E6070: Perturbation Theory for a ring in an Electric Field

E6070: Perturbation Theory for a ring in an Electric Field
Submitted by: Jacob Hanouna and Sarit Nagar
The problem:
A particle with mass M and charge e is placed in a one dimensional ring with radius R.
A homogenous electric field E is being induced parallel to the ring’s plane.
(1)
(2)
(3)
(4)
(5)
Write the Hamiltonian H(θ, p) of the system.
What are the symmetries the system has without/with the perturbation.
Write the matrical presentation of the Hamiltonian with the suitable basis for the perturbated system.
Calculate the ground state energy n = 0 up to the second order in the perturbation.
Calculate the excited states energies n > 1 up to the second order in the perturbation.
The solution:
(1) The Hamiltonian is:
Ĥ =
p̂2
p̂2
+ V̂ (θ) =
+ eER cos θ
2m
2m
(2) Without the perturbation the Hamiltonian has symmetry with respect to rotations and reflections.
In the presence of the electric field, only the reflection remains.
(3) We will use the basis that complies with the reflection symmetry:
|n = 0i 7→ √
1
1
7−→ √
2π
2πR
1
1
cos(kn x) 7−→ √ cos(nθ)
π
πR
1
1
sin(kn x) 7−→ √ sin(nθ)
|n, −i 7→ √
π
πR
|n, +i 7→ √
Notice that describing the basis using the relation x = Rθ changes the normalization as mentioned
on the right side.
There is a degeneracy between the even states and the odd states, though they are not ”coupled”
to each other. Therefore, the Hamiltonian will be of the form:

ε0

ε1


ε2


H 7→ 





where εn =
2
kn
2M
=
..
.
ε1
ε2

(+)
(+)
V00 V01
V02
 (+)
(+)
(+)
 V10
V12
  (+) V11
(+)
(+)
 V
V21
V22
  20
  .
..
..
 +  ..
.
.
 
  (−)
0
0
 V10
  (−)
 V20
0
0

..
..
..
..
.
.
.
.

n2
.
2M R2
1
(−)
(−)
···
···
···
..
.
V01
0
0
..
.
···
···
..
.
V11
V12
(−)
(−)
V21
V22
..
..
.
.
(−)
V02
0
0
..
.
(−)
···
···
···
..
.










··· 

··· 

..
.
V00 = hn = 0|V (θ)|n = 0i = 0
(+)
V0m
(+)
Vnm
eER
= hn = 0|V (θ)|m, +i = √
2π
Z
0
2π
eER
cos(θ) cos(mθ)dθ = √ δm,1
2
Z
eER 2π
cos(nθ) cos(θ) cos(mθ)dθ
= hn, +|V (θ)|m, +i =
π 0
Z 2π
eER
=
[cos((n − m − 1)θ) + cos((n − m + 1)θ)] dθ +
4π
0
Z 2π
+
[cos((n + m − 1)θ) + cos((n + m + 1)θ)] dθ
0
eER
eER
=
[δn,m+1 + δn,m−1 + δn,−m+1 + δn,−m−1 ] =
[δn,m+1 + δn,m−1 ]
2
2
The last transition is due to the fact that n, m > 0.
Z
eER 2π
= hn = 0|V (θ)|m, −i = √
cos(θ) sin(mθ)dθ = 0
2π 0
Z
eER 2π
eER
(−)
Vnm = hn, −|V (θ)|m, −i =
sin(nθ) cos(θ) sin(mθ)dθ =
[δn,m+1 + δn,m−1 ]
π 0
2
√




2
0
0
·
·
·
0
0
0
0
0
·
·
·
√
 2 0 1 0 · · ·
0 0 1 0 · · · 




 0
(−)
eER 0 1 0 1 · · ·
1 0 1 · · ·
= eER
V
=




n,m
2 
2 


0
0
1
0
·
·
·
0
0
1
0
·
·
·




.. .. .. .. . .
..
..
.. .. . .
.
.
. . . .
.
.
. .
(−)
V0m
(+)
Vn,m
[0]
[1]
(4) Notice that En = εn and En = Vnn = 0 Therefore, the ground state energy up to the second
order in the perturbation is simply
[2]
E0 = E0 =
X
m6=0
|V0m |2
0−
m2
2M R2
= −2M e2 E 2 R4
(+)
(−)
(5) For the excited states n, m > 1 we have Vn,m = Vn,m
En[2]
X |Vnm |2
M e2 E 2 R 4
1
1
M e2 E 2 R 4
=
=
+
=
εn − εm
2
n2 − (n − 1)2 n2 − (n + 1)2
4n2 − 1
m6=n
Therefore the energy up to second correction is:
En =
n2
M e2 E 2 R 4
+
2M R2
4n2 − 1
2