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Suppose I have a 1D box of length $L$ and I put $n\gg1$ fermions in it at zero temperature. Using the energy levels for a particle in a box, the total energy is

$$E\sim \frac{n^3}{L^2}$$

If I have $k$ such copies of these boxes, the energy is $kE$. But next suppose I put these boxes end-to-end and drop the walls, so I now have a single box of length $kL$ and with $kn$ fermions. The energy is

$$E_k \sim \frac{(kn)^3}{(kL)^2} = kE$$

So putting the boxes together and dropping the walls didn't change anything.

The "pressure", $dE_k/d(kL)$, (or "tension", whatever you want to call it in 1D) is the same as for a single box, $dE/dL$. In other words, degeneracy pressure is intensive.

Other than calculating it, is there some reason this should be so? Is it just a lucky coincidence? Because the quantum particles "feel" the entire volume of the box instead of being localized, it seems I would want to think about the entire system at once, and pressure would instead depend on the size of the system, but it doesn't. Why not? Is there some lesson here I should learn about quantum statistical mechanics more generally?

Mark Eichenlaub
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    The fact that energy is extensive and pressure intensive is true also in classical physics. So why are you surprised to find the same result in quantum physics? I'm just curious. – valerio Jul 18 '17 at 06:46
  • @valerio92 Quoting the question: "Because the quantum particles "feel" the entire volume of the box instead of being localized, it seems I would want to think about the entire system at once, and pressure would instead depend on the size of the system" – Mark Eichenlaub Jul 24 '17 at 10:49

2 Answers2

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Semiclassically, quantum particles in $n$ dimensions take up a volume $h^n$ of phase space. Therefore, if you keep the position-space density constant, the volume of momentum space per particle stays the same. That means that the characteristic momenta of the particles is intrinsic. Since pressure goes with energy density, this implies the pressure is intrinsic as well.

knzhou
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    This reduces the question to something more general, but still feels a bit like begging the question. Why should quantum particles take up a volume of phase space per particle that's independent of the number of particles? (Perhaps everyone but me already understands this...) – Mark Eichenlaub Jul 18 '17 at 02:19
  • @MarkEichenlaub Would you accept "the WKB approximation says $\int p dx = nh$"? I asked the same question once too, and apparently it's a pretty deep general feature of semiclassical QM. – knzhou Jul 18 '17 at 02:28
  • Ahhh, that gives me a lot to think about. Thanks, Kevin! – Mark Eichenlaub Jul 18 '17 at 02:34
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First of all, I don't think this is true in general. Specifically, for systems with sufficiently long-range interactions, like $V\sim r^{-d}$ or longer range, the Fermi pressure will no longer be intensive. Indeed, the energy density will not be intensive for any particle statistics, and as a result the thermodynamic limit does not exist for such systems.

So a more general form of this question would be to ask: why does the thermodynamic limit exist for non-interacting particles? But I guess it is pretty much the definition of "non-interacting" that the particles don't change their energies when you bring several of them together!

Rococo
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