# When creating term symbols, how do you know if the angular momentum $L$ is antisymmetric of symmetric?

For example I'm trying to get the term symbol of $$(1s)^{2}(2s)^{2}(2p)^2$$ . In the answers they state the following:

The combination of angular momenta $$L_1 = L_2 = 1$$ gives $$L = 2$$ (symmetric), $$L = 1$$ (antisymmetric) and $$L = 0$$ (symmetric). This must be combined with the spin wave function of opposite symmetry, thus $$^1D_2, ^3P_{0, 1, 2}$$ and $$^1S_0.$$

I totally understand this, except for how they assign symmetric and antisymmetric to the angular momenta. In the previous exercise I only had $$L = 0$$ and they said it was symmetric and antisymmetric. So how do I know if the angular momentum is symmetric or antisymmetric?

The orbital angular momentum quantum number is just a number so it cannot be symmetric or antisymmetric.

Also, in this context you should specify that the "symmetry" is with respect to particle re-labelling/exchange. For Pauli exclusions etc.

The actual proof is (I think) quite mathematical with Slater determinants, but the general rule of thumb is that:

For odd numbers for the total angular momentum $$(L= 1,3,5, ...)$$ the spatial wavefunction is antisymmetric upon particle exchange.

You can test it though.
I worked out the angular momentum states of your possible total $$L$$ systems.

In the following, on the LHS there will be the basis in the total (composite) angular momentum $$|L, m_L\rangle$$, while on the RHS there will be the state in the individual angular momentum basis $$|\ell, m_\ell\rangle$$. Each term on the RHS is a tensor product between particle $$A$$ and particle $$B$$, so $$|1,0\rangle |1,-1\rangle$$ means "$$|1,0\rangle_A \otimes|1,-1\rangle_B$$ ".
In this specific case $$\ell = 1$$, always, as both electrons are the in $$p$$ orbital.

# $$L= 0$$:

\begin{align} \left|0,0\right> &= \frac1{\sqrt3} \left( \big|1,1\big>\big|1,-1\big> ~~+~~ \big|1,-1\big>\big|1,1\big> ~~-~~ \big|1,0\big>\big|1,0\big> \right) \end{align}

If you swap labels $$A\leftrightarrow B$$, each term stays exactly the same, so $$|0,0\rangle$$ is symmetric.

# $$L =1$$:

\begin{align} \lvert1,1\rangle & = \frac{1}{\sqrt 2} \left( \lvert 1,0\rangle\lvert1,1\rangle - \lvert1,1\rangle\lvert1,0\rangle \right) \\ \lvert1,0\rangle & = \frac{1}{\sqrt 2} \left( \lvert1,-1\rangle\lvert1,1\rangle - \lvert1,1\rangle\lvert1,-1\rangle \right) \\ \lvert1,-1\rangle & = \frac{1}{\sqrt 2} \left( \lvert1,0\rangle\lvert1,-1\rangle - \lvert1,-1\rangle\lvert1,0\rangle\right) \end{align}

If you swap labels $$A\leftrightarrow B$$, each line gets an overall minus sign, so $$|1,*\rangle$$ is antisymmetric.

# $$L=2$$:

You can check every case yourself, but for instance:

\begin{align} \left|2,0\right> &= \frac1{\sqrt6} \left( \big|1,1\big>\big|1,-1\big> ~~+~~ \big|1,-1\big>\big|1,1\big> ~~+~~ \sqrt4\cdot \big|1,0\big>\big|1,0\big> \right) \end{align}

If you swap labels $$A\leftrightarrow B$$, each term stays exactly the same, so $$|2,0\rangle$$ is symmetric.