# Eigenvalues of time dependent Hamiltonian

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

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): $$\Big[-\frac{\hbar^2}{2m}\nabla^2+V({\bf r},t) \Big]\psi({\bf r})\varphi(t)=i\hbar\frac{\partial}{\partial t}[\psi({\bf r})\varphi(t)]$$

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

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

• why exactly can you not separate $r$ and $t$? $r$ is not a position, it's a coordinate so it doesn't depend on $t$.... Mar 2, 2017 at 0:30
• 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 Mar 2, 2017 at 1:13
• of course... I should have realized this is what you meant... my bad. Sorry. Mar 2, 2017 at 1:26
• Have you reviewed Dirac's approach? Mar 2, 2017 at 2:03
• 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/… Mar 2, 2017 at 4:16