D.59 The canonical probability distribution

This note deduces the canonical probability distribution. Since the derivations in typical textbooks seem crazily convoluted and the made assumptions not at all as self-evident as the authors suggest, a more mathematical approach will be followed here.

Consider a big system consisting of many smaller subsystems $A,B,\ldots$ with a given total energy $E$. Call the combined system the collective. Following the same reasoning as in derivation {D.58} for two systems, the thermodynamically stable equilibrium state has shelf occupation numbers of the subsystems satisfying

\begin{eqnarray*}
& \displaystyle
\frac{\partial \ln Q_{\vec I_A}}{\partial ...
... {\vphantom' E}^{\rm p}_{s_B} = 0\\
& \displaystyle
\ldots
\end{eqnarray*}

where $\epsilon_2$ is a shorthand for 1/${k_{\rm B}}T$.

An individual system, take $A$ as the example, no longer has an individual energy that is for certain. Only the collective has that. That means that when $A$ is taken out of the collective, its shelf occupation numbers will have to be described in terms of probabilities. There will still be an expectation value for the energy of the system, but system energy eigenfunctions $\psi^{\rm S}_{q_A}$ with somewhat different energy ${\vphantom' E}^{\rm S}_{q_A}$ can no longer be excluded with certainty. However, still assume, following the fundamental assumption of quantum statistics, {N.23}, that the physical differences between the system energy eigenfunctions do not make (enough of) a difference to affect which ones are likely or not. So, the probability $P_{q_A}$ of a system eigenfunction $\psi^{\rm S}_{q_A}$ will be assumed to depend only on its energy ${\vphantom' E}^{\rm S}_{q_A}$:

\begin{displaymath}
P_{q_A} = P({\vphantom' E}^{\rm S}_{q_A}).
\end{displaymath}

where $P$ is some as yet unknown function.

For the isolated example system $A$, the question is now no longer “What shelf numbers have the most eigenfunctions?” but “What shelf numbers have the highest probability?” Note that all system eigenfunctions $\psi^{\rm S}_{q_A}$ for a given set of shelf numbers $\vec{I_A}$ have the same system energy ${\vphantom' E}^{\rm S}_{\vec{I}_A}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\sum_{s_A}I_{s_A}{\vphantom' E}^{\rm p}_{s_A}$. Therefore, the probability of a given set of shelf numbers $P_{\vec{I}_A}$ will be the number of eigenfunctions with those shelf numbers times the probability of each individual eigenfunction:

\begin{displaymath}
P_{\vec I_A} = Q_{\vec I_A} P({\vphantom' E}^{\rm S}_{\vec I_A}).
\end{displaymath}

Mathematically, the function whose partial derivatives must be zero to find the most probable shelf numbers is

\begin{displaymath}
F = \ln\left(P_{\vec I_A}\right)
- \epsilon_{1,A}\bigg(\sum_{s_A} I_{s_A} - I_A\bigg).
\end{displaymath}

The maximum is now to be found for the shelf number probabilities, not their eigenfunction counts, and there is no longer a constraint on energy.

Substituting $P_{\vec{I}_A}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $Q_{\vec{I}_A}P({\vphantom' E}^{\rm S}_{\vec{I}_A})$, taking apart the logarithm, and differentiating, produces

\begin{displaymath}
\frac{\partial \ln Q_{\vec I_A}}{\partial I_{s_A}}
+ \fr...
...vec I_A}} {\vphantom' E}^{\rm p}_{s_A}
- \epsilon_{1,A} = 0
\end{displaymath}

That is exactly like the equation for the shelf numbers of system $A$ when it was part of the collective, except that the derivative of the as yet unknown function $\ln(P_A)$ takes the place of $-\epsilon_2$, i.e. $\vphantom0\raisebox{1.5pt}{$-$}$1/${k_{\rm B}}T$. It follows that the two must be the same, because the shelf numbers cannot change when the system $A$ is taken out of the collective it is in thermal equilibrium with. For one, the net energy would change if that happened, and energy is conserved.

It follows that ${\rm d}\ln{P}$$\raisebox{.5pt}{$/$}$${{\rm d}}{\vphantom' E}^{\rm S}_{\vec{I}_A}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$1$\raisebox{.5pt}{$/$}$${k_{\rm B}}T$ at least in the vicinity of the most probable energy ${\vphantom' E}^{\rm S}_{\vec{I}_A}$. Hence in the vicinity of that energy

\begin{displaymath}
P({\vphantom' E}^{\rm S}_A) = \frac{1}{Z_A} e^{-{\vphantom' E}^{\rm S}_A/{k_{\rm B}}T}
\end{displaymath}

which is the canonical probability. Note that the given derivation only ensures it to be true in the vicinity of the most probable energy. Nothing says it gives the correct probability for, say, the ground state energy. But then the question becomes “What difference does it make?” Suppose the ground state has a probability of 0. followed by only 100 zeros instead of the predicted 200 zeros? What would change in the price of eggs?

Note that the canonical probability is self-consistent: if two systems at the same temperature are combined, the probabilities of the combined eigenfunctions multiply, as in

\begin{displaymath}
P_{AB} = \frac{1}{Z_AZ_B} e^{-({\vphantom' E}^{\rm S}_A+{\vphantom' E}^{\rm S}_B)/{k_{\rm B}}T}.
\end{displaymath}

That is still the correct expression for the combined system, since its energy is the sum of those of the two separate systems. Also for the partition functions

\begin{displaymath}
Z_AZ_B =\sum_{q_A}\sum_{q_B} e^{-({\vphantom' E}^{\rm S}_{q_A}+{\vphantom' E}^{\rm S}_{q_B})/{k_{\rm B}}T} = Z_{AB}.
\end{displaymath}