HOMEWORK ASSIGNMENT 7: Solutions

PHYS852 Quantum Mechanics II, Spring 2010
HOMEWORK ASSIGNMENT 7: Solutions
Topics covered: addition of three angular momenta, degenerate perturbation theory
~1 and S
~2 , and one spin-1 particle, described
1. Consider a system of two spin-1/2 particles, described by S
0
0
~
~
~
~
by S3 . Let S = S1 + S2 . What are the allowed values of s ? For each allowed value, give the allowed
m0 values. Use Clebsch Gordan coefficients to express the |s0 m0 m3 i states in terms of the |m1 m2 m3 i
states.
~=S
~1 + S
~2 + S
~3 . What are the allowed values of the quantum number s? For each s value, list
Let S
the alllowed m-values. Express the states |s0 , s, mi as linear superpositions of the |s0 m0 m3 i states,
and then as linear superpositions of the |m1 m2 m3 i states.
Notation:
For s = 2, we use | ⇑i for m = 2, | ↑i for m = 1, |0i for m = 0, | ↓i for m = −1, and | ⇓i for m = −2;
For s = 1, we use | ↑i for m = 1, |0i for m = 0, and | ↓i for m = −1;
For s = 1/2, we use | ↑i for m = 1/2, and | ↓i for m = −1/2.
The allowed values of s0 are 0, 1
For s0 = 0, the only allowed value for m0 is 0.
For s0 = 1, the allowed values for m0 are −1, 0, 1.
The |s0 m0 m3 i states follow the standard singlet/triplet forms:
|s0 m0 m3 i = |m1 m2 m3 i
|1 ↑ m3 i = | ↑↑ m3 i
1
|10m3 i = √ (| ↑↓ m3 i+| ↓↑ m3 i)
2
|1 ↓ m3 i = | ↓↓ m3 i
1
|00m3 i = √ (| ↑↓ m3 i − | ↓↑ m3 i)
2
For
For
For
For
For
s0 = 0, the allowed value of s is 1.
s0 = 1, the allowed values of s are 0, 1, 2.
s = 0, the allowed value of m is 0.
s = 1, the allowed values of m are −1, 0, 1.
s = 2, the allowed values of m are −2, −1, 0, 1, 2.
1
(1)
(2)
(3)
(4)
(5)
|s0 smi =
|s0 m0 m3 i
|12 ⇑i =
|1 ↑↑i
|12 ↑i =
|120i =
|12 ↓i =
|12 ⇓i =
|11 ↑i =
|110i =
|11 ↓i =
|100i =
|01 ↑i =
|010i =
|01 ↓i =
= |m1 m2 m3 i
= | ↑↑↑i
1 √
√1 (|1 ↑ 0i + |10 ↑i)
=
( 2| ↑↑ 0i + | ↑↓↑i + | ↓↑↑i)
2
2
√
√
1
√1 (|1 ↑↓i+2|100i+|1 ↓↑i) = √ (| ↑↑↓i+ 2| ↑↓ 0i+ 2| ↓↑ 0i+| ↓↓↑i)
