D.61 Chem­i­cal po­ten­tial in the dis­tri­b­u­tions

The fol­low­ing con­vo­luted de­riva­tion of the dis­tri­b­u­tion func­tions comes fairly straightly from Baier­lein [4, pp. 170-]. Let it not de­ter you from read­ing the rest of this oth­er­wise very clearly writ­ten and en­gag­ing lit­tle book. Even a nonengi­neer­ing au­thor should be al­lowed one mis­take.

The de­riva­tions of the Maxwell-Boltz­mann, Fermi-Dirac, and Bose-Ein­stein dis­tri­b­u­tions given pre­vi­ously, {D.57} and {D.58}, were based on find­ing the most nu­mer­ous or most prob­a­ble dis­tri­b­u­tion. That im­plic­itly as­sumes that sig­nif­i­cant de­vi­a­tions from the most nu­mer­ous/prob­a­ble dis­tri­b­u­tions will be so rare that they can be ig­nored. This note will by­pass the need for such an as­sump­tion since it will di­rectly de­rive the ac­tual ex­pec­ta­tion val­ues of the sin­gle-par­ti­cle state oc­cu­pa­tion num­bers $\iota$. In par­tic­u­lar for fermi­ons, the de­riva­tion will be solid as a rock.

The mis­sion is to de­rive the ex­pec­ta­tion num­ber $\iota_n$ of par­ti­cles in an ar­bi­trary sin­gle-par­ti­cle state $\pp{n}////$. This ex­pec­ta­tion value, as any ex­pec­ta­tion value, is given by the pos­si­ble val­ues times their prob­a­bil­ity:

$$\iota_n = \sum_q i_n P_q$$

where $i_n$ is the num­ber of par­ti­cles that sys­tem en­ergy eigen­func­tion $\psi^{\rm S}_q$ has in sin­gle-par­ti­cle state $\pp{n}////$, and $P_q$ the prob­a­bil­ity of the eigen­func­tion. Since ther­mal equi­lib­rium is as­sumed, the canon­i­cal prob­a­bil­ity value $e^{-{\vphantom' E}^{\rm S}_q/{k_{\rm B}}T}$$\raisebox{.5pt}{$/$}$​$Z$ can be sub­sti­tuted for $P_q$. Then, if the en­ergy ${\vphantom' E}^{\rm S}_q$ is writ­ten as the sum of the ones of the sin­gle par­ti­cle states times the num­ber of par­ti­cles in that state, it gives:

$$\iota_n = \frac{1}{Z} \sum_q i_n e^{-(i_1{\vphantom' E}^{\... ...}_n +i_{n+1}{\vphantom' E}^{\rm p}_{n+1}+\ldots)/k_{\rm B}T}.$$

Note that $i_n$ is the oc­cu­pa­tion num­ber of sin­gle-par­ti­cle state $\pp{n}////$, just like $I_s$ was the oc­cu­pa­tion num­ber of shelf $s$. Deal­ing with sin­gle-par­ti­cle state oc­cu­pa­tion num­bers has an ad­van­tage over deal­ing with shelf num­bers: you do not have to fig­ure out how many sys­tem eigen­func­tions there are. For a given set of sin­gle-par­ti­cle state oc­cu­pa­tion num­bers $\vec\imath$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vert i_1,i_2,\ldots\rangle$, there is ex­actly one sys­tem en­ergy eigen­func­tion. Com­pare fig­ures 11.2 and 11.3: if you know how many par­ti­cles there are in each sin­gle-par­ti­cle state, you know every­thing there is to know about the eigen­func­tion de­picted. (This does not ap­ply to dis­tin­guish­able par­ti­cles, fig­ure 11.1, be­cause for them the num­bers on the par­ti­cles can still vary for given oc­cu­pa­tion num­bers, but as noted in chap­ter 11.11, there is no such thing as iden­ti­cal dis­tin­guish­able par­ti­cles any­way.)

It has the big con­se­quence that the sum over the eigen­func­tions can be re­placed by sums over all sets of oc­cu­pa­tion num­bers:

$$\begin{eqnarray*} \lefteqn{\iota_n = \frac{1}{Z} \underbrace{\sum_{i_1}\sum_{i... ...m p}_n +i_{n+1}{\vphantom' E}^{\rm p}_{n+1}+\ldots)/k_{\rm B}T} \end{eqnarray*}$$

Each set of sin­gle-par­ti­cle state oc­cu­pa­tion num­bers cor­re­sponds to ex­actly one eigen­func­tion, so each eigen­func­tion is still counted ex­actly once. Of course, the oc­cu­pa­tion num­bers do have to add up to the cor­rect num­ber of par­ti­cles in the sys­tem.

