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I wish to calculate the values of the orbital angular momentum of excited electronic states, using Molpro. So $\langle L_x \rangle$, $\langle L_y \rangle$, and $\langle L_z \rangle$. (Specifically for my case, I'm looking for the projection on the z-axis in a diatomic molecule. But I guess the answer should apply generally for all projections.) This should ideally be done at the CASSCF or MRCI level. But feel free to share procedures if they exist in other programs too.

I can find one mention of these operators in the documentation, called LOP taken together, or LX, LY, LZ, individually. However they are listed with parity -1, and: "Expectation values are only nonzero for symmetric operators (parity=1)". Still, the electronic state must have some angular momentum. For instance, in CASSCF I can calculate the $\langle L_x^2 \rangle$, $\langle L_y^2 \rangle$, and $\langle L_z^2 \rangle$ with the card EXPEC2,LXX,LYY,LZZ. But then the sign of the projected angular momentum is lost, and I wish to compare signs for the different excited states.

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    $\begingroup$ Have you taken into account spin-orbit coupling (SOC)? The angular momentum expectation value of a real wavefunction is always 0, and as long as you don't take into account SOC and there isn't a magnetic field, the wavefunctions can always be taken as real. So you can only expect a non-zero expectation value of angular momentum if you turn on SOC $\endgroup$
    – wzkchem5
    Jul 8 at 9:58
  • $\begingroup$ Did you try LX and confirm that you got zero? $\endgroup$ Jul 9 at 10:09
  • $\begingroup$ Thanks @wzkchem5! I have indeed not turned on spin-orbit coupling. So this question is based on a misconception. I will leave it up for others who might be inclined to make the same mistake. Cheers! $\endgroup$ Jul 11 at 9:32
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"However they are listed with parity -1, and: "Expectation values are only nonzero for symmetric operators (parity=1)"."

This matches the comment by wzkchem5 which suggests that those expectation values, like $\left\langle \hat{L}_x \right\rangle$ which you correctly found to be listed as -1 in the parity column of the table in the link you provided, will be zero. It may be true that $\left\langle \hat{L}^2_x \right\rangle$ is not zero, because there's no longer cancellation of positive and negative parts.

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    $\begingroup$ Thanks @Nike! This question was indeed based on a misconception (as you know, without spin-orbit coupling the wave-function is taken to be real, and $\langle L_x \rangle$ is zero). I will leave the question for others not to make my mistake. Cheers! $\endgroup$ Jul 11 at 9:32

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