Feedback â Homework 6
Feedback — Homework 6 You submitted this homework on Sun 12 May 2013 11:25 PM EDT (UTC -0400). You got a score of 11.00 out of 11.00.
Here are some identities and definitions that were used during week 6. They may be useful in completing this assignment. Notation: Cartesian coordinates are designated by x 1 , x 2 , x 3 , and the unit vectors in those directions by ^ ^ ^ e1 , e2 , e3
. We designate the position operator for a particle by r t
define the radial coordinate, r
=
R__t Ğ__t. The corresponding momentum operators are : : r
ℏ i
xk
We suppress the use of ^ to designate operators. Quantities like 2
= L
t
, and
r
^ pk =
L
^ ^ ^ = e1 x1 + e2 x 2 + e3 x3
t t t
t
t and
r , p, L = r × p
Ğ t are assumed here to be operators unless stated otherwise. L
Einstein summation convension: When one encounters an expression with repeated indices, such as a i b i then the expression is to be summed over the values i = 1, 2, and 3 of that index. For example, ai bi
º
a1 b1 + a2 b2 + a3 b3
.
Levi-Civita symbol: The Levi-Civita symbol P ijk is defined for the values 1, 2, and 3 of each of its indices. For an even permutation of (123), i.e. (123),(231) or (312), we have
P For
(ijk)
ijk
= +1.
an odd permutation of (123), i.e. (213),(132) or (321), we have
P Otherwise, P ijk
= 0
ijk
=
J
1.
.
Kronecker delta: # ij
= 1
if i
= j
... Otherwise, # ij
Commutation relationships: Definition of commutator
[p j , x k ] = p j x k
J J
[x a p j , x k ] = x a p j x k
J
^ ^ ^^ [ A, B] = AB
J
^^ BA =
= 0
.
^ ^ [ B, A] ℏ x k pj =
i
#
jk
ℏ x k xa pj =
i
#
jk x a
(ijk)
[L a , x b ] = L a x b
[ Lt , p u ] = L t p u
[ L u , L v ] = L u Lv
J J
x b L a = iℏ
P
P
J
p u L t = iℏ
L v L u = iℏ
P
abc
xc
tuv p v
uvw L w
Question 1 Derive the contraction identity for
P
abc
P
cde
. Which of the following expressions is correct?
Your Answer
P
abc
P P
abc
abc
P
P
P
P
abc
=
cde
cde
=
cde
P
=
cde
=
Score
# # ab
de
# # ae
# # ad
bd
be
# # ad
J# # J# # J# # ac
be
ad
ae
+
be
be
1.00
bd
# # ae
Explanation
bd
Total
1.00 / 1.00
Question Explanation We may cyclicly permute the indices of the first Levi-Civita symbol to obtain:
P
cab
P
cde
=
# # ad
be
J#
ae
#
bd
Question 2 By r t
Ğ t we mean p
x i pi
. What is the commutator
Your Answer
Ğt tĞ t tĞ t
[L 1 , r
[L 1 , r
[
1,
t
p ] = iℏL 2
p ] = iℏx 2 p 3
] = 0
[L 1 , r
t
Ğt
p]
?
Score
Explanation
[L 1 , r
[L 1 , r
Ğt tĞ t t
p] = 0
1.00
p ] = iℏ(x 1 + p 2 /ℏ)
Total
1.00 / 1.00
Question Explanation We will make use of the linearity of the commutator and the identity: [A, BC] = [A, B]C + B[A, C]
both of which can be simply shown. Then using the Einstein summation convention, we have: [ L 1 , r i p i ] = [L 1 , r i ]p i + r i [ L 1 , p i ] = iℏ
P
1ik r k p i
+ iℏ
P
1ik r i p k
= iℏ
P
1ik
(r k p i
J
rk pi ) = 0
Question 3 Same as Question 2 but with
L
instead of L 1 Which is correct?
