Feedback â Homework 4
Feedback — Homework 4 You submitted this homework on Mon 29 Apr 2013 1:13 PM EDT -0400. You got a score of 6.00 out of 6.00.
Question 1 Suppose we have a potential V (x)
4
= ax
2
+ bx
+ c,
with
a, b, c
positive, and imagine that we are examining a low energy
bound state of this system. Expressing the low energy approximation of this potential as a harmonic oscillator, what is its characteristic frequency, ω? Your Answer
Score
Explanation
− − − ω = √
4πa c
−− ω = √
2b a
−− ω = √
✔
2b
1.00
m
−− − ω = √
a
cm
Total
1.00 / 1.00
Question Explanation For small x, this can be approximated as V (x)
2
≈ bx
+ c =
1 2
2
2
mω x
+ c,
where
2
ω
=
2b m
.
Question 2 †
^ , and annihilation, a ^, operators, how can we express the momentum operator in terms of Recalling our definition of the creation, a
these? Your Answer
Score
− − − ^ = √ℏω a ^ p
†
^ = ℏωa a ^ p − − − − ^ = i√ p
mℏω 2
(^ a
− − − − ^ = i√ p
mℏω 2
^ (a
†
†
✔
− ^ a)
1.00
^) + a
Total
1.00 / 1.00
Question 3 2
2 3=2
3
Explanation
Calculate the expectation value of the kinetic, 2T^3
=
^ p
2
2 2m 3 , and potential, 2V^3 = 2 2 1
2
2
^ mω x
3, energy operators in the n-th
excited state of the harmonic oscillator, |n3. (You may notice that there is a relationship between the two, regardless of what n is. This is a manifestation of a result known as the Virial theorem.)
You should not run into any integrals while calculating this.
Your Answer
2T^3 =
ℏω
2V^3 =
ℏω
2 3=
ℏω
2V^3 =
ℏω
2T^3 =
ℏω
2 3=
ℏω
2T^3 =
ℏω
2V^3 =
ℏω
2T^3 =
ℏω
2V^3 =
ℏω
4
4
Score
n
(2n + 1)
✔ ^ T
4
4
Explanation
1.00
(2n + 1)
(2n + 1)
4
^ V
4
2
4
4
4
(2n + 1)
(2n + 1)
n
n
2T^3 =
ℏωn
2V^3 =
ℏω(n + 1/2)
Total
1.00 / 1.00
Question Explanation ^ We recall that a
− − − = √
− − − − ^ = √ x
ℏ
2mω
2ℏ
i
^ + x
√2mωℏ
^. p
Using this we can write
^ x
and
^ p
as:
†
^ + a ^ ) (a
− − − −
^ = i√ p
mω
mωℏ 2
^ (a
†
^ ). − a
So,
2T^3 = 2n| −
1
mωℏ
2m
2
†
2
[(a ^ )
2
†
+ (a ^)
†
−a ^ a ^ − a ^a ^ ]|n
3=
ℏω 4
2n|(a ^
†
†
^ + a a ^a ^ )|n
3=
ℏω 4
2n|(2a ^
†
^ + 1)|n a
3=
ℏω 4
(2n + 1)
Likewise,
2V^3 = 2n|
1 2
2
mω
ℏ 2mω
†
2
^ ) [(a
2
^) + (a
†
†
^ a ^ + a ^a ^ ]|n + a
3=
ℏω 4
^ 2n|(a
†
†
^ +a ^a ^ )|n a
3=
ℏω 4
^ 2n|(2a
†
^ + 1)|n a
3=
ℏω 4
(2n + 1)
Question 4 What is the wavefunction of the first excited state of the quantum harmonic oscillator, |13? (You do not need to solve the Schrödinger equation) Your Answer
Score 1/4
ψ (x) = ( 1
mω πℏ
)
Explanation
2
exp(−
2mωx ℏ
)
ψ (x) = 0 1
− − − −
1/4
ψ (x) = ( 1
mω πℏ
)
√
2mω ℏ
1/4
ψ (x) = ( 1
mω πℏ
)
✔
2
x exp(−
mωx 2ℏ
)
1.00
2
exp(−
mωx 2ℏ
)
Total
1.00 / 1.00
Question Explanation We know that |n
3=
+ 1
^ a
†
√n+1
|n
− − −
3, and a ^
†
= √
mω 2ℏ
^ − x
i 2mωℏ
^. p
So,
