Have a question from my class which I'm struggling with.

We have a particle $m$ with wavefunction $φ=Ax \exp(-ax)$ when $x≥0$, and $φ=0$ otherwise.

We are asked to show that the double derivative $φ''=(a²-2a/x)φ.$ So far so good.

Then "Supposing that the particle is an eigenstate of the hamiltonian, determine the potential V(x) and the total energy E of this state.

I've worked through this and got the Schrödinger eq into the form:

$$-\hbar^2/2m (a^2-2a/x)=E-V$$

After this I'm not sure what to do. How can I separate energy and potential with the information I've been given ?


closed as off-topic by ZeroTheHero, Yashas, tpg2114 May 11 at 20:16

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – ZeroTheHero, Yashas, tpg2114
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ Welcome! Please not that unfortunately the community does not favour answering homework-like questions and yours is precisely of that kind. $\endgroup$ – ZeroTheHero May 11 at 13:06

The separation you want is not possible - you've done as much as you can in this problem without additional constraints.

The reason is that the potential energy, as always, is only defined up to the addition of a global constant, whose only effect would be to shift the eigenvalues by an equal amount. Since that transformation has no effect on the eigenfunctions, the latter cannot be used to break that ambiguity. Thus, knowledge of an eigenfunction is sufficient information to recover the function $V(x) - E$ (the potential minus the corresponding eigenvalue), but not to separate those two objects.

To do that, you need some additional, external criterion, which you can use to fix the zero value of the potential energy. This will depend on the configuration, and there is no guarantee that there will exist a convenient canonical choice. For example, if the potential is bounded from below (as with the harmonic oscillator) it can be convenient to set that bound at zero; similarly, a common choice is to set the zero at the value at infinity, if that exists.

If you do see a convenient choice, then provide a brief argument for why it makes sense, and move on.


Not the answer you're looking for? Browse other questions tagged or ask your own question.