11.3 How Many Sys­tem Eigen­func­tions?

The fun­da­men­tal ques­tion from which all of quan­tum sta­tis­tics springs is a very ba­sic one: How many sys­tem en­ergy eigen­states are there with given generic prop­er­ties? This sec­tion will ad­dress that ques­tion.

Of course, by de­f­i­n­i­tion each sys­tem en­ergy eigen­func­tion is unique. Fig­ures 11.1-11.3 give ex­am­ples of such unique en­ergy eigen­func­tions for sys­tems of dis­tin­guish­able par­ti­cles, in­dis­tin­guish­able bosons, and in­dis­tin­guish­able fermi­ons. But try­ing to get ac­cu­rate data on each in­di­vid­ual eigen­func­tion just does not work. That is much too big a chal­lenge.

Quan­tum sta­tis­tics must sat­isfy it­self by fig­ur­ing out the prob­a­bil­i­ties on groups of sys­tem eigen­func­tions with sim­i­lar prop­er­ties. To do so, the sin­gle-par­ti­cle en­ergy eigen­states are best grouped to­gether on shelves of sim­i­lar en­ergy, as il­lus­trated in fig­ures 11.1-11.3. Do­ing so al­lows for more an­swer­able ques­tions such as: “How many sys­tem en­ergy eigen­func­tions $\psi^{\rm S}_q$ have $I_1$ out of the $I$ to­tal par­ti­cles on shelf 1, an­other $I_2$ on shelf 2, etcetera?” In other words, if $\vec{I}$ stands for a given set of shelf oc­cu­pa­tion num­bers $(I_1,I_2,I_3,\ldots)$, then what is the num­ber $Q_{\vec{I}}$ of sys­tem eigen­func­tions $\psi^{\rm S}_q$ that have those shelf oc­cu­pa­tion num­bers?

That ques­tion is an­swer­able with some clever math­e­mat­ics; it is a big thing in var­i­ous text­books. How­ever, the sus­pi­cion is that this is more be­cause of the neat math­e­mat­ics than be­cause of the ac­tual phys­i­cal in­sight that these de­riva­tions pro­vide. In this book, the de­riva­tions are shoved away into {D.56}. But here are the re­sults. (Drums please.) The sys­tem eigen­func­tion counts for dis­tin­guish­able par­ti­cles, bosons, and fermi­ons are:

$$Q^{\rm {d}}_{\vec I} = I! \prod_{{\rm all} s}\frac{N_s^{I_s}}{\Big(I_s\Big)!}$$ (11.1)

$$Q^{\rm {b}}_{\vec I} = \prod_{{\rm all} s} \frac{\Big(I_s+N_s-1\Big)!}{\Big(I_s\Big)!\Big(N_s-1\Big)!}$$ (11.2)

$$Q^{\rm {f}}_{\vec I} = \prod_{{\rm all} s} \frac{\Big(N_s\Big)!}{\Big(I_s\Big)!\Big(N_s-I_s\Big)!}%$$ (11.3)

where $\Pi$ means the prod­uct of all the terms of the form shown to its right that can be ob­tained by sub­sti­tut­ing in every pos­si­ble value of the shelf num­ber $s$. That is just like $\Sigma$ would mean the sum of all these terms. For ex­am­ple, for dis­tin­guish­able par­ti­cles

$$Q^{\rm {d}}_{\vec I} = I! \frac{N_1^{I_1}}{\Big(I_1\Big)!}... ..._3}}{\Big(I_3\Big)!} \frac{N_4^{I_4}}{\Big(I_4\Big)!} \ldots$$

where $N_1$ is the num­ber of sin­gle-par­ti­cle en­ergy states on shelf 1 and $I_1$ the num­ber of par­ti­cles on that shelf, $N_2$ the num­ber of sin­gle-par­ti­cle en­ergy states on shelf 2 and $I_2$ the num­ber of par­ti­cles on that shelf, etcetera. Also an ex­cla­ma­tion mark in­di­cates the fac­to­r­ial func­tion, de­fined as

$$n! = \prod_{{\underline n}=1}^n {\underline n}= 1 \times 2 \times 3 \times \ldots \times n$$

For ex­am­ple, 5! = 1 $\times$ 2 $\times$ 3 $\times$ 4 $\times$ 5 = 120. The eigen­func­tion counts may also in­volve 0!, which is de­fined to be 1, and $n$! for neg­a­tive $n$, which is de­fined to be in­fin­ity. The lat­ter is es­sen­tial to en­sure that the eigen­func­tion count is zero as it should be for fermion eigen­func­tions that try to put more par­ti­cles on a shelf than there are states on it.

This sec­tion is mainly con­cerned with ex­plain­ing qual­i­ta­tively why these sys­tem eigen­func­tion counts mat­ter phys­i­cally. And to do so, a very sim­ple model sys­tem hav­ing only three shelves will suf­fice.

