Wavepacket expansion coefficients

[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...