# Why can we say that at zero absolute temperature is there only one accesible state?

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?

• If zero-termperature quantum states can be degenerate, then it's not true. But that requires some kind of unbroken symmetry. Feb 3 '19 at 20:10
• Possible duplicate: physics.stackexchange.com/q/109941/50583 and its linked questions Feb 3 '19 at 20:16
• @ACuriousMind not the same question, thelast answer only takes he numbe of miscrostates as 1 but does not explain why Feb 3 '19 at 20:27
• Feb 3 '19 at 20:38
• Related (possible duplicate?): physics.stackexchange.com/questions/260978/… Feb 4 '19 at 18:14

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.