D.57 Number of system eigenfunctions

This note derives the number of energy eigenfunctions $Q_{\vec{I}}$ for a given set $\vec{I}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $(I_1,I_2,I_3,\ldots)$ of shelf occupation numbers, $I_s$ being the number of particles on shelf number $s$. The number of single-particle eigenfunctions on shelf number $s$ is indicated by $N_s$.

Consider first the case of distinguishable particles, referring to figure 11.1 for an example. The question is how many different eigenfunctions can be created with the given shelf numbers. What are the ways to create different ones? Well, the first choice that can be made is what are the $I_1$ particles that go on shelf 1. If you pick out $I_1$ particles from the $I$ total particles, you have $I$ choices for particle 1, next there are $I-1$ choices left for particle 2, then $I-2$ for particle 3. The total number of possible ways of choosing the $I_1$ particles is then

\begin{displaymath}
I \times (I{-}1) \times (I{-}2) \times \ldots \times (I{-}I_1{+}1)
\end{displaymath}

However, this overestimates the number of variations in eigenfunctions that you can create by selecting the $I_1$ particles: the only thing that makes a difference for the eigenfunctions is what particles you pick to go on shelf 1; the order in which you chose to pick them out of the total set of $I$ makes no difference. If you chose a set of $I_1$ particles in an arbitrary order, you get no difference in eigenfunction compared to the case that you pick out the same particles sorted by number. To correct for this, the number of eigenfunction variations above must be divided by the number of different orderings in which a set of $I_1$ particles can come out of the total collection. That will give the number of different sets of particles, sorted by number, that can be selected. The number of ways that a set of $I_1$ particles can be ordered is

\begin{displaymath}
I_1! = I_1 \times (I_1-1) \times (I_1-2) \times
\ldots \times 3 \times 2 \times 1;
\end{displaymath}

there are $I_1$ possibilities for the particle that comes first in the sorted set, then $I_1-1$ possibilities left for the particle that comes second, etcetera. Dividing the earlier expression by $I_1!$, the number of different sets of $I_1$ particles that can be selected for shelf 1 becomes

\begin{displaymath}
\frac{I \times (I-1) \times (I-2) \times \ldots \times (I-...
...) \times (I_1-2) \times
\ldots \times 3 \times 2 \times 1}.
\end{displaymath}

But further variations in eigenfunctions are still possible in the way these $I_1$ particles are distributed over the $N_1$ single-particle states on shelf 1. There are $N_1$ possible single-particle states for the first particle of the sorted set, times $N_1$ possible single-particle states for the second particle, etcetera, making a total of $N_1^{I_1}$ variations. That number of variations exists for each of the individual sorted sets of particles, so the total number of variations in eigenfunctions is the product:

\begin{displaymath}
N_1^{I_1} \frac{I \times (I-1) \times (I-2) \times \ldots ...
...) \times (I_1-2) \times
\ldots \times 3 \times 2 \times 1}.
\end{displaymath}

This can be written more concisely by noting that the bottom of the fraction is per definition $I_1!$ while the top equals $I!$$\raisebox{.5pt}{$/$}$$(I-I_1)!$: note that the terms missing from $I!$ in the top are exactly $(I-I_1)!$. (In the special case that $I$ $\vphantom0\raisebox{1.5pt}{$=$}$ $I_1$, all particles on shelf 1, this still works since mathematics defines 0! = 1.) So, the number of variations in eigenfunctions so far is:

\begin{displaymath}
N_1^{I_1} \frac{I!}{I_1!(I-I_1)!}.
\end{displaymath}

The fraction is known in mathematics as “I choose $I_1$.”

Further variations in eigenfunctions are possible in the way that the $I_2$ particles on shelf 2 are chosen and distributed over the single-particle states on that shelf. The analysis is just like the one for shelf 1, except that shelf 1 has left only $I-I_1$ particles for shelf 2 to chose from. So the number of additional variations related to shelf 2 becomes

\begin{displaymath}
N_2^{I_2} \frac{(I-I_1)!}{I_2!(I-I_1-I_2)!}.
\end{displaymath}

The same way the number of eigenfunction variations for shelves 3, 4, ... can be found, and the grand total of different eigenfunctions is

\begin{displaymath}
N_1^{I_1} \frac{I!}{I_1!(I-I_1)!} \times
N_2^{I_2} \frac...
..._3} \frac{(I-I_1-I_2)!}{I_3!(I-I_1-I_2-I_3)!} \times
\ldots
\end{displaymath}

This terminates at the shelf number $S$ beyond which there are no more particles left, when $I-I_1-I_2-I_3-\ldots-I_B$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. All further shelves will be empty. Empty shelves might just as well not exist, they do not change the eigenfunction count. Fortunately, there is no need to exclude empty shelves from the mathematical expression above, it can be used either way. For example, if shelf 2 would be empty, e.g. $I_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, then $N_2^{I_2}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 and $I_2!$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, and the factors $(I-I_1)!$ and $(I-I_1-I_2)!$ cancel each other. So the factor due to empty shelf 2 becomes multiplying by one, it does not change the eigenfunction count.

