# Total angular momentum of N identical bosons

1. Jun 6, 2009

### wdlang

assume that we have N spin-1 bosons all on the same spatial orbit.

The problem is that what values the total angular momentum can be?

I am puzzled by this problem for a long time

maybe a bit permutation group theory is needed?

i guess this type of problem is well solved

Is there any book i can consult?

2. Jun 7, 2009

3. Jun 7, 2009

4. Jun 8, 2009

### Avodyne

I assume that there is no orbital angular momentum, and so we are just keeping track of spin.

First of all, if the particles were distinguishable, we could assign $m=-1,0,+1$ to each particle (where $m$ is the eigenvalue of $S_z$ for that particle). Then there would be $3^N$ possible states.

If the particles are bosons, however, then we are allowed to keep only the totally symmetric part of the state. Thus the state is labeled just by the total numbers of +1's, 0's, and -1's, which I will call $n_+, n_0, n_-$, with $n_++n_0+n_-=N$. For example, with $N=3$ and $n_+=2, n_0=1, n_-=0$, the only allowed state is

$${\textstyle{1\over\sqrt3}}(|{+1},+1,0\rangle+|{+1},0,+1\rangle+|0,+1,+1\rangle).$$

The total number of completely-symmetric states is given by the number of ways to partition $N$ among $n_+, n_0, n_-$. This number is

$$\sum_{n_-=0}^N\sum_{n_0=0}^N\sum_{n_+=0}^N \delta_{n_++n_0+n_-,\,N} = \sum_{n_-=0}^N\sum_{n_0=0}^{N-n_-}1 = \sum_{n_-=0}^N N+1-n_- = {\textstyle{1\over2}}(N+1)(N+2).$$

Suppose we want a state with total $S_z=+N$. The only way to get it is to have each individual $S_z=+1$, or equivalently $n_+=N, n_0=0, n_-=0$. This state must be part of a multiplet with total spin quantum number $s=N$. There are $2N+1$ states in this multiplet, with z-component quantum number ranging from $m=-N$ to $m=+N$.

Now suppose we want a state with total $S_z=N-1$. The only way to get it is to have $n_+=N-1, n_0=1, n_-=0$. However, we already have one state with $m=N-1,$ the one with $s=N.$ So the one state with $m=N-1$ must be the one in the $s=N$ multiplet. Therefore, there is no state with $s=N-1.$ If there was, there would have to be a second, different state with $m=N-1.$

Now suppose we want a state with total $S_z=N-2$. There are two ways to get it: $n_+=N-2, n_0=2, n_-=0$ and $n_+=N-1, n_0=0, n_-=1$. One of these states must be the one with $s=N$ and $m=N-2,$ and the other must be a state with $s=N-2$ and $m=N-2.$ There are $2(N-2)+1=2N-1$ states in this multiplet.

Now suppose we want a state with total $S_z=N-3$. There are two ways to get it: $n_+=N-3, n_0=3, n_-=0$ and $n_+=N-2, n_0=1 , n_-=1$. However, we already have two states with $m=N-3,$ the one with $s=N$ and the one with $s=N-2.$ So there is no room for another state with $m=N-3,$ and so there is no state with $s=N-3$.

And so on. We ultimately find $s=N,N-2,N-4,\ldots$, down to 1 or 0, are the allowed values.

We can check this by counting states. We already counted the total number of completely symmetric states, with the result ${\textstyle{1\over2}}(N+1)(N+2).$ Now let's count them according to the allowed values of $s.$

If $N$ is even, the allowed values of $s$ are $0,2,4,\ldots,N$, and the total number of states is

$$\sum_{s=0,2,4,\ldots}^N2s+1 = {\textstyle{1\over2}}(N+1)(N+2),$$

which agrees with the other way of counting. If $N$ is odd, the allowed values of $s$ are $1,3,5,\ldots,N$, and the total number of states is

$$\sum_{s=1,3,5,\ldots}^N2s+1 = {\textstyle{1\over2}}(N+1)(N+2),$$

which again agrees.

So we're done. If $N$ is even, the allowed values of $s$ are $0,2,4,\ldots,N,$ and if $N$ is odd, the allowed values of $s$ are $1,3,5,\ldots,N.$ In both cases, each value of $s$ occurs once.

5. Jun 11, 2009

### Avodyne

There is a minor mistake above (and for some reason I don't get an "edit" button to fix it): I wrote $2(N-2)+1=2N-1$ instead of $2(N-2)+1=2N-3$.