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 ${\vphantom' E}^{\rm S}_{q_A}$:

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

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:

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

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

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

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

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

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

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

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

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

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

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