# QM: with V = -x

1. Jul 2, 2010

### gn0m0n

1. The problem statement, all variables and given/known data
This is from an old exam I'm studying from. It goes:

"A particle of mass m is traveling in one dimension under the influence of a potential

V = -Fx

where F is a known constant. Find the energy spectrum and wavefunctions.

Hint: You may want to work in momentum space, but do not have to."

2. Relevant equations

Schrodinger eqn, other fundamental quantum relations

3. The attempt at a solution

I solved for the momentum space wave function and got an exponential solution ~ exp[(i/hF)(p^3-E*p)], I think, or something much like that. But how can I find an energy spectrum when I don't seem to have any boundary conditions? It does not say anything about, say, an infinite potential for x < 0 or anything.

I also tried using

d<p>/dt= <-dV/dx>

but that didn't seem to help.

It seems like the energy could be anything. Could that be all I need to say? And what about the wave functions?

It just seems like a weird problem. I'm wondering if there could have been a typo or something obvious I'm missing. I can post more of my work if it will help.

2. Jul 2, 2010

### vela

Staff Emeritus
You got it. No boundary conditions means no restriction on the allowed energies, similar to the case of a free particle.

The potential corresponds to a particle in a constant force field. Classically, the energy would just allow you to locate where the turning point was. Quantum mechanically, that's where the wave function should go from being oscillatory to decaying. I haven't worked it out, but I'd expect that changing E just shifts the wave function left or right.

Since you have the momentum-space wave function, you can Fourier transform it to obtain the position-space wave function.

You might want to look up Airy functions as well.

3. Jul 2, 2010

### bjnartowt

If V = V(x) = F*x, this seems like a linear potential, not that of a free particle.

Is that not so? If so: a constant force (gradient of potential) would be on the particle, rather like the quantum picture of a particle ascending a hill against a gravitational potential. No?

4. Jul 2, 2010

### vela

Staff Emeritus
Yes, you're right. They're different potentials. My point was that you don't get quantized energy levels with V=-Fx because the particle is unbounded, just like you don't get quantized energy levels with V=0 because again the particle is unbounded. The potential in both cases does not confine the particle to a finite region of space.

5. Jul 2, 2010

### bjnartowt

Oh, I see. You said "SIMILAR" to a free particle. Sorry about that: details are not my strong suit.