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I don't know if my question has sense at all but while doing my homework there appeared in my mind this question. Say, for a particle in a box, the confinement makes that the wave number k is discrete, depending on integer numbers $\textbf{n}=(n_{x}, n_{y},n_{z})$. If De Broglie's formula holds, it doesn't mean that momentum $\textbf{p}=\hbar \textbf{k}$ is also discrete?

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  • $\begingroup$ Are you talking about a particle in a 3-dimensional box? $\endgroup$
    – Tachyon209
    Apr 9 '20 at 14:16
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No. You can't use the De Broglie formula here because De Broglie's formula is true only for a free particle $(V(x)=0$ everywhere$)$.

For the infinite square well, we can write down the energy eigenvalue equation as: $$\frac{-\hbar^2}{2m}\frac{\partial^{2}\psi(x)}{\partial x^2}+V(x)\psi(x)=E\psi(x)$$ such that $V(x)=0 $ inside the box and $V(x)\rightarrow\infty$ outside. Now, clearly we can see that the Energy operator $\hat{E}$ doesn't commute with $\hat p$ (although they do commute for a free particle potential.) If you try to use the definition of momentum operator $ \hat p_x=(\hbar/i)\partial_x $ on the stationary solutions of infinite well, you'll see that none of them is an eigenfunction of $\hat p$.

The probable reason for your confusion here is naming the constant $2mE/\hbar^2$ as $\textbf k$. Here, $\textbf k$ is not the wavenumber as we had for a free particle. Just name it as something else, and logically try to argue if it still makes sense to call it the same wavenumber. The definition $\textbf p =\hbar\textbf k$ doesn't extend here.

Also, if you are thinking about three dimensions, you can extend the argument very similarly to higher dimensions by symmetry.

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  • $\begingroup$ Ok I got it, I agree with you. $\endgroup$
    – Amadeus
    Apr 11 '20 at 16:00
  • $\begingroup$ Cool. If there's still any doubt, feel free to ask. $\endgroup$
    – Tachyon209
    Apr 11 '20 at 16:01
  • $\begingroup$ But, for example, for a free particle if we simply impose periodic boundary conditions without confinement, the wave number takes again discrete values. What happens there? $\endgroup$
    – Amadeus
    Apr 11 '20 at 16:03
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    $\begingroup$ Sure, I agree with you when you say that for infinite square well I can't use De Broglie's formula since it is true just for a free particle. But if we take the free particle Hamiltonian and impose periodic boundary conditions in a volume V but without confinement, say phi(L)=phi(0) in one axis and generalizing it to higher dimensions, the wave number will take discrete values, just as in the infinite well (same problem but with confinement). Following your answer then De Broglie formula will be still applicable and momentum p takes discrete values. Am I wrong? $\endgroup$
    – Amadeus
    Apr 11 '20 at 16:29
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    $\begingroup$ Yes you said it well. Thanks $\endgroup$
    – Amadeus
    Apr 11 '20 at 18:04
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Yes. But remember, you can think of the wavefunction in quantum physics like a wave in the configuration space of the system, that is isomporphic to the physical space only for a single particle case. Momentum is discrete. Actually, 1 of the 1st quantum observed effects was the discreteness of the energy spectrum, that depends on momentum and electrostatic pontential $ V\vec(r) $ for the electron in Hydrogen atom in an approximated electrostatic potencial within its interactions with the positive charge in nucleus.

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Yes it is applicable. Note that the expectation value of momentum is zero for any stationary solution in a box, so $\langle \psi | \frac{\hbar}{i} \vec \nabla |\psi\rangle = 0$. Still $ \psi | \vec p |\psi = 0$, without the integration over space is the Noether momentum distribution belonging to the Schrödinger lagrangian ${\cal L} = i\hbar \psi^* \dot \psi = -\hbar^2 \vec \nabla \psi^* \cdot \vec \nabla \psi +qV\psi^* \psi$ .

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