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We don't get eigenstates of momentum when we operate momentum operator in the wave function of particle in a 1D box problem yet we say momentum is quantized in this situation. Why is it so?

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    $\begingroup$ You do get eigenstates when you impose periodic boundary conditions. $\endgroup$ Commented Jun 4, 2014 at 15:34
  • $\begingroup$ Related: physics.stackexchange.com/q/45498/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Oct 28, 2016 at 13:25
  • $\begingroup$ Momentum isn’t quantized for the particle in a box. $\endgroup$
    – tparker
    Commented Jul 11, 2022 at 16:52

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Taking the standard $[0,L]$ problem, eigenfunction and energy eigenvalues are: $$ \varphi_n=\sqrt{\frac{2}{L}}\sin\frac{n\pi x}{L}, \qquad E_n=\frac{\hbar^2\pi^2n^2}{2mL^2}. $$ This means that stationary 1D box systems (e.g. insulated ones) only admit states with a discrete set of possible energies, as above.

Now, that as far energy is concerned. What about linear momentum? We know that by observing momentum we force the system to collapse into a momentum eigenstate, which would be of the form: $$ \phi_p = \frac{1}{\sqrt{2\pi\hbar}}e^{ipx/\hbar}, $$ where $p$ is continuous. Indeed, the probability for a measurement performed on an insulated 1D box system in the $n$-th energy level to produce the momentum value p is given by: $$ |\langle\phi_p|\varphi_n\rangle|^2. $$ However, we can think of the wave vector $k$ namely: $$ k=\frac{n\pi}{L} $$ which is of course quantized, and then invoking the analogy between this writing and the usual theory of harmonic waves write: $$ p=\hbar k = \frac{n\pi\hbar}{L}, $$ such formula looks reasonable, since $$ E=\frac{p^2}{2m}=\frac{\hbar^2\pi^2n^2}{2mL^2} $$ and we recover the correct formula for the energy levels. However, that expression for $p$ is just based on analogy and does not actually mean that the momentum measurement on the $n$-th energy eigenstate will yield $n\pi\hbar/L$ as a result.

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Recall the Heisenberg principle which tells us that a we cannot know both position and momentum at the same time $$\Delta x \Delta p \geq \hbar/2$$ For a particle, we have actually at most $\Delta x = L$, so that we cannot in principle have a sharp momentum state.

Fortunately, there are eigenstates which are eigenstates of $\hat{P}^2$, that is, states superposed of eigenvalues $p$ and $-p$. However, the gap in between them cannot be arbitrary small, so the momentum $p,-p$ gets quantized as $p_n = \pm \frac{n h}{2 L}$.

That is, we find particles only with discrete (quantized) momenta $\pm p_n$ but once measuring the momentum, we actually did not collapse the wave function on a $p_n$ state but on the $\pm p_n$ superposition and in a second measurement we cannot be sure about the sign of the momentum. The classical picture of a particle bouncing of elastically of the walls can be of help to understand this.

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  • $\begingroup$ The energy eigenstates for the particle in a box aren’t eigenstates of $P^2$. They have a continuous spread of momenta. $\endgroup$
    – tparker
    Commented Jul 11, 2022 at 16:54
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I change totally my answer and acknowledges that indeed the momentum does not have stationary eigenvalues in this problem. That is because there is no invariance of the system by spatial translation and therefore translational momentum is not conserved. This implies for a start that it does not commute with the hamiltonian and that thus hamiltonian eigenstates are not momentum eigenstates.

The physical reason is pretty easy to grasp. A stationary state in a box involves a quantum particle bouncing back and forth against the walls of the box. This means that if you start with an initial state which is a momentum eigenstate with momentum $p$, it will reach a wall of the box within a time of the order of $mL/p$ with $m$ the mass and $L$ the size of the box.

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  • $\begingroup$ Hm. You say momentum is quantized in a 1D box, but if you take the fourier transform of the energy eigenstates $\psi(x)$ you get a continuous range of momenta. So what do you mean by momentum is quantized? $\endgroup$
    – user35033
    Commented Apr 29, 2014 at 9:54
  • $\begingroup$ I don't get it...the wave functions in a box are $\psi_n(x) \propto e^{ik_nx}$ with $k_n = \frac{2\pi n}{L}$ and the eigenvalues of $P$ are then $p_n = \hbar k_n$. If this is not quantized, I wonder how you call it. $\endgroup$
    – gatsu
    Commented Apr 29, 2014 at 10:34
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    $\begingroup$ The eigenfunctions of momentum are $exp(ikx)$ but these functinos don't satisfy the boundary conditions of the particle in a box. Operate with $p$ on $\psi(x)=\theta(x)\theta(1-x)\sin(n \pi x / L)$ -- you don't get an eigenvalue. $\endgroup$
    – user35033
    Commented Apr 29, 2014 at 16:59
  • $\begingroup$ My bad, I am too used to the treatment in statistical mechanics.... $\endgroup$
    – gatsu
    Commented Apr 29, 2014 at 19:35
  • $\begingroup$ Both {cos(kx),sin(kx)} and {exp(ikx),exp(-ikx)} form orthogonal bases. If you impose periodic boundary conditions on Aexp(ikx)+Bexp(-ikx) you still get sin(pinx/L) (from Euler's identity), and k=2pi*n/L are the momentum eigenvalues. Is this helpful? $\endgroup$ Commented Jul 5, 2014 at 20:05
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Solving the schrodinger equation gives us: $$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}=E\psi\:\:\:\xrightarrow{\kappa^2\:\equiv\:2mE/\hbar^2}\:\:\: \frac{d^2\psi}{dx^2}=-\kappa^2\psi\:\:\:\rightarrow\:\:\:\psi(x)=Ae^{i\kappa x}+Be^{-i\kappa x}$$

