# Free particle in one dimension

1. May 23, 2014

### angass

Hi,
I´m not sure if my way of tackling a question, probably it's a trivial problem, but it's important for me to get it right so any help will be greeted.
The question is as follows:

Problem: consider a particle in a one-dimensional system. The wave function ψ(x) is as follows:
ψ(x)= 0 for (-∞<x<0),
ψ(x)= 1/√a for (0<x<a)
ψ(x)= 0 for (a<x<∞)

i) if the kinetic energy is measured what is the most probable value?
ii) which values of the momentum can never be found when measuring it?

My reasoning:
i) I use the operator for kinetic energy: K= -$\frac{h^2}{2m}$$\frac{∂^2}{∂x^2}$
which when applied: ∫ψ(x)*Kψ(x)dx gives me zero.
If this was right I assume the most probable value of the kinetic energy is zero.
ii) I have no clue whatsoever what the question means.

Thanks and forgive my bad english, regards from Spain!

2. May 23, 2014

### Staff: Mentor

Hi angass, welcome to PF!

Is that correct? Because this is not a continuous function, and therefore not a valid wave function.

3. May 27, 2014

### rigetFrog

This looks suspiciously like a griffiths or leibowitz homework problem....

You want to take express psi(x) in a fourier basis (as a sum of e^i(kx-wt)). Doing the fourier transform will be a simple integral from 0 to a with 1/a^1/2 as a constant.

i) You're forgetting the edges at 0 and a. The derivative is not zero there. Once you express this function as fourier expansion, you'll see you don't get zero anymore.

ii) apply the definition of p = i hbar d/dx to the fourier transform. The answer should drop out. I expect some of the coefficients for specific k terms in the fourier transform will be zero. (the fourier basis is an eigen value of the momentum and kinetic energy operator)

4. May 28, 2014

### Staff: Mentor

Same here mate.

To the OP if it is we have a homework section.