2
Your Answer [L
[L
2
2
[L
[L
,r
,r
2
2
Score
Ğ t J tĞ t tĞ t tĞ t tĞ t t
p] =
2
ℏ r
Explanation
p
p ] = iℏL 2
,r
p] = 0
,r
p] = r
2
+ p
1.00
2
Total
1.00 / 1.00
Question Explanation Once again: [L
2
, r i p i ] = r i [L
2
, p i ] + [L
2
, r i ]p i
Now, [L
2
P
2, r i ] = [L j L j , r i ] = L j [L j , r j ] + [L j , r j ]L j = L j iℏ
Likewise: [L
2
, p i = L j iℏ
P
jik p k
P
+ iℏ
jik p k
Lj
jik
r k + iℏ
P
jik r k L j
We then have: [L
2
P
, r i p i ] = iℏ
P
jik r i L j p k
+ iℏ
jik
P
r i p k L j + iℏ
jik L j r k
P
p i + iℏ
jik r k L j p i
We now relabel i õ k in the third and fourth terms and swap those two indices in the LeviCivita symbol to give: iℏ
P
jik
(r i L j p k + r i p k L j
J
J
Lj r i pk
r i L j p k ) = iℏ
P
ijk
[L j , r i p k ]
Then as in problem one:
P
[ Lj , r i p k ] = iℏ
jil
P
r l p k + iℏ
ri pl
jkl
So we arrive at:
J In the second term we rename
J
ℏ
2
P P jik
i
=
J
õ
J
jil ( r l p k
ℏ
2
P
jik
(
P
jil r l p k
+
P
jkl r i p l )
k
r k pl ) =
2
ℏ (3 r k p k
J
J
2
ℏ (
3 rk pk
# # ii
J
J# #
kl
il
ik
)( r l p k
J
r k pl )
ri pi + r i pi ) = 0
Question 4 The following questions deal with a particle of mass M moving in an isotropic three-dimensional harmonic oscillator potential, ^ H =
J
ℏ
2
2M
?
2
+ V ( x1 , x 2 , x 3 ) =
J
ℏ
2
2M
?
2
+
M
8
2
2 2
2
2
(x 1 + x 2 + x 3 ).
This potential has been used frequently during the past 15 years to trap ultracold atoms. The energy eigenfunctions of this system can be expressed as products
ç
ç
ç
|n 1 (x 1 ) |n 2 ( x 2 ) |n 3 (x 3 ) ,
where |n k (x k ) >=
1
_ R__ nk !
M
(a k )
nk
|0 >,
with
a
M k
=
m 8_ _ __ √ 2ℏ
xk
J
i
____ 8 _ℏ_
√ 2m
pk
being the creation operator introduced in Lecture 7, part 2, and |0> being the ground state of a
one-dimensional harmonic oscillator, which is given by the Gaussian function discussed in Lecture 10. A state
©
(x 1 , x 2 , x 3 )
odd parity if
©
is said to have even parity if
(x 1 , x 2 , x 3 ) =
What is the energy, E n n 1
2 n3
J J J J ©
(
x1 ,
of the state
x 2,
©
x3)
(x 1 , x 2 , x 3 ) =
E n1n2 n3 =
J J J x1 ,
ç
ç
|n 1 (x 1 ) |n 2 ( x 2 ) |n 3 (x 3 )
ℏ
x 2,
x3)
and
ç?
Score
Explanation
8
2
8
3ℏ 2
E n1n2 n3 = ℏ
E n1n2 n3 =
(
.
