# Does a particle in a spherically symmetric infinite square well potential exert a force on the inner and outer shell barrier?

For a particle in the potential:

$$V(r) = \begin{cases} 0 & \text{a < r < b}\\ \infty & \text{otherwise.} \end{cases}$$

Does this guy in the ground-state exert a force on the shells a and b? I don't think so because at ground-state both the centrifugal term of the radial equation and the angular momentum of the particle is zero. What say you brethren?

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What do you mean by force? Like an electromagnetic force? –  santa claus Dec 15 '12 at 1:08
Particles don't like to be trapped and exert a sort of pressure on their boundaries. The force tries to push the boundaries out such that the particle is free. –  Synthetic Dec 15 '12 at 1:37

Yes, the particle in the ground state will exchange momentum and exert a pressure on the walls of the container in which it is enclosed. This can be inferred from the fact that the ground state energy will increase when the container shrinks adiabatically in size. The pressure follows from equating the energy increase to the integral of $P dV$.
In case of the annular container that you mention, it is simple to see that the ground state energy increases when the annular gap $b-a$ decreases. The fact that in this example the container is spherically symmetric with vanishing angular momentum (and hence zero centrifugal barier) is just a distraction. A particle doesn't need angular momentum to exchange momentum with the walls.
Yes, $P(V)$ represents the pressure and $dV$ the change in volume. –  Johannes Dec 15 '12 at 1:54
Ok. My math is a bit weak I guess. I'm having trouble translating this into anything worthwhile. I have a volume $$V = 4 \pi r*(b-a)$$ and energy $$E_0$$. How would I calculate the pressure? Thanks for the answer by the way. It makes sense when you see the change in ground-state energy. –  Synthetic Dec 15 '12 at 2:16
Is this a homework problem? A few hints: the correct equation for the volume is $V = 4 \pi (b^2-a^2)$, and it is easy to calculate the ground state energy by separation of variables (you will find an energy that is a function of $b-a$). –  Johannes Dec 15 '12 at 2:23
Oops... That should be $V = (4/3) \pi (b^3-a^3)$. The equation $V = 4 \pi r^2 (b-a)$ ($r=(b-a)/2$) simplifies the math, but works only for thin shells ($b-a << r$). –  Johannes Dec 15 '12 at 2:34