# Number of individual states with the same occupation numbers

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1. Oct 27, 2017

### WeiShan Ng

1. The problem statement, all variables and given/known data
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 $n_k$?

2. Relevant equations
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
$$\frac{N!}{n_1!n_2! \dots n_k! \dots}$$

3. The attempt at a solution
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?

2. Oct 28, 2017

### DeathbyGreen

So a couple of things:

1) I'm not sure about your problem statement. You say that $k$ is the number of particles in a given state. This should be $n_k$. In occupation number representation the number operator $n_k$ returns the number of particles in the $k^{th}$ momentum state.

2) Are you using bosonic particles or fermionic particles? Because the statistics change depending on which; for example you can have two fermions in a state (spin up and down), and exchanging them is non trivial and creates a different wave function. For bosonic particles you can have infinitely many in a state (infinite dimensional Hilbert space) and swapping them leaves the wavefunction unchanged.