```notes13-- Interactions of electrons with an electromagnetic field
We will use Gaussian units, then the electric and magnetic fields satisfy Maxwell
equations

B  0

  E  4

 1 B
x E 
0
c t

 1 E 4 
x B 

j
c t
c

In terms of potentials  and A , the Schrodinger equation for the interaction of an
electron with charge -e and mass  with the EM field is



  / i   e / c Ar, t  2

 r, t 
 e ( r, t )  r, t   i

2
t


where


B  xA
(13.1)


1 A
 
and E  
c t
13-1. Gauge transformation of the Schrodinger equation
Note that under the gauge transformation

1 f
'   
A'  A  f ,
where f=f(x,t)
c t
the electric and magnetic field remain invariant.
For the Schrodinger equation to remain invariant under the Gauge transformation, the
wave function transforms as
' ( x, t )  ( x, t )ei ( x ,t )
(13.2)
where


e
( x, t ) 
f ( x, t )
c
For the time-independent fields, we can write
 ( r, t )  e iEt /  ( r )
If we choose the gauge

  Ar   0
we then get the time-independent equation
(13.3)
(13.4)

2 2
ie 
e2
 
A   
A2  e ( r )  E
2
2
c
2c
(13.5)
13.2. Constant magnetic field
For a constant magnetic field B, we can take

1 
A  r B
2
The Schrodinger equation takes the form
(13.6)
2 2
ie 
e2
 
A   
A2  e ( r )  E
2
c
2c 2
Manipulating the 2nd term above,

(13.7)
ie  
ie  
r  B    
B  r  
2c
2 c
e   
e  

B  r   
B  L
2 c
i
2 c
(13.8)
Manipulating the 3rd term,
 2
e2   2
e2
e2 B 2 2
2 2
r

B


r
B

r

B


x  y 2  (13.9)
2
2
2
8c
8c
8c







We will note that for ordinary magnetic field (13.8) and (13.9) are quite small, and thus
the two terms can be treated as perturbation. (13.8) is for the normal Zeeman effect, and
(13.9) is for the quadratic Zeeman effect.
For a free electron in a static magnetic field, the electron is confined by the potential
given by (13.9). Note that this is a 2D harmonic potential, and the allowed energy levels
are given by equal spacing according to the harmonic oscillator solutions. These levels
are called Landau levels.
Small B-field-- normal Zeeman effect
If B-field is small, equation (13.7) is solved by the perturbation theory
with
H1 
e
2 c
BL z  L Lz
where
L 
e
2 c
B
is the cyclotron frequency.
If the unperturbed Hamiltonian has spherical symmetry, the eigenstates are given by
unm such that the perturbation energy for such a state is given by
E n1m   L m
This is called normal Zeeman effect. Note that in the current model, we have not
included the spin of the electron. This will be done later.
The perturbation H1 can be viewed as the interaction of a magnetic dipole due to the
orbital motion of an electron in a magnetic field. The dipole moment is given by



e 
e
ML 
L
g L L  gL
2mc
2mc
where we have used m for the mass of the particle. The second equality is to show the
general relation between the magnetic dipole moment and angular momentum. In the
present case, g L =1. The last equality shows that  is an elementary unit of the magnetic
dipole moment. The last expression can be generalized to any particle and the orbital
angular momentum can be replaced by the spin. In that case, the g can take different
values.

Using M L , the interaction energy of such a magnetic dipole in a B field is given by
 
e
H1   M L  B 
BL z
2mc
If the B-field is along the z-direction.
In the case of electron spin, the magnetic dipole moment is given by


e
MS 
gS S
2mc
where

 ... ) is the g-factor for the electron, taken to be 2 for most of our
2
purpose. It is a fundamental property of an electron and its value can be calculated using
quantum electrodyamics (QED).
g S =2 (1+
13.3. Flux Quantization and the Aharanov-Bohm effect
The gauge invariance of the electromagnetic fields has a number of important real
physical consequences. We talk about two here. The other important one is the quantum
Hall effect which is a little bit more complicated.
(a) Flux quantization
From (13.1), the time-independent equation is
1 
e 
(   A( r )) 2 ( r )  V ( r )( r )  E( r )
2 i
c
This equation can be solved by writing
( r )  0 ( r )e  i ( e / c ) f ( r )
where
(13.10)
(13.11)
1  2
( ) 0 ( r )  V ( r )0 ( r )  E0 ( r )
2 i
and
r

  
f ( r )   dr '  A( r ' )
P
The line integrals along two different paths would have the difference
r
  
d
r
 '  A(r ' ) -
 

 
  
  
d
r
'

A
(
r
'
)


'
x
A
(
r
'
)

d
S

B
=
d
r
'

A
(
r
'
)


  dS  

1
2
r
S

This if the line integral encloses an area where the magnetic flux is zero, f (r ) obtained
from the two paths are identical. If the magnetic field is not identically zero, then for the
wavefunction to be single-valued, the phase factor from the two different paths should be
equal to 2n  , or
e
  2 n
c
and  
2c
n, n  0,1,2,...
e
(13.12)
This is the magnetic flux quantization-- a direct consequence of Gauge invariant.
Example: Consider a magnet having the shape of a donut. At normal temperature the
magnet is a normal metal and the magnetic flux lines can penetrate the hole as well as the
metal. At low temperature where the metal becomes a superconductor. For the
superconductor, the so-called Meissner effect says that the magnetic field lines cannot
penetrate the superconductor, i.e., B=0 inside the superconductor. The flux quantization
rule says that the total magnetic flux penetrating inside the hole should be governed by eq.
(13.12). Actual experiment shows that the electronic charge e should be replaced by 2e.
The reason is that the charge carrier in a superconductor is a Cooper pair-- two electrons


with momentum p and  p coupled together to form an elementary unit.
Aharanov-Bohm effect:
Schematically, electrons following the two paths shown in the diagram will have
interference fringes depending on the total magnetic flux enclosed. The additional phase
e
 , in addition to the phase from the path difference. By
difference is given by
c
changing the magnetic field the interference fringe will differ.
Homework 13
13.1. Let us estimate if the 2nd term in (13.7) can indeed be treated as perturbations.
Consider the 2p state of atomic hydrogen (neglect spin). Estimate the ratio of the energy
from the second term with respect to the 2p state energy for a magnetic field strength of
100 Gauss. Be careful that we are using the Gaussian units here so make sure you have
the B-field in that units.
13.2. A particle of mass M is attached to a rigid massless rod of length R whose other end
is fixed at the origin. The rod is fixed to rotate about its fixed point.
(a) Show that the Hamiltonian is given by

L2
1   2
H

( Rxp )
2I 2I
where I=MR2.
(b) If the particle carries charge q, and the rotor is placed in a constant magnetic field B
along the z-direction, what is the modified Hamiltonian? (Follow the example in this
chapter)
(c) For small B, what is the energy spectrum?
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