# Gauge invariance of Fermi's golden rule

I am having some issues with gauge invariance of Fermi's golden rule. Say we have a system Hamiltonian for a particle in an electric field and some additional potential $V$ with $$H=(p-A(t))^2+\Phi(t)+V,$$ where $p$ is the momentum operator and the classical $A$ and $\Phi$ should reproduce a given classical time dependent electric field $E(t)$.Treating the electric field as a perturbation of the particle in the potential $V$ we have the unperturbed Hamiltonian $H_0=p^2+V$ and a time dependent perturbation Hamiltonian \begin{align}H_P=pA(t)+A(t)p+A^2(t)+\Phi(t).\end{align} If I treat this problem with time dependent perturbation theory I should be able to use Fermi's Golden rule with the matrix elements $\|\langle n'|H_P|n\rangle\|^2$. If I choose different gauges $H_P$ depends on this choice of gauge. This is the point which is not quiet clear to me. Will the results calculated by Fermi's golden rule be the same, independent of gauge or is this formula not gauge invariant? And if not, what would be the right/best choice of gauge using the formula? If it is gauge invariant, how do I see this?

• If you change $H_P$ and keep the same functions $\psi_n$, the expression $\|\langle n'|H_P|n\rangle\|^2$ will generally change. Perhaps there is some clever way to change $\psi_n$'s so this expression does not change, but I do not see it. – Ján Lalinský Jan 5 '15 at 23:42
• This is an old issue and in the literature there are cases where wrong conclusions were obtained by not realising that both operators and states have to be transformed. You may be interested in a paper by Aharonov and Au, Phys. Rev. A 20 (1979) 1553. – Urgje Jan 6 '15 at 11:36
• Thank you, the paper was very interesting and helpful. The matrix elements are equal, if one respects energy conservation. – Daniel Feb 7 '15 at 17:59