So, on a recent quantum mechanics exam (undergrad QM 1) I was given a problem which asked: "an electron is confined to the interior of a hollow spherical cavity of radius R with impenetrable walls. Find an expression for the pressure exerted on the walls of the cavity by the electron in its ground state."
My initial approach (get ready for what I'm sure is some broken logic, sorry in advance): I started with the classical expression for force (Newton's 2nd) F=dp/dt and, noticed, that the Force on the wall is just the radial component of F, so, P=Fr/A=(1/A)dpr/dt, where A=4πR2 but since the system is conservative… Fr=(∂V/∂r)=dpr/dt. Now taking the expectation values of everything I get… $\langle P\rangle$=$\langle F$r$\rangle$=$\langle∂V/∂r\rangle$=d<pr>/dt. Now , this is the point in the problem where I started going in circles because since the particle is in a stationary state and V=0 inside the well I couldn't think of how to find any expectation value which I could get pressure from that wouldn't come out to be 0. I just moved on at this point and didn't do the problem because I was obviously missing something and didn't have time to sit there and really ponder the problem.
Post exam I was thinking that maybe I am supposed to use the first law of thermo or something and act like I am creating the system and thus must do work on the enviroment to "make room for it" but alas that did not yield fruit either. So, I went and asked my professor about the problem and they sent me the following: "The average force the electron exerting radially on the wall is equal to the change of the energy with respect to R , <-dV/dR>
since inside the cavity there is no other potential, the energy of the electron is
it will equal the=<-dH/dR>=-d< H >/dR= -dE/dR."
I don't really understand, first of all, why the partial derivative is w.r.t R rather than r for one, secondly, why the potential being zero somehow lets us replace V with the Hamiltonian (total energy) and how the change of the total energy of an energy eigenstate w.r.t the radius, which, in reality is constant somehow gives us a force on the cavity wall. I'm sorry for being so stupid/dense right now, this is my first semester of QM and I've been doing well so far but this is the first problem I've seen that asks for what force a particle exerts on something else (up until now I've actually never even seen a quantum problem that involves directly with forces at all except indirectly in the form of potential energy functions/operators when finding bound states and/or finding the spin wavefunction for an electron in an external magnetic field so I don't really see what's going on here as I don't have much experience dealing with this sort of a problem quantum mechanically).
If anyone could provide any insight that would be highly appreciated. Also, I don't feel like I'm great at communication to be honest so if anything is unclear and needs rephrasing just ask. Thanks again!
As a disclaimer, I haven't really posted much on here so if my post breaks any rules or something please let me know (this is not a current assignment or anything, rather its me trying to understand something from a previous assignment that I don't understand fully, hopefully, in a generalizable way that leads to a deeper understanding of the theory/model)
Best Answer
We have for pressure $P$ the thermodynamic identity
$$ P=-\frac{dU}{dV}\bigg|_S $$
Where $U$ is the interal energy, $V$ is volume and $S$ is entropy. Of course, these are ensemble quantities which are not even defined for a single particle. We can perform a formal calculation: replacing $U=E_n$, the nth eigen-energy, and ignoring $S$ we have
$$ P_n=-\frac{d}{dV}E_n $$
In the nth energy eigenstate, with $l=0$, $E_n=\frac{n^2\pi^2}{2mR^2}$, $\hbar=1$. Note that $R=R(V)$ since $V=\frac{4}{3}\pi R^3$. Using the chain rule
$$ P_n=\frac{n^2\pi^2}{m}R^{-3}\frac{dR}{dV}=\frac{n^2 \pi}{4mR^5} $$
It should be stressed that this calculation is by formal analogy rather than from a microscopic thermodynamic theory. It is also not a deduction purely from single particle quantum mechanics.
If we let $P=F/A$, with force $F$ and area $A$, the above does agree with what your prof said. Again, note that force is not even defined in quantum mechanics.