How to Determine Expansion Coefficients for a Wavepacket in a Periodic Box?

[raintrek]

I'm trying to get my head around the idea of expansion coefficients when describing a wavefunction as

\Psi(\textbf{r}, t) = \sum a_{n}(t)\psi_{n}(\textbf{r})

As I understand it, the expansion coefficients are the a_{n} s which include a time dependence and also dictate the probability of obtaining an eigenvalue whereby \sum |a_{n}(t)|^{2} = 1. I also understand that the expectation values of operators can be given as function of the a_{n}(t) coefficients given the orthonormality in the eigenfunctions, whereby <H> = \sum |a_{n}(t)|^{2} E_{n}.



If I'm looking at the wavepacket:

\psi(x) = \sqrt{\frac{2}{L}}sin(\frac{\pi x}{L})

How would I determine the expansion coefficients of the wavepacket in the basis states \psi_{n}(x) for the particle in the periodic box, length L? I'm completely confused about the terminology here.

Any help/explanation would be massively appreciated

 

[malawi_glenn]

you fist find the corresponding eigenfunctions \phi (x)for the particle in the periodic box, length L? Then you do this:

a_n = \int \psi ^*(x) \phi _n(x) dx

i.e

\psi (x) = \sum a_n \phi _n(x)

wave functions are normalised here.

So now find the eigenfunction for a box with length L, and do the integral.

 

[raintrek]

I thought that the eigenfunctions \psi(x)were already specific by
\psi(x) = \sqrt{\frac{2}{L}}sin(\frac{\pi x}{L})?

 

[malawi_glenn]

ok your post was not clear.

You state that your wave packed was:
\psi(x) = \sqrt{\frac{2}{L}}sin(\frac{\pi x}{L})

But that is the wave function for the groud state for the box with length L.

The eigenfunctions are altough:

\phi _n(x) = \sqrt{\frac{2}{L}}sin(\frac{n \pi x}{L})

So IF your wave function was \psi(x) = \sqrt{\frac{2}{L}}sin(\frac{\pi x}{L}), then it is trivial to find the expansion coefficients in the basis \phi _n(x)

 

[raintrek]

Sorry, I should have probably just transcribed the question as it's written here:

What are the expansion coefficients of the wavepacket \psi(x) = \sqrt{\frac{2}{L}}sin(\frac{\pi x}{L}) in the basis states \psi_{n}(x) of a particle in periodic box of size L?

I thought that maybe I'd need to use this relation:

a_{n}(t) = \int \psi^{*}_{m}(r) \Psi(r,t) dV

But that gives me a sin² integral which seems very involved for the question...