Recently I had a quiz in my physics class and I feel like the professor made a mistake on the solution for it. Yes, I already have the answer to this question. I am not trying to get people's answers for homework or anything. But I would like to hear people responses to this. The question was:

Which of the following conditions are necessary restrictions imposed on the solutions of the one dimensional Schrodinger equation? (Choose all the apply)
1. The wave function must be continuous.
2. The wave function needs to be normalized.
3. The wave function must converge to zero as x tends to positive and negative infinity.
4. The wave is zero at the boundary conditions

I know that the first three are correct because they are right out of the book. But my problem lies with the last one. So far we have only dealt with a particle in a rigid box and in the derivation, the wave function was zero at the boundary conditions. That's why I said 4 was correct too. This was before we studied finite potential wells where the wave function exponentially decays at the boundaries. So with knowing our limits to the material for the quiz, do you guys think it is fair that he take points off for saying the wave function zero at the boundary condition when it really is zero at the boundary with what we were studying?

  • 1
    $\begingroup$ Nice question ... until you get to the part where you turn it into a plea for justice. There will be a lot more such oversights in future courses so develop a thick skin. Physicists are far more imperfect than you'd imagine. $\endgroup$
    – user346
    Apr 17, 2011 at 20:57
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    $\begingroup$ As much as I appreciate you putting in the effort to ask a well-stated question, @Greg, it runs afoul of item #3 (and perhaps #5) on the What should I not ask here section of the FAQ: "there is no actual problem to be solved: 'I’m curious if other people feel like I do.'" Besides, it's not about physics, it's about grading. If you asked why choice 4 is incorrect, then that would be a perfectly valid physics question. (You could take this up in the chat room if you want, we're less particular about being on topic there) $\endgroup$
    – David Z
    Apr 17, 2011 at 22:25

2 Answers 2


I would say it's definitely fair. You're not supposed to just take (4) for granted in the rigid box case - you were probably expected to understand why (4) holds in the rigid box case, and if you did, you would see why (4) doesn't have to hold in the general case.

A particle inside a rigid box can never be found outside the box. So the wavefunction is zero everywhere outside the box, but non-zero generally inside the box. Now since the wavefunction is continuous everywhere, this necessarily means that it has to be zero at the boundary of the box.

This means that if (as in the case for a finite potential well) the wavefunction does not have to be zero outside the well (because of tunelling), then continuity does not require it to be zero at the boundary, so clearly (4) is false in general.

In short, the thing that causes (4) to be true in the rigid box is the absence of a wavefunction outside the box. Since you wouldn't expect this to be true in general, (4) need not hold true in general.


Regarding (4), you are asking wether it was fair to have asked a question which employed stuff you presumably hadn't covered. That is a question regarding academic and testing practice. However, to answer how you can know by just knowing the theory you knew:

For the infinite potential well the hamiltonian has the potential term $$\hat{H}\psi(r) = -\frac{\hbar^2}{2m}\nabla^2 \psi(r) - V(r) \psi(r)$$ Because $V(x)=\infty$ at the boundary, the only way you can stop the hamiltonian from blowing up is by taking $\psi=0$. It becomes obvious hereon, that if the potential does not blow up at the boundary, there is no need for the wavefunction to be zero.

Note: Regarding (3), although correct, is necessary but not sufficient. It lies in the requirement for square integrability (or belogning to $L_2$) $$\int|\psi(x,t)|^2 dx\lt \infty$$ Infact, a wavefunction that depends on one spatial coordinate $x$ must fall off at least as fast as $\frac{1}{\sqrt{x}}$. So just being zero at infinity is not enough.


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