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.


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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.