Note that various factors cancel in the eigenfunction count above, it simplifies to the final expression

\begin{displaymath}
Q^{\rm {d}}_{\vec I} = I! \frac{N_1^{I_1}}{I_1!} \times
...
...^{I_2}}{I_2!} \times
\frac{N_3^{I_3}}{I_3!} \times
\ldots
\end{displaymath}

Mathematicians like to symbolically write a product of indexed factors like this using the product symbol $\Pi$:

\begin{displaymath}
Q^{\rm {d}}_{\vec I} = I! \prod_{{\rm all\ }s} \frac{N_s^{I_s}}{I_s!}.
\end{displaymath}

It means exactly the same as the written-out product.

Next the eigenfunction count for fermions. Refer now to figure 11.3. For any shelf $s$, it is given that there are $I_s$ particles on that shelf, and the only variations in eigenfunctions that can be achieved are in the way that these particles are distributed over the $N_s$ single-particle eigenfunctions on that shelf. The fermions are identical, but to simplify the reasoning, for now assume that you stamp numbers on them from 1 to $I_s$. Then fermion 1 can go into $N_s$ single-particle states, leaving $N_s-1$ states that fermion 2 can go into, then $N_s-2$ states that fermion 3 can go into, etcetera. That produces a total of

\begin{displaymath}
N_s \times (N_s-1) \times (N_s-2) \times \ldots \times (N_s-I_s+1)
= \frac{N_s!}{(N_s-I_s)!}
\end{displaymath}

variations. But most of these differ only in the order of the numbers stamped on the fermions; differences in the numbers stamped on the electrons do not constitute a difference in eigenfunction. The only difference is in whether a state is occupied by a fermion or not, not what number is stamped on it. Since, as explained under distinguishable particles, the number of ways $I_s$ particles can be ordered is $I_s!$, it follows that the formula above over-counts the number of variations in eigenfunctions by that factor. To correct, divide by $I_s!$, giving the number of variations as $N_s!$$\raisebox{.5pt}{$/$}$$(N_s-I_s)!I_s!$, or “$N_s$ choose $I_s$.” The combined number of variations in eigenfunctions for all shelves then becomes

\begin{displaymath}
Q^{\rm {f}}_{\vec I} =
\frac{N_1!}{(N_1-I_1)!I_1!} \time...
... \ldots
= \prod_{{\rm all\ }s} \frac{N_s!}{(N_s-I_s)!I_s!}.
\end{displaymath}

If a shelf is empty, it makes again no difference; the corresponding factor is again one. But another restriction applies for fermions: there should not be any eigenfunctions if any shelf number $I_s$ is greater than the number of states $N_s$ on that shelf. There can be at most one particle in each state. Fortunately, mathematics defines factorials of negative integer numbers to be infinite, and the infinite factor $(N_s-I_s)!$ in the bottom will turn the eigenfunction count into zero as it should. The formula can be used whatever the shelf numbers are.

Figure D.3: Schematic of an example boson distribution on a shelf.
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
\begin{picture}(...
...{\circle{10}}
\put(125,10){\rule{.2pt}{.2pt}}
\end{picture}
\end{figure}

Last but not least, the eigenfunction count for bosons. Refer now to figure 11.2. This one is tricky, but a trick solves it. To illustrate the idea, take shelf 2 in figure 11.2 as an example. It is reproduced in condensed form in figure D.3. The figure merely shows the particles and the lines separating the single-particle states. Like for the fermions, the question is, how many ways can the $I_s$ bosons be arranged inside the $N_s$ single-particle states? In other words, how many variations are there on a schematic like the one shown in figure D.3? To figure it out, stamp identifying numbers on all the elements, particles and single-state separating lines alike, ranging from 1 to $I_s+N_s-1$. Following the same reasoning as before, there are $(I_s+N_s-1)!$ different ways to order these numbered objects. As before, now back off. All the different orderings of the numbers stamped on the bosons, $I_s!$ of them, produce no difference in eigenfunction, so divide by $I_s!$ to fix it up. Similarly, all the different orderings of the single-particle state boundaries produce no difference in eigenfunction, so divide by $(N_s-1)!$. The number of variations in eigenfunctions possible by rearranging the particles on a single shelf $s$ is then $(I_s+N_s-1)!$$\raisebox{.5pt}{$/$}$$I_s!(N_s-1)!$. The total for all shelves is

\begin{eqnarray*}
Q^{\rm {b}}_{\vec I} & = &
\frac{(I_1+N_1-1)!}{I_1!(N_1-1)...
...
& = & \prod_{{\rm all\ }s} \frac{(I_s+N_s-1)!}{I_s!(N_s-1)!}.
\end{eqnarray*}