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.57}. But here are the results. (Drums please.) The system eigenfunction counts for distinguishable particles, bosons, and fermions are:

 $\displaystyle Q^{\rm {d}}_{\vec I}$ $\textstyle =$ $\displaystyle I! \prod_{{\rm all}\ s}\frac{N_s^{I_s}}{\Big(I_s\Big)!}$  (11.1)
 $\displaystyle Q^{\rm {b}}_{\vec I}$ $\textstyle =$ $\displaystyle \prod_{{\rm all}\ s}
\frac{\Big(I_s+N_s-1\Big)!}{\Big(I_s\Big)!\Big(N_s-1\Big)!}$  (11.2)
 $\displaystyle Q^{\rm {f}}_{\vec I}$ $\textstyle =$ $\displaystyle \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

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

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

\begin{displaymath}
n! = \prod_{{\underline n}=1}^n {\underline n}= 1 \times 2 \times 3 \times \ldots \times n
\end{displaymath}

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.
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
\begin{picture}(...
...82,21,t'$\pp11////$'
\PB210,21,t'$\pp12////$'}
\end{picture}
\end{figure}

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.
\begin{figure}
\centering
{}%
\setlength{\unitlength}{1pt}
\begin{pic...
.../I$}}
\put(278,68){\makebox(0,0)[l]{$I_3/I$}}
\end{picture}
\end{figure}

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

\begin{displaymath}
\psi^{\rm S}_1 =
\pp1/{\skew0\vec r}_1//z1/ \pp1/{\skew0...
...//z2/ \pp1/{\skew0\vec r}_3//z3/ \pp1/{\skew0\vec r}_4/p/z4/.
\end{displaymath}

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,

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

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

\begin{displaymath}
4 \times 3 \times 8 \times 3 \times 3 = 864
\end{displaymath}

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

\begin{displaymath}
{\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
\end{displaymath}

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.
\begin{figure}
\centering
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\begin{pic...
...110,-13)(140.3,0){3}{\makebox(0,0)[b]{$I_2/I$}}
\end{picture}
\end{figure}

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.