Fig­ure 11.4: Il­lus­tra­tive small model sys­tem hav­ing 4 dis­tin­guish­able par­ti­cles. The par­tic­u­lar eigen­func­tion shown is ar­bi­trary.

The first ex­am­ple is il­lus­trated in quan­tum-me­chan­i­cal terms in fig­ure 11.4. Like the other ex­am­ples, it has only three shelves, and it has only $I$ $\vphantom0\raisebox{1.5pt}{$=$}$ 4 dis­tin­guish­able par­ti­cles. Shelf 1 has $N_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 sin­gle-par­ti­cle state with en­ergy ${\vphantom' E}^{\rm p}_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 (ar­bi­trary units), shelf 2 has $N_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ 3 sin­gle-par­ti­cle states with en­ergy ${\vphantom' E}^{\rm p}_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2, (note that 3 $\vphantom0\raisebox{1.1pt}{$\approx$}$ $2\sqrt{2}$), and shelf 3 has $N_3$ $\vphantom0\raisebox{1.5pt}{$=$}$ $4\sqrt{4}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 8 sin­gle-par­ti­cle states with en­ergy ${\vphantom' E}^{\rm p}_3$ $\vphantom0\raisebox{1.5pt}{$=$}$ 4. One ma­jor de­fi­ciency of this model is the small num­ber of par­ti­cles and states, but that will be fixed in the later ex­am­ples. More se­ri­ously is that there are no shelves with en­er­gies above ${\vphantom' E}^{\rm p}_3$ $\vphantom0\raisebox{1.5pt}{$=$}$ 4. To mit­i­gate that prob­lem, for the time be­ing the av­er­age en­ergy per par­ti­cle of the sys­tem eigen­func­tions will be re­stricted to no more than 2.5. This will leave shelf 3 largely empty, re­duc­ing the ef­fects of the miss­ing shelves of still higher en­ergy.

Fig­ure 11.5: The num­ber of sys­tem en­ergy eigen­func­tions for a sim­ple model sys­tem with only three en­ergy shelves. Po­si­tions of the squares in­di­cate the num­bers of par­ti­cles on shelves 2 and 3; dark­ness of the squares in­di­cates the rel­a­tive num­ber of eigen­func­tions with those shelf num­bers. Left: sys­tem with 4 dis­tin­guish­able par­ti­cles, mid­dle: 16, right: 64.

Now the ques­tion is, how many en­ergy eigen­func­tions are there for a given set of shelf oc­cu­pa­tion num­bers $\vec{I}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(I_1,I_2,I_3)$? The an­swer, as given by (11.1), is shown graph­i­cally in the left graph of fig­ure 11.5. Darker squares in­di­cate more eigen­func­tions with those shelf oc­cu­pa­tion num­bers. The oblique line in fig­ure 11.5 is the line above which the av­er­age en­ergy per par­ti­cle ex­ceeds the cho­sen limit of 2.5.

Some ex­am­ple ob­ser­va­tions about the fig­ure may help to un­der­stand it. For ex­am­ple, there is only one sys­tem eigen­func­tion with all 4 par­ti­cles on shelf 1, i.e. with $I_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ 4 and $I_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $I_3$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0; it is

$$\psi^{\rm S}_1 = \pp1/{\skew0\vec r}_1//z1/ \pp1/{\skew0\v... ...2//z2/ \pp1/{\skew0\vec r}_3//z3/ \pp1/{\skew0\vec r}_4/p/z4/.$$

This is rep­re­sented by the white square at the ori­gin in the left graph of fig­ure 11.5.

As an­other ex­am­ple, the dark­est square in the left graph of fig­ure 11.5 rep­re­sents sys­tem eigen­func­tions that have shelf num­bers $\vec{I}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(1,2,1)$, i.e. $I_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, $I_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2, $I_3$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1: one par­ti­cle on shelf 1, two par­ti­cles on shelf 2, and one par­ti­cle on shelf 3. A com­pletely ar­bi­trary ex­am­ple of such a sys­tem en­ergy eigen­func­tion,

$$\pp3/{\skew0\vec r}_1//z1/ \pp1/{\skew0\vec r}_2//z2/ \pp4/{\skew0\vec r}_3//z3/ \pp8/{\skew0\vec r}_4//z4/,$$

is the one de­picted in fig­ure 11.4. It has par­ti­cle 1 in sin­gle-par­ti­cle state $\pp3////$, which is on shelf 2, par­ti­cle 2 in $\pp1////$, which is on shelf 1, par­ti­cle 3 in $\pp4////$ which is on shelf 2, and par­ti­cle 4 in $\pp8////$, which is on shelf 3. But there are a lot more sys­tem eigen­func­tions with the same shelf oc­cu­pa­tion num­bers; in fact, there are

