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It's a reasonably standard fact (though not nearly well-known enough) that the momentum operator in the infinite square well is a very problematic beast (as explained e.g. in this and this answers).

In the comments under this answer, mike stone proposed an interesting look at how this might work:

[the] question can be formulated more physically by asking what happens in finite depth square well on the entire line. Then a self-adjoint momentum operator exists (but does not commute with the Hamiltonian) and the overlaps do give the possible momenta, [and] one can then explore what happens as the depth increases.

I think that's an excellent way to look at it, so: what are the momentum properties of the finite-well eigenstates, and how do they change in the limit of a very tall well? How do they mesh with the properties of the (possible self-adjoint extensions of the) momentum operator on the infinite square well, and its interactions with the hamiltonian over the restricted-interval version?

Emilio Pisanty
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  • As a side note, one might be also interested in the related the problem of self-adjointness of the Hamiltonian in an infinite well as discussed in https://arxiv.org/pdf/quant-ph/0103153.pdf, where it is shown that $H=p^2/2m$ is not self-adjoint in this case. Also related is an extension of the above discussion available from http://fma.if.usp.br/~piza/artigos/001401ajp_Chico.pdf – ZeroTheHero Oct 10 '17 at 17:54
  • @ZeroTheHero Yeah, I looked for good references for that earlier today and didn't find any as easily as I'd've liked. We're probably overdue a good canonical question, with solid and comprehensive references, on the status of the momentum operator in the infinite square well. – Emilio Pisanty Oct 10 '17 at 18:03
  • There's a lot of related literature that cites Bonneau so Gsearching for this paper in GScholar is a good first step. I think the explicit case of the momentum operator is discussed in Chap.6 of the QM book by Capri. – ZeroTheHero Oct 10 '17 at 18:43
  • @ZeroTheHero Yeah, the information definitely exists - I'm just saying that this site ought to have a concise, clear and complete summary. – Emilio Pisanty Oct 10 '17 at 20:19
  • Amen to that!!! – ZeroTheHero Oct 10 '17 at 20:28
  • @ZeroTheHero The paper you cited does not show that $H$ is not self-adjoint in that case, it is only demostrated that there are functions you can formally define $H$ on, but do not belong to the domain of a self-adjoint extension. It also uses some of the mathematical language in a very questionable way. If you are interested, I would rather recommend one of the classics, e.g. Reed/Simon, especially part 2 on self-adjointness. Hall's "QM for Mathematicians" is also not a bad start. – Daniel Oct 31 '17 at 13:47
  • @EmilioPisanty I am not entirely sure what you are actually asking. What do you mean by "momentum properties", and what kind of "transition" between the two systems (finite and infinite well) are you looking for? Keep in mind that the Hilbert spaces of both systems are different, one is $L^2(\mathbb{R})$, the other is $L^2(0,L)$, so it's a bit tricky to compare them directly. Also, $p$ is technically not an observable for the infinite square well, since the boundary conditions of the self-adjoint extensions do not correspond to the physical situation. – Daniel Oct 31 '17 at 13:54
  • @Daniel I'm asking for the limiting properties of the momentum-space wavefunctions of the finite-well eigenfuctions as the well depth tends to infinity. – Emilio Pisanty Oct 31 '17 at 14:22
  • Well, if you carry out the limit of the Fourier transform, you will get something well-defined, since what you will be doing technically is equivalent to extending the infinite-well solution by zero on $\mathbb{R}$ and then performing a Fourier transform, which certainly makes sense from a physical point of view. The tricky thing is that mathematically speaking, for $p$ you are treating the eigenfunction as a $H^1(\mathbb{R})$-function and for $H$ as a $H^2_0(0,L)$-function (otherwise $H$ is ill-defined). Does this go in the right direction? – Daniel Oct 31 '17 at 15:53
  • @Daniel Frankly, you lost me somewhat on the terminology there. – Emilio Pisanty Oct 31 '17 at 15:56
  • However, I don't expect that the Fourier transform of the infinite-well eigenfunctions to contain all the relevant information. The finite-well eigenfunctions are continuously differentiable but the infinite-well ones have a discontinuous derivative, which means that their momentum-space representation decays to zero more slowly as $p\to\infty$. I would like to see a quantitative description of those properties. – Emilio Pisanty Oct 31 '17 at 16:00
  • I guess that means I wasn't pointing in the right direction ;), but for the record, it should've been the Sobolev space $H^1_0(0,L)\cap H^2(0,L)$ instead of $H^2_0(0,L)$. Considering the $p$-representations of the eigenfunctions and their asymptotic behaviour - shouldn't that be a pretty straightforward calculation? From the part inside the well there should come exactly one order less for the tall-well-limit, while the rest should converge pretty fast for $p\rightarrow\infty$. – Daniel Oct 31 '17 at 16:30
  • @Daniel Yes, I don't expect this to be too hard a calculation. – Emilio Pisanty Oct 31 '17 at 16:31
  • It seems to me that all the discussion relies upon a misunderstanding. It is false that with a finite square well the generic wavefunctions tend to vanish inside the barrier when the height of the barrier increases. It only happens to eigenfunctions of the Hamiltonian. The other operators like the momentum are not affected by this phenomenon. So I think it is very difficult to grasp some intuition on the failure to define a momentum operator in the infinite square well through this way. – Valter Moretti Nov 16 '17 at 14:17
  • Also the idea that boundary conditions $\psi(0)=\psi(L)=0$ matter in the infinite well for all possible vawefunctions is wrong. These conditions are valid for the Hamiltonian eigenfunctions but the Hilbert space of the infinite well also includes vawefunctions like the constant one $\psi(x) = 1/\sqrt{L}$. – Valter Moretti Nov 16 '17 at 14:22

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