We can use the Hamiltonian as a starting point for path integrals. This adds
another to our list of temporal quantization formalisms. And we can derive the
Feynman path integral from the Hamiltonian path integral, thus showing all four
formalisms equivalent.
Development
After
[Schulman-1981],
[Kleinert-2004]
As before, simpler using Lorentz gauge, as
p
and
A
can be passed through each other:
Hamiltonian:
Where the operators are:
The kernel is (formally):
We want to get every operator next to one of its eigenfunctions, whether on left
or right. Some slicing will be required.
Break up the path from start with
N
cuts into
N+1
pieces by inserting value for one:
Getting:
Identifying:
Look at one piece:
Break Hamiltonian up into two pieces:
Break up exponential into two pieces, error of order
squared:
Insert another expression for one between the two pieces:
Getting for the infinitesimal kernel:
This implies one integration between each pair of
x
's. Counting the
x
's at the ends, we have N+1
p
integrations, one more than the number of
x
integrations.
Given:
We have for the integrand of the infinitesimal kernel:
Measure is:
Result is:
We have a midpoint rule for
A
, without explicitly inserting it.
Rewrite as Hamiltonian:
Hamiltonian path integral has integrals over
p
and
q
; Feynman path integral only over
q
; integrate out the
p
integrals to derive the Feynman path integral and close the
circle.
Use kernel from above.
A typical
p
integral is:
With definition:
Get:
Expand argument of exponential:
Normalization constant
C
defined by:
The
term almost cancels against the
term above:
Difference between vector potential at sequential slices:
Difference between time/space coordinates of order square root of
:
(
see derivation of
Schrödinger equation above) so difference in vector potential terms of order
:
Since this is already being multiplied by a factor of
, it is of order
squared, and drops out.
Therefore we get the previous coordinate space kernel back:
As in
Explicit
expression.