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Quantum thermodynamics speaks of probability distributions of observables. These are mysterious and complicated quantities that arise from measurement - a complicated process.

Question

Has no one spoken of a wavefunction thermodynamics? What I mean by this is that in classical thermo, we could distinguish members of an ensemble via its configuration $(\vec x_i , \vec p_i)$. We spoke of probabilities on these configurations $P[(\vec x_i , \vec p_i)]$. We can also uniquely characterize a quantum system via the collection of complex amplitudes $(c_i)$. Why does no one care about $P[(c_i)]$?

Motivation

It seems to me that if one cares about the particulars of the dynamics that is really going on, one would need to know the amplitudes very well, they determine how the system will evolve due to interference, so it would be important to speak of their probabilities as opposed to observable probabilities.

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    $\begingroup$ There is such a thing. It's called the density matrix. $\endgroup$ Jun 4, 2018 at 21:58
  • $\begingroup$ My guess is that it's because most of the states you derive in statistical mehcanics require the paritition function. Maybe since the partition function simply "counts" all possible configuration states, it doesn't care about quantum interference of these states....hmm..but what would the partition function for hong-ou-mandal be? would there be 2 states or 4? $\endgroup$ Jun 4, 2018 at 22:03
  • $\begingroup$ Have you tried a google search for quantum statistical mechanics? $\endgroup$ Jun 5, 2018 at 0:19
  • $\begingroup$ @Enrique Mendez please clarify if you are talking about the density matrix, which seems to be precisely what you're talking about. $\endgroup$ Jun 5, 2018 at 4:47
  • $\begingroup$ So the density matrix is different I think. The density matrix is an average over pure states $\sum_i p_i |i><i|$, it's a summed or averaged quantity over the probability distribution. It is not $P[ (c_i) ]$. I do grant that you can pull out $p_i$ out but that's dependent on the basis you choose. Presumably there would be a unique $P[ (c_i)]$. $\endgroup$ Jun 5, 2018 at 16:55

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