Now con­sider first the case of $I$ iden­ti­cal bosons. For them the oc­cu­pa­tion num­bers may have val­ues up to a max­i­mum of I:

$$\begin{eqnarray*} \lefteqn{\iota_n = \frac{1}{Z} \underbrace{\sum_{i_1=0}^I\su... ...p}_n +i_{n+1}{\vphantom' E}^{\rm p}_{n+1}+\ldots)/{k_{\rm B}}T} \end{eqnarray*}$$

One sim­pli­fi­ca­tion that is im­me­di­ately ev­i­dent is that all the terms that have $i_n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 are zero and can be ig­nored. Now ap­ply a trick that only a math­e­mati­cian would think of: de­fine a new sum­ma­tion in­dex $i_n'$ by set­ting $i_n$ $\vphantom0\raisebox{1.5pt}{$=$}$ $1+i_n'$. Then the sum­ma­tion over $i_n'$ can start at 0 and will run up to $I-1$. Plug­ging $i_n$ $\vphantom0\raisebox{1.5pt}{$=$}$ $1+i_n'$ into the sum above gives

$$\begin{eqnarray*} \lefteqn{\iota_n = \frac{1}{Z} \underbrace{\sum_{i_1=0}^I\ld... ...m p}_n +i_{n+1}{\vphantom' E}^{\rm p}_{n+1}+\ldots)/k_{\rm B}T} \end{eqnarray*}$$

This can be sim­pli­fied by tak­ing the con­stant part of the ex­po­nen­tial out of the sum­ma­tion. Also, the con­straint in the bot­tom shows that the oc­cu­pa­tion num­bers can no longer be any larger than $I-1$ (since the orig­i­nal $i_n$ is at least one), so the up­per lim­its can be re­duced to $I-1$. Fi­nally, the prime on $i_n'$ may as well be dropped, since it is just a sum­ma­tion in­dex and it does not make a dif­fer­ence what name you give it. So, al­to­gether,

$$\begin{eqnarray*} \lefteqn{\iota_n = \frac{1}{Z} e^{-{\vphantom' E}^{\rm p}_n/{... ...m p}_n +i_{n+1}{\vphantom' E}^{\rm p}_{n+1}+\ldots)/k_{\rm B}T} \end{eqnarray*}$$

The right hand side falls apart into two sums: one for the 1 in $1+i_n$ and one for the $i_n$ in $1+i_n$. The first sum is es­sen­tially the par­ti­tion func­tion $Z^-$ for a sys­tem with $I-1$ par­ti­cles in­stead of $I$. The sec­ond sum is es­sen­tially $Z^-$ times the ex­pec­ta­tion value $\iota_n^-$ for such a sys­tem. To be pre­cise

$$\iota_n = \frac{1}{Z} e^{-{\vphantom' E}^{\rm p}_n/{k_{\rm B}}T} Z^- \left[1 + \iota_n^-\right]$$

This equa­tion is ex­act, no ap­prox­i­ma­tions have been made yet.

The sys­tem with $I-1$ par­ti­cles is the same in all re­spects to the one for $I$ par­ti­cles, ex­cept that it has one less par­ti­cle. In par­tic­u­lar, the sin­gle-par­ti­cle en­ergy eigen­func­tions are the same, which means the vol­ume of the box is the same, and the ex­pres­sion for the canon­i­cal prob­a­bil­ity is the same, mean­ing that the tem­per­a­ture is the same.

But when the sys­tem is macro­scopic, the oc­cu­pa­tion counts for $I-1$ par­ti­cles must be vir­tu­ally iden­ti­cal to those for $I$ par­ti­cles. Clearly the physics should not change no­tice­ably de­pend­ing on whether 10$\POW9,{20}$ or 10$\POW9,{20}$ + 1 par­ti­cles are present. If $\iota_n^-$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\iota_n$, then the above equa­tion can be solved to give:

$$\iota_n = 1\left/\left[\frac{Z}{Z^-} e^{{\vphantom' E}^{\rm p}_n/{k_{\rm B}}T} - 1 \right]\right.$$

The fi­nal for­mula is the Bose-Ein­stein dis­tri­b­u­tion with

$$e^{-\mu/{k_{\rm B}}T} = \frac{Z}{Z^-}$$

Solve for $\mu$:

$$\mu = - k_{\rm B}T \ln\left(\frac{Z}{Z^-}\right) = \frac{ - k_{\rm B}T\ln(Z)+k_{\rm B}T\ln(Z^-)}{I - (I-1)}$$

