PHY4605–Spring 2005
Problem Set 2, Solutions
Due: Jan. 24, 2005
1. Gauge invariance
.
(a)
B0 = ∇ × A0 = ∇ × A + ∇ × (∇λ) = ∇ × A = B
(since ∇ × (∇φ) = 0)
µ ¶
∂A
∂A0
∂
∂Λ
0
0
−
E = −∇φ −
= −∇φ + ∇
− (∇Λ)
∂t
∂t
∂t
∂t
∂A
= −∇φ −
=E
∂t
(b) First let’s apply the operator p − qA0 to Ψ0 :
[−i~∇ − qA − q(∇Λ)] eiqΛ/~ Ψ = q(∇Λ)eiqΛ/~ Ψ − i~eiqΛ/~ ∇Ψ
− qAeiqΛ/~ Ψ − q(∇Λ)eiqΛ/~ Ψ
= −i~eiqΛ/~ ∇Ψ − qAeiqΛ/~ Ψ
We need the square of the operator applied to Ψ0 , so do it again:
[−i~∇ − qA − q(∇Λ)]2 eiqΛ/~ Ψ
= (−i~∇ − qA − q(∇Λ))
¡
¢
× −i~eiqΛ/~ ∇Ψ − qAeiqΛ/~ Ψ
µ
¶
iq
2
iqΛ/~
iqΛ/~ 2
= −~
(∇Λ · ∇Ψ)e
+e
∇Ψ
~
+ i~q(∇ · A)eiqΛ/~ Ψ − q 2 (A · ∇Λ)eiqΛ/~ Ψ
+ iq~eiqΛ/~ A · (∇Ψ) + iq~eiqΛ/~ (A · ∇Ψ) + q 2 A2 eiqΛ/~ Ψ
+ iq~eiqΛ/~ (∇Λ · ∇Ψ) + q 2 (A · ∇Λ)eiqΛ/~ Ψ
¡
= eiqΛ/~ −~2 ∇2 Ψ + i~q(∇ · A)Ψ + 2iq~(A · ∇Ψ) + q 2 A2 Ψ
¢
− iq~∇Λ · ∇Ψ − q 2 (A · ∇Λ)Ψ + iq~(∇Λ · ∇Ψ + q 2 (A · ∇Λ)Ψ
£
¤
= eiqΛ/~ (−i~∇ − qA)2 Ψ
Therefore
·
¸
·
¸
1
1
∂Λ
2
0 2
0
0
iqΛ/~
(−i~∇ − qA ) + qφ Ψ = e
(−i~∇ − qA) + qφ − q
Ψ
2m
2m
∂t
¸
·
∂Λ
∂Ψ
iqΛ/~
−q Ψ
= e
i~
∂t
∂t
∂ ¡ iqΛ/~ ¢
∂Ψ0
= i~
QED
e
Ψ = i~
∂t
∂t
1
2. Harmonic oscillator in B-field
a) B=0 From problem set 4, we know the solution to the 3D simple harmonic
oscillator. If φn (x) are the energy eigenfunctions of the 1D simple harmonic
oscillator we find,
ψn1 n2 n3 (x, y, z) = φn1 (x)φn2 (y)φn3 (z) n1 , n2 , n3 = 0, 1, 2, · · ·
and
µ
Enkm
3
= ~ω
+ n1 + n2 + n3
{z
}
2 |
(1)
¶
(2)
N
How the ψn1 n2 n3 are related to the angular momentum eigenfunctions ψnlm is
not clear, although we worked out a few simple cases. [Since it is a central force
problem, however, we know we may write ψnlm = Rn (r)Ylm (θ, φ) where only
the Rnl (r)’s will differ from the hydrogen-atom ones!] Since the eigenenergies
must be the same regardless of the labelling, we must have,
2n + l = n1 + n2 + n3 ≡ N
(3)
where the right hand side can be any integer, as can the left hand side if n and
l are integers. We need only check for degeneracies. For the cartesian case it
was
−p
N N
X
X
1
3
ν=
1 = N2 + N + 1
(4)
2
2
p=0 q=0
(see problem set 4 solutions of last semester)
N ν
0 1
1 3
2 6
3 10
4 15
How many allowed energies correspond to fixed N=2n+l?
Check for cases: e.g. N = 3 combinations
n
1
0
l degeneracy 2l + 1
1
3
3
7
for total of 3+7=10 states. For the general case sum (2l + 1) over all possible
2
n:
¶
¸
N/2 · µ
X
ν=
2 N − 2n + 1
n=0
| {z }
µ
=
l
N
+1
2
¶µ
¶
2N + 1 − 4
N/2
X
n
n=0
¶µ
¶
µ
¶
1N N
N
+ 1 2N + 1 − 4
+1
=
2
22 2
µ
¶µ
¶
N
1
3
=
+ 1 N + 1 = N2 + N + 1
2
2
2
µ
(5)
Note principal q − n. n is unrelated to the hydrogen-atom “n”: no restriction
on the size of l depending on n.
b) 3D simple harmonic oscillator in a magnetic field has the following Hamiltonian
µ
¶2
1
e
1
~2 2
ie~
ie~
e2
1
2 2
p − A + mω r = −
∇ +
A·∇+
∇·A+
A2 + mω 2 r2
2
2m
c
2
2m
2mc
2mc
2mc
2
µ
¶
ie~
weakfield
A·∇+∇·A
(6)
' HSHO +
2mc
¡
¢
¡
¢
Now for a constant
field take A = 12 r × B.¤ Note ∇ · Aψ = A · ∇ψ + ∇ · A ψ.
£
But ∇ · A = − 12 B · ∇
× r} − r| · ∇{z× B} = 0. So
| {z
0
0sinceB=const.
µ
¶
µ
¶
ie~
ie~
ie~
ie~
A·∇+∇·A ψ =
A·∇ ψ =
r × B · ∇ψ = −
B × r · ∇ψ
2mc
mc
2mc
2mc
¡
¡
¢¢
e
e
=
B · r × − i~∇ ψ =
B · Lψ.
(7)
2mc
2mc
N.B.: At the last step we could interchange · and × since ∇i never acted on ri :
(B × r · ∇) = ²ijk Bj rk ∇i = Bj ²jki rk ∇i .
c) Spectrum of 3 lowest energy states
i) Ground state in zero field N = 2n + l = 0 only possible combination n = 0,
l = 0 non-degenerate ν = 1
∆H '
e
e
B · L̂ =
Bz Lz
2mc
2mc
Lz ψN =0 = 0
∆EN =0 = 0
ii) First excited state N = 1
3
(8)
(9)
(10)
+~
Only n = 0, l = 1 allowed, degeneracy ν = 3 for Lˆz ψN =1 = 0 ψN =1 .
−~
∆EN =1 =
e
Bz m~ m = 1, 0, −1
2mc
iii) 2nd excited state N = 2
e
l=2
n = 0, l = 2 → ∆EN
=2 = 2mc Bz m~ m = 2, 1, 0, −1, −2 (degeneracy 5)
l=0
n = 1, l = 0 → ∆EN
=2 = 0 (non-degenerate)
4
(11)