# Time-Dependent Perturbation Theory & Completeness

1. May 29, 2015

### Runei

Hello!

I just want to make sure that I have understood the following argument the correct way:

For a given quantum system we take the hamiltonian to be a time-independent (and soluble) part, and a time-dependent part.

$\hat{H} = \hat{H_0} + H'(t)$

Now, the solutions to the unperturbed system are given by

$\Psi_n(x,t) = \psi_n(x) e^{-iE_nt/\hbar}$

And any solution to the system can be written as

$\Psi(x,t) = \sum\limits_n c_n \psi_n(x) e^{-iE_nt/\hbar}$

Argument:
When we go ahead and introduce the perturbed system, we can still write the solution to the system as a sum of the eigenfunctions of the unperturbed system, since these eigenfunctions represent a basis to Hilbert-space. And the solution to the perturbed system must also belong to Hilbert-space. Therefore:

$\Psi'(x,t) = \sum\limits_n c_n(t) \psi_n(x) e^{-iE_nt/\hbar}$

The expansion coefficients in this case will depend on time, since the full hamiltonian now also depends on time, but the central argument is that we can still write the solution we are looking for, as an sum of the solutions to the unperturbed system.

Is this correct or am I missing something essential? :)

2. May 29, 2015

### andresB

Yes, you can expand any vector using the eigenvectors of the unperturbed hamiltonian since they (are supposed at least) form complete basis for the Hilbert space.

3. May 29, 2015

### Runei

Thanks a lot, mate! :)