Determine the potential V(x) from the Hamiltonian

1. Sep 29, 2016

Milsomonk

1. The problem statement, all variables and given/known data
Assuming psi is an eigenstate of the Hamiltonian (TISE) and that E=0, determine the potential V(x) appearing in the Hamiltonian.

2. Relevant equations
Time Independant Schrodinger Equation - 1 Dimensional (x)

I am given the wavefunction psi = N/(1+x^2)
I have found the normalization coefficient N, but this cancels in the calculation.

3. The attempt at a solution
Hey guys,
So I have this question which i'm a little unsure on, so far I've taken these steps.
1. wrote down the TISE, since E=0 i set the right hand side to zero.
2. next I moved V(x) over to the left, cancelled the minus sign on each side.
3. So now I have (Hbar^2/2m)*(second derivative of psi wrt x)= v(x) psi. I divided both sides by psi, carried out the defferentiation and simplified.
4. I have checked my calculus and simplifications in mathematica and they are correct. Basically for V(x) I have Hbar^2/m * (function of x), im wondering if this is the write form for a potential and whether my method was sound. It seems odd to me to have the planck constant and mass involved in the potential, but maybe its fine? Am I one the right lines with my technique for answering the question?

Any insights would be really appreciated :)

2. Sep 29, 2016

TSny

Of course we would need to see you answer in order to know if it is correct, but your method appears to be correct.
Often, the potential does not have Planck's constant, h, or the mass of the particle, m (e.g., the hydrogen atom or the harmonic oscillator). But in these cases, the wavefunction will contain h and m. For such systems, what would you expect to happen to h and m if you carried through your procedure to derive V from psi?

3. Sep 29, 2016

Staff: Mentor

Your approach sounds correct.

This is a very weird problem, because the wave function and the resulting potential don't make sense from a units point of view, unless the x is is some adimensional length.

4. Sep 29, 2016

Milsomonk

Thanks! awesome answers :) I feel a bit more confident in my answer now. I guess i'd expect them to cancel if h and m were contained in the wavefunction, leaving a potential purely dependant on x.

I'm happy to here that it is a weird problem, I thought that as well but didn't know if I was just missing something. I guess this particular problem is more of a mathematical excersise than a particularly physical one.

Thanks again guys :)