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In my Quantum mechanics 1 lecture the professor proofed that the wave function in one dimension has to be continuous as long as the potential is "well behaved". My question is whether the wave function still has to be continuous for an arbitrary potential, which is arbitrarily "wild". Do you know a proof that it has to be continuous/does not have to be continuous or some arguments why it should/should not be the case?

I understand that from a physical standpoint, the wavefunction must be continuous, because else we will get an infinite kinetic energy, which is unphysical. But in physics generally all potentials are continuous as well when you look at the smallest dimensions.

But I still pose this question because I think it is somewhat different from the other questions which have been asked on this topic. What this question is about is the purely mathematical point of view, if we take the whole physics out of it. If one assumes there can be arbitrarily "wild" potentials (for example a potential which has a delta function at x if x is rational and is zero if x is irrational), does the wave function still have to be continuous?

Luis
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  • Possible duplicates: https://physics.stackexchange.com/q/19667/2451 , https://physics.stackexchange.com/q/262671/2451 , https://physics.stackexchange.com/q/164524/2451 , https://physics.stackexchange.com/q/594123/2451 , https://physics.stackexchange.com/q/149001/2451 and links therein. – Qmechanic Dec 15 '21 at 19:32
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    @Qmechanic I think the dupe targets don't necessarily address the question. They talk about smoothness, and certainly, if the wave function is smooth then it's also continuous, but if a wave function is not smooth, it is not necessarily not continuous. Furthermore, about half the answers in the dupe targets disagree with one another, but they are all at $0$ or upvoted, and some of the most relevant material is buried in comments. A person Googling the same question who is directed to the dupe targets will have a hard time getting an answer. – user400188 Dec 16 '21 at 00:23
  • Hi Luis. In order to reopen this post (v5) consider to be specific: Do you mean the time-dependent or the time-independent Schroedinger equation? The different answers are mainly because of this distinction. – Qmechanic Dec 16 '21 at 03:02

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