6
6
√
1
√1 (|10 ↓i + |1 ↓ 0i)
=
(|
↑↓↓i
+
|
↓↑↓i
+
2| ↓↓ 0i)
2
2
|1 ↓↓i
= | ↓↓↓i
1 √
√1 (|1 ↑ 0i − |10 ↑i)
=
( 2| ↑↑ 0i − | ↑↓↑i − | ↓↑↑i)
2
2
1
√1 (|1 ↑↓i − |1 ↓↑i)
= √ (| ↑↑↓i − | ↓↓↑i)
2
2
√
1
√1 (|10 ↓i − |1 ↓ 0i)
= (| ↑↓↓i + | ↓↑↓i − 2| ↓↓ 0i)
2
2
√
1 √
√1 (|1 ↑↓i−|100i+|1 ↓↑i)
√ ( 2| ↑↑↓i − | ↑↓ 0i − | ↓↑ 0i + 2| ↓↓↑i)
=
3
6
1
|00 ↑i
= √ (| ↑↓↑i − | ↓↑↑i)
2
1
|000i
= √ (| ↑↓ 0i − | ↓↑ 0i)
2
1
|00 ↓i
= √ (| ↑↓↓i − | ↓↑↓i)
2
2
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
(15)
(16)
(17)
(18)
Calculations:
0
S− |12 ⇑i = (S−
+ S3− )|1 ↑↑i = (S1− + S2− + S3− )| ↑↑↑i
p
√
√
√
2 · 3−2· ↑|12 ↑i = 1 · 2−1 · 0|10 ↑i+ 1 · 2−1 · 0|1 ↑ 0i = | ↓↑↑i+| ↑↓↑i+ 1 · 2 − 1 · 0| ↑↑ 0i
√
√
2|12 ↑i = 2(|10 ↑i + |1 ↑ 0i) = | ↓↑↑i + | ↑↓↑i + 2| ↑↑ 0i
√
1
1
|12 ↑i = √ (|10 ↑i + |1 ↑ 0i) = (| ↓↑↑i + | ↑↓↑i + 2| ↑↑ 0i).
2
2
√
1
1
0
S− |12 ↑i = (S−
+ S3− ) √ (|10 ↑i + |1 ↑ 0i) = (S1− + S2− + S3− ) (| ↓↑↑i+| ↑↓↑i+ 2| ↑↑ 0i
2
2
√
√
√
6|120i = |1 ↓↑i + 2|100i + |10 ↓i = (| ↓↓↑i+ 2| ↓↑ 0i+ 2| ↑↓ 0i + | ↑↑↓i)
√
√
1
1
|120i = √ (|1 ↓↑i + 2|100i + |10 ↓)i) = √ (| ↓↓↑i+ 2| ↓↑ 0i+ 2| ↑↓ 0i + | ↑↑↓i)
6
6
1
1 √
0
S− |11 ↑i = (S−
+S3− ) √ (|1 ↑ 0i−|10 ↑i) = (S1− +S2− +S3− ) ( 2| ↑↑ 0i−| ↑↓↑i−| ↓↑↑i)
2
2
√
2|110i = |1 ↑↓i − |1 ↓↑i = | ↑↑↓i − | ↓↓↑i
1
1
|110i = √ (|1 ↑↓i − |1 ↓↑i) = √ (| ↑↑↓i−| ↓↓↑i)
2
2
1
1
0
S− |110i = (S−
+ S3− ) √ (|1 ↑↓i − |1 ↓↑i) = (S1− + S2− + S3− ) √ (| ↑↑↓i − | ↓↓↑i)
2
2
√
√
1
2|11 ↓i = |10 ↓i − |1 ↓ 0i = √ (| ↓↑↓i + | ↑↓↑i − 2| ↓↓ 0i)
2
√
1
1
|11 ↓i = √ (|10 ↓i − |1 ↓ 0i) = (| ↓↑↓i + | ↑↓↑i − 2| ↓↓ 0i)
2
2
3
(19)
(20)
(21)
(22)
(23)
(24)
(25)
(26)
(27)
(28)
(29)
(30)
(31)
2. Derive general expressions for the energy-shifts up to third-order and state vectors up to second-order
in degenerate perturbation theory.
As described in the lecture notes, solving the first- and second-order equations gives:
Enm = En(0) + λvnm − λ2
X Vnmn0 m0 Vn0 m0 nm
∆n0 n
0
(32)
n 6=n
m0



