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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?

Qmechanic
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Caspertijmen1
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2 Answers2

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For two particles with the same angular momentum $\ell_1=\ell_2=\ell$, the permutation symmetry follows immediately from the symmetries of the Clebsch-Gordan coefficients: $$ C^{LM}_{\ell m_1;\ell m_2}=(-1)^{2\ell+L} C^{LM}_{\ell m_2;\ell m_1} $$ so that (in accordance with the answer of @Superciocia), the symmetric states have $L$ even and the antisymmetric ones has $L$ odd. This holds for any $\ell$.

The situation is much more complicated if you have $3$ or more particles. For instance, in the coupling of $3$ states with $\ell=1$, there are three sets of states with total $L=1$. One of them is symmetric but the other two have mixed symmetry.

One of the sets with mixed symmetry comes from coupling the first and second particles to $L_{12}=0$, then coupling this to the third to get $L=1$. It’s easy to see the resulting states are neither symmetric nor antisymmetric (although they are antisymmetric under the exchange $1\leftrightarrow 2$.)

The symmetric states with $L=1$ are in fact linear combinations of $L_{12}=1$ and $L_{12}=2$ states.

ZeroTheHero
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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.

SuperCiocia
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