[NOTES/SM-04001] The Canonical Ensemble

We begin with a an ensemble of systems which is in thermal equilibrium with a reservoir at the some temperature is called a {\tt canonical ensemble}. The distribution function \(f(\vec{r}_1,\vec{r}_2,\cdots,\vec{v}_1,\cdots,\vec{v}_N)\) is a function of energy. \begin{equation} f(\vec{r}_1,\vec{r}_2,\cdots,\vec{v}_1,\cdots,\vec{v}_N) \equiv f(E(\vec{r}_1,\vec{r}_2,\cdots,\vec{v}_1,\vec{v_2})) \end{equation} For discrete energies, let $P(E_j)$ be the probability that a system is in a state with energy $E_j$.

 Consider two systems $A$ and $B$, which are in thermal equilibrium the a heat reservoir. Let \(P_A(E_B)\) and \(P_B(E_B)\), repectively, denote the probabilities that the system \(A\) is in state with energy \(E_A\) and that the system \(B\) is in state with energy \(E_B\).

Note that probability that system being in an {\tt energy level} $E$ is not the same as the system being in a {\it particular micro state} with energy $E$. Very many micro states correspond to a macro state of a system with energy \(E\).

The energy of composite system $A+B$ is equal to $E_A+E_B$. $P_{A+B}(E_{A+B})=$ is the probability that the composite system is in a particular state with energy $E_{A+B}$. $$ P_{A+B}(E_A+E_B)= P_A(E_A)P_B(E_B) $$ In most cases of interest $E_{A+B}=E_A+E_B$ interaction energy is small compared to the individual values $E_A$ and $E_B$.

We note that the two events

1. System $A$ is in a particular state with energy $E_A$, and
2. System $B$ is in a particular state with energy $E_B$.

are independent.

There are both in thermal equilibrium with heat reservoir and can exchange energy with the heat bath. Presence of $B$ does not affect the exchange of energy of system $A$ with the heat reservoir. The state, in which system $A$ is in, is no way affected by system $B$ \begin{equation} P_{A+B}(E_A+E_B) = P_A(E_A)P_B(E_B). \end{equation} Equating the derivatives of \(P_{A+B}(E_A+E_B)\) w.r.t. \(E_A\) and w.r.t. \(E_B\), we get \begin{align*} P^\prime_A(E_A)P_B(E_B)=&P_A(E_A)P^\prime_B(E_B)\\ \frac{P^\prime_A(E_A)}{P_A(E_A)} = & \frac{P^\prime_B(E_B)}{P_B(E_B)} = -\beta. \end{align*} $E_A$ and $E_B$ can be any values hence the way above equation can be satisfied is that both sides are equal to a constant $\beta$ which does not depend on $E_A$ or $E_B$. Also $\beta$ cannot depend on properties of systems $A$ and $B$. It can depend only on quantity, or property, that is common between $A$ and $B$. Hence $\beta$ must depend on reservoir, and hence temperature $T$ of the reservoir. Thus we get $$ P_A(E_A) = C_Ae^{-\beta E_A} $$

Thus the probability that a system, which is in thermal equilibrium with a reservoir at temperature $T$, will be found in a particular state with energy $E$ is
$$ P(E) = C\exp(-\beta E) $$
 \(P(E)\) is the probability of a particular macro state or micro state with energy.Here it is important to remember that  \(P(E)\) is not the probability that the system be found in energy level \(E\). If \(\Omega(E)\) is the degeneracy of energy level \(E\), there are \(\Omega(E)\) microstates with energy \(E\). Therefore, the probability that the system be found in energy level \(E\) is \(\Omega(E) e^{-\beta E}\). In macroscopic systems $\Omega(E)$ increases rapidly  with energy $E$.

Demanding that probability of all events added together must be equal to unity, we have \begin{equation} \sum_{k}P(E_k)= \sum_k C e^{-\beta E_k} = 1 \rightarrow C \sum_ke^{-\beta E_k} =  1 . \end{equation}
The constant \(C\) is now fixed and the probability \(P(E_k)\) is written as \begin{equation} P(E_k) = Z^{-1}e^{-\beta E_k},\end{equation} where
\begin{equation} Z = \sum_{\text{micro states } k} e^{-\beta E_k},\label{EQ05} \end{equation} is called the canonical partition function. Here the sum \(k\) is over all micro states.

This sum can also be written as 
\begin{equation}Z = \sum_{\text{Energy levels  }\  E} \Omega(E) e^{-\beta E}.\end{equation}

For continuous energies, the  distribution function will be of the form
\begin{align*} f(E) =& C^\prime e^{-\beta E}.\\ f(E) =& Z^{-1} D_N \int d^r_1..\int d^3r_N...  \end{align*}
where a constant \(D_N\) with dimension \(L^{-3N} [V]^{-3N}\) has been introduced to make the integration volume element dimensionless. \begin{equation} {\rm dim} [D_N] \times L^{3N}(\text{Vel})^{3N} = \text{dimensionless}. \end{equation} The condition that the sum of all probabilities be equal to one becomes \begin{equation} D_N\iiint_V d^3r \iiint_{\infty}^\infty d^3v_1\cdots d^3v_N f(E)=1 \end{equation} Introduce a dimensionless function of \(\beta\) by \begin{align*} Z&=D_N\iint d^3r_1\cdots d^3r_N\iint d^3v_1\cdots d^3N_N e^{-\beta E} \end{align*}
\(Z\) is called the canonical partition function or Zustandssumme.  The distribution function can be written in terms of the partition function  as \begin{equation} f(E) = Z^{-1} D_N e^{-\beta E}. \end{equation} The numerical factor $Z$  ensures that the sum of all probabilities is equal to one.

We have seen that the probability, for the systems to be in a \textit{particular} micro state with energy \(E\), is $Ce^{-\beta E}$. Therefore, the probability that the system has energy $E$ is given by

\begin{equation} \begin{pmatrix} \text{no. of microstates}\\ \text{with energy $E$}\end{pmatrix}\times e^{-\beta E} \end{equation}

An energy level consists of all states with given energy. Thus the probability of system being in a level $E$ is $$\sum_{E_k=E} P(E_k) = \Omega(E_k)P(E_k),$$ where  $\Omega(E)$ is the degeneracy {\it i.e.} the number of states with energy $E$. In macroscopic systems $\Omega(E)$ increases rapidly  with energy $E$.

We have obtained an expression for the probability that the system will be in energy  level \(E\). The answer is in terms of an unknown parameter \(\beta\) which will turn out to be related to the temperature of the body.

One way to make this identification is compute partition function of an ideal gas and the internal energy of ideal gas and compare the answer with the result from equipartition of energy.