X

λ2 X Vnmn0 m0 

|nmi = |nm(0) i 
1
−

2 0
∆n0 n  00
m 6=m n 6=n
m000
n 6=n
m0

+
X
|nm
0(0)
m0 6=m
X Vn0 m0 nm00 Vnm00 n000 m000 Vn000 m000 nm

0 m0 nm 
+
V
n
2

vnm
00 m ∆n000 n
000


X
 X Vnm0 n00 m00 Vn00 m00 nm 

vnm
Vn0 m0 nm
0 0(0) 


i λ
1+λ
|n m i − λ
+
vnm0 m ∆n00 n
∆n0 n
∆n0 n
00
0
n 6=n
m00
n 6=n
m0

+ λ2
X

Vn0 m0 n00 m00 −
n00 6=n
m00

X Vn0 m0 nm000 Vnm000 n00 m00

00 00
 Vn m nm 
vnm000 m
∆n0 n ∆n00 n 
000
(33)
m 6=m
To proceed, we need to use the third-order equations to find the third-order energy-shift and the
second-order correction to the |nm0(0) i states.
The third-order equation is:
(3)
(2)
(1)
(H0 − En(0) )|nm(3) i = −V |nm(2) i + Enm
|nm(0) i + Enm
|nm(1) i + Enm
|nm(2) i
(34)
(3)
Hitting from the left with hnm(0) | and solving for Enm gives
(3)
(2)
(1)
Enm
= hnm(0) |V |nm(2) i − Enm
hnm(0) |nm(1) i − Enm
hnm(0) |nm(2) i
X
=
Vnmn0 m0 hn0 m0(0) |nm(2) i
n0 6=n
m0
=
X
X
n0 6=n n00 6=n,n0
m0
m00
+
Vnmn0 m0 Vn0 m0 n00 m00 Vn00 m00 nm
∆n0 n ∆n00 n
X Vnmn0 m0 vn0 m0 Vn0 m0 nm
X Vnmn0 m0 Vn0 m0 nm vnm
−
2
∆n0 n
∆2n0 n
0
0
n 6=n
m0
−
X X
n 6=n
m0
X Vnmn0 m0 Vn0 m0 nm000 Vnm000 n00 m00 Vn00 m00 nm
vnm000 m ∆n0 n ∆n00 n
000
n0 6=n n00 6=n m 6=m
m00
m0
4
(35)
Hitting from the left with hnm0(0) | and solving for hnm0(0) |nm(2) i gives
hnm0(0) |V |nm(2) i Enm (2)
−
hnm0(0) |nm(1) i
vnm
vnm
X Vnm0 n00 m00 Vn00 m00 nm
= −
∆2n00 n
00
hnm0(0) |nm(2) i =
n 6=n
m00

+
X 


n00 6=n
m00
−
X

X
n000 6=n,n00
m000
X
n00 6=n n000 6=n m
m00
m000
−
X
Vnm0 n00 m00 Vn00 m00 n000 m000 Vn000 m000 nm Vnm0 n00 m00 vn00 m00 Vn00 m00 nm 

+

vnm ∆n00 n ∆n000 n
vnm ∆2n00 n
X Vnm0 n00 m00 Vn00 m00 nm0000 Vnm0000 n000 m000 Vn000 m000 nm
vnm vnm0000 m ∆n00 n ∆n000 n
0000
6=m
X Vnm0 n00 m00 Vn00 m00 nm Vnmn000 m000 Vn000 m000 nm
vnm ∆n00 n vnm0 m ∆n000 n
000
n00 6=n n 6=n
m000
m00
5
(36)
3. Consider a system described in the basis {|1i, |2i, |3i, |4i}

1 0 0
 0 1 0
H0 = ~ω0 
 0 0 2
0 0 0
by the bare Hamiltonian

0
0 

0 
2
(37)
Let the system be perturbed by the operator

0
 2
V = ~g 
 1
1
2
0
1
1
1
1
0
2

1
1 

2 
0
(38)
First, determine the ‘good basis’ for degenerate perturbation theory, then compute the eigenvalues
and eigenvectors of H = H0 + V to second-order in perturbation theory.
The bare Hamiltonian has two degenerate subspaces. The set {|1i, |2i} spans the space corresponding
(0)
(0)
to E1 = ~ω0 , and the set {|3i, |4i} spans the space corresponding to E2 = 2~ω0 .
By inspection, we see that
0 2
(39)
V1 = V2 =
2 0
There fore the good basis states are
|11(0) i =
|12(0) i =
|21(0) i =
|22(0) i =
1
√ (|1i − |2i)
2
1
√ (|1i + |2i)
2
1
√ (|3i − |4i)
2
1
√ (|3i + |4i)
2
(40)
(41)
(42)
(43)
with first-order energies given by
In this basis we have
v11 = −2~g
(44)
v12 = 2~g
(45)
v21 = −2~g
(46)
v22 = 2~g
(47)