Your Answer E n1n2 n3 =
©
8
(n 1 + n 2 + n 3 +
8
3ℏ 2
3 2
1.00
)
n1 n2 n3
Total
1.00 / 1.00
Question Explanation The form of this hamiltonian is such that when we convert it to raising and lowering operators we will have: ^ H = ℏ
8
M
^ ^ (a 1 a 1 +
1 2
M
^ ^ + a2 a2 +
1 2
M
^ ^ + a3 a3 +
1
)
2
From the definitions of these operators, we can see that they each only act on the corresponding state. Thus each get:
M
^ ^ ai ai
acts a number operator on the state
E = ℏ
8
(n 1 + n 2 + n 3 +
3
|n i
ç and thus we
)
2
Question 5 What is the parity of the state Your Answer Even Odd
ç
ç
|n 1 (x 1 ) |n 2 ( x 2 ) |n 3 (x 3 )
ç? Score
Explanation
Even if n 1
+ n2 + n3
is even, odd if n 1
+ n2 + n3
is odd
1.00
Can't be determined from the information given Total
1.00 / 1.00
Question Explanation M
^ The ground state has even parity as it is just the product of gaussians. We note that as a is a k
M
^ ^ linear combination of x^k and p , both of which have odd parity, a must also have odd parity. k k So our state will have even parity of we have an even number of creation operators and odd otherwise. In terms of the occupation numbers this gives. Means the state has even parity if n 1 + n 2 + n 3 is even and the state is odd if n 1 + n 2 + n 3 is odd.
Question 6 There are a number of different states of the form |n 1 (x 1 )ç|n 2 ( x 2 )ç|n 3 (x 3 )ç that have the same energies. Do all states of the same energy have the same parity? Your Answer
Score
Explanation
Can't be determined from the information given Yes
1.00
No Total
1.00 / 1.00
Question Explanation For states to have the same energy, n 1 + n 2 + n 3 must be the same for all of them, and thus by the previous problem they must have the same parity.
Question 7 ^
For which values of k does [L k , H ] Your Answer
^ = Lk H
J
^ H Lk = 0
Score
? Explanation
k = 3
only 1.00
k = 1, 2, 3
None Total
1.00 / 1.00
Question Explanation ^
Clearly L k commutes with the constant part of the hamiltonian, so using the summation M
convention we only need to compute: [L k , ℏ8 ^
drop the
ℏ
8 for now and restore it at the end.
^ ^ aj aj] = ℏ
^ ^ [L k , a j ] =
P
= iℏ
m 8_^ _ __ √ 2ℏ
[√ 2ℏ
M
M
^ ^ ^ ^ ^ ^ ( a j [L k , a j ] + [ L k , a j ]a j )
i
^ [L k , x j ] +
_____ 8_
√ 2ℏm
m _ _8__ kjl
8
^ ^ [L k , pj ]
i
xl +
2ℏm ____ 8_ √_
pl
We will
P
= iℏ ]
kjl
^ al
Likewise: M ^ ^ [ L k , a j ] = iℏ
P
^
M
kjl a l
So our expression becomes: iℏ
Exchanging j
õ
l
P
^
M ^M
kjl (a j
al
M
^ ^ + al aj )
in the second term we obtain: iℏ
P
M M
^ ^ al
kjl (a j
J
M
^ ^ aj al ) = 0
So: ^ ^ [L k , H ] = 0
Question 8 Questions 8-10. The following questions deal with a particle of mass M moving in a different three-dimensional harmonic oscillator potential than used in Questions 4-7. This has also been used to trap ultracold atoms. It is given by
M V(
1,
2,
3)
=
2
(
+
+ 4
).
M V (x 1 , x 2 , x 3 ) =
8
2 2
2
2
( x 1 + x 2 + 4 x 3 ).
2
What is the ground state energy, E 0 , of this system? Your Answer E 0 = 3ℏ
E0 =
ℏ
Explanation
8
8
2
E0 = ℏ
8
E 0 = 2ℏ
E0 =
Score
3ℏ
8
1.00
8
2
Total
1.00 / 1.00
Question Explanation This is quite similar to the previous problem if we take the x 3 oscillator to have a frequency of 2 8 . When going to the raising and lowering operators, everything goes through the same except we need to keep in mind this different frequency. So the hamiltonian becomes: ^ H = ℏ
8
M
^ ^ (a 1 a 1 +
= ℏ
8
1 2
M
^ ^ + a2 a2 +
M
1 2
) + 2ℏ
M
8
M
^ ^ (a 3 a 3 +
1 2
)
M
^ ^ ^ ^ ^ ^ ( a 1 a 1 + a 2 a 2 + 2 a 3 a 3 + 2)
The ground state energy is then: 2ℏ
8
Question 9 ^
For which values of i for this new potential does [L i , H ] Your Answer
Score
i = 1, 2, 3
i = 3
only
1.00
^ = LiH
J
^ H Li = 0
?