$$4 \times 3 \times 8 \times 3 \times 3 = 864$$

such eigen­func­tions, since there are 4 pos­si­ble choices for the par­ti­cle that goes on shelf 1, times a re­main­ing 3 pos­si­ble choices for the par­ti­cle that goes on shelf 3, times 8 pos­si­ble choices $\pp5////$ through $\pp12////$ for the sin­gle-par­ti­cle eigen­func­tion on shelf 3 that that par­ti­cle can go into, times 3 pos­si­ble choices $\pp2////$ through $\pp4////$ that each of the re­main­ing two par­ti­cles on shelf 2 can go into.

Next, con­sider a sys­tem four times as big. That means that there are four times as many par­ti­cles, so $I$ $\vphantom0\raisebox{1.5pt}{$=$}$ 16 par­ti­cles, in a box that has four times the vol­ume. If the vol­ume of the box be­comes 4 times as large, there are four times as many sin­gle-par­ti­cle states on each shelf, since the num­ber of states per unit vol­ume at a given sin­gle-par­ti­cle en­ergy is con­stant, com­pare (6.6). Shelf 1 now has 4 states, shelf 2 has 12, and shelf 3 has 32. The num­ber of en­ergy states for given shelf oc­cu­pa­tion num­bers is shown as grey tones in the mid­dle graph of fig­ure 11.5. Now the num­ber of sys­tem en­ergy eigen­func­tions that have all par­ti­cles on shelf 1 is not one, but 4$\POW9,{16}$ or 4 294 967 296, since there are 4 dif­fer­ent states on shelf 1 that each of the 16 par­ti­cles can go into. That is ob­vi­ously quite lot of sys­tem eigen­func­tions, but it is dwarfed by the dark­est square, states with shelf oc­cu­pa­tion num­bers $\vec{I}$ $\vphantom0\raisebox{1.5pt}{$=$}$ (4,6,6). There are about 1.4 10$\POW9,{24}$ sys­tem en­ergy eigen­func­tions with those shelf oc­cu­pa­tion num­bers. So the $\vec{I}$ $\vphantom0\raisebox{1.5pt}{$=$}$ (16,0,0) square at the ori­gin stays lily-white de­spite hav­ing over 4 bil­lion en­ergy eigen­func­tions.

If the sys­tem size is in­creased by an­other fac­tor 4, to 64 par­ti­cles, the num­ber of states with oc­cu­pa­tion num­bers $\vec{I}$ $\vphantom0\raisebox{1.5pt}{$=$}$ (64,0,0), all par­ti­cles on shelf 1, is 1.2 10$\POW9,{77}$, a tremen­dous num­ber, but to­tally hu­mil­i­ated by the 2.7 10$\POW9,{138}$ eigen­func­tions that have oc­cu­pa­tion num­bers $\vec{I}$ $\vphantom0\raisebox{1.5pt}{$=$}$ (14,27,23). Tak­ing the ra­tio of these two num­bers shows that there are 2.3 10$\POW9,{61}$ en­ergy eigen­func­tions with shelf num­bers $(14,27,23)$ for each eigen­func­tion with shelf num­bers $(64,0,0)$. By the time the sys­tem reaches, say, 10$\POW9,{20}$ par­ti­cles, still less than a mil­limol, the num­ber of sys­tem en­ergy eigen­states for each set of oc­cu­pa­tion num­bers is as­tro­nom­i­cal, but so are the dif­fer­ences be­tween the shelf num­bers that have the most and those that have less. The tick marks in fig­ure 11.5 in­di­cate that for large sys­tems, the dark­est square will have 40% of the par­ti­cles on shelf 2, 37% on shelf 3, and the re­main­ing 23% on shelf 1.

These gen­eral trends do not just ap­ply to this sim­ple model sys­tem; they are typ­i­cal:

The num­ber of sys­tem en­ergy eigen­func­tions for a macro­scopic sys­tem is as­tro­nom­i­cal, and so are the dif­fer­ences in num­bers.

An­other trend il­lus­trated by fig­ure 11.5 has to do with the ef­fect of sys­tem en­ergy. The sys­tem en­ergy of an en­ergy eigen­func­tion is given in terms of its shelf num­bers by

$${\vphantom' E}^{\rm S}=I_1{\vphantom' E}^{\rm p}_1+I_2{\vphantom' E}^{\rm p}_2+I_3{\vphantom' E}^{\rm p}_3$$

so all eigen­func­tions with the same shelf num­bers have the same sys­tem en­ergy. In par­tic­u­lar, the squares just be­low the oblique cut-off line in fig­ure 11.5 have the high­est sys­tem en­ergy. It is seen that these shelf num­bers also have by far the most en­ergy eigen­func­tions:

The num­ber of sys­tem en­ergy eigen­func­tions with a higher en­ergy typ­i­cally dwarfs the num­ber of sys­tem eigen­func­tions with a lower en­ergy.

