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So I recently attended some lectures at a summer school on basic equilibrium statistical mechanics and couldn't really understand the following slide: enter image description here

The part I don't really get is how you change from $p^2$ (which I guess really is $<p^2>$) to $(\Delta p)^2$. The justification given was that because the average momentum is zero by symmetry, then the variance is $<p^2>$. However, I don't see why the variance of the energy of a single particle within a many-particle system is the same thing as the typical uncertainty in momentum arising due to quantum mechanics. I mean it seems to be equating the "classical" uncertainty of saying we only know the probability distribution of momenta because this is a large system of particles with the fundamental uncertainty in the momenta of a single particle's wavefunction that comes from quantum mechanics.

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    $\begingroup$ By definition, $\Delta p=p-\langle p \rangle$. And, as you mentioned, $\langle p \rangle$ is zero by symmetry. $\endgroup$ – pathintegral Aug 25 '16 at 15:55
  • $\begingroup$ OK but let's pretend that everything is completely classical - this is still how you would define the "uncertainty" because you are describing a system of many particles with a probability distribution. I don't see any reason why it should be the same thing as the typical fundamental uncertainty that arises due to the need to describe particles with a wavefunction in quantum mechanics. $\endgroup$ – Henry Aug 26 '16 at 11:06
  • $\begingroup$ I think this slide is concerned with quantum uncertainty of momentum of a single particle only. No statistical uncertainty yet. $\endgroup$ – pathintegral Aug 26 '16 at 13:51
  • $\begingroup$ No saying that $<p>=0$ is saying that statistically the average is zero, you can't say this applies to a typical quantum mechanical wave function in the system (or at least I can't see the justification for doing so) $\endgroup$ – Henry Aug 26 '16 at 15:34

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