Explanation
None Total
1.00 / 1.00
Question Explanation Reusing the work from problem 7, we need to calculate: M
M
M
^ ^ ^ ^ ^ ^ [L k , a 1 a 1 + a 2 a 2 + 2 a 3 a 3 ]
M
M
i
3
^ ^ ^ a ^ ] = [L k , a ai] + [Lk , a 3
Where we have used the result of question 7. We thus only need to calculate this one commutator: M
P
^ ^ [ Lk , a 3 a 3 ] = iℏ
M
k3j
M
^ ^ ^ ^ (a 3 a j + a j a 3 )
We may evaluate this then for different values of k . ⎧ ⎪ ⎨ ⎪
J
M
M
^ ^ ^ ) iℏ( a a2 + ^ a a 3 3
2
M ^ ^
M ^ ^
iℏ( a 3 a 1 + a 1 a 3 )
⎩0
So only for
k = 3
do we have
^ ^ [L k , H ] = 0
k = 1 k = 2 k = 3
.
Question 10 Do all states of this new Hamiltonian that have the same energy also have the same parity? Your Answer
Score
Explanation
Can't be determined from the information given No
1.00
Yes Total
1.00 / 1.00
Question Explanation The criteria for being even or odd parity are the same as for the isotropic case. With that in mind, here is a counter example: (1, 1, 0) and (0, 0, 1) have the same energy but the first is even parity while the second is odd parity.
Question 11 We return to the "quantum bouncing ball" problem discussed in the first part of Lecture 6, and apply it to the problem of a neutron subject to the Earth's gravitational field, which is reflected perfectly by a mirror. The one-dimensional time-independent Schrödinger equation that describes this problem is ^ H =
J
ℏ
2
:
2
2M
:
©
x
Determine the lowest-energy wavefunction
(x) + Mgx
2
©
0 (x)
©
(x) = E
©
(x)
from the properties of the Airy function .
Use the following parameters: g
= acceleration due to Earth's gravity = 9.8 m/s2
M
J
= mass of neutron = 1.6749 × 10
ℏ = 1.0546 × 10
J
34
Js
k =
27
kg
Boltzmann constant =
J
1.3806 × 10
23
J/K
Calculate the ground state energy, E 0 . To report it in units that are relevant to modern ultracold neutron sources, convert this energy to an equivalent temperature, T
= E 0 /k
. Enter your
answer as an integer value of nanokelvin, rounded to the nearest nanokelvin. You entered: 16
Your Answer 16
Score
Total
Explanation
1.00 1.00 / 1.00
Question Explanation First we make the change of variables x =
0
+ E
Mg
where 2
Mg ℏ
1/3
2
= (
2
)
With this definition: d
2
dx
M
2
=
2
g
2
d
2
(
) d
2
0
2
Our equation then becomes:
J J
ℏ
0 =
2
d
©
2M
=
dx 2
Let us now divide both sides by
J
2
2
2
+ (Mgx
d
©
2
d
2
E)
2
0
©
0
+
©
to obtain:
2
ℏ g M 0 =
2
ℏ g M 2
J
2
3
d d
2
0
©
2
J0
©
=
d d
2
0
©
2
J0
©
So we obtain the differential equation for the Airy function. The ground state will be the lowest energy state for which (x = 0) = 0 . x = 0 corresponds to ©
0O
=
J
E
. We want the Airy function Ai as our solution and in order to match our boundary conditions we have: (x) Ai( 0 ) where 0 O is a zero of the Airy function. So the ground state value is 0 O = a0 where a0 2.338 is the first zero of the Airy function. The ground state energy is then: ©
U
¡J
E0 =
J
a0
And the associated temperature is:
T0 =
J
a0
= k
J
2
Mg ℏ
a0 k
2
(
1/3
2
)
Plugging in numerical values, we obtain: T0
¡
1.632 × 10
J
8
K
¡
16 nK
Here are some identities and definitions that were used during week 6. They may be useful in completing this assignment. Notation: Cartesian coordinates are designated by x 1 , x 2 , x 3 , and the unit vectors in those directions by ^ ^ ^ e1 , e2 , e3
. We designate the position operator for a particle by r t
define the radial coordinate, r
=
R__t Ğ__t. The corresponding momentum operators are : : r
ℏ i
xk
We suppress the use of ^ to designate operators. Quantities like 2
= L
t
, and
r
^ pk =
L
^ ^ ^ = e1 x1 + e2 x 2 + e3 x3
t t t
t
t and
r , p, L = r × p
Ğ t are assumed here to be operators unless stated otherwise. L
Einstein summation convension: When one encounters an expression with repeated indices, such as a i b i then the expression is to be summed over the values i = 1, 2, and 3 of that index. For example, ai bi
º
a1 b1 + a2 b2 + a3 b3
.
