D.58 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.57} 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 I_...
...n_2 {\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}}
+ \frac...
...{\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}