# Addition of Angular Momentum for identical particles

This is the problem I'm trying to understand:

Consider two particles with spin 1 without orbital angular momentum. If they are distinguishable, from the rule of addition of angular momentum applied to spin, we'll have states of total spin $j=0,1,2$. If we have, however, identical particles which are the possible states?

In textbooks, the addition of angular momentum is never treated in terms of distinguishable and identical particles, at least I don't recall it. The way I would approach this problem is to acknowledge that the possible total spin would be $j=0,1,2$ and then, from the state $|-1\rangle |-1\rangle$, I would use the ladder operator to build all the other four states compatible with $j=2$.

How do I build the states for $j=0,1$? The state $|j=1,m_{j}=-1\rangle$ must be built from the same states as $|j=2,m_{j}=-1\rangle$, that is, $|-1\rangle |0\rangle$ and $|0\rangle |-1\rangle$. So how are they any different?

Thank you very much.

## Answers and Replies

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Orodruin
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What is important for distinguishable particles is that the total state is (anti-)symmetric (depending on Bose/Fermi statistics). You need to take the spin state into account as well as the spatial wave function.