Expansion coefficients of a wave packet

[rwooduk]

Homework Statement

What are the expansion coefficients of a wavepacket [tex]\Psi (x) = \sqrt{\frac{2}{L}}sin \frac{\pi x}{L}[/tex] in the basis Ψ_n(x) of a particle in a periodic box of size L?

Homework Equations

[tex]\Psi (r,t) = {\sum_{n}^{}} a_{n}(t) \Psi _{n}(r)[/tex]

The Attempt at a Solution

[tex]\left \langle \Psi _{m}| \Psi \right \rangle= {\sum_{n}^{}} a_{n}(t)\left \langle \Psi _{m}| \Psi_{n} \right \rangle[/tex]

all zero except for m=n therefore

[tex]a_{n}(t)=\left \langle \Psi _{n}| \Psi \right \rangle[/tex]

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.

 

[Orodruin]

What are the basis functions for your potential?

 

[rwooduk]

What are the basis functions for your potential?

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

[tex]\Psi _{n}=\sqrt{\frac{1}{L}} exp(\frac{i2\pi nx }{L})[/tex]

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

thanks for the reply!

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 Σa_n that's on the right hand side.

 

[Orodruin]

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

 

[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 dependence in either wavefunction, so why am I using a_n(t).

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

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

but then expand

[tex]\Psi (x) = \sqrt{\frac{2}{L}}sin \frac{\pi x}{L}[/tex]

[tex]\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})[/tex]

so I get

[tex]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[/tex]

this can't be correct?

any further help on this would be appreciated, I still don't really understand what the question is asking to be honest.