Why Is the Scalar Potential Ignored in Coulomb Gauge?

[eoghan]

Hi there,
I'm studying the interaction of one electron atom with an electromagnetic field. In every textbook the starting point is the hamiltonian of the system containing the scalar potential and the vector potential. But then the scalar potential is ignored and I don't understand why.
I've read that in the coulomb gauge I can choose the scalar potential to be 0, why??

 

[vela]

In electromagnetism, you're allowed to perform the gauge transformation

\textbf{A} \rightarrow \textbf{A}'= \textbf{A}+\nabla\lambda

\Phi \rightarrow \Phi'= \Phi - \partial_t \lambda

The situation you're describing is a combination of the Lorenz and Coulomb gauges. In the Lorenz gauge, you have

-\partial_t \Phi + \nabla \cdot \textbf{A} = 0

The Lorenz gauge is only a partial gauge fixing, so you still have the freedom to do an additional gauge transformation. In particular, if you choose the gauge function λ such that

\Phi = \partial_t \lambda

you will have, after the gauge transformation, \Phi' = 0 and \nabla\cdot\textbf{A}'=0, which is the Coulomb gauge.

 

[eoghan]

Thank you!