# Three particle composite angular momentum states

I'm working on this problem to determine the ground state angular momenta ($$S,L,J$$) for nitrogen using Hund's rules, and I want to see if the total orbital angular momentum $$L=2$$ is possible (the states have to be antisymmetric). In this case, the three $$2p$$ electrons all have $$l=1$$.

To my understanding, if the "top of the ladder" ($$L = M$$) of the composite states with $$L=2$$ is symmetric/antisymmetric, then the whole collection of states with $$L=2$$ is symmetric/antisymmetric. Hence, we can examine the composite state $$|22\rangle$$.

I am unable to find a Clebsch-Gordan table for 3 particles, so I just added up the first two electrons, and then added the third to the composite state of the two.

Since the third electron has $$l=1$$, the composite state of the first two electrons can have either $$L_{12}=1$$ or $$L_{12}=2$$ in order to obtain that final $$L=2$$. With reference to the Clebsch-Gordan table, if $$L_{12} = 1$$, then $$|22\rangle = |11\rangle_{12}|11\rangle_3 = \frac1{\sqrt 2}(|11\rangle_1 |10\rangle_2 - |10\rangle_1 |11\rangle_2 )|11\rangle_3,$$ and if $$L_{12} = 2$$, then $$|22\rangle = \sqrt{\frac23}|22\rangle_{12}|10 \rangle_3 - \sqrt{\frac13}|21\rangle_{12}|11 \rangle_3 = \sqrt{\frac23}|11\rangle_1|11\rangle_2|10 \rangle_3 - \sqrt{\frac13}\sqrt{\frac12}(|11\rangle_1 |10\rangle_2 + |10\rangle_1 |11\rangle_2)|11\rangle_3.$$ I noticed that not only are these states not antisymmetric, they are not even symmetric, meaning that if I put the three electrons in the $$|22\rangle$$ state, exchanging two of the electrons will change the wave function. How could that be if electrons are identical? Otherwise, what went wrong with my procedure, and what is the proper way of adding the angular momenta of three particles?

## 2 Answers

your calculations are correct, however your beginning assumption is not: the $$L=2$$ states do not have do be specifically symmetric or antisymmetric. Looking at the representation $$1\otimes 1 \otimes 1 =(2\oplus 1 \oplus 0)\otimes 1 = 3\oplus 2\cdot2 \oplus 3\cdot1 \oplus 0$$, you can see that you can form $$2$$ representations of $$L=2$$. This hints that combined, at each value of $$L_z$$, the $$2$$ dimensional space is a standard representation of $$S_3$$. This is why the two states you calculated in the case $$L_z=2$$ are not symmetric.

In order to get the correct fully antisymmetric wave function of the three electrons, you need to combine these states with the $$s=1$$ states of spin (if you're familiar with particle physics, this is a bit how you combine the wave functions of the quarks to get the spin 1/2 baryons like the nucleons).

For the final resolution of your problem, I would recommend using different states though. Setting $$|L_{z1}L_{z2}L_{z3}\rangle_o$$ the orbital angular momentum states (no ambiguity since $$L_1=L_2=L_3=1$$) and similarily $$|s_{z1}s_{z2}s_{z3}\rangle_s$$ for the spin part (no ambiguity since $$s_1=s_2=s_3=1/2$$), it is usually more simple to consider: $$|(12)\rangle_o = \frac{1}{\sqrt 2}(|101\rangle_o-|011\rangle_o)$$ $$|(23)\rangle_o = \frac{1}{\sqrt 2}(|110\rangle_o-|101\rangle_o)$$ $$|(31)\rangle_o = \frac{1}{\sqrt 2}(|011\rangle_o-|110\rangle_o)$$ which are linear combinations of the two states you identified in the $$L=L_z=2$$ subspace. They may not be orthogonal and independent, but they make the symmetrisation process more transparent.

Hope this helps and please tell me if you find some mistakes.

There’s a number of subtle elements to your questions.

You have just discovered that the theory of permuting $$3$$ objects is more complicated (and more interesting) that permuting just $$2$$. Indeed, for $$3$$ objects, there are 2-dimensional spaces where the basis states do not necessarily transform back to a multiple of themselves after every permutation. By contrast with 2 objects states are either symmetric or antisymmetric (or combination of those).

You are dealing with the angular part but there’s also a spin part. Combining three spins will also produce repeated values of total $$S$$ and you would need to combine those with the spatial part to get a totally antisymmetric state.

It is for such a fully antisymmetric state that electrons are indistinguishable. In fact, if the particles are indistinguishable, the many-particle state must return to a multiple of itself so it will come as no surprise to learn that this fully antisymmetric state is a sum of products of partially symmetric spatial and partially symmetric spin states.

See for instance

Kaplan IG. The Pauli exclusion principle. Can it be proved?. Foundations of Physics. 2013 Oct;43(10):1233-51.

for a discussion of indistinguishability, and a specific application to 3-particles.

Note that you cannot have a fully antisymmetric spin state with three spin-1/2 particles, so in turn, this means spin-statistics prevents the occurrence of fully symmetric spatial wavefunction (which would have $$L=3$$).