Why can we rewrite the wave function like so:

[EngageEngage]

Hi, I've been working on writing the wave function in terms of momentum eigenfunctions. The only problem I have with the derivation is the last step, which allows me to write:

$$\Psi(x) = \int^{\infty}_{-\infty} \phi(p)u_{p}(x)dp$$
where
$$u_{p}(x) = \frac{1}{\sqrt{2\pi\hbar}}e^{\frac{ipx}{\hbar}}$$
What allows me to do so? I haven't found a way to prove this. I know that I can write the wave function like so:
$$\Psi(x) = \sum c_{n}u_{n}$$
as long as i have a complete basis, but how do I go from this to the integral? Or is this irrelevant? his is on Griffiths QM Ed2 on page 104, but all he says is "Any (square-integratable) function f(x) can be written in the form [what i showed above]" I haven't found anywhere where he proves this or expects one to prove this in the problems.

Can someone please tell me how to get started on proving this? If anyone can help me out with this one I would appreciate it greatly.

 

[Dick]

The u_p are a complete basis. This is the subject of Fourier analysis. If u_n are complete you have a sum over the integers n. The index p is continuous. So you have to move from a sum to an integral to add them all up. What exactly do you have to PROVE?

 

[EngageEngage]

I think you just answered my question. I spaced that p is continuous. So that's why I can rewrite the sum like the integral with no troubles then. Its nothing i have to prove, I just didn't understand why I could go to the integral from the sum ( i was thinking that p was quantized for some reason), but with p being continuous this makes sense. Thanks a lot for the help!

 

[Domnu]

One more thing... remember that when you try to find $$\phi(p)$$, you have to change the sum to an integral as well... i.e.

$$\phi(p) = \int_{-\infty}^{\infty} \psi(x) u_p^*(x) dx$$

 

[EngageEngage]

Thanks for the help. Yeah I see that one because it is just the Fourier transform and inverse from there.