# How to solve the time-dependent Schrödinger equation

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Hi

This is a general question about the possibilities to solve the time dependent Schrödinger equation:

My goal is to solve this PDE for $f:[-1,1] \times \mathbb{R}\rightarrow \mathbb{C}$  $$\partial_t f(x,t) = -\partial_x^2 f(x,t) + g(t)V(x)f(x,t).$$

So I am assuming that we have a time-dependent potential that consists of a function $g \in L^2 \cap C^{\infty}$ and a potential $V \in C^{\infty}$

I would consider this PDE to be solved if I get two ODEs just depending on either $x$ or $t$. ( Though separation of variables does not work directly a good substitution or integral transform could do it)

Furthermore, I want to specify the wavefunction $f(.,t_0)$ at a particular point in time $t_0$ as a boundary condition. The other boundary conditions shall be time-independent.

As an example, we could look at the PDE where $f(.,0)$ is given and at some point in time $t_0>0$ we apply a short potential pulse $V$ to the wavefunction.

$$\partial_t f(x,t) = -\partial_x^2 f(x,t) + \delta(t-t_0)V(x)f(x,t).$$

Can we say anything in this general setting about ways to solve this PDE.

Of course, perturbation theory or any standard ways of approximations are not interesting to me.

If anything about my question is unclear, please let me know.

asked Jun 23, 2014
edited Jun 23, 2014

Have you tried to express the general solution via the retarded Green's function to see the general properties of the solution? After a pulse force, the solution becomes a new superposition of the "old" Hamiltonian eigenfunctions, I guess.

In a "pulse" case, the "perturbation" part can be expressed via $f(x,t_0)$, i.e., via known function (because of delta-function). So the exact explicit solution exists.

In case of arbitrary $g(t)$, the superposition of eigenfunctions will have time-dependent coefficients.

@Vladimir: I don't understand how your answers helps me to find the solutions? Do you know how to get the solution (in the particular / general ) case? If not, I think it is better if you delte your answer and post it as a comment.

When $V(x)=0$, your solution is a superposition like $f(x,t)=\sum_n A_n e^{-\lambda_n t}\sin(k_n x)$ where $A_n$ can be found from the initial condition. The Green's function can also be expressed via a spectral sum (but not only!):$$G(x,x'|t,t')=\sum_n e^{-\lambda_n (t-t')}\sin(k_n x) \sin(k_n x'),$$ so the integral with the retarded Green's function can be calculated and new coefficients can be found in case of a "pulse" potential.

In a general case you can obtain an infinite system of algebraic equations for new coefficients, i.e., no explicit analytical solution.

Are you  aware of the existence of a general solution? I know of solutions only in two limits -- the sudden and the adiabatic cases. Why are you restricting time to take values in $[-1,1]$? Is $g[\pm1=0$?

No, I am not aware of the existence. I am restricting myself to the set $[-1,1]$ cause I want to treat a Hamiltonian on a compact set (for mathematical reasons). $g$? $g$ depends on time, so why should it be specified on $\pm 1$? (In case you mean $V$, the answer is also no.)

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