# Exchange symmetry and addition of angular momentum

1. Dec 1, 2014

### thegreenlaser

In case it's relevant, the context of my question is finding the allowed states of an atom. For example, given a nitrogen atom with (1s)2(2s)22p3, how do we find the possible states in terms of total orbital angular momentum L, total spin S, and total angular momentum J = L + S. It seems that the solutions to problems like this rely on determining whether given pairs of (L,S) lead to completely antisymmetric states or not. Only the antisymmetric states are allowed because of the symmetrization postulate.

Now, my question: say we add together angular momenta $J = J_1 + \cdots + J_n$ and get a resulting basis $\left| J, M \right\rangle$. Now, is it true that the exchange symmetry of $\left| J, M \right\rangle$ depends only on J? For example, if I know that one state with J = 4 is completely antisymmetric, can I be sure that all states with J = 4 are completely antisymmetric with respect to interchange? (I don't know if it makes a difference, but if it does, I'm only really interested in cases where $j_1, \ldots, j_n$ are either all integers or all half-integers.)

It seems that this is true for the case of adding two angular momenta (though I don't really know how to prove it). The states with $J = J_{max}$ will all be symmetric, the states with $J = J_{max} - 1$ will be antisymmetric, etc. The key point though is that states with the same $J$ will have the same exchange symmetry, even if $M$ is different. Is there a good justification for this, and does it hold for addition of three or more angular momenta?

2. Dec 2, 2014

### DrDu

That's a complex question. In general a state with a given J can be both symmetric and anti-symmetric, especially if the electrons occupy different orbitals, like 1p and 2p.

3. Dec 2, 2014

### thegreenlaser

I just realized my notation was a bit confusing, so I want to clarify. The J in the second half of my original post is not the J = L+S from the first half.

To put my question another way, if I add together orbital angular momenta $L = L_1 + \cdots + L_N$, will the symmetry of the resulting states $\left| L, M_L \right\rangle$ depend only on $L$ and not on $M_L$?

Similarly, if I add together spins $S = S_1 + \cdots + S_N$, will the symmetry of the resulting states $\left| S, M_S \right\rangle$ depend only on $S$ and not on $M_S$?

I ask because I see a lot of statements like "All the S = 1 spin states are symmetric and all the L = 1 orbital states are antisymmetric, so S = 1, L = 1, J = 0, 1, 2 are allowed states." But the whole thing seems to rely on the fact that the symmetry of the total spin states depends only on S and the symmetry of the total orbital angular momentum states depends only on L.

4. Dec 2, 2014

### thegreenlaser

I think I've maybe figured it out. Basically, I suspect that the lowering operator $\hat{L}_- = \hat{L}_{1-} + \cdots + \hat{L}_{N-}$ preserved symmetry. So if $\hat{L}_-$ operates on a completely antisymmetric state, the result is antisymmetric. If it operates on a completely symmetric state, the result is symmetric. If it operates on a partially symmetric state, the result is partially symmetric. I don't know how to prove this, but I'm pretty sure that it's true from squinting at it for a bit.

If the above statement is true, then my question is answered easily. If $\left| L, L \right\rangle$ is antisymmetric, then so is $\hat{L}_- \left| L, L \rightr\rangle \propto \left| L, L - 1 \right\rangle$. Repeated operations of $\hat{L}_-$ shows that $\left| L, L - 2 \right\rangle, \left| L, L-3 \right\rangle, \cdots, \left| L, -L \right\rangle$ are all antisymmetric. Similarly for non-antisymmetric states and for $S$.

The result, then, is that the symmetry of $\left| L, M_L; S, M_S \right\rangle$ depends only on $L, S$ and not on $M_L, M_S$.

Last edited: Dec 2, 2014
5. Dec 3, 2014

### DrDu

I now remember a more formal proof that the permutation group commutes with the tensor reps of the general linear group and its subgroups like the rotation group. Hamermesh's book on group theory discusses this at length.

6. Dec 3, 2014

### thegreenlaser

Ugh... I just realized that a small algebra error had been throwing me off this whole time. It's actually not hard at all to show that $\hat{L}_{\pm} = \hat{L}_{1\pm} + \cdots \hat{L}_{N\pm}$ and $\hat{S}_{\pm} = \hat{S}_{1\pm} + \cdots \hat{S}_{N\pm}$ commute with the interchange operators $\hat{P}_{ij}$. Then of course the raising and lowering operators will not change the symmetry of a state, and $\left| L, M_L; S, M_S \right\rangle$ must all have the same symmetry given the same $L$ and $S$.