There is a question in SE about the fact that the Hamiltonian isn't invariant under the EM gauge transformations. I wanted to ask about its consequences here.
I know that in general, the Hamiltonian and wave functions would transform under the gauge transformation in such a way that the Schrödinger equation remains true. Specifically, if under a gauge transformation $H$ goes to $\tilde{H}$ and $\psi$ goes to $\tilde{\psi}$, We still have: $$ \tilde{H} \tilde{\psi}=i\hbar\partial_t\tilde{\psi}$$
Also, I know under a time-independent gauge transformation, if $\psi$ is a eigenstate of Hamiltonian with energy $E$, the $\tilde{\psi}$ would be the eigenstate of $\tilde{H}$ with the same energy. So the energy spectrum of Hamiltonian is invariant under the time-independent gauge transformations. But what confuses me, are the time-dependent gauge transformations. It can be seen that under the gauge transformation $$(\phi,\mathbf{A})\to(\phi+\partial_t \lambda,\mathbf{A}-\nabla \lambda)\qquad (1),$$ the eigenvalue equation for the Hamiltonian transforms to: $$\tilde{H}\tilde{\psi}=E\tilde{\psi}+\partial_t\lambda(x,t) \tilde{\psi}\qquad (2)$$ The derivation of this relation is appended at the end of the question. So as it can be seen $\tilde{\psi}$ isn't anymore the eigenstate of the Hamiltonian and it seems that the energy spectrum would also change. I mean that even the energy differences would change under this gauge transformation.
This somehow looks odd to me. Take for example the hydrogen atom. On one hand, we know that the $13.6$ eV can be measured in experiment, or the energy differences between various level can be observed with spectroscopy techniques, but on the other hand they are somehow dependent on the gauge we choose. Take into account that gauge transformations doesn't represents a physical change in the system like changing the observer, but it's just a reformulation of the same problem. It seems that even the orbital shapes would change under time-dependent gauge transformations.
Is the above reasoning correct? Is energy deference an observable physical quantity or not? or maybe only the eigenvalues of time-independent Hamiltonians represents the energy levels of the system and we should look at time-dependent Hamiltonians as just the time evolution generator and do not relate its spectrum to energies of the system. Can someone clarify the situation?
Proof of Eq. 2:(I have set $\hbar=e=1$ for simplicity)
Suppose that $H=\frac{(p-A)^2}{2m}+\phi$ is the systems Hamiltonian in the original gauge and $\psi$ is one of its eigenvectors with energy $E$, so in the position representation: $$H\psi=[\frac{(-i\nabla-A)^2}{2m}+\phi]\psi(x)=E\psi(x),\qquad (3)$$ Now consider the gauge transformation which is given by Eq. (1), under which the Hamiltonian transforms to: $$\tilde H=\frac{(p-A+\nabla \lambda)^2}{2m}+\phi+\partial_t \lambda$$ and also the $\psi$ wave function would go to: $$\tilde \psi(x,t)=\exp{[-i\lambda(x,t)]}\psi(x,t)$$ It is easy to see that: $$(-i\nabla-A+\nabla \lambda)\exp{[-i\lambda(x,t)]}\psi(x,t)=\exp{[-i\lambda(x,t)]}(-i\nabla-A)\psi(x,t),\qquad (4)$$ If we operate the $\tilde H$ on the $\tilde \psi$ form left and use Eq. 4 twice: $$\tilde H \tilde \psi =\exp[-i\lambda(x,t)]\Big(\frac{(-i\nabla-A)^2}{2m}+\phi\Big)\psi(x,t) +\partial_t \lambda\exp[-i\lambda(x,t)]\psi(x,t)$$ but the expression in the big parentheses is just the original Hamiltonian $H$ and according to Eq. 3, it can be replace by $E$, So: $$\tilde H\tilde \psi=E\tilde\psi +(\partial_t\lambda)\tilde\psi $$ which completes the proof.
