Is there any instance where the microcanonical, canonical and grand canonical ensembles give equal results?

For most systems, the derived macroscopic thermal quantities from anyone of the formulations should have a same expression. Lets examine the simplest two-level system, this example appeared in many statistical text books, being treated in details for both microcanonical and canonical ensembles.

The two-level non-interaction system assumes particle can exists in one of the two levels:

begin{align} text{ Level 2 } --- & & text{Number } = N_2 & & text{ Energy } = epsilon text{ Level 1 } --- & & text{Number } = N_1 & & text{ Energy } = 0 end{align}

Since this system has no vloume information. We can only find the average energy $U(T) = langle Erangle$, and the exprssion of entropy $S(T)$.

The canonical ensemble is the simplist one, Lets do it first: given total partical number $N$ and temperature $T$:

One-partical partition function $$ Z_1 = e^{-beta 0} + e^{-beta epsilon} = 1 + e^{-beta epsilon}. $$

Average occupation numbers: $$ N_1 = N frac{e^{-beta 0}}{e^{-beta 0} + e^{-beta epsilon}} = N frac{1}{1 + e^{-beta epsilon}} $$ $$ N_2 = N frac{e^{-beta epsilon}}{e^{-beta 0} + e^{-beta epsilon}} = N frac{e^{-beta epsilon}}{1 + e^{-beta epsilon}} = N frac{1}{1 + e^{beta epsilon}} $$ Therefore $$ tag{1} U = langle Erangle = N_2 epsilon = frac{N epsilon}{1 + e^{beta epsilon}} $$

Helmhotz free energy:

$$tag{2} F = - K T ln Z_N =- NKT ln(1 + e^{-beta epsilon}) . $$

Thus entropy $$ tag{3} S = frac{U-F}{T} = frac{N}{T}frac{epsilon}{1 + e^{beta epsilon}} + NK ln(1 + e^{-beta epsilon}). $$

Micro-canonical ensemble: given N, E

$$ N_2 = frac{E}{epsilon} $$ $$ N_1 = N - frac{E}{epsilon} $$

The microcanonical configurational number: $$ Gamma(N, E) = frac{N!}{frac{E}{epsilon}! left(N -frac{E}{epsilon} right)!} $$ $$ ln Gamma(N, E) = N ln frac{Nepsilon}{Nepsilon-E} + frac{E}{epsilon} ln frac{Nepsilon-E}{E} $$

The entropy $S$ $$ tag{4} S = K ln Gamma(N, E) =NKln frac{Nepsilon}{Nepsilon-E} + Kfrac{E}{epsilon} ln frac{Nepsilon-E}{E} $$

The temperature $frac{1}{T} = frac{partial S}{partial E}]_{E=U}$

$$tag{5} frac{1}{T} = frac{ K }{epsilon} ln frac{Nepsilon-U}{U} $$

Invert the above equation to find $U(T)$ $$tag{6} U = N frac{epsilon}{1 + e^{frac{epsilon}{K T}} } $$ The same as Eq. (1).

Substitute Eq.(5) and Eq.(6) into Eq.(4) $$ tag{7} S = NK ln left( 1 + e^{frac{epsilon}{K T}} right) + frac{U}{T} = NK ln left( 1 + e^{frac{epsilon}{K T}} right) + frac{N}{T} frac{epsilon}{1 + e^{frac{epsilon}{K T}} } $$

Eq. (7) is the same as Eq. (3).

 Grand Canonical Ensemble: given chemical potential and temperature 

For the system having partical number $N$ and total energy $E$, the probability is proportional to $exp(betamu N - beta E)$, and $N_2 = E / epsilon$, $N=N_1+N_2$. And the multiplicity $frac{N!}{N_1! N_2!}$ $$ Z = sum_{N,E} frac{N!}{N_1! N_2!} expleft(betamu N - beta E right) = sum_{N, N_2} frac{N!}{(N-N_2)! N_2!} expleft{betamu (N-N_2) + beta N_2 (mu-epsilon) right} $$

$$ = sum_{N=0,1,2..} sum_{N_2=0}^N frac{N!}{(N-N_2)! N_2!} exp^{N-N_2}left(betamuright) exp^{N_2}left{beta (mu- epsilon) right} $$ $$ = sum_{N=0,1,2..} left{ e^{betamu} + e^{beta(mu-epsilon)} right}^N = frac{1}{left{ 1-e^{betamu} - e^{beta(mu-epsilon)} right} }. $$ $$ ln Z = - ln left{ 1 - e^{betamu} - e^{beta(mu-epsilon)} right} . $$ The grand potential $Omega = -K T ln Z$ $$ Omega = -KTln Z = KTln left{ 1 - e^{betamu} - e^{beta(mu-epsilon)} right}. $$

The average number $bar{N} = langle N rangle$: $$ bar{N} = -frac{partial Omega}{partial mu}=frac{e^{betamu} +e^{beta(mu-epsilon)} }{ 1 - e^{betamu} - e^{beta(mu-epsilon)}} equiv bar{N}_1 + bar{N}_2. $$ Where $bar{N}_1/bar{N}_2 = e^{beta epsilon}$.

The average anergy $U$ $$ U -mu bar{N} = -frac{partial ln Z}{partial beta} = -frac{ mu e^{betamu} +(mu -epsilon )e^{beta(mu-epsilon)} }{ 1 - e^{betamu} - e^{beta(mu-epsilon)}} = -mu bar{N} + epsilon bar{N}_2. $$ $$ tag{8} U = bar{N}_2 epsilon = frac{bar{N}epsilon}{1+e^{betaepsilon}} . $$ Eq. (8) is same as Eq.(1) from canonocal ensemble and Eq.(6) from microcanonical ensemble.

The entropy $$ S = -frac{partial Omega}{partial T} = -K ln left{ 1 - e^{betamu} - e^{beta(mu-epsilon)} right} + KT frac{ mu e^{betamu} +(mu -epsilon )e^{beta(mu-epsilon)} }{ 1 - e^{betamu} - e^{beta(mu-epsilon)}} frac{-1}{KT^2} $$ The first term resemble the grand potentail, and the second term $U - mubar{N}$ $$ S = -frac{Omega}{T} + frac{U-mubar{N}}{T} =frac{U-Omega-mubar{N}}{T}=frac{U-F}{T} $$ This relation resemble Eq.(3). The direct comparison between this entropy with Eq.(3) and Eq.(7) serves a charllenging exercise.