Levi-Civita symbol: The Levi-Civita symbol P ijk is defined for the values 1, 2, and 3 of each of its indices. For an even permutation of (123), i.e. (123),(231) or (312), we have
P For
(ijk)
ijk
= +1.
an odd permutation of (123), i.e. (213),(132) or (321), we have
P Otherwise, P ijk
= 0
ijk
=
J
1.
.
Kronecker delta: # ij
= 1
if i
= j
... Otherwise, # ij
Commutation relationships: Definition of commutator
[p j , x k ] = p j x k
J J
[x a p j , x k ] = x a p j x k
J
^ ^ ^^ [ A, B] = AB
J
^^ BA =
= 0
.
^ ^ [ B, A] ℏ x k pj =
i
#
jk
ℏ x k xa pj =
i
#
jk x a
(ijk)
[L a , x b ] = L a x b
[ Lt , p u ] = L t p u
[ L u , L v ] = L u Lv
J J
x b L a = iℏ
P
P
J
p u L t = iℏ
L v L u = iℏ
P
abc
xc
tuv p v
uvw L w
Question 1 Derive the contraction identity for
P
abc
P
cde
. Which of the following expressions is correct?
Your Answer
P
abc
P P
abc
abc
P
P
P
P
abc
=
cde
cde
=
cde
P
=
cde
=
Score
# # ab
de
# # ae
# # ad
bd
be
# # ad
J# # J# # J# # ac
be
ad
ae
+
be
be
1.00
bd
# # ae
Explanation
bd
Total
1.00 / 1.00
Question Explanation We may cyclicly permute the indices of the first Levi-Civita symbol to obtain:
P
cab
P
cde
=
# # ad
be
J#
ae
#
bd
Question 2 By r t
Ğ t we mean p
x i pi
. What is the commutator
Your Answer
Ğt tĞ t tĞ t
[L 1 , r
[L 1 , r
[
1,
t
p ] = iℏL 2
p ] = iℏx 2 p 3
] = 0
[L 1 , r
t
Ğt
p]
?
Score
Explanation
[L 1 , r
[L 1 , r
Ğt tĞ t t
p] = 0
1.00
p ] = iℏ(x 1 + p 2 /ℏ)
Total
1.00 / 1.00
Question Explanation We will make use of the linearity of the commutator and the identity: [A, BC] = [A, B]C + B[A, C]
both of which can be simply shown. Then using the Einstein summation convention, we have: [ L 1 , r i p i ] = [L 1 , r i ]p i + r i [ L 1 , p i ] = iℏ
P
1ik r k p i
+ iℏ
P
1ik r i p k
= iℏ
P
1ik
(r k p i
J
rk pi ) = 0
Question 3 Same as Question 2 but with
L
instead of L 1 Which is correct?
