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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 $|\uparrow\rangle |\uparrow\rangle$ 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 $|\Psi\rangle $ is symmetric under particle exchange and state $|\Phi\rangle$ is such that $\langle \Psi | \Phi \rangle=0$, can one show that $|\Phi\rangle$ is antisymmetric?

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

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  • $\begingroup$ Is there a general argument which shows that singlet, triplet, quintet, etc. states should be alternatingly symmetric and antisymmetric? $\endgroup$ Oct 10 '20 at 18:24
  • $\begingroup$ Yes. Look at symmetry of CG coefficient under permutation of two indices: $C^{LM}_{\ell_1m_1;\ell_2m_2}=(-1)^{\ell_1+\ell_2-L}C_{\ell_2m_2;\ell_1m_1}^{LM}$. Thus, for any $\ell_1=\ell_2=\ell$, the different $L$ states will alternate between symmetric and antisymmetric as you go from $L=2\ell, 2\ell-1,\ldots, 0$. $\endgroup$ Oct 10 '20 at 21:46
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You are overthinking it; you didn't do this in class?

For your specific example, you see it. Ignoring normalizations, $ |\uparrow\rangle |\uparrow\rangle$ is lowered by $J_-$ to $|\uparrow\rangle | 0\rangle+ |0\rangle |\uparrow\rangle$, and so on, as you indicated. But this state is orthogonal to $|\uparrow\rangle | 0\rangle - |0\rangle |\uparrow\rangle$, which is antisymmetric, so it is annihilated by $J_+$, so it's the highest spin state of the triplet.

Lowering it once and twice will likewise produce antisymmetric states.

The singlet, $|\uparrow\rangle | \downarrow\rangle + | \downarrow\rangle|\uparrow\rangle - |0\rangle | 0\rangle$, will be symmetric.

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  • $\begingroup$ How did you get the form of the singlet state? I know that it should be a linear combination of $|\uparrow\rangle | \downarrow\rangle,\ | \downarrow\rangle|\uparrow\rangle\ ,|0\rangle | 0\rangle$ but how did you get the coefficients? The way we'd do it in class would be either by orthogonality with another state or by using that fact that operating on it with the lowering/raising operator gives the zero vector. Is there a more efficient way? $\endgroup$ Oct 10 '20 at 18:23
  • $\begingroup$ Also, my question was rather whether there is a general argument that shows that singlet, triplet quintet etc. states are alternatingly symmetric and antisymmetric. $\endgroup$ Oct 10 '20 at 18:29
  • $\begingroup$ 1. Both methods work; the second is more organized basic linear algebra. 2. Again, as indicated, if you descend from the highest M of Jmax state, the lower rung will allow construction of the state of J=Jmax-1 , orthogonal to Jmax, so will have the opposite symmetry; the one below, similarly, etc... There are plethysm arguments, but it is so trivial to just do them by inspection as here, so you'd certainly check them this way. There are formulas for the most general case. $\endgroup$ Oct 10 '20 at 19:17

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