Is there a minimum value different from zero for the uncertainty in momentum for a particle in a box? 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.
 A: 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.