2
Your Answer [L
[L
2
2
[L
[L
,r
,r
2
2
Score
Ğ t J tĞ t tĞ t tĞ t tĞ t t
p] =
2
ℏ r
Explanation
p
p ] = iℏL 2
,r
p] = 0
,r
p] = r
2
+ p
1.00
2
Total
1.00 / 1.00
Question Explanation Once again: [L
2
, r i p i ] = r i [L
2
, p i ] + [L
2
, r i ]p i
Now, [L
2
P
2, r i ] = [L j L j , r i ] = L j [L j , r j ] + [L j , r j ]L j = L j iℏ
Likewise: [L
2
, p i = L j iℏ
P
jik p k
P
+ iℏ
jik p k
Lj
jik
r k + iℏ
P
jik r k L j
We then have: [L
2
P
, r i p i ] = iℏ
P
jik r i L j p k
+ iℏ
jik
P
r i p k L j + iℏ
jik L j r k
P
p i + iℏ
jik r k L j p i
We now relabel i õ k in the third and fourth terms and swap those two indices in the LeviCivita symbol to give: iℏ
P
jik
(r i L j p k + r i p k L j
J
J
Lj r i pk
r i L j p k ) = iℏ
P
ijk
[L j , r i p k ]
Then as in problem one:
P
[ Lj , r i p k ] = iℏ
jil
P
r l p k + iℏ
ri pl
jkl
So we arrive at:
J In the second term we rename
J
ℏ
2
P P jik
i
=
J
õ
J
jil ( r l p k
ℏ
2
P
jik
(
P
jil r l p k
+
P
jkl r i p l )
k
r k pl ) =
2
ℏ (3 r k p k
J
J
2
ℏ (
3 rk pk
# # ii
J
J# #
kl
il
ik
)( r l p k
J
r k pl )
ri pi + r i pi ) = 0
Question 4 The following questions deal with a particle of mass M moving in an isotropic three-dimensional harmonic oscillator potential, ^ H =
J
ℏ
2
2M
?
2
+ V ( x1 , x 2 , x 3 ) =
J
ℏ
2
2M
?
2
+
M
8
2
2 2
2
2
(x 1 + x 2 + x 3 ).
This potential has been used frequently during the past 15 years to trap ultracold atoms. The energy eigenfunctions of this system can be expressed as products
ç
ç
ç
|n 1 (x 1 ) |n 2 ( x 2 ) |n 3 (x 3 ) ,
where |n k (x k ) >=
1
_ R__ nk !
M
(a k )
nk
|0 >,
with
a
M k
=
m 8_ _ __ √ 2ℏ
xk
J
i
____ 8 _ℏ_
√ 2m
pk
being the creation operator introduced in Lecture 7, part 2, and |0> being the ground state of a
one-dimensional harmonic oscillator, which is given by the Gaussian function discussed in Lecture 10. A state
©
(x 1 , x 2 , x 3 )
odd parity if
©
is said to have even parity if
(x 1 , x 2 , x 3 ) =
What is the energy, E n n 1
2 n3
J J J J ©
(
x1 ,
of the state
x 2,
©
x3)
(x 1 , x 2 , x 3 ) =
E n1n2 n3 =
J J J x1 ,
ç
ç
|n 1 (x 1 ) |n 2 ( x 2 ) |n 3 (x 3 )
ℏ
x 2,
x3)
and
ç?
Score
Explanation
8
2
8
3ℏ 2
E n1n2 n3 = ℏ
E n1n2 n3 =
(
.
Your Answer E n1n2 n3 =
©
8
(n 1 + n 2 + n 3 +
8
3ℏ 2
3 2
1.00
)
n1 n2 n3
Total
1.00 / 1.00
Question Explanation The form of this hamiltonian is such that when we convert it to raising and lowering operators we will have: ^ H = ℏ
8
M
^ ^ (a 1 a 1 +
1 2
M
^ ^ + a2 a2 +
1 2
M
^ ^ + a3 a3 +
1
)
2
From the definitions of these operators, we can see that they each only act on the corresponding state. Thus each get:
M
^ ^ ai ai
acts a number operator on the state
E = ℏ
8
(n 1 + n 2 + n 3 +
3
|n i
ç and thus we
)
2
Question 5 What is the parity of the state Your Answer Even Odd
ç
ç
|n 1 (x 1 ) |n 2 ( x 2 ) |n 3 (x 3 )
ç? Score
Explanation
Even if n 1
+ n2 + n3
is even, odd if n 1
+ n2 + n3
is odd
1.00
Can't be determined from the information given Total
1.00 / 1.00
Question Explanation M
^ The ground state has even parity as it is just the product of gaussians. We note that as a is a k
M
^ ^ linear combination of x^k and p , both of which have odd parity, a must also have odd parity. k k So our state will have even parity of we have an even number of creation operators and odd otherwise. In terms of the occupation numbers this gives. Means the state has even parity if n 1 + n 2 + n 3 is even and the state is odd if n 1 + n 2 + n 3 is odd.
