D.25 Num­ber of bo­son states

For iden­ti­cal bosons, the num­ber is $I+N-1$ choose $I$. To see that think of the $I$ bosons as be­ing in­side a se­ries of $N$ sin­gle par­ti­cle-state boxes. The idea is as il­lus­trated in fig­ure D.2; the cir­cles are the bosons and the thin lines sep­a­rate the boxes. In the pic­ture as shown, each term in the group of states has one bo­son in the first sin­gle-par­ti­cle func­tion, three bosons in the sec­ond, three bosons in the third, etcetera.

Fig­ure D.2: Bosons in sin­gle-par­ti­cle-state boxes.

Each pic­ture of this type cor­re­sponds to ex­actly one sys­tem state. To fig­ure out how many dif­fer­ent pic­tures there are, imag­ine there are num­bers writ­ten from 1 to $I$ on the bosons and from $I+1$ to $I+N-1$ on the sep­a­ra­tors be­tween the boxes. There are then $(I+N-1)!$ ways to arrange that to­tal of $I+N-1$ ob­jects. (There are $I+N-1$ choices for which ob­ject to put first, times $I+N-2$ choices for which ob­ject to put sec­ond, etcetera.) How­ever, the $I!$ dif­fer­ent ways to or­der the sub­set of bo­son num­bers do not pro­duce dif­fer­ent pic­tures if you erase the num­bers again, so di­vide by $I!$. The same way, the dif­fer­ent ways to or­der the sub­set of box sep­a­ra­tor num­bers do not make a dif­fer­ence, so di­vide by $(N-1)!$.

For ex­am­ple, if $I$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 and $N$ $\vphantom0\raisebox{1.5pt}{$=$}$ 4, you get 5!$\raisebox{.5pt}{$/$}$​2!3! or 10 sys­tem states.