Considering a two level system with energies $ 0 $ and $ \epsilon$, we write out the single particle partition function with ease to be, also N-particle partition function for non-interacting particles. $$ Z_1 = 1+e^{-\beta\epsilon}, \qquad Z_N = (1+e^{-\beta\epsilon})^{N} $$

Further, we can deduce from $F=-kT\ln Z_N$, that $$ F = -NkT\ln (1+e^{-\beta\epsilon}) $$ and the entropy deduced from this, $$ S = -\frac{\partial F}{\partial T} = -Nk\ln (1+e^{-\beta\epsilon})-\frac{N\epsilon}{T}\frac{1}{1+e^{-\beta\epsilon}} $$

However, now if we work this system out in microcanonical ensemble, the total no. of microstates can be written out to be (using the constraint $n_1+n_2 = N$), $$ \Omega = \sum_{n_1 = 0}^{N}\frac{N!}{(N-n_1)!n_1!} = 2^N $$

From this, if we calculate the entropy, $S= k\ln \Omega \implies S = Nk\ln 2$. This is completely independent of temperature and entirely different fro the one we derived using Canonical Partition function.

Where does the definition of temperature come in the microcanonical formalism for this problem ??

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    $\begingroup$ You need to fix the total energy in your microcanonical computation. $\endgroup$ Commented Sep 16, 2014 at 6:51

1 Answer 1


The microcanonical ensemble is the (maximum entropy) probability distribution for a given specified total energy. What you've calculated is actually the maximum entropy distribution with no constraints on the energy, which is the same as the canonical distribution at infinite temperature ($\beta = 0$).

To correctly calculate the microcanonical entropy, first observe that in this system, the total energy $E=\epsilon n_1$, or $n_1 = E/\epsilon$, as long as $E$ is an integer multiple of $\epsilon$.

Since the total energy completely determines $n_1$, we can calculate the number of states in the microcanonical ensemble just as you did, but without the summation over $n_1$:

$$\Omega(E) = \frac{N!}{(N-E/\epsilon)!(E/\epsilon)!}$$

This is only defined when $E$ is an integer multiple of $\epsilon$, but in an appropriate limit of large $N$ and small $\epsilon$ you should be able to use Stirling's approximation to write $S(E) = \log \Omega(E)$ as if it were a continuous function of $E$.

Once this is done, the temperature enters through the usual thermodynamic definition, $$ \frac{1}{T} = \frac{\partial S}{\partial E}, $$ which ought to be fairly straightforward to calculate.

I hope this helps, and sorry for not working through all the details myself.


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