Download Lagrangian and Hamiltonian forms of the Electromagnetic Interaction

Lagrangian and Hamiltonian forms of the Electromagnetic Interaction
(version 2.0 : Nov 20, 2008)
Quantum mechanics is formulated in terms of of energies and potentials rather
than forces, and the Hamiltonian formulation provides the basic framework. This
describes electromagnetic interactions in terms of the vector and scalar potential rather
than the E and B fields.
I.
The Lagrangian for electromagnetic interactions.
The electromagnetic Lagrangian is given by
e
L  12 m 2  e   A
(1.1)
c
which leads to the correct equations of motion in the usual way. [I’m using “ e ” as the
symbol for charge, rather than q, to avoid confusion with generalized coordinates. The
charge of the electron is e .] Hamilton’s equations for this Lagrangian become (using
implied summation for repeated indices):
d  L  L
0


dt  q j  q j
(1.2)
d
e  
e


 m j  Aj    e j   j k Ak    0
dt 
c  
c

 m
d j
dt
  e j 
e dAj e
  j k Ak 
c dt
c
(1.3)
But since A j in this equation is a function of the particle’s position, the total time
derivative of A j depends on the particle’s velocity in addition to any explicit time
dependence in the field. Thus
dAj
Aj Aj xk
Aj



 k  k  Aj
dt
t xk t
t
Substituting this into (1.3) gives
m
d j

1 Aj  e
 e  j 
   j k Ak   k k  Aj 
dt
c t  c 

(1.4)
(1.5)
Using the “BAC-CAB” rule, A  ( B  C )  B( A C )  ( B A)C , this becomes
d

1 Aj 
(1.6)
m j  e  j 
  e     A j
dt
c t 

Since the expressions in brackets are just the EM fields expressed in terms of  and A ,
we have derived the force equation from the Lagrangian:
m
d
e
 eE    B
dt
c
(1.7)
[Note added in proof:
The derivation of (1.6) from (1.5) can be done somewhat more rigorously using
the definition of cross-product,
3
A  B   Ai B j ijk kˆ  Ai B j ijk kˆ
(implied summation on repeated indices)
(1.8)
i , j 1
This also provides a good exercise in managing indices. It’s easier if we work backwards
from the third term in (1.6):
    A   i   A  ijk kˆ  i   A   j   ijk kˆ


j




 A

 A

(1.9)
 i     j   ijk kˆ  i     j   ijk kˆ
 x

 x

 A A 

i  i  k   ijk kˆ
 xk xi 
Note that in the last line, the first term corresponds to   k ,   i and the second term to
  i,   k . These are the only possibilities, since ijk must all be different (otherwise
 ijk  0 ). The student should convince himself/herself that the last line of (1.9) is indeed
identical to the last term in (1.6); this may not be immediately obvious because the
dummy indices are all different.]
2.
The Hamiltonian for electromagnetic interactions.
In the Hamiltonian formulation, we define the Hamiltonian in terms of the
Lagranian by
H   pi qi  L
where the canonical momentum pi is defined by
L
pi 
qi
(2.1)
(2.2)
and all quantities in (2.1) are expressed in terms of the conjugate variables  qi , pi  . In
the case of the electromagnetic Lagrangian, this leads to
L
e
pi 
 mi  Ai
(2.3)
i
c
Note that pi is different from mi , but that is not really surprising. When the coordinate
system is not Cartesian, the canonical momentum may not have the dimensions of a
linear momentum (e.g., angular momentum). And when velocity-dependent potentials
are used, then even in Cartesian coordinates the canonical momentum may be different
from the usual mechanical momentum.
The Hamiltonian can be derived from the Lagrangian following the usual
prescription:
L
H  qj
 L  qj pj  L
q j
(2.4)
H is always expressed a function of the generalized momenta p j , so we must replace  j
in L with p j using (2.3). In the case of the electromagnetic Hamiltonian, this yields
2
1 
e 
 pk  Ak   e
c 
k 2m 
The equations of motion can be derived in the usual way:
2
H
  1 
e  
1
e 
xj 

p

A


k  
 k
 pk  Ak   jk
p j
p j  2m 
c  
m
c 
1
e 

 p j  Aj 
m
c 
which is equivalent to (2.3), and
2

H
  1 
e 
pj  


 pk  Ak   e 
x j
x j  2m 
c 

H 
1
e   e Ak
   pk  Ak  
m
c   c x j


  e
x j

(2.5)
(2.6)
(2.7)
To put this into the more familiar form of equation (1.7), we need to convert from p j
back to  j using (2.3) [on both sides of the equation, and being careful of indices] :
m
d j
dt
 m


 e A
1
 mk   k
m
 c x j


  e
x j

 
 e Ak
e dAj
 k 
 c x
c dt
j



  e
x j

e dAj
c dt
d j
dt
 1  A x Aj    e Ak 

 e   j  k
  e
    k
t xk    c x j 
x j
 c  t
 1 Aj   e  Ak Aj
 e 


  k 
 c t x j  c  x j xk
which is the same as equation (1.5).



(2.8)