Fig­ure 11.6: Num­ber of en­ergy eigen­func­tions on the oblique en­ergy line in the pre­vi­ous fig­ure. (The curves are math­e­mat­i­cally in­ter­po­lated to al­low a con­tin­u­ously vary­ing frac­tion of par­ti­cles on shelf 2.) Left: 4 par­ti­cles, mid­dle: 64, right: 1,024.

Next as­sume that the sys­tem has ex­actly the en­ergy of the oblique cut-off line in fig­ure 11.5, with zero un­cer­tainty. The num­ber of en­ergy eigen­states $Q_{\vec{I}}$ on that oblique line is plot­ted in fig­ure 11.6 as a func­tion of the frac­tion of par­ti­cles $I_2$$\raisebox{.5pt}{$/$}$​$I$ on shelf 2. (To get a smooth con­tin­u­ous curve, the val­ues have been math­e­mat­i­cally in­ter­po­lated in be­tween the in­te­ger val­ues of $I_2$. The con­tin­u­ous func­tion that in­ter­po­lates $n!$ is called the gamma func­tion; see the no­ta­tions sec­tion un­der ! for de­tails.) The max­i­mum num­ber of en­ergy eigen­states oc­curs at about $I_2$$\raisebox{.5pt}{$/$}$​$I$ $\vphantom0\raisebox{1.5pt}{$=$}$ 40%, cor­re­spond­ing to $I_3$ $\vphantom0\raisebox{1.5pt}{$=$}$ 37% and $I_1$ $\vphantom0\raisebox{1.5pt}{$=$}$23%. This set of oc­cu­pa­tion num­bers, $(I_1,I_2,I_3)$ = (0.23,0.40,0.37)$I$, is called the most prob­a­ble set of oc­cu­pa­tion num­bers. If you pick an eigen­func­tion at ran­dom, you have more chance of get­ting one with that set of oc­cu­pa­tion num­bers than one with a dif­fer­ent given set of oc­cu­pa­tion num­bers.

To be sure, if the num­ber of par­ti­cles is large, the chances of pick­ing any eigen­func­tion with an ex­act set of oc­cu­pa­tion num­bers is small. But note how the spike in fig­ure 11.6 be­comes nar­rower with in­creas­ing num­ber of par­ti­cles. You may not pick an eigen­func­tion with ex­actly the most prob­a­ble set of shelf num­bers, but you are quite sure to pick one with shelf num­bers very close to it. By the time the sys­tem size reaches, say, 10$\POW9,{20}$ par­ti­cles, the spike be­comes for all prac­ti­cal pur­poses a math­e­mat­i­cal line. Then es­sen­tially all eigen­func­tions have very pre­cisely 23% of their par­ti­cles on shelf 1 at en­ergy ${\vphantom' E}^{\rm p}_1$, 40% on shelf 2 at en­ergy ${\vphantom' E}^{\rm p}_2$, and 37% on shelf 3 at en­ergy ${\vphantom' E}^{\rm p}_3$.

Since there is only an in­cred­i­bly small frac­tion of eigen­func­tions that do not have very ac­cu­rately the most prob­a­ble oc­cu­pa­tion num­bers, it seems in­tu­itively ob­vi­ous that in ther­mal equi­lib­rium, the phys­i­cal sys­tem must have the same dis­tri­b­u­tion of par­ti­cle en­er­gies. Why would na­ture pre­fer one of those ex­tremely rare eigen­func­tions that do not have these oc­cu­pa­tion num­bers, rather than one of the vast ma­jor­ity that do? In fact, {N.23},

It is a fun­da­men­tal as­sump­tion of sta­tis­ti­cal me­chan­ics that in ther­mal equi­lib­rium, all sys­tem en­ergy eigen­func­tions with the same en­ergy have the same prob­a­bil­ity.

So the most prob­a­ble set of shelf num­bers, as found from the count of eigen­func­tions, gives the dis­tri­b­u­tion of par­ti­cle en­er­gies in ther­mal equi­lib­rium.

This then is the fi­nal con­clu­sion: the par­ti­cle en­ergy dis­tri­b­u­tion of a macro­scopic sys­tem of weakly in­ter­act­ing par­ti­cles at a given en­ergy can be ob­tained by merely count­ing the sys­tem en­ergy eigen­states. It can be done with­out do­ing any physics. What­ever physics may want to do, it is just not enough to off­set the vast nu­mer­i­cal su­pe­ri­or­ity of the eigen­func­tions with very ac­cu­rately the most prob­a­ble shelf num­bers.