The momentum operator in one dimensional quantum mechanics is: $$\hat p_x=\frac{\hbar}{i}\frac{d}{dx} $$ and we can imagine creating an eigenvalue-eigenfunction system $$\hat p_x\psi = p_x\psi.$$ As a student of ODE, I see here a Sturm-Liouville problem if we let $\lambda=-i\hbar p_x$ then we can say $\psi'+\lambda\psi=0$. If we then imposed boundary conditions of a 1-D particle in a box such that $\psi(0)=\psi(L)=0$, then this generates a Sturm-Liouville problem which should have eigenfunctions of $ \psi_n(x)=\sin(\frac{n\pi x}{L}) $ and eigenvalues of $\lambda_n=\frac{n^2\pi^2}{L^2}$. However, when we calculate the expectation value of $\hat p $ we get $0$. $$$$How are these solutions physically self-consistent with each other? I am confused because the solution to a PIB in an infinite well is $\sqrt\frac{2}{L}\sin(\frac{n\pi x}{L})$ so the eigenfunction system makes sense, but my eigenvalues don't seem to.
2 Answers
Note also that the expectation value of momentum for any real eigenfunction is zero.
Also, while the accepted answer by @Andrew does give you the right intuition as to why the average momentum is zero, there is a slight subtlety due to the fact that the wavefunction isn't $\sin(px)$ over all space, but rather $\sin(px)$ within the box and zero outside it. As a result, the possible values of momentum aren't just $\pm p$, but rather have some spread, as I discuss in this answer. That being said, the distribution of possible momenta is symmetric about $p=0$, and so @Andrew's argument still holds.
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Note that $\sin(px)=\frac{1}{2i}\left(e^{ipx}-e^{-ipx}\right)$ is a superposition of a state with positive momentum and a state with negative momentum. The average momentum is $p-p=0$.
Your square well problem is a special case of this. Mathematically, the boundary conditions mean that you only get $\sin$ solutions, and not complex exponentials. Physically, a particle in a well can't have any net average momentum in the positive or negative direction, or else it would leave the well.
As Philip pointed out in another answer, there is a small subtlety in that the real wavefunction is $\theta(x)\theta(L-x)\sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)$ (where $\theta(x)=1$ for $x>0$ and $0$ for $x<0$), which is not purely a superposition of 2 complex exponentials. However since it is real, the wavefunction is still of the form $\psi=\phi+\phi^\star$, where $\phi$ is a superposition of positive momentum modes $\phi \sim \sum e^{ipx}, p>0$, and so $\psi$ has zero average momentum.
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