answer

Score

Explanation

x ¨ = −mqEx

x ¨ =

✔

qE

1.00

m

x ¨ = −mE

x ¨ = −E

Total

1.00 / 1.00

Question Explanation This is of the same form as in the lectures S

= ∫ dt [

1 2

principle of least action gives us the equation of motion If we compute

dV dx

2

mx ˙

− V (x)] ,

mx ¨ = −

dV dx

we obtain:

−qE .

Dividing both sides of the equation of motion by m we obtain: x ¨

=

qE m

with

V (x) =

1 2

2

V0 sin (kx).

(Netwon's Second Law).

We know then that the