− − − − − 1/4 − − − − 2 mω ℏ d mω mωx ψ 1 (x) = ^ a ψ 0 (x) = [√ x − √ ]( ) exp(− ) 2ℏ 2mω dx πℏ 2ℏ †
1/4
mω = (
) πℏ
− − − − − − − − − 2 mω ℏ mω mωx √ [√ x + x ] exp(− ) 2ℏ 2mω ℏ 2ℏ 1/4
mω = (
) πℏ
− − − − − − − − 2 mω mω mωx x + √ x] exp(− ) 2ℏ 2ℏ 2ℏ
[√
1/4
mω = (
)
− − − − − 2 2mω mωx x exp(− ) ℏ 2ℏ
√
πℏ
Question 5 In the context of the model of a crystal we considered in the lectures, why does the ground state of a crystal break continuous translation symmetry? Your Answer
Score
The ground state of a crystal is translationally invariant. The breaking of translational invariance is a consequence of Goldstone's theorem. A crystal has a periodic structure, but it is not uniform in space. Translating by some distance
✔
1.00
which is not a multiple of the lattice spacing gives a state which is different from that of the original state. There is a term in the Hamiltonian of the lattice which breaks the translation symmetry. Total
1.00 / 1.00
Explanation
Question 6 Which of the following is most closely related to Goldstone's theorem? Your Answer
Score
The formation of a Fermi surface. The existence of optical phonons. The spontaneously broken translational symmetry of the ground state of crystal implies the
✔
1.00
existence of gapless excitations, which turn out to be phonons. The phase space trajectory of a harmonic oscillator traces out a circle. Total
1.00 / 1.00
Explanation
Question 1 Suppose we have a potential V (x)
4
= ax
2
+ bx
+ c,
with
a, b, c
positive, and imagine that we are examining a low energy
bound state of this system. Expressing the low energy approximation of this potential as a harmonic oscillator, what is its characteristic frequency, ω? Your Answer
Score
Explanation
− − − ω = √
4πa c
−− ω = √
2b a
−− ω = √
✔
2b
1.00
m
−− − ω = √
a
cm
Total
1.00 / 1.00
Question Explanation For small x, this can be approximated as V (x)
2
≈ bx
+ c =
1 2
2
2
mω x
+ c,
where
2
ω
=
2b m
.
Question 2 †
^ , and annihilation, a ^, operators, how can we express the momentum operator in terms of Recalling our definition of the creation, a
these? Your Answer
Score
− − − ^ = √ℏω a ^ p
†
^ = ℏωa a ^ p − − − − ^ = i√ p
mℏω 2
(^ a
− − − − ^ = i√ p
mℏω 2
^ (a
†
†
✔
− ^ a)
1.00
^) + a
Total
1.00 / 1.00
Question 3 2
2 3=2
3
Explanation
Calculate the expectation value of the kinetic, 2T^3
=
^ p
2
2 2m 3 , and potential, 2V^3 = 2 2 1
2
2
^ mω x
3, energy operators in the n-th
excited state of the harmonic oscillator, |n3. (You may notice that there is a relationship between the two, regardless of what n is. This is a manifestation of a result known as the Virial theorem.)
You should not run into any integrals while calculating this.
