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I'm learning quantum mechanics on my own. I've known that energy is quantized and I've started wondering about temperature. From thermodynamics we have: $$U=\frac{3}{2}NkT $$ (for ideal gas, of course)
Both U and N aren't continous, so i think T shouldn't be, too. Is that formula correct also for quantum mechanics?
I'm really sorry because of my language, I'm still working on it.

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2 Answers 2

up vote 7 down vote accepted

No. Temperature is a concept of statistical mechanics and only exists in the limit of large numbers of objects. So it's an average. And since the number of particles is (typically) not known exactly, the possible temperature values (due to the quantization of energy) also cannot be known exactly.

That's the best case. More generally, a temperature bath can have any real temperature and so temperature is not quantized.

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In the canonical case it's clear that $T$, if coming from outside, can be continous (the probability $0<p<1$ for some random state is continous too, after all). However, in the microcanonical case of a closed system, where $1/T:=\partial S/\partial E$, where $S$ depends on a counting quantity (some trace over an operator which counts different energies where $[E,E+\Delta]$ gives some energy interval, say $\Gamma (E)$), my question would be what guaranties you that there is a smooth counting quantity, such that the derivative makes sense. –  NikolajK Jul 22 '12 at 23:05
Excuse me, what canonical and microcanonical cases are? And what does $\Gamma(E) $ mean? –  user10768 Jul 24 '12 at 12:45

I think this is energy of system in macroscopic scale and in macroscopic scale the energy is not quantized.

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