While studying the microcanonical ensemble, the entropy definition requires that at T=0 there is only one accesible state so that the entropy, S=0. Why is it true?
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2If zero-termperature quantum states can be degenerate, then it's not true. But that requires some kind of unbroken symmetry. – Peter Shor Feb 03 '19 at 20:10
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Possible duplicate: https://physics.stackexchange.com/q/109941/50583 and its linked questions – ACuriousMind Feb 03 '19 at 20:16
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@ACuriousMind not the same question, thelast answer only takes he numbe of miscrostates as 1 but does not explain why – miguel pedraza Feb 03 '19 at 20:27
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Related? https://physics.stackexchange.com/questions/299331/explicit-nontrivial-examples-in-quantum-mechanics-ground-state-degeneracy – dmckee --- ex-moderator kitten Feb 03 '19 at 20:38
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Related (possible duplicate?): https://physics.stackexchange.com/questions/260978/the-correct-statement-of-the-third-law-of-thermodynamics – valerio Feb 04 '19 at 18:14
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It is not true.
It is a typical misconception originating from a careless connection between statistical mechanics and thermodynamics. Thermodynamic behavior can be recovered from the ensemble approach to Statistical Mechanics only looking at the behavior of entropy, free energies, energy per particle or per unit volume at the so-called thermodynamic limit (by the way, only at the thermodynamic limit results derived in different ensembles become equivalent).
As a consequence, in order to have $$ s(u,v)=0 $$ where $u=U/N$, $v=V/N$, and $s=S/N$, at the state where $\left( \frac{\partial s}{\partial u} \right)_u^{-1}=0$ (which is a statement equivalent to the third law) it is not necessary that $S=0$, but it is enough that it would not diverge with the size as fast as $N$. For example, a logarithmic divergence with $N$ of the ground state degeneracy would be perfectly compatible with the Planck's form of the third principle.
