State orthogonal to a symmetric state must be antisymmetric

Question

I am working out the exchange symmetry of the eigenstates of the total angular momentum operator of a system of two spin-1 bosons.
I know that there must be a quintet, triplet, and a singlet state.
The highest state of the quintet is $|uparrowrangle |uparrowrangle$ and is symmetric under particle exchange.
Applying the lowering operator in this state generates the entire quintet. Since the lowering operator is symmetric itself, all states in the quintet are symmetric.
Using the orthogonality of states, one can deduce the highest state of the triplet. Successively applying the lowering operator generates the entire triplet.
My problem is: How does one show that the triplet states are antisymmetric, given that the quintet states are symmetric.
Phrased more generally perhaps, if the state $|Psirangle $ is symmetric under particle exchange and state $|Phirangle$ is such that $langle Psi | Phi rangle=0$, can one show that $|Phirangle$ is antisymmetric?

ZeroTheHero · Answer

In fact as stated
$$langle Phivert Psirangle =0 tag{1}
$$ does not imply that one is symmetric and the other antisymmetric.  The coupling $1otimes 1=2oplus 1oplus 0$ and the state with $L=0$ is orthogonal to all the $L=2$ states yet it is also symmetric, v.g. with $vert Phirangle=vert 2,0rangle$ and $vertPsirangle=vert 0,0rangle$ then both are symmmetric but (1) still,holds.

Cosmas Zachos · Answer

State orthogonal to a symmetric state must be antisymmetric

2 Answers

Add your own answers!

Ask a Question