Example

As an example of the equivalence between the microcanonical and canonical ensembles, consider the calculation of the internal energy $E$ in a system of N two-level particles. The goal of this pedagogical example is to show that the ensemble average internal energy is the same when computed according to the canonical or microcanonical ensembles.

Microcanonical ensemble: In a system of N two-level particles (e.g., N spins that can be up or down) each particles can be assumed to be either in the ground state with energy equal to zero, or in the excited state with energy $\epsilon$. The total internal energy is

$$E = m \epsilon = \sum_{j=1}^N n_j \epsilon,$$ (50)

where $n_j=0, 1$ and $m$ is the number of particels with energy $\epsilon$. The number of possible states $\Omega(E)$ with energy $E = m \epsilon$,

$$\Omega(E) =(_m^N) = \frac{N!}{m!(N-m)!},$$ (51)

determines the entropy of the system according to Eq. (46),

$$S = k \Omega(E),$$ (52)

and, therefore, the average temperature $T$ of the system according to Eq. (39),

$$\frac{1}{T}= \frac{\partial S}{\partial E} \Bigg )_N = \frac{\par......rtial m}{\partial E} \Bigg )_N =\frac{k}{\epsilon} \frac{\partial}{\partial m}( N!- m!- (N-m)!),$$ (53)

since according to the Stirling formula,

$$N! \approx N N- N,$$ (54)

ln$m! \approx m$   ln$m - m$, etc. Therefore,

$$\frac{1}{T}= \frac{k}{\epsilon}(0-lnm-1+1 + ln(N-m)+ \frac{(N-m)}{(N-m)} -1),$$ (55)

or

$$\frac{1}{T}= \frac{k}{\epsilon} \Bigg (\frac{N-m}{m} \Bigg) =\frac{k}{\epsilon} \Bigg (\frac{N}{m} -1 \Bigg ).$$ (56)

Thus,

$$\beta \epsilon = \Bigg (\frac{N}{m} -1 \Bigg) \Rightarrow (\beta \epsilon ) +1 =\frac{N}{m},$$ (57)

and, therefore, the internal energy $E$ is obtained as follows,

$$\boxed{\frac{m}{N}= \frac{1}{1+\text{exp}(\beta \epsilon)}}\Rightarrow \boxed{E= \frac{N \epsilon}{1+\text{exp}(\beta \epsilon)}}$$ (58)

Canonical Ensemble: The partition function of the canonical ensemble is

$$Z = \prod_{j=1}^N \sum_{n_j=0}^1 e^{-\beta \epsilon n_j} = (1+e^{-\beta \epsilon})^N,$$ (59)

Therefore,

$$Z = N (1+e^{-\beta \epsilon}),$$ (60)

and

$$E= -\frac{\partial \text{ln} Z}{\partial \beta} \Bigg )_{N,V} = \frac{N \epsilon }{1+ e^{\beta \epsilon}},$$ (61)

which coincides with Eq. (58).