physical chemistry - Molpro - Connection between atomic term symbols and system symmetry

Let's say I have a carbon atom with the electron configuration $1s^22s^22p^2$. We know, that using atomic term symbols we can describe the following states:

$${}^3P_0, {}^3P_1, {}^3P_2, {}^1D_2, {}^1S_0$$

Now I'd like to compute their energies with Molpro. But how should I specify those terms using just symmetry point groups?

I saw something similar using state-averaged multi computation from several different symmetries. But, I don't know, how to derive them just from the term symbol?

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My attempt

I was thinking about writing the state wavefunction as the linear combination of microstates and specifying each of then separately.

I.e. let's have the state ${}^1S_0$. According to this answer it can be described like this:

$$|L = 0, M_L = 0\rangle = \frac{1}{\sqrt 3} \left( |m_{l1}= 1, m_{l2} = -1\rangle + |m_{l1}= -1, m_{l2} = 1\rangle - |m_{l1}= 0, m_{l2} = 0\rangle \right)$$

I presume, that I can choose any of the available point groups in Molpro for 1 atom. Let's say I'll choose $D_{2h}$ point group.

Then the orbitals' symmetry can be described like this:

• $1s$ ($A_g$)


• $2s$ ($A_g$)


• $2p_x$ ($B_{3u}$)


• $2p_y$ ($B_{2u}$)


• $2p_z$ ($B_{1u}$)

According to this, the symmetry of the state $\left| m_{l1} = 1, m_{l2} = -1 \right>$ should be $$(A_g)^2 \otimes (A_g)^2 \otimes (B_{3u})^1 \otimes (B_{1u})^1 = B_{2g}$$

and the corresponding wavefunction specification in Molpro: wf,6,6,0.

The state $\left| m_{l1}=-1, m_{l2} = 1 \right>$ posses the same symmetry, i.e. $B_{2g}$ and so the Molpro wavefunction specification stays the same wf,6,6,0.

Finally, the state $\left| m_{l1} = 0, m_{l2} = 0 \right>$ has the symmetry

$$(A_g)^2 \otimes (A_g)^2 \otimes (B_{2u})^2 = A_g$$, i.e. the wavefunction will be specified like wf,6,1,0 and multi will be called with both of the wavefunctions together

{multi;
 occ,2,1,1,0,1,0,0,0;
 core,1,0,0,0,0,0,0,0;
 wf,6,6,0;
 wf,6,1,0;
}

Is this the correct approach?

Answer

You are trying to use the wf card to pass individual configurations. This is wrong in 2 ways:

1. The wf card does not specify which configurations to use, it specifies the symmetry of the wave function (or electronic state).


2. You don't need to tell Molpro which configurations to use. It will figure this out on its own based on the symmetry of the wave function you are requesting. All you need is the IRREP of the $^1S_0$ state, which is totally symmetric.

{multi; occ,2,1,1,0,1,0,0,0; core,1,0,0,0,0,0,0,0; wf,6,1,0; }

However, this will give you different a state than expected. Add expec2,lxx,lyy,lzz; and check $\langle L^2\rangle$ in the output. It should be 6, telling us the $^1D_1$ state is lower in energy than $^1S_0$. So we need to request more states of that symmetry (use the state card). Also, we should make sure the $^1D_1$ is described appropriately, as it might negatively affect the targeted $^1S_0$ state. So you should add all of the 5 degenerate $^1D_1$ components. I leave that as an exercise for you (see also below).

Some further note about your configuration $(a_g)^2\otimes(a_g)^2\otimes(b_{3u})^1\otimes(b_{1u})^1=B_{2g}$ (Note that the symmetry of orbitals is commonly denoted by small letters, while capital letters are used for electronic states.)

• Only IRREPs of same symmetry mix, so this one will not contribute to $^1S_0$.


• It is breaking the spherical symmetry of the atom. This can only happen if there is some external perturbation, e.g. a magnetic or electric field.


• It is however, one component of a degenerate (different) state. But you need to consider all the degenerate components to preserve symmetry. This is an issue of the symmetry reduction from $O_h$ to $D_{2h}$. For example you can calculate the $^3P$ state with

{multi; occ,2,1,1,0,1,0,0,0; core,1,0,0,0,0,0,0,0; wf,6,4,2; wf,6,6,2; wf,6,7,2; }

Check that you get 3 exactly degenerate $p$ orbitals, and 3 exactly degenerate electronic state energies. If you would request each state separately, by running multi 3 times with only one wf card each, you will get 3 degenerate state energies, but the $p$ orbitals won't be degenerate.