The fi­nal frac­tion is a dif­fer­ence quo­tient ap­prox­i­ma­tion for the de­riv­a­tive of the Helmholtz free en­ergy with re­spect to the num­ber of par­ti­cles. Now a sin­gle par­ti­cle change is an ex­tremely small change in the num­ber of par­ti­cles, so the dif­fer­ence quo­tient will be to very great ac­cu­racy be equal to the de­riv­a­tive of the Helmholtz free en­ergy with re­spect to the num­ber of par­ti­cles. And as noted ear­lier, in the ob­tained ex­pres­sions, vol­ume and tem­per­a­ture are held con­stant. So, $\mu$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\left(\partial{F}/\partial{I}\right)_{T,V}$, and (11.39) iden­ti­fied that as the chem­i­cal po­ten­tial. Do note that $\mu$ is on a sin­gle-par­ti­cle ba­sis, while $\bar\mu$ was taken to be on a mo­lar ba­sis. The Avo­gadro num­ber $I_{\rm A}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 6.022 1 10$\POW9,{26}$ par­ti­cles per kmol con­verts be­tween the two.

Now con­sider the case of $I$ iden­ti­cal fermi­ons. Then, ac­cord­ing to the ex­clu­sion prin­ci­ple, there are only two al­lowed pos­si­bil­i­ties for the oc­cu­pa­tion num­bers: they can be zero or one:

$$\iota_n = \frac{1}{Z} \underbrace{\sum_{i_1=0}^1\ldots\sum... ..._n +i_{n+1}{\vphantom' E}^{\rm p}_{n+1}+\ldots)/{k_{\rm B}}T}$$

Again, all terms with $i_n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 are zero, so you can set $i_n$ $\vphantom0\raisebox{1.5pt}{$=$}$ $1+i_n'$ and get

$$\begin{eqnarray*} \lefteqn{\iota_n = \frac{1}{Z} \underbrace{\sum_{i_1=0}^1\ld... ...p}_n +i_{n+1}{\vphantom' E}^{\rm p}_{n+1}+\ldots)/{k_{\rm B}}T} \end{eqnarray*}$$

But now there is a dif­fer­ence: even for a sys­tem with $I-1$ par­ti­cles $i_n'$ can still have the value 1 but the up­per limit is zero. For­tu­nately, since the above sum only sums over the sin­gle value $i_n'$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, the fac­tor $(1+i_n')$ can be re­placed by $(1-i_n')$ with­out chang­ing the an­swer. And then the sum­ma­tion can in­clude $i_n'$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 again, be­cause $(1-i_n')$ is zero when $i_n'$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. This sign change pro­duces the sign change in the Fermi-Dirac dis­tri­b­u­tion com­pared to the Bose-Ein­stein one; the rest of the analy­sis is the same.

Here are some ad­di­tional re­marks about the only ap­prox­i­ma­tion made, that the sys­tems with $I$ and $I-1$ par­ti­cles have the same ex­pec­ta­tion oc­cu­pa­tion num­bers. For fermi­ons, this ap­prox­i­ma­tion is jus­ti­fied to the gills, be­cause it can be eas­ily be seen that the ob­tained value for the oc­cu­pa­tion num­ber is in be­tween those of the sys­tems with $I-1$ and $I$ par­ti­cles. Since no­body is go­ing to count whether a macro­scopic sys­tem has 10$\POW9,{20}$ par­ti­cles or 10$\POW9,{20}$ + 1, this is truly as good as any the­o­ret­i­cal pre­dic­tion can pos­si­bly get.

But for bosons, it is a bit trick­ier be­cause of the pos­si­bil­ity of con­den­sa­tion. As­sume, rea­son­ably, that when a par­ti­cle is added, the oc­cu­pa­tion num­bers will not go down. Then the de­rived ex­pres­sion over­es­ti­mates both ex­pec­ta­tion oc­cu­pa­tion num­bers $\iota_n$ and $\iota_n^-$. How­ever, it could at most be wrong, (i.e. have a fi­nite rel­a­tive er­ror) for a fi­nite num­ber of states, and the num­ber of sin­gle-par­ti­cle states will be large. (In the ear­lier de­riva­tion us­ing shelf num­bers, the ac­tual $\iota_n$ was found to be lower than the Bose-Ein­stein value by a fac­tor $(N_s-1)$$\raisebox{.5pt}{$/$}$​$N_s$ with $N_s$ the num­ber of states on the shelf.)

If the fac­tor $Ze^{{\vphantom' E}^{\rm p}_1/{k_{\rm B}}T}$$\raisebox{.5pt}{$/$}$​$Z^-$ is one ex­actly, which def­i­nitely means Bose-Ein­stein con­den­sa­tion, then $i_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $1+i_1^-$. In that case, the ad­di­tional par­ti­cle that the sys­tem with $I$ par­ti­cles has goes with cer­tainty into the ground state. So the ground state bet­ter be unique then; the par­ti­cle can­not go into two ground states.