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A particle bound in an infinite potential wall at $x=0$ will apply a force on the wall. For a plane wave and imagining it as a fluid bouncing off the reflection wall at $x=0$, find the force in terms of $\phi'(0)$, the derivative of wave function at $x=0$.

We know, for a reflecting fluid, $\textrm{force }= \frac{dp}{dt}$, where $p$ is momentum. Classically, $p=mv$. In our case, $v$ is $\hbar k/m$, the velocity of the plane wave. Instead of mass, we have $|\phi|^2dx$, the probability of finding the massive particle in an interval $dx$. So $p= \frac{\hbar k}{m}|\phi|^2 dx$. Then, force $f=\frac{dp}{dt} =\frac{d(|\phi|^2 dx \hbar k/m)}{dt}$

By the continuity equation, $\frac{d(|\phi|^2)}{dt}=-\frac{dj}{dx}$, where $j$ is probability current = $\hbar k/m |\phi|^2$. So $f=\frac{dp}{dt}=\left(\frac{\hbar k}{m}\right)^2 \frac{d(|\phi|^2)}{dx} = \frac{\hbar k}{m} 2 \phi \phi' dx$.

But for infinite potential, $\phi(0)=0$, so this gives 0. So I have an unwanted $\phi$ and a pesky $dx$ in my problem. I'm stumped for days, I'd appreciate all help. I'm fairly sure I should be getting an answer of force ~ $\phi'(0)^2$.

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Your answer gets derailed at the point where you try to apply the two classical equations $$p=mv\textrm{ and }F=\frac{dp}{dt}$$ to this (very) quantum problem. To see why this is fundamentally flawed, notice that the plane-wave situation you consider (not exactly a plane wave, though: it's two counter-propagating plane waves) is a stationary state, so that all time derivatives are zero.

You then make a couple of slips that get you badly off track. Equating $p=\frac{\hbar k}{m}|\phi|^2 dx$ can't be right since it's a finite quantity equal to an infinitesimal. Though in this particular situation the equation $j=v|\phi|^2$ kind-of holds (instead of the actual current, $j=\frac{\hbar}{2im}\left(\phi^* \frac{d\phi}{dx}-\phi \frac{d\phi^*}{dx}\right)$), here you must add the currents of left- and right-going particles to get $j=0$.

The right approach to this problem needs a fundamental re-think of what we mean by force in this context, and indeed in all of quantum mechanics, where energy is a far more fundamental concept than force (which becomes merely the gradient of the potential). Suppose the wall slipped to the left a small amount $\Delta x$, over a long time. What would be the work done by the particle on the wall? This work would then have to come from the particle's (purely kinetic) energy. How would that change with the slight enlargement of the box?

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To answer the question you posed at the very end. One can use f=dE/dl,It works for energy eigenstates, but we soon run into problems because energy is not well defined for superposition states. –  Prathyush Nov 5 '12 at 7:00
    
Yes, that is quite right. That just tells you that you need to be careful by what you mean by force! You can define it to be $F=\frac{d}{dL}\langle E\rangle$ and that makes a definite statement about whoever is pushing the wall being a classical entity. You can also define an operator $\hat{F}=\frac{d}{dL}\hat{E}$ (easiest path: on each basis state and then by linearity) and then you can ask quantum mechanical questions on $\hat{F}$ on the particle - or even on a second, slower (entangled?) quantum system whose coordinate is $L$. –  Emilio Pisanty Nov 5 '12 at 14:16
    
Perhaps it is relevant to ask how does one measure the force on the wall? –  Prathyush Nov 5 '12 at 15:34
    
Yes. And these considerations ought to wait until one has a definite answer (out of the several reasonable ones). –  Emilio Pisanty Nov 5 '12 at 16:36
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