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Suppose I want to describe an arbitrary state of a quantum particle in a box of side $L$. The relevant eigenmodes are those of standing waves, namely $$ \left<x|n\right>=\sqrt{\frac{2}{L}}\cdot \sin \left(\frac{n\pi x}{L}\right)$$ In this basis, the operate $\hat{p}^2$ is diagonal by construction, so every state has a definite energy.

But suppose I want to construct this operator from the matrix elements of $\hat{p}$? These are ($\hbar=1$) $$\left<m|\hat{p}|n \right>=-\frac{2imn((-1)^{m+n}-1)}{L\cdot(m^2-n^2)}$$ which vanishes for $m$ and $n$ with the same evenness, making this an off-diagonal "checkered" matrix.

I tried taking a finite portion of this matrix and squaring it in order to obtain $\hat{p}^2$, but from what I could tell, this yields an on-diagonal checkered matrix, contrary to the construction that it would be strictly diagonal.

I assume this has something to do with the fact that the wavefunctions obtained by applying $\hat{p}$ do not satisfy the same boundary conditions as the basis functions (do not vanish). Would this problem be solved I had taken the the infinite matrix in its entirety before squaring? Or is this entire problem (decomposing $\hat{p}$ in this basis) ill-defined?

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The Hilbert space for a particle in a box has no associated momentum operator (as a momentum operator implies that the state space is invariant under space translations). So your attempt do define one leads to artifacts.

[Edit] In general, the Hamiltonian must be a densely defined operator on the physical Hilbert space. But a square well potential is too singular as an operator on the space of all square integrable functions on $R$ (i.e., no densely defined subspace is mapped by it into the Hilbert space), hence its Hilbert space cannot be the standard Hilbert space.

[Edit2] Note that the usual formula for the momentum is not a canonical momentum on the restricted space of wave functions defined only in the box, as it fails to satisfy either the commutation relation or does not preserve the boundary conditions.

The appropriate replacement for the momentum operator is the mode counting operator $\hat n$ with $\hat n|n\rangle=n|n\rangle$; see also the derivation of Qmechanic.

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Shouldn't the momentum operator exist anyway, and any invariance of the Hamiltonian just imply that it (meaning its expected value) is conserved? – Benji Remez Nov 30 '12 at 19:29
@BenjiRemez: see my edited answer. It is not about any invariance, it is about being able to define the translation as an operator in the same space. The translated potential lives in a different Hilbert space! – Arnold Neumaier Nov 30 '12 at 19:35
Alright, so here's a followup question. I agree that the mode counting operator produces equivalent information. But my problem is that these states do no distinguish between positive and negative momenta (they don't specify a sign). So, if I were to put wave packet in the box with some momentum, moving along, say, the positive $x$ axis, how would such a packet be decomposed in the basis of the eigenstates? And how would it differ if the initial momentum had been negative? – Benji Remez Nov 30 '12 at 19:52
@BenjiRemez: There is no momentum, so the question is moot. Standing waves have no direction, so there sohouldn't be a sign. – Arnold Neumaier Dec 2 '12 at 14:04
@ArnoldNeumaier: In no bound state the momentum has a certain value, but it does not mean it does not exist. Rather, it fluctuates. One can calculate its "variation" $\sqrt{\langle n|\hat p ^2|n\rangle - \langle n|\hat p|n\rangle ^2} $, can't one? – Vladimir Kalitvianski Dec 2 '12 at 16:15

Applying $\hat p$ to a Hamiltonian eigenfunction gives another state indeed, but this state can be decomposed in the eigenstates of the Hamiltonian. Numerically it is good everywhere except, maybe, the boundary points, so for integration purposes such a decomposition (superposition) is good. You did not obtain a diagonal matrix exclusively because you took a finite matrix for $\hat p$.

EDIT: Yes, at $t=0$ you can form a wave packet moving in a certain direction from "plane waves" $exp(i k_n x)$, as usual, and then you can replace each momentum eigenfunction $exp(i k_n x)$ with a linear combination of sines (Hamiltonian eigenfunctions $\psi_m (x)$ with the corresponding time exponentials $exp(−i E_m t),\; t>0)$. In particular, if you choose the largest possible wave packet with a certain momentum $\psi(x,t=0)\propto exp(-ik_n x),\;0<x<L$, then in later moments it will not keep its form because of reflections and the momentum will not conserve.

EDIT 2: Any standing wave $\psi(x)$ on an interval $(0,L)$ can be decomposed into a superposition of the momentum eigenfunctions $ exp(\pm|k_n|x)$. In our case the superposition contains only two running waves with certain momenta: $\sin(k_n x)=(e^{ik_n x}-e^{-ik_n x})/(2i)$, so the momentum operator exists and there is no problem here.

