Gauge Transformation Quantum Electrodynamics

1. Apr 17, 2015

Juan Carlos

It's well known when if we are working on problems related to particles in presence of an electromanetic field, the way we state the problem can be done using the next Hamiltonian:
$H=\dfrac{(p-\frac{e}{c}A)^2}{2m} +e \phi$ where the only condition for A is: $\vec{\nabla } \times \vec{A} =\vec{B}$
So we have this "gauge freedom" under (of course) the above condition .
What happens when are studying two or more partcles in presence of the electromagnetic field.

Naturally:
$H=\dfrac{(p_{1}-\frac{e}{c}A)^2}{2m} +\dfrac{(p_{2}-\frac{e}{c}A)^2}{2m} +e \phi$ where the only condition for A is: $\vec{\nabla } \times \vec{A} =\vec{B}$

The question is: can we relax the condition of one unique gauge for all the particles?

Just saying:
can we select two different gauges? obviously with the condition : $\vec{\nabla } \times \vec{A} =\vec{B}$

I've been working on this, I satetd two different gauges and I can say that there is no difference in terms between the equations of motion having the same gauge. (Hamilton)

In particular I've done the same for the correspoding quantum problem obtainig the heisemberg equations, having the same result: no difference.

I'm interested on the quantum problem because having two different gauges could provoque loosing the symmetry of a Hamiltonian given.

2. Apr 19, 2015

Einj

When you fix a gauge you do that for the electromagnetic potential, which is only one in your problem. This shouldn't be influenced by how many particles you have. What do you mean by fixing two different gauges?