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Lately, I have been studying QM more deeply and I just discovered how many important subtleties the 'well-known' particle in an infinite potential well hides, which are precious for extending knowledge on the subject. I know there are a lot of interesting questions about quantum mechanics on finite real interval on this website (I think I have read all of them), nevertheless, there are some puzzling aspects that I would like to address.

In ordinary QM, i.e. on the whole real line $\mathbb{R}$, we are able to make the uncertainty $\Delta p$ in momentum arbitrarily small, asymptotically going to zero. Consequently, the initial wave function will approach (asymptotically, again) a plane wave. This is always true, even if the plane waves do not belong to the Hilbert space, nor to the domain of the operator $[\hat{x},\hat{p}]$, where $\hat{x}$ is the usual position operator.

Now, when we are dealing with a particle in an infinity square well, that is with a QM implemented on a finite interval of the real line, a straightforward and maybe naive calculation shows the following: $$ \Delta x \Delta p \geq \hbar/2 \Rightarrow \Delta p^{min}\geq \frac{\hbar}{2 \Delta x^{max}}= \frac{\hbar}{2L}$$

where $L$ is the length of the interval on which our particle lives.

This seems to suggest that there is a minimum value for the uncertainty $\Delta p$ in momentum and hence that it is not possible to make $\Delta p$ arbitrarily small.

The existence of eigenstates of the momentum operator $\hat{p}$ (which must be defined carefully in order to be self-adjoint, see e.g. What's the deal with momentum in the infinite square well?) does not help, in my opinion.

Indeed these eigenstates violate the uncertainty principle and do not belong to the domain of the commutator $[\hat{x},\hat{p}]$ (see e.g. from here: How can I solve this quantum mechanical "paradox"?). This can be regarded as something similar to what happens when the configuration space is the whole real line (the "ordinary" plane waves do not belong to the physical Hilbert space, as said above), nevertheless while in this case we can approach asymptotically the plane waves, that is the states with $\Delta p =0 $, in the "compact" case this is not possible anymore if it is true that our physical states can reach a finite value, different from zero, of $\Delta p$.

Hence my questions: is all of this correct? If so, should not this mean that $p-representation$ is not available in this case? If not, what is wrong with this argument?

Thank you all in advance.

RH_ss
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  • The more we assume, the more we can derive. But if we assume too much, things get inconsistent. I worry maybe some assumptions here are questionable. Do you assume or derive that momentum operator is not just $-i\hbar \partial_x$, but is something different due to boundaries? If so, why? Boundaries are fictitious and are always realized through high potential walls, so we can always go back to infinite coordinate domain like in standard exposition of QT and then $p_x = -i\hbar \partial_x$ and the uncertainty relation is satisfied. – Ján Lalinský Apr 27 '22 at 14:09
  • In my argument the momentum operator is still the differential operator $-i \hbar \partial _{x}$, with particular boundary conditions in order to assure self-adjointness. Of course, we can always go back to the infinite case, but I'm asking why in the compact case (or the infinite potential well, if you want) seems that some particularities arise concerning the Heisenberg uncertainty principle. – RH_ss Apr 27 '22 at 14:26
  • Well this seems mostly about functional analysis, which I don't understand. I guess it's possible that theorems (uncertainty relation) valid in infinite domain need not be valid in finite domain. Maybe it would be good to ask this also on a math forum. There is very little physics in this question; from physics standpoint, we just assume that operators represent measurable things, that hermitian operators eigenfunctions form a basis, and other usual rules. But this may very well fail mathematically in some scenarios, like when restricting everything to a finite domain,which is very unphysical. – Ján Lalinský Apr 27 '22 at 14:41
  • Thanks for the tip. Functional analysis is for sure one of the tools which should be used to explain this situation. Nevertheless, I partially have to disagree with you about the physical content of the question: even if it is true that this system is unrealistic, deeply understanding what is really going on with momentum can help shed light on more complex situations. At least in my opinion. – RH_ss Apr 27 '22 at 14:59
  • I agree getting better understanding of mathematics is benefitial. Restricting model to a finite domain sometimes simplifies and sometimes complicates things, so one can't just dismiss it generally. Whether it's a good idea depends on details. – Ján Lalinský Apr 27 '22 at 15:07

1 Answers1

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As long as you define $\hat p$ as a self-adjoint operator, the spectral theorem applies, so you do have a $p-$representation. In the square well, the usual definition of $\hat p$ is $-i\partial_x$ on $\mathcal C^1$ spatial wavefunctions satisfying periodic boundary conditions and makes it essentially self-adjoint. You can check that the usual Fourier basis used in Fourier series is a complete eigenbasis of $\hat p$, so you do have a $p-$representation.