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I'm doing this indeed for numerical purposes. So technically speaking, since the matrix elements of $\hat{p}$ vanish when the difference between $m$ and $n$ increases, would taking a larger finite matrix result in the off-diagonal elements of $\hat{p}^2$ getting smaller? – Benji Remez Nov 30 '12 at 19:33
Yes, I think so. – Vladimir Kalitvianski Nov 30 '12 at 20:16

The operator is well-defined for instance in the position basis the operator corresponding to $\Psi(x) = \langle x | n \rangle$ is given by the ordinary form $\hat{p} = -i \hbar (d/dx)$ [*].

What happens is that $\Psi(x)$ are not eigenfunctions of the momentum operator $\hat{p}$ and, therefore, you cannot find a spectral decomposition of the operator in that basis (and, as a consequence, you cannot find a diagonal matrix).

Notice that $\Psi(x)$ are eigenfunctions of the operator $\hat{p}^2$ and it is possible to obtain the spectral decomposition

$$\hat{p}^2 = \sum p_n^2 | n \rangle \langle n |$$

with the corresponding eigenvalues $p_n^2 = n^2h^2/4L^2$.

As a final note, it is not possible to obtain the matrix $\mathbf{p}$ from the matrix $\mathbf{p}^2$. First because the sign of $p$ is not defined and second because $\pm p$ does not correspond to a measurement of the momentum.

[*] By simplicity I am assuming one dimension.

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A geometric approach can be achieve via an orbifold construction.

I) Start with a free 1D quantum particle $\psi(x)=\psi(x+2\pi R)$ on a circle $S^1$ of circumference $2\pi R=2L$. The momentum $p=n\frac{\hbar }{R}$ is then quantized, $n\in\mathbb{Z}.$ The Hilbert space $H$ has momentum basis $(| n\rangle)_{n\in\mathbb{Z}}$, where

$$\hat{p} | n\rangle~=~n\frac{\hbar }{R}| n\rangle, \qquad \langle n| m\rangle~=~\delta_{n,m}, \qquad n,m\in\mathbb{Z}. $$

II) The orbifold
$$I ~=~ S^1/ \mathbb{Z}_2~=~ S^1/\sim$$ is an interval of length $L$. A position $x \sim -x$ is identified with the opposite position $x\in S^1$ via a discrete $\mathbb{Z}_2$-group action. To implement the two Dirichlet boundary conditions


we impose that the wave function should be odd $\psi (x) = -\psi(-x)$. This kills half of the previous Hilbert space. The $\mathbb{Z}_2$-reduced Hilbert space $H/\sim~$ has basis $(| n\rangle_{\sim})_{n\in\mathbb{N}}$, where

$$|n \rangle_{\sim}~:=~\frac{|n\rangle - |-n\rangle}{\sqrt{2}}=-|-n \rangle_{\sim}, \qquad {}_{\sim}\langle n| m\rangle_{\sim}~=~\delta_{n,m}- \delta_{n,-m}, \qquad n,m\in\mathbb{Z}.$$

We can implement the orbifold construction through a parity operator $P$,

$$P\hat{x}P~=~-\hat{x}, \qquad P\hat{p}P~=~-\hat{p}, \qquad P^2~=~{\bf 1}, \qquad P^{\dagger}~=~P.$$

A ket state $|\psi \rangle~\sim 0 $ is identified with zero if it is even, $P|\psi \rangle= |\psi \rangle$. The new basis $|n \rangle_{\sim}$ is odd, $P|n \rangle_{\sim}=-|n \rangle_{\sim}$. The absolute value $\hat{|p|}$ of the momentum operator is a $P$-even (unbounded) operator. It is diagonal in the $| n\rangle_{\sim}$ basis, $$ \hat{|p|}~ | n\rangle_{\sim}~=~ |n|\frac{\hbar }{R}| n\rangle_{\sim},\qquad {}_{\sim}\langle n| ~\hat{|p|}~|m\rangle_{\sim}~=~ |n| \frac{\hbar }{R}~ {}_{\sim}\langle n| m\rangle_{\sim}\qquad n,m\in\mathbb{Z}. $$

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Introducing multivalued operators would be a novelty in quantum mechanics, with unexplored consequences (but probably leading to inconsistency). Indeed, your final formula for the action of $\hat p$ is not what you get wehn you substitute the original definitions on the left hand side. – Arnold Neumaier Dec 2 '12 at 17:34
I updated the answer. – Qmechanic Dec 2 '12 at 18:16

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