Your Answer
2T^3 =
ℏω
2V^3 =
ℏω
2 3=
ℏω
2V^3 =
ℏω
2T^3 =
ℏω
2 3=
ℏω
2T^3 =
ℏω
2V^3 =
ℏω
2T^3 =
ℏω
2V^3 =
ℏω
4
4
Score
n
(2n + 1)
✔ ^ T
4
4
Explanation
1.00
(2n + 1)
(2n + 1)
4
^ V
4
2
4
4
4
(2n + 1)
(2n + 1)
n
n
2T^3 =
ℏωn
2V^3 =
ℏω(n + 1/2)
Total
1.00 / 1.00
Question Explanation ^ We recall that a
− − − = √
− − − − ^ = √ x
ℏ
2mω
2ℏ
i
^ + x
√2mωℏ
^. p
Using this we can write
^ x
and
^ p
as:
†
^ + a ^ ) (a
− − − −
^ = i√ p
mω
mωℏ 2
^ (a
†
^ ). − a
So,
2T^3 = 2n| −
1
mωℏ
2m
2
†
2
[(a ^ )
2
†
+ (a ^)
†
−a ^ a ^ − a ^a ^ ]|n
3=
ℏω 4
2n|(a ^
†
†
^ + a a ^a ^ )|n
3=
ℏω 4
2n|(2a ^
†
^ + 1)|n a
3=
ℏω 4
(2n + 1)
Likewise,
2V^3 = 2n|
1 2
2
mω
ℏ 2mω
†
2
^ ) [(a
2
^) + (a
†
†
^ a ^ + a ^a ^ ]|n + a
3=
ℏω 4
^ 2n|(a
†
†
^ +a ^a ^ )|n a
3=
ℏω 4
^ 2n|(2a
†
^ + 1)|n a
3=
ℏω 4
(2n + 1)
Question 4 What is the wavefunction of the first excited state of the quantum harmonic oscillator, |13? (You do not need to solve the Schrödinger equation) Your Answer
Score 1/4
ψ (x) = ( 1
mω πℏ
)
Explanation
2
exp(−
2mωx ℏ
)
ψ (x) = 0 1
− − − −
1/4
ψ (x) = ( 1
mω πℏ
)
√
2mω ℏ
1/4
ψ (x) = ( 1
mω πℏ
)
✔
2
x exp(−
mωx 2ℏ
)
1.00
2
exp(−
mωx 2ℏ
)
Total
1.00 / 1.00
Question Explanation We know that |n
3=
+ 1
^ a
†
√n+1
|n
− − −
3, and a ^
†
= √
mω 2ℏ
^ − x
i 2mωℏ
^. p
So,
− − − − − 1/4 − − − − 2 mω ℏ d mω mωx ψ 1 (x) = ^ a ψ 0 (x) = [√ x − √ ]( ) exp(− ) 2ℏ 2mω dx πℏ 2ℏ †
1/4
mω = (
) πℏ
− − − − − − − − − 2 mω ℏ mω mωx √ [√ x + x ] exp(− ) 2ℏ 2mω ℏ 2ℏ 1/4
mω = (
) πℏ
− − − − − − − − 2 mω mω mωx x + √ x] exp(− ) 2ℏ 2ℏ 2ℏ
[√
1/4
mω = (
)
− − − − − 2 2mω mωx x exp(− ) ℏ 2ℏ
√
πℏ
Question 5 In the context of the model of a crystal we considered in the lectures, why does the ground state of a crystal break continuous translation symmetry? Your Answer
Score
The ground state of a crystal is translationally invariant. The breaking of translational invariance is a consequence of Goldstone's theorem. A crystal has a periodic structure, but it is not uniform in space. Translating by some distance
✔
1.00
which is not a multiple of the lattice spacing gives a state which is different from that of the original state. There is a term in the Hamiltonian of the lattice which breaks the translation symmetry. Total
1.00 / 1.00
Explanation
Question 6 Which of the following is most closely related to Goldstone's theorem? Your Answer
Score
The formation of a Fermi surface. The existence of optical phonons. The spontaneously broken translational symmetry of the ground state of crystal implies the
✔
1.00
existence of gapless excitations, which turn out to be phonons. The phase space trajectory of a harmonic oscillator traces out a circle. Total
1.00 / 1.00
Explanation
