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The question is as follow:

A state of a system of many noninteracting particles can be specified by listing which particle is in which of the accessible single particle states. In each microscopic state we can identify the number of particles in a given single particle state $k$. This number is called the occupation number, of state k and is denoted by $n_k$.
How many individual microscopic states have the same set of occupation numbers $n_k$?


Solution given:

The number of states with the same set of $n_k$ is the number of ways the N particles can be distributed in groups of $n_k$ each. It is the combinatorial factor that expresses the fact that all N particles can be interchanged, but interchanges of particles within each group do not produce new states. Hence the number is $$\frac{N!}{n_1!n_2! \dots n_k! \dots}$$

So I tried using this equation to see if I can find the correct answer. I took a system of $N=13$ particles $$ \begin{array}{|c|c|c|c|c|c|} \hline k & 1 & 2 & 3 & 4 & 5 \\ n_k & 3 & 2 & 1 & 4 & 3 \\ \hline \end{array} $$
where $k$ are the individual microscopic states and $n_k$ is their occupation numbers respectively.

And when I applied the equation, I get $$\frac{13!}{3!2!1!4!}=21621600$$ which is so far off from the number expected (2). I must have understand the equation wrong. Can someone please explain to me how exactly does this equation work?

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  • $\begingroup$ You forgot another $3!$ factor from the denominator, the formula should be $13!/(3! 2! 4! 3!)$, but apart from that what is your issue/problem with this answer? $\endgroup$ – Zoltan Zimboras Oct 28 '17 at 7:06
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    $\begingroup$ It turns out I misunderstood the statement. At first I thought "number of states with the same set of $n_k$ " means "how many k will have the same number of $n_k$", now I realize it actually means the combinatorics of the states... $\endgroup$ – WeiShan Ng Oct 28 '17 at 9:26
  • $\begingroup$ @WeiShang Ng: Good, then this is settled! Right? $\endgroup$ – Zoltan Zimboras Oct 28 '17 at 9:55
  • $\begingroup$ @ZoltanZimboras: Yup. $\endgroup$ – WeiShan Ng Oct 28 '17 at 10:12
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Apart from the fact that you forgot another $3!$ factor from the denominator, the answer is correct. That is, the number of states in a $N=13$ particle system with the same set of $n_k$-s that you described is equal to: $$ \frac{13!}{3!2!1!4!3!}=3603600. $$

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