Your analysis falls down because the uncertainty principle fails, especially on the $\hat p$ eigenvectors. This may seem counterintuitive because the commutator is still the same and $\Delta x,\Delta p$ are well-defined, so the formal proof you must know should hold. The problem, as always, is domain issues, here because $\hat x |p\rangle$ ($|p\rangle$ is an $\hat p$ eignevector of eigenvalue $p$) is not in the domain of $\hat p$ (no periodic boundary conditions). For the uncertainty principle to work on looser domain assumptions, the exponentiated version must hold. This is no the case here, as $e^{is\hat p}$ is not your usual translation operator, but makes the wavefunction loop back due to the choice of the domain.

For more information, check out Quantum Theory for Mathematicians by BC Hall, which includes the above argument in more detail.

Of course, your reasoning works if you place yourself now on the entire line, but I think you weren't interested in this loophole. In this case, the Heisenberg uncertainty holds, and you extend the wavefunctions to the entire line with $1_{[-L/2,L/2]}$. It is pretty obvious that restricted by this constraint, you can never approach asympotically plane waves, so no $p-$resolution. In particular, our previous eigenvectors become $sinc$ functions in momentum space, so $\Delta p=\infty$.

Hope this helps and tell me if you find some mistakes.

LPZ
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  • Part 1. Thanks for your comment. This was exactly the answer I was looking for. I was sure something had to be wrong with the uncertainty relation in this case. Nevertheless, I have some concerns. I understand your argument about $\hat{x} \rangle{p}$ not being in the domain of $\hat{p}$, but - if I am not wrong - this is always true unless at the boundaries the wave function is zero. In this case the wave function should be in the domains of $\hat{x}\hat{p}$ and $\hat{p}$. – RH_ss Apr 27 '22 at 15:50
  • Part 2. While this can never happen for an eigenstate of $\hat{p}$ (which were the ones you were talking about), since they cannot be never zero at the endpoints (indeed they are correctly ruled out from the HUP domain), this in principle could happen for some other state. So there should be states which correctly lie in the HUP domain. So, for these states should not hold the statement about the minimum value of $\Delta p$? – RH_ss Apr 27 '22 at 15:54
  • Just to be sure, I think you meant at the end of part 1 the domains of $\hat p$ and $\hat p \hat x$. Yes in this case the uncertainty inequality applies when the wavefunction is in their intersection. For example, it is the case for $1_{[-L/4,L/4]}(x)$ or $x(L-x)$ (after normalization) and you can explicitly do the calculations. Word of caution for $\Delta p$, expand it in Fourier series rather than staying in real space with $-i\partial_x$, especially in the second example, as it won’t work. – LPZ Apr 27 '22 at 16:27
  • Sorry the second example should have been $1-(2x/L)^2$ – LPZ Apr 27 '22 at 17:34
  • Yes, sorry, I meant $\hat{p}\hat{x}$ of course. I am not sure I have understood what you meant to say in the last lines of the comment. Just to be sure, if this does not bother you, can you clarify? Thanks – RH_ss Apr 27 '22 at 19:33
  • No problem. To calculate $\Delta p^2$, you can either stay in position space and use: $\Delta p^2=\int \psi^*(-i\partial_x-\langle p \rangle)^2\psi dx$ with a careful limiting procedure when you’re at the boundary of the domain, or go to momentum space $\Delta p^2=\sum_p(p-\langle p \rangle)^2|\phi |^2$ which is more convenient as you have a Hilbert basis. – LPZ Apr 27 '22 at 19:55
  • Let us continue this discussion in chat. – RH_ss Apr 27 '22 at 20:25