1
 0
V = 2~g 
 0
0
0
1
0
1
0
0
1
0

0
1 

0 
1
(48)
From this, we see that states |11(0) i and |21(0) i are not coupled to any other states by V . Thus,
beyond the first-order energy-shift, they are not perturbed by V .
6
Thus the problem maps onto a single non-degenerate problem in the {|12(0) i, |22(0) i} Hilbert space,
for which we have
1 0
(49)
H0 = ~ω0
0 2
and
V = 2~g
0 1
1 0
(50)
The states to second-order are therefore
|11i = |11(0) i
(51)
2g
2g 2
(0)
|12i ≈ |12 i 1 − 2 − |22(0) i
ω0
ω0
(52)
|21i = |21(0) i
(53)
2g 2
(0) 2g
(0)
|22i ≈ |12 i + |22 i 1 − 2
ω0
ω0
(54)
The energies to second-order are
E11 = ~ω0 − 2~g
E12 ≈ ~ω0 + 2~g −
(55)
4~g 2
ω0
E21 = 2~ω0 − 2~g
E22 ≈ 2~ω0 + 2~g +
7
(56)
(57)
4~g 2
ω0
(58)
4. Prove that in order to predict the resonance-frequency shifts of a perturbed quantum system to
second-order, one needs only to compute the energy levels to second-order, and that this implicitly
requires computing the eigenstates only to first-order.
Then show that in order to compute the shift in the expectation-value of an arbitrary operator to
second-order, one must compute the eigenstates to second-order.
The resonance frequency shifts are the differences between the energy eigenvalues, which means
they can be computed to second-order directly from the second-order energies. The second-order
(2)
energy corrections are given in terms of the first-order eigenstates via En = −hn(0) |V |n(1) i.
To compute the expectation value of an operator to second-order, we need to evaluate
hAi = hψ(t)|A|ψ(t)i
X
c∗m cn hm|A|niei(Em −En )t/~
=
mn
(59)
expanding to 2nd order gives
h
X
hAi =
nc∗m cn hm(0) |A|n(0) i + λ hm(1) |A|n(0) i + hm(0) |A|n(1) i
m
i
+λ2 hm(0) |A|n(2) i + hm(1) |A|n(1) i + hm(2) |A|n(0) i
h
i
(0)
(0)
(1)
(1)
(2)
(2)
i Em −En +λ Em −En +λ2 Em −En
t/~
× e
which shows that the second-order eigenstates are required.
8
(60)
5. Consider a pair of identical harmonic oscillators, described by the Hamiltonian
H0 = ~ω(A†1 A1 + A†2 A2 + 1).
(61)
Determine the energy eigenvalues and degeneracies of the two lowest energy levels of H0 .
Let the system be perturbed by the operator
V = ~λ(A†1 A†2 + A†1 A2 + A†2 A1 + A1 A2 ).
(62)
For the states comprising the two lowest bare energy levels, determine the full energy eigenvalues of
H = H0 + V to second-order and the eigenstates to first-order in λ.
(0)
The lowest energy level is the ground state, E1 = ~ω, with non-degenerate eigenstate
|1(0) i = |00i,
(63)
(0)
The first excited energy level is E2 = 2~ω, with d2 = 2. The degenerate subspace is spanned by the
states {|01i, |10i}.
in basis {|01i, |10i}, we have
V2 = ~λ
0 1
1 0
(64)
For the ground-state, we have
(1)
E1 = 0
and
(1)
|1
∞
X
i=−
n1 ,n2 =0
(n1 ,n2 )6=(0,0)
|n1 n2 i
(65)
hn1 n2 |V |00i
~ω(n1 + n2 )
(66)
with
p
p
hn2 n2 |V |m1 m2 i = ~λ
(m1 +1)(m2 +1)δn1 ,m1 +1 δn2 ,m2 +1 + m2 (m1 +1)δn1 ,m1 +1 δn2 ,m2 −1
p
√
+
m1 (m2 +1)δn1 ,m1 +1 δn2 ,m2 −1 + m1 m2 δn1 ,m1 −1 δn2 ,m2 −1
(67)
this becomes
λ
|11i
2ω
(68)
λ
~λ2
h00|V |11i = −
2ω
2ω
(69)
|1(1) i = −
with the second-order energy then given by
(2)
E1 = −
For the n = 2 subspace, the ’good basis’ is
1
|21(0) i = √ (|01i − |10i)
2
(70)
1
|22(0) i = √ (|01i − |10i)
2
(71)
9
The first-order energies are
(1)
E21 = −~λ
(72)
(1)
E22 = ~λ
(73)
The first-order eigenstates are
|21(1) i = −|12i
h12|V |21(0) i
h21|V |21(0) i
− |21i
2~ω
2~ω
λ
(|12i − |21i)
2ω
λ
(|12i + |21i)
|22(1) i = −
2ω
= −
(74)
so that the second-order energies are
(2)
E21
= h21(0) |V |21(1) i
~λ2
= − √ (h01| − h10|)A1 A2 |(|12i − |21i)
2 2ω
~λ2
= −
ω
(2)
= h22(0) |V |22(1) i
~λ2
= −
ω
E22
(75)
(76)
We have then
E1 ≈ ~ω −
~λ2
2ω
(77)
~λ2
ω
~λ2
≈ 2~ω + ~λ −
ω
E21 ≈ 2~ω − ~λ −
(78)
E22
(79)
and
|1i ≈ |00i −
|21i ≈
|22i ≈
λ
|11i
2ω
1
√ (|01i − |10i) −
2
1
√ (|01i + |10i) −
2
10
(80)
λ
(|12i − |21i)
2ω
λ
(|12i + |21i)
2ω
(81)
(82)