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

$$I \times (I{-}1) \times (I{-}2) \times \ldots \times (I{-}I_1{+}1)$$

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

$$I_1! = I_1 \times (I_1-1) \times (I_1-2) \times \ldots \times 3 \times 2 \times 1;$$

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

$$\frac{I \times (I-1) \times (I-2) \times \ldots \times (I-... ...) \times (I_1-2) \times \ldots \times 3 \times 2 \times 1}.$$

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:

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

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:

$$N_1^{I_1} \frac{I!}{I_1!(I-I_1)!}.$$

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

$$N_2^{I_2} \frac{(I-I_1)!}{I_2!(I-I_1-I_2)!}.$$

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

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

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

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

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

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

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

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

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

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

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.

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

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