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All the literature says that the physically meaningful solutions to the Schrödinger equation in an infinite potential well must fulfill the boundary condition that the wave function is $0$ at the walls of the well, otherwise the wave function wouldn't be continuous.

But what differentiates an infinite potential well from a bounded universe with the dimensions of the potential well, where the wave function isn't even defined outside of the "well"/universe? In such a case, the wave function could very well be continuous without being $0$ at the boundary.

What I'm looking for is either an explanation why the case of an infinite potential well is actually different from a bounded universe, or an explanation why the wave function would also have to satisfy the boundary condition in a bounded universe.

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    $\begingroup$ Maybe it is the scale? A micron versus 14 billion light-years? That sets them apart by 32 orders of magnitude. $\endgroup$
    – my2cts
    Feb 23, 2021 at 18:20
  • $\begingroup$ Read my2cts: a these scales there's no quantum physics, not by a looooooong shot! ;-) $\endgroup$
    – Gert
    Feb 23, 2021 at 18:21
  • $\begingroup$ There's probably an answer to be written here about how self-adjointness of differential operators (essential to the mathematical foundations of QM) depends on the boundary conditions used. But I'm headed into office hours and can't do so myself just now. $\endgroup$ Feb 23, 2021 at 18:46
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    $\begingroup$ A bounded universe with no boundary conditions would be like a loose rope. Any wave would just roll off and be gone. $\endgroup$
    – A. P.
    Feb 23, 2021 at 20:13
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/113012/2451 and links therein. $\endgroup$
    – Qmechanic
    Feb 23, 2021 at 20:26

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There is no a-priori sense in having an infinitely large potential, you can't calculate with infinite numbers (except for some very special mathematical considerations). So, the physical meaning that is given to such a potential must be implicitely contained in the accompanying text that explains the infinite potential well. An this meaning is commonly: "the particle is not allowed to penetrate the outside of the well". Or in mathematical terms: $\psi(x<0)=0$ and $\psi(x>L)=0$.

So you are absolutely right in suspecting that the infinite potential well can be considered equivalent to a bounded universe (with the presumably small dimension of the well, of course), as long as your definition of a bounded universe implies impenetrable boundaries. Note, however, that this is not the only way a (conceptual) universe can be bounded. You could also demand a finite universe with periodic boundary conditions $\psi(0)=\psi(L)$, which would be inequivalent to the infinite potential well.

If you find the boundary conditions of the infinite potential well rather unexpected, think about the similar finite potential well for a moment. This has everywhere "well"-defined potential, so clearly no math crash here. As the closer analysis shows, the walls of the finite well can actually be penetrated. If the energy $E$ of an eigen-state is above the upper potential value $V_{out}$, the solution outside the well is purely oscillatory, as is the case for the inside. But, if the energy is below the upper potential value (so that classically the particle would not be able to be outside), the solution becomes damped oscillatory, with the damping coefficient being proportional to the "missing" energy. So the more energy the particle is missing with respect to the outside potential, the shorter the distance until the wave function has attenuated to $1/e$ or beyond. This is actually the basis of the effect of "quantum tunneling".

Now you can imagine what happens, if you let $V_{out}\to\infty$: the energy of an arbitrary given eigen-state will less and less likely be above $V_{out}$, but more and more likely be below it. So it will be damped more and more strongly. Moreover, the damping goes to infinity because the state is missing more and more energy to the boundary of the well. In the limit, you can consider the wave function as being damped to zero within an infinitely short distance, which is nothing else than what you expect from the infinite potential well. Note, that this is not meant as a rigorous mathematical treatment, but just a mental picture, or as you have desired: a physical justification.

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    $\begingroup$ I like your last section the most. The infinite potential well is mostly used an idealisation of the finite potential well with a high outer potential, so it makes sense to have it fulfill boundary conditions which are obtained as the limit of the boundary conditions of the finite well. $\endgroup$ Feb 23, 2021 at 21:22
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All the literature says that the physically meaningful solutions to the Schrödinger equation in an infinite potential well must fulfill the boundary condition that the wave function is 0 at the walls of the well, otherwise the wave function wouldn't be continuous.

That's not the reason at all: there are plenty potentials $V(\mathbf{r})$ for particles with 'soft' boundaries that generate continuous wave functions (think quantum harmonic oscillator, e.g.)

For the infinite well, the boundary conditions $\psi(0)=0$ and $\psi(L)=0$ arise from the simple fact that the potentials at these boundaries is $+\infty$ and so no particle can penetrate these potential walls. It's like a prisoner in a cell: he can move inside the cell but he can never escape!

But what differentiates an infinite potential well from a bounded universe [...]

In a word: scale. Even at very modest scales quantum physics reverts back to Newtonian physics. The Universe is so ginormously large that thinking in terms of bounded wave functions is absurd and a complete waste of time.

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  • $\begingroup$ About the first paragraph: I didn't mean that to be continuous, all wave functions need to satisfy the given boundary condition, but that to be continuous, the wave function of the infinite potential well needs to satisfy the condition. Which is certainly true: if we assume that the wave function is 0 outside the well, then it must be 0 at the boundaries, too, to be continuous. $\endgroup$ Feb 23, 2021 at 19:03
  • $\begingroup$ Maybe a better way to phrase my question: say the well goes from $x=-a$ to $x=a$. Why do we assume that the wave function is a function $\psi:\mathbb R\to\mathbb R$, instead of $\psi:(-a,a)\to\mathbb R$? In the first case, boundary conditions apply, in the second case they don't, since there wouldn't be any boundaries to speak of. $\endgroup$ Feb 23, 2021 at 19:08
  • $\begingroup$ Why would scale matter in an empty universe? The Schrödinger equation doesn't have a size limit. Of course there are some spontaneous collapse models around, which postulate dynamics in addition to the Schrödinger equation, but none of them has been experimentally confirmed. $\endgroup$
    – A. P.
    Feb 23, 2021 at 19:53
  • $\begingroup$ @A.P. There's a reason why QP applies to the microscopic only. See the Correspondence principle: en.wikipedia.org/wiki/…. In a Universe wave functions bound by the boundaries of the Universe all trajectories become Newtonian. $\endgroup$
    – Gert
    Feb 23, 2021 at 20:27
  • $\begingroup$ @Gert The Correspondence Principle describes how classical mechanics arises as a fringe case of quantum mechanics, but it doesn't mean quantum mechanics stops working there. One can still solve for eigenfunctions of the Hamiltonian. $\endgroup$
    – A. P.
    Feb 23, 2021 at 21:05

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