Question 6 There are a number of different states of the form |n 1 (x 1 )ç|n 2 ( x 2 )ç|n 3 (x 3 )ç that have the same energies. Do all states of the same energy have the same parity? Your Answer
Score
Explanation
Can't be determined from the information given Yes
1.00
No Total
1.00 / 1.00
Question Explanation For states to have the same energy, n 1 + n 2 + n 3 must be the same for all of them, and thus by the previous problem they must have the same parity.
Question 7 ^
For which values of k does [L k , H ] Your Answer
^ = Lk H
J
^ H Lk = 0
Score
? Explanation
k = 3
only 1.00
k = 1, 2, 3
None Total
1.00 / 1.00
Question Explanation ^
Clearly L k commutes with the constant part of the hamiltonian, so using the summation M
convention we only need to compute: [L k , ℏ8 ^
drop the
ℏ
8 for now and restore it at the end.
^ ^ aj aj] = ℏ
^ ^ [L k , a j ] =
P
= iℏ
m 8_^ _ __ √ 2ℏ
[√ 2ℏ
M
M
^ ^ ^ ^ ^ ^ ( a j [L k , a j ] + [ L k , a j ]a j )
i
^ [L k , x j ] +
_____ 8_
√ 2ℏm
m _ _8__ kjl
8
^ ^ [L k , pj ]
i
xl +
2ℏm ____ 8_ √_
pl
We will
P
= iℏ ]
kjl
^ al
Likewise: M ^ ^ [ L k , a j ] = iℏ
P
^
M
kjl a l
So our expression becomes: iℏ
Exchanging j
õ
l
P
^
M ^M
kjl (a j
al
M
^ ^ + al aj )
in the second term we obtain: iℏ
P
M M
^ ^ al
kjl (a j
J
M
^ ^ aj al ) = 0
So: ^ ^ [L k , H ] = 0
Question 8 Questions 8-10. The following questions deal with a particle of mass M moving in a different three-dimensional harmonic oscillator potential than used in Questions 4-7. This has also been used to trap ultracold atoms. It is given by
M V(
1,
2,
3)
=
2
(
+
+ 4
).
M V (x 1 , x 2 , x 3 ) =
8
2 2
2
2
( x 1 + x 2 + 4 x 3 ).
2
What is the ground state energy, E 0 , of this system? Your Answer E 0 = 3ℏ
E0 =
ℏ
Explanation
8
8
2
E0 = ℏ
8
E 0 = 2ℏ
E0 =
Score
3ℏ
8
1.00
8
2
Total
1.00 / 1.00
Question Explanation This is quite similar to the previous problem if we take the x 3 oscillator to have a frequency of 2 8 . When going to the raising and lowering operators, everything goes through the same except we need to keep in mind this different frequency. So the hamiltonian becomes: ^ H = ℏ
8
M
^ ^ (a 1 a 1 +
= ℏ
8
1 2
M
^ ^ + a2 a2 +
M
1 2
) + 2ℏ
M
8
M
^ ^ (a 3 a 3 +
1 2
)
M
^ ^ ^ ^ ^ ^ ( a 1 a 1 + a 2 a 2 + 2 a 3 a 3 + 2)
The ground state energy is then: 2ℏ
8
Question 9 ^
For which values of i for this new potential does [L i , H ] Your Answer
Score
i = 1, 2, 3
i = 3
only
1.00
^ = LiH
J
^ H Li = 0
?
