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I'm thinking about the possibility of a discontinuous solution of the Schrödinger equation and what is needed to get such an object.

It should be a condition to the potential, I think. If the potential is bounded everywhere a discontinuous wavefunction does not make sense because the second derivative will be singular and so the left side of the equation while the right side ist still bounded. So the only possibility for such solutions should be a singularity in the potential. Is this correct so far? Does there exit any potential which results in a discontinuous solution?

If I'm thinking about the particle in a box problem: Why isn't it possible that this solution becomes discontinuous at the boundary points?

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    $\begingroup$ Oh, copy&paste fail. I meant this question and its linked questions. $\endgroup$ – ACuriousMind Aug 9 '16 at 12:21
  • $\begingroup$ Comment to the post (v2): Are you talking about TDSE or TISE? $\endgroup$ – Qmechanic Aug 9 '16 at 12:27
  • $\begingroup$ More related: physics.stackexchange.com/q/38181/2451 , physics.stackexchange.com/q/186323/2451, physics.stackexchange.com/q/262671/2451 and links therein. $\endgroup$ – Qmechanic Aug 9 '16 at 12:36
  • $\begingroup$ Usually the wave function, in order to appear in the Schrödinger equation, must be differentiable in some domains, hence continuous. Unless you allow for some ill-defined distributional or discontinuous potential or the like on the right hand side of the equation. $\endgroup$ – gented Aug 9 '16 at 14:35
  • $\begingroup$ I found many useful hints related to my question but I have still the feeling that they only raise the subject. Is there a more precise answer for this infinite potentials? $\endgroup$ – Core2016 Aug 9 '16 at 17:59
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Certain Hamiltonian operators can actually lead to solutions that have discontinuities. Since there are derivatives (from a momentum $P$) which can cause difficulties, these points have to be treated by multiplications by a position variable, e.g., $Q$, where $[Q,P] = i\hbar$. For example, if $P^2$ is replaced by the Hermitian combination $(QP+PQ)^2/4$, then the Schroedinger equation solution can be discontinuous at $x = 0$, where $Q|x\rangle = x|x\rangle$. An example of such discontinuous Schroedinger equation solutions appears in Section VI of this paper.

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It doesn't make sense to have a discontinuous wave function solution to S.E. That is because of the Born interpretation of it. A probability distribution doesn't work with this singularities. However the first derivative it's related with the momentum of the particle and this can be discontinuous when there is an infinite potential. A discontinuity would represent violation of momentum conservation, but infinite potentials don't exist in real life.

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  • $\begingroup$ But in general we only assume that the wavefunction is square-integrable. And not all L² functions are continuous? $\endgroup$ – Core2016 Aug 9 '16 at 18:04
  • $\begingroup$ All $L^{2}$ must be continuous, or the integral over all space would diverge. $\endgroup$ – Javier Vazquez Sep 3 '16 at 2:03

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