0

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.

Stoby
  • 520
  • 1
    Related: What's the deal with momentum in the infinite square well? – Emilio Pisanty Feb 24 '21 at 21:17
  • 1
    #Sigh#. Asking people to not use uncommon abbreviations or acronyms without defining them is a fruitless task. But Il try to knock one apple off of the tree here. What the heck is PID? and please define abbreviations and acronyms on first use. What defines "uncommon"? I can't define it, but I know it when I see it. :-) Choose wisely and conservatively. – garyp Feb 26 '21 at 13:06
  • @garyp its PIB, particle in a box, pretty common in quantum chemistry, where my training is, but this is a physics group, so I understand the confusion, and I will fix. Thanks – Stoby Feb 26 '21 at 20:35
  • If you see it as an eigenvalue problem for P and the „physical” boundary conditions $\psi (0) = \psi (L) = 0$, you obtain $\psi_n (x) \equiv 0$. – DanielC Feb 26 '21 at 22:18

2 Answers2

2

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.

Philip
  • 11,393
1

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.

Andrew
  • 48,573