Explanation
None Total
1.00 / 1.00
Question Explanation Reusing the work from problem 7, we need to calculate: M
M
M
^ ^ ^ ^ ^ ^ [L k , a 1 a 1 + a 2 a 2 + 2 a 3 a 3 ]
M
M
i
3
^ ^ ^ a ^ ] = [L k , a ai] + [Lk , a 3
Where we have used the result of question 7. We thus only need to calculate this one commutator: M
P
^ ^ [ Lk , a 3 a 3 ] = iℏ
M
k3j
M
^ ^ ^ ^ (a 3 a j + a j a 3 )
We may evaluate this then for different values of k . ⎧ ⎪ ⎨ ⎪
J
M
M
^ ^ ^ ) iℏ( a a2 + ^ a a 3 3
2
M ^ ^
M ^ ^
iℏ( a 3 a 1 + a 1 a 3 )
⎩0
So only for
k = 3
do we have
^ ^ [L k , H ] = 0
k = 1 k = 2 k = 3
.
Question 10 Do all states of this new Hamiltonian that have the same energy also have the same parity? Your Answer
Score
Explanation
Can't be determined from the information given No
1.00
Yes Total
1.00 / 1.00
Question Explanation The criteria for being even or odd parity are the same as for the isotropic case. With that in mind, here is a counter example: (1, 1, 0) and (0, 0, 1) have the same energy but the first is even parity while the second is odd parity.
Question 11 We return to the "quantum bouncing ball" problem discussed in the first part of Lecture 6, and apply it to the problem of a neutron subject to the Earth's gravitational field, which is reflected perfectly by a mirror. The one-dimensional time-independent Schrödinger equation that describes this problem is ^ H =
J
ℏ
2
:
2
2M
:
©
x
Determine the lowest-energy wavefunction
(x) + Mgx
2
©
0 (x)
©
(x) = E
©
(x)
from the properties of the Airy function .
Use the following parameters: g
= acceleration due to Earth's gravity = 9.8 m/s2
M
J
= mass of neutron = 1.6749 × 10
ℏ = 1.0546 × 10
J
34
Js
k =
27
kg
Boltzmann constant =
J
1.3806 × 10
23
J/K
Calculate the ground state energy, E 0 . To report it in units that are relevant to modern ultracold neutron sources, convert this energy to an equivalent temperature, T
= E 0 /k
. Enter your
answer as an integer value of nanokelvin, rounded to the nearest nanokelvin. You entered: 16
Your Answer 16
Score
Total
Explanation
1.00 1.00 / 1.00
Question Explanation First we make the change of variables x =
0
+ E
Mg
where 2
Mg ℏ
1/3
2
= (
2
)
With this definition: d
2
dx
M
2
=
2
g
2
d
2
(
) d
2
0
2
Our equation then becomes:
J J
ℏ
0 =
2
d
©
2M
=
dx 2
Let us now divide both sides by
J
2
2
2
+ (Mgx
d
©
2
d
2
E)
2
0
©
0
+
©
to obtain:
2
ℏ g M 0 =
2
ℏ g M 2
J
2
3
d d
2
0
©
2
J0
©
=
d d
2
0
©
2
J0
©
So we obtain the differential equation for the Airy function. The ground state will be the lowest energy state for which (x = 0) = 0 . x = 0 corresponds to ©
0O
=
J
E
. We want the Airy function Ai as our solution and in order to match our boundary conditions we have: (x) Ai( 0 ) where 0 O is a zero of the Airy function. So the ground state value is 0 O = a0 where a0 2.338 is the first zero of the Airy function. The ground state energy is then: ©
U
¡J
E0 =
J
a0
And the associated temperature is:
T0 =
J
a0
= k
J
2
Mg ℏ
a0 k
2
(
1/3
2
)
Plugging in numerical values, we obtain: T0
¡
1.632 × 10
J
8
K
¡
16 nK
