Download E5664: Spin-orbit interaction

E5664: Spin-orbit interaction
Submitted by: Snir Cohen
The problem:
An electron of mass M and charge e is located in a one-dimensional ring of radius R. The electron
has a spin-orbit interaction with the electric field that creates a charge Q that is located in the
center of the ring. It is recommended to use units of ~ =c= 1.
(1) Write the Hamiltonian H(x, p, S) of the electron.
(2) What are the eigenenergies of the electron?
(3) What is the velocity of the electron in each one of the energy states?
(4) What is the magnetic flux Φ through the center of the ring that will eliminate the effect of Q
on an electron that has spin up?
Express all the answers by the given data alone.
The solution:
(1) The general Hamiltonian of a spin-orbit interaction:
H=
p2
+ HSO
2M
The spin-orbit term:
HSO = −
e
~ = − eQ pSz
(E × p~) · S
2
2M
2M 2 R2
Where the electric field
E=
Q
R2
Hence
1
H=
2M
eQ
2
p −
pSz
M R2
We complete the square and get
2
2 2 2
1
eQ
1
e Q Sz
H=
Sz −
p−
2
2M
2M R
2M 4M 2 R4
(2) From periodic boundary conditions:
k=
2πn
n
=
, n = integer
L
R
Where
L = 2πR
For an electron
m=±
1
2
1
Hence the eigenenergies are:
En,±
1
=
2M
n
eQ
∓
R 4M R2
2
−
e2 Q2
32M 3 R4
(3) The definition of the velocity operator is implied by the ”rate of change fomula”:
eQ
1
2
p −
pSz
v̂ = i[H, x̂] , H =
2M
M R2
1
eQ
eQ
1
2
v̂ = i ·
[p̂ , x̂] −
−2ip̂ + i ·
Ŝz [p̂ , x̂] = i ·
Ŝz
2M
M R2
2M
M R2
eQ
1
p̂ −
⇒ v̂ =
Ŝz
M
2M R2
Hence the velocities are:
1 n
eQ
vn,± =
∓
M R 4M R2
(4) Spin up:
m=
1
2
The effect of a magnetic flux through the center of the ring:
En,+
1
=
2M
n
eQ
eΦ
−
−
2
R 4M R
L
2
+ const
eΦ
eΦ
=
L
2πR
We demand
−
eQ
eΦ
−
=0
4M R2 2πR
So
Φ=−
πQ
2M R
2