Is there any instance where the microcanonical, canonical and grand canonical ensembles give equal results? In certain circumstances; when energy fluctuation effects are minimal, the microcanonical and canonical ensembles give the same results.
Then, under different circumstances the canonical and grand canonical ensembles give the same results.
Can there be any instances where the microcanonical, canonical, and grand canonical ensembles all give equal results? If so, what would the circumstances be?
 A: For most systems, the derived macroscopic thermal quantities from any one of the formulations should have the same expression. Let's examine the simplest two-level system, which appears in many statistical text books, being treated in detail for both microcanonical and canonical ensembles.
The two-level non-interacting system assumes particles can exist 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 volume information, we can only express the average energy $U(T) = \langle E\rangle$, and the entropy $S(T)$, as functions of temperature.
The canonical ensemble is the simplest one. Let's do it first: given total particle number $N$ and temperature $T$:
One-particle 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
$$ \tag{1}
U = \langle E\rangle = N_2 \epsilon =  \frac{N \epsilon}{1 + e^{\beta \epsilon}}
$$
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:
$$
   \Gamma(N, E) = \frac{N!}{\frac{E}{\epsilon}! \left(N -\frac{E}{\epsilon} \right)!}
$$
$$
   \ln \Gamma(N, E) = N \ln \frac{N\epsilon}{N\epsilon-E} + \frac{E}{\epsilon} \ln \frac{N\epsilon-E}{E}
$$
The entropy $S$
$$ \tag{4}
S = K \ln \Gamma(N, E) =NK\ln \frac{N\epsilon}{N\epsilon-E} + K\frac{E}{\epsilon} \ln \frac{N\epsilon-E}{E}
$$
The temperature $\frac{1}{T} = \frac{\partial S}{\partial E}]_{E=U}$
$$\tag{5}
\frac{1}{T} = \frac{ K }{\epsilon} \ln \frac{N\epsilon-U}{U} 
$$
Invert the above equation to find $U(T)$
$$\tag{6}
U =   N \frac{\epsilon}{1 + e^{\frac{\epsilon}{K T}} }
$$
This is 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 particle number $N$ and total energy $E$, the probability is proportional to $\exp(\beta\mu N - \beta E)$, and $N_2 = E / \epsilon$, $N=N_1+N_2$. And the multiplicity $\frac{N!}{N_1! N_2!}$
$$
Z = \sum_{N,E} \frac{N!}{N_1! N_2!} \exp\left(\beta\mu N - \beta E \right) = \sum_{N, N_2} \frac{N!}{(N-N_2)! N_2!} \exp\left\{\beta\mu (N-N_2) + \beta N_2 (\mu-\epsilon) \right\}
$$
$$
 = \sum_{N=0,1,2..}   \sum_{N_2=0}^N \frac{N!}{(N-N_2)! N_2!}  \exp^{N-N_2}\left(\beta\mu\right)  \exp^{N_2}\left\{\beta (\mu- \epsilon) \right\}
$$
$$
 = \sum_{N=0,1,2..}  \left\{ e^{\beta\mu} + e^{\beta(\mu-\epsilon)} \right\}^N\\
 = \frac{1}{\left\{ 1-e^{\beta\mu} - e^{\beta(\mu-\epsilon)} \right\} }.
$$
$$
 \ln Z = - \ln \left\{ 1 - e^{\beta\mu} - e^{\beta(\mu-\epsilon)} \right\} .
$$
The grand potential $\Omega = -K T \ln Z$
$$
\Omega = -KT\ln Z = KT\ln \left\{ 1 - e^{\beta\mu} - e^{\beta(\mu-\epsilon)} \right\}.
$$
The average number $\bar{N} = \langle N \rangle$:
$$
  \bar{N} = -\frac{\partial \Omega}{\partial \mu}=\frac{e^{\beta\mu} +e^{\beta(\mu-\epsilon)} }{ 1 - e^{\beta\mu} - e^{\beta(\mu-\epsilon)}}
\equiv \bar{N}_1 + \bar{N}_2.
$$
Where $\bar{N}_1/\bar{N}_2 = e^{\beta \epsilon}$.
The average energy $U$
$$
  U -\mu \bar{N} = -\frac{\partial \ln Z}{\partial \beta} = -\frac{ \mu e^{\beta\mu} +(\mu -\epsilon )e^{\beta(\mu-\epsilon)} }{ 1 - e^{\beta\mu} - e^{\beta(\mu-\epsilon)}} = -\mu \bar{N} + \epsilon \bar{N}_2.
$$
$$ \tag{8}
  U = \bar{N}_2 \epsilon = \frac{\bar{N}\epsilon}{1+e^{\beta\epsilon}} .
$$
Eq. (8) is same as Eq.(1) from the canonical ensemble and Eq.(6) from the microcanonical ensemble.
The entropy
$$
  S = -\frac{\partial \Omega}{\partial T} = -K \ln \left\{ 1 - e^{\beta\mu} - e^{\beta(\mu-\epsilon)} \right\} + KT \frac{ \mu e^{\beta\mu} +(\mu -\epsilon )e^{\beta(\mu-\epsilon)} }{ 1 - e^{\beta\mu} - e^{\beta(\mu-\epsilon)}} \frac{-1}{KT^2}
$$
The first term resemble the grand potential, and the second term $U - \mu\bar{N}$
$$
  S = -\frac{\Omega}{T} + \frac{U-\mu\bar{N}}{T} =\frac{U-\Omega-\mu\bar{N}}{T}=\frac{U-F}{T}
$$
This relation resemble Eq.(3). The direct comparison between this entropy with Eq.(3) and Eq.(7) serves a challenging exercise.
