= Boltzmann distribution in a four-state system
{c}
{title2=$Z=(1+q)^2$}
Consider two distinguishable, noninteracting <two-level systems>, each with <energies> $0$ and $\Delta>0$. Their four <microstates> $(0,0),(0,1),(1,0),(1,1)$ have total <energies> $0,\Delta,\Delta,2\Delta$. Thus the three <energy> <macrostates> have <state degeneracies> $1,2,1$. This is a stipulated finite model, not a measured material.
Set $q=e^{-\beta\Delta}$. The four individual <Boltzmann factors> are $1,q,q,q^2$, so the <partition function> and aggregate <probabilities> are
$$
Z=1+2q+q^2=(1+q)^2,\qquad
\bigl(P(0),P(\Delta),P(2\Delta)\bigr)=\frac{(1,2q,q^2)}{(1+q)^2}.
$$
Each of the two middle <microstates> has <probability> $q/Z$, while their shared <energy> <macrostate> has <probability> $2q/Z$.
At the <temperature> satisfying $\Delta=k_BT\ln2$, the <Boltzmann factor> is $q=1/2$ and the <partition function> is $Z=9/4$. Therefore
$$
\bigl(p_{00},p_{01},p_{10},p_{11}\bigr)=\left(\frac49,\frac29,\frac29,\frac19\right),\qquad
\bigl(P(0),P(\Delta),P(2\Delta)\bigr)=\left(\frac49,\frac49,\frac19\right).
$$
The middle <energy> is as likely as the <ground state> <energy> because its <state degeneracy> is twice as large. Yet each individual middle <microstate> is only half as likely as the <ground state> <microstate>. The diagram places the four <energies> beside their individual <probabilities>, then shows the sums within each <macrostate>.
\Image[/boltzmann-microstates-and-degeneracy.png]
{title=Four <microstates> and three <energy> <macrostates> at a gap of $k_BT\ln2$}
The <expected value> of <energy> follows directly by weighting the three possible <energies>:
$$
\langle E\rangle=0\,P(0)+\Delta P(\Delta)+2\Delta P(2\Delta)
=\frac{2\Delta q+2\Delta q^2}{(1+q)^2}
=\frac{2\Delta q}{1+q}.
$$
At $q=1/2$ this is $2\Delta/3$. An individual <energy> measurement still returns $0$, $\Delta$ or $2\Delta$; the <expected value> describes the <ensemble average>, rather than a fourth possible measured <energy>.
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