# Free particle wave function confusion.

1. Aug 12, 2016

### Oz123

• Member reminded to use the homework template for posts in the homework sections of PF.
Hi! I'm currently studying Griffith's fantastic book on QM, and I'm confused for a bit about the wave function for a free particle.
Here's what I think so far; for a free particle, there are no stationary states, so therefore we can't solve the SE with
ψ(x)=Aeikx+Be-ikx

That is, we can't write a discrete sum. But We can have solutions as:

ψ(x,t)=∫dkφ(k)ei(kx-ωt)

I don't know if my understanding is correct, so please tell me so. Now, I assume that this understanding is correct and get to the question: If the solutions can only be the latter, then why was the solution from the book for the scattering states in the delta function potential a sum of stationary states and not the continuous sum? Also, why is it the same for the bound states if we are solving for the free particle when x<0 and x>0? Is it because it has a potential at x=0?

2. Aug 13, 2016

### Orodruin

Staff Emeritus
These functions do solve the Schrödinger equation, but they are not normalisable and therefore not actually in the relevant Hilbert space of square integrable functions.

Generally, in scattering theory, you will look at an in-state of definite momentum. Of course, the actual physical state is a superposition of such states and not a plane wave. However, in many cases, looking at just an incoming plane wave solution is a sufficiently accurate description.

In the case of a scattering potential, you will often have both bound and free states. The bound states correspond to energy levels with an energy lower than the energy at $\pm\infty$ and are generally discrete while the free scattering states show a continuous spectrum. In both cases you have to find the solutions to the Schrödinger equation in all of space.