With the boundary condition and normalization, the solution takes the form \begin{align} \psi_{n}(x)&=A\sin(\kappa x)=\sqrt{\frac{2}{L}}\sin\left(\frac{n\pi}{L}x\right)\\ \\ E&=\kappa^2\hbar^2/2m=\frac{n^2\hbar^2\pi^2}{2mL^2} \end{align}

This wavefunction is not the eigenfunction of momentum operator. \begin{align} \hat{P}\psi(x) &= -i\hbar\frac{\partial}{\partial x}\psi(x) = -\sum_{n=1}^{\infty}i\hbar c_{n}\frac{\partial}{\partial x}\psi_{n}(x)=-\sum_{n=1}^{\infty}i\hbar c_{n}\sqrt{\frac{2}{L}}\frac{n\pi}{L}\cos\left(\frac{n\pi}{L}x\right)\\ \hat{P}^2\psi(x) &= -\hbar^2\frac{\partial^2}{\partial x^2}\psi(x)=\sum_{n=1}^{\infty}\hbar^2c_{n}\sqrt{\frac{2}{L}}\left(\frac{n\pi}{L}\right)^2\sin\left(\frac{n\pi}{L}x\right)=\left(\frac{n\hbar\pi}{L}\right)^2\psi(x)\end{align}

Since energy eigenstates are not momentum eigenstates implies $[H, P]\ne0$ and thus we can not simultaneously find all eigenstates of both $\hat{H}$ and $\hat{P}$. However $\hat{P}^2$ does have an eigenstate with eigenvalue $(n\hbar\pi/L)^2$. Nevertheless, it would be naive to directly associate $p=\hbar\kappa$ to assert that momentum is quantized. To quote wiki

In this sense, it is quite dangerous to call the number $k$ a wavenumber, since it is not related to momentum like "wavenumber" usually is. The rationale for calling k the wavenumber is that it enumerates the number of crests that the wavefunction has inside the box, and in this sense it is a wavenumber. This discrepancy can be seen more clearly below, when we find out that the energy spectrum of the particle is discrete (only discrete values of energy are allowed) but the momentum spectrum is continuous (momentum can vary continuously) and in particular, the relation $E=\frac{p^{2}}{2m}$ for the energy and momentum of the particle does not hold. As said above, the reason this relation between energy and momentum does not hold is that the particle is not free, but there is a potential V in the system, and the energy of the particle is $E = T + V$ , where T is the kinetic and V the potential energy.

But there are solutions like this : $$\psi_{n}(x)= -A\theta(x)\theta(L-x)e^{\frac{n\pi}{L}x}$$

with normalization $A=\frac{1}{L}$ But this is momentum eigenstate with complex eigenvalue $$\hat{P}\psi_{n}(x)= -i\hbar\frac{n\pi}{L}\psi_{n}(x)$$

Just like this post implies about momentum operator. There are other solutions as well with different eigenvalues of the momentum operator. But from this, it is clear that even though $\hat{H}$ and $\hat{P}$ share some common eigenstate. They don't share them all and thus it may not be possible to always diagonalize them. However, $\hat{H}$ and $\hat{P}^2$ can be diagonalized simultaneously. Thus the interpretation of $p=\hbar\kappa$ is wrong. The parameter $\kappa$ is not wavenumber and so does not imply quantization of momentum.

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    $\begingroup$ The logic in your answer breaks at the point of "The solution takes the form ... this has no momentum eigenstate". The solution you put there is just the energy eigenvalue solution. But there are many more solutions to the TDSE, which are all linear combinations of the type of solution you have written down. So this does not prove that there is no momentum eigenstate, only that the energy eigenstates are not simultaneous energy and momentum eigenstates. $\endgroup$ Commented Jul 11, 2022 at 11:38
  • $\begingroup$ @doublefelix Thanks for correcting me $\endgroup$ Commented Jul 11, 2022 at 14:16

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