Physics Asked by TomS on November 15, 2020
Suppose we have two-particle states
$$ |0rangle^{(j)} = frac{1}{sqrt{2}} big(|+m,-mrangle – |-m,+mrangle big) $$
based on one-particle states in the SU(2) $j$-representation, where I use the notation
$$ |+m,-mrangle = |j,+mrangle otimes |j,-mrangle $$
with arbitrary $j = 1/2, 1, 3/2, ldots$ and $m le j$.
A rotation $ R(alpha,beta,gamma) $ acts as *)
$$ R(alpha,beta,gamma) , |+m,-mrangle = R(alpha,beta,gamma) , |j,+mrangle otimes R(alpha,beta,gamma) ,|j,-mrangle $$
For singlet states the rotations reduce to the identity.
Is there a simple way to see that for the above mentioned states $|0rangle^{(j)}$ without going through the calculation of matrix elements, i.e. Wigner D-functions for each representation j?
*)
$$ R(alpha,beta,gamma) = e^{-ialpha J_z} e^{-ibeta J_y} e^{-igamma J_z} $$
$$ J_i = J_i^{(1)} otimes 1^{(2)} + 1^{(1)} otimes J_i^{(2)} $$
$$ implies ; R(alpha,beta,gamma) = R^{(1)}(alpha,beta,gamma) otimes R^{(2)}(alpha,beta,gamma) $$
The rotation matrices are but exponentials of linear combinations of generators J, both at the tensor factor level and the coproduct level. So if you see that all three coproduct generators in the ladder decomplexified basis, $J_z,J_{pm}$, annihilate your state, for any j, you have seen that their exponential amounts to the identity for that state. Conversely, if some (combinations of) generators do not annihilate your state, then there is a departure from the identity, in general, e.g. for a small angle, and your state is not a singlet.
Now, indeed, for your state, $$ J_z Big( |+m,-mrangle - |-m,+mrangle Big ) =Big((m-m) |+m,-mrangle -(-m+m) |-m,+mrangle Big ) =0. $$
However, examine the action of $J_+$ on it, but take j=2, m=1, for simplicity. $$ J_+ Big( |1,-1rangle - |-1,1rangle Big ) = Big( 2 |2,-1rangle + sqrt{ 6 } |1, 0rangle - sqrt{ 6 } | 0, 1rangle - 2 | -1, 2rangle Big )neq 0. $$ This is not a singlet.
Correct answer by Cosmas Zachos on November 15, 2020
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