Consider the Schrödinger equation $H\Psi=i\hbar\frac{\partial \Psi}{\partial t}({\bf r},t)$. The hamiltonian $H$ is: \begin{equation} H=-\frac{\hbar^2}{2m}\nabla^2+V({\bf r},t) \end{equation} And the equation is: \begin{equation} \Big[-\frac{\hbar^2}{2m}\nabla^2+V({\bf r},t) \Big]\Psi({\bf r},t)=i\hbar\frac{\partial \Psi}{\partial t}({\bf r},t) \end{equation}

If $V$ is time-independent, then $H$ too. In this case the wave function can be factored as $\Psi({\bf r},t)=\psi({\bf r})e^{-i\omega t}$, and then $H\Psi=E\Psi$, so the eigenvalues of $H$ are the energies $E$.

If $V$ is time-dependent, we can't separate the variables ${\bf r}$ and $t$ like before, but $H$ is still an hermitian operator and it must have certain eigenvalues.

My question is: What are the eigenvalues of $H$ when it depends on time? (and if they are the energies, why?)

EDIT: If I try to separate the variables, I have $\Psi({\bf r},t)=\psi({\bf r})\varphi(t)$, putting this in Schrödinger equation (this is the method of separation of variables): \begin{equation} \Big[-\frac{\hbar^2}{2m}\nabla^2+V({\bf r},t) \Big]\psi({\bf r})\varphi(t)=i\hbar\frac{\partial}{\partial t}[\psi({\bf r})\varphi(t)] \end{equation}

\begin{equation} -\varphi(t)\frac{\hbar^2}{2m}\nabla^2\psi({\bf r})+V({\bf r},t)\psi({\bf r})\varphi(t)=i\hbar\psi({\bf r})\frac{\partial \varphi}{\partial t}(t) \end{equation} Dividing by $\Psi({\bf r},t)=\psi({\bf r})\varphi(t)$:

\begin{equation} -\frac{\hbar^2}{2m}\frac{1}{\psi({\bf r})}\nabla^2\psi({\bf r})+V({\bf r},t)=i\hbar\frac{1}{\varphi(t)}\frac{\partial \varphi}{\partial t}(t) \end{equation} But I can't continue with the method, because the left side depends on $t$ (it doesn't depend on ${\bf r}$ only).

  • $\begingroup$ why exactly can you not separate $r$ and $t$? $r$ is not a position, it's a coordinate so it doesn't depend on $t$.... $\endgroup$ Commented Mar 2, 2017 at 0:30
  • $\begingroup$ I can't apply the method of separation of variables to factor the wavefunction in two wavefunctions (one depending only on $r$ and the other depending only on $t$), I edited the question whit some details $\endgroup$
    – slaaidenn
    Commented Mar 2, 2017 at 1:13
  • $\begingroup$ of course... I should have realized this is what you meant... my bad. Sorry. $\endgroup$ Commented Mar 2, 2017 at 1:26
  • $\begingroup$ Have you reviewed Dirac's approach? $\endgroup$ Commented Mar 2, 2017 at 2:03
  • 1
    $\begingroup$ You could solve the differential equation if you have to, but a better method would be to expand the states into a linear superposition of stationary States. This approach is given here in detail. ocw.mit.edu/courses/chemistry/… $\endgroup$
    – Sumant
    Commented Mar 2, 2017 at 4:16


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