# Interchange symmetry for states with identical particles

I was reading this web page about interchange symmetry for states with identical particles here:

http://quantummechanics.ucsd.edu/ph130a/130_notes/node317.html

The article states that the highest total angular momentum state will always be symmetric under interchange and the next highest state is antisymmetric.

So my questions is:

Why does the symmetry of the angular momentum wave function alternate? How to prove it?

• Relevant things to google or otherwise investigate: Permutation group, young tableaux, representation theory and young tableaux, young tableaux and angular momentum. You might find the following useful: hep.caltech.edu/~fcp/math/groupTheory/young.pdf – joshphysics Jun 10 '14 at 6:53

Let there be two identical particles with angular momentum operators $$J_1, J_2$$ having the value $$j$$.
A state in the joined angular momentum basis will be $$|J, m\rangle$$
A state in the separated angular momentum basis will be $$|m1, m2\rangle$$
We shall observe the state with the highest azimuthal component $$m = j + j = 2j$$
$$|J=2j,m=2j\rangle = |m_1=j,m_2=j\rangle$$
It is easy to see that the state is symmetrical under interchange. The $$J_-$$ operator is commuting with the permutation operator, hence every $$|J=2j,m\rangle$$ state is also symmetrical under interchange.
Next, we know that $$|J=2j-1,m=2j-1\rangle = const * (|m_1=j,m_2=j-1\rangle - |m_1=j-1,m_2=j\rangle)$$ We can see that this next state with lower total angular is anti-symmetrical, and because of the previous statement every $$|2j-1,m\rangle$$ state is also symmetrical under interchange.
We can repeat this process down to the desired value of $$J$$