# Expansion coefficients of a wave packet

1. Dec 6, 2014

### rwooduk

1. The problem statement, all variables and given/known data
What are the expansion coefficients of a wavepacket $$\Psi (x) = \sqrt{\frac{2}{L}}sin \frac{\pi x}{L}$$ in the basis Ψn(x) of a particle in a periodic box of size L?

2. Relevant equations
$$\Psi (r,t) = {\sum_{n}^{}} a_{n}(t) \Psi _{n}(r)$$

3. The attempt at a solution
$$\left \langle \Psi _{m}| \Psi \right \rangle= {\sum_{n}^{}} a_{n}(t)\left \langle \Psi _{m}| \Psi_{n} \right \rangle$$

all zero except for m=n therefore

$$a_{n}(t)=\left \langle \Psi _{n}| \Psi \right \rangle$$

so I have a term for the coefficients but how do I apply it to the specific wavefunction? espcially as it has no n in it.

Any help as always very much appreciated. Thanks.

2. Dec 6, 2014

### Orodruin

Staff Emeritus
What are the basis functions for your potential?

3. Dec 6, 2014

### rwooduk

hmm do you mean the basis wavefunctions? if it's a periodic box then

$$\Psi _{n}=\sqrt{\frac{1}{L}} exp(\frac{i2\pi nx }{L})$$

so i substitute this into the above an(t) equation?

edit i expanded the exponential in terms of cos and sin but it gets a bit complicated, also not sure what to do with the Σan thats on the right hand side.

Last edited: Dec 6, 2014
4. Dec 6, 2014

### Orodruin

Staff Emeritus
It will be more fruitful to expand the sine in terms of exponentials ...

5. Jan 4, 2015

### rwooduk

Apologies for the late reply. I'm getting really lost with this one, I'm really not sure my method is correct as there is no time dependance in either wavefunction, so why am I using an(t).

If I use:
$$a_{n}(t)=\left \langle \Psi _{n}| \Psi \right \rangle$$
with

$$\Psi _{n}=\sqrt{\frac{1}{L}} exp(\frac{i2\pi nx }{L})$$
$$\Psi (x) = \sqrt{\frac{2}{L}}sin \frac{\pi x}{L}$$

but then expand

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

$$\Psi (x) = \sqrt{\frac{2}{L}}sin \frac{\pi x}{L} = \sqrt{\frac{2}{L}}\frac{1}{2i}(exp\frac{in\pi}{L}- exp \frac{-in\pi}{L})$$

so I get

$$a_{n}(t) = \int_{0}^{L} \sqrt{\frac{1}{L}} exp(\frac{i2\pi nx }{L})\sqrt{\frac{2}{L}}\frac{1}{2i}(exp\frac{in\pi}{L}- exp \frac{-in\pi}{L}) dx$$

this cant be correct?

any further help on this would be appreciated, I still dont really understand what the question is asking to be honest.