Time Varying Potential, series solution

Question

Suppose we have a time varying potential $$\left( -\frac{1}{2m}\nabla^2+ V(\vec{r},t)\right)\psi = i\partial_t \psi$$ then I want to know why is the general solution written as $\psi = \displaystyle\sum_n a_n(t)\phi_n(\vec{r})e^{-iE_n t} $ Particularly, why do we get a time dependent coefficient $a_n(t)$. This confuses me because when we have a time independent potential, then we use variable separation and usual method to get the general solution $$\psi = \displaystyle\sum_n a_n\phi_n(\vec{r})e^{-iE_n t}$$ However, the time varying counterpart cannot be reduced this way by variable seperation.

EDIT: I could not find a free preview of the book I am using, however, the lectures here for example, use the same solution.

Answer 1

The basis functions $\phi_n(\vec{r})$ and the energies $E_n$ are the solutions of the stationary Schrödinger equation: $$ \left( -\frac{1}{2m}\nabla^2+ V_0(\vec{r})\right)\phi_n(\vec{r}) = E_n \phi_n(\vec{r}) $$ If the Hamiltonian depends on time one even can not write this equation. But the set of functions $\phi_n(\vec{r})$ is a full basis in the Hilbert space. So one always can expand any function (from this space) over this basis.

The snapshot of the wavefunction $\psi(\vec{r},t)$ at the moment $t$ is just a function of coordinates and the element of this Hilbert space. So we can expand it: $$ \psi(\vec{r},t) = \sum_n b_n(t) \phi_n(\vec{r}) $$ If the Hamiltonian do not depend on time the expansion coefficients can be easily derived from the general Schrödinger equation (the one with the time derivative): $$ b_n(t) = a^{(0)}_n e^{-iE_nt} $$

In the case of time-dependent potential these coefficients are usually considered as unknown functions of time: $$ b_n(t) = a_n(t) e^{-iE_nt} $$

The perturbation theory is used to find the approximation for these functions.

Answer 2

1) What OP is looking at is known as time-dependent perturbation theory. Here the energies $E_n$ are eigenvalues for the unperturbed time-independent Hamiltonian $H^{(0)}$. The full Hamiltonian is

$$H(t)~=~H^{(0)}+V(t).$$

2) Imagine for a second that the potential $V$ is time-independent and commutes with $H^{(0)}$. Let $v_n$ be the eigenvalues of $V$. In the time independent case, the wavefunction solution is then of the form

$$\psi(t,\vec{r}) ~=~ \displaystyle\sum_n c_n\phi_n(\vec{r})e^{-i(E_n+v_n) t} ~=~ \displaystyle\sum_n \left(c_n e^{-iv_nt}\right) \phi_n(\vec{r})e^{-iE_n t}.$$

3) For general time-dependent perturbations $V(t)$, it is hence natural to expect that the coefficients $a_n(t)$ in the eigenfunction expansion

$$\psi(t,\vec{r}) ~=~ \displaystyle\sum_n a_n(t)\phi_n(\vec{r})e^{-iE_n t} $$

could depend on time $t$, cf. OP's question(v2). Here $\phi_n(\vec{r})$ denote eigenfunctions for the unperturbed problem,

$$ H^{(0)}\phi_n(\vec{r})~=~E_n\phi_n(\vec{r}).$$

Answer 3

It is simply a matter of definition. If the time-dependent coefficients can be reasonably found from the Scroedinger equation, then it is a solution for your time-dependent wave function. It is introducing new variables $a_n (t)$ and determining them from the exact equation. It can always be done.