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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.

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  • $\begingroup$ Perhaps I didn't understand your question. From what I understood you seems to be contradicting yourself. So if the particular transformation preserves Schrodinger eqn, now since Schrodinger eq is by definition an eigen equation, it will definitely have eigen values(equivalent to diagonalising the matrix). But my question is something different, is your transformation still keeps the Hamiltonian Hermitian? I cant see that explicitly. $\endgroup$ Commented Mar 9, 2021 at 8:10
  • $\begingroup$ Is $V(t)$ unitary? $\endgroup$
    – Noiralef
    Commented Mar 9, 2021 at 9:59
  • $\begingroup$ @TheImperfectCrazy I am wondering if there is a prescription to actually compute the eigenvalues of the transformed Hamiltonian. H' is Hermitian, you can show it e.g. by expressing V as $\exp{i A}$, with A a Hermitian operator. $\endgroup$
    – ablagi
    Commented Mar 11, 2021 at 17:49
  • $\begingroup$ @Noiralef Yes, let's assume it is. I edited the question, thank you. $\endgroup$
    – ablagi
    Commented Mar 11, 2021 at 17:49
  • $\begingroup$ If $V$ is unitary, then $H'$ is Hermitian and can thus always be diagonalized, as you just wrote. However, I am not aware of an easy way to determine the eigenvectors of $H'$ starting with those of $H$. $\endgroup$
    – Noiralef
    Commented Mar 11, 2021 at 22:36

2 Answers 2

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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.

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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.

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