Hamiltonian eigenvalues in a transformed reference frame

Question

Under a time-dependent unitary transformation $V(t)$ of the state vectors $|{\psi}\rangle$

\begin{equation} |\psi'(t)\rangle = V(t) |\psi(t)\rangle \end{equation}

The Hamiltonian $H(t)$ has to transform as

\begin{equation} H' = V H V^{\dagger} - i \hbar V \dot{V}^{\dagger} \end{equation}

to preserve the form of the Schrödinger equation. $H'$ is Hermitian, and as such can be diagonalized.

If the original Hamiltonian has instantaneous eigenvectors $\{|n(t)\rangle\}_n$ of eigenvalues $E_n(t)$ such that

\begin{equation} H(t)|n(t)\rangle = E_n(t) |n(t)\rangle, \end{equation}

can one also find eigenvector and eigenvalues of the transformed Hamiltonian $H'$ starting from those of $H$?

I know that, in particular cases, it is possible to do that. I am wondering if there is a scheme to do it in the general case.

Answer 1

If you know the time dependent eigenvectors and eigenvalues of your original Hamiltonian, at any given time you could try using ordinary perturbation theory on the Hamiltonian $$V^\dagger H' V= H + i \hbar {V}^{\dagger}\dot{V}$$ treating the first part as your free theory and the second part as a perturbation.

In general, since $V$ and $\dot{V}$ are arbitrary, I don't think you can do any better than this. If it's a finite Hilbert space with $N$ vectors your perturbation is an arbitrary generator of $U(N)$ which means it's an arbitrary Hermitian matrix, so the form of the perturbation is completely general.

Answer 2

If we know eigenvectors of a Hamiltonian, we know a transformation that diagonilizes this Hamiltonian (a matrix, each column of which is an eigenvector). This transformation, like any other operator, can be transformed to the new basis, where it will be diagonalizing the new Hamiltonian.