Rising and lowering operators in a molecule

[Malamala]

Hello! If I have a molecule in Hund case a, I can write its wavefunction (electronic + rotational) as $|e,\Lambda,\Omega,S>|J,\Sigma,\Lambda,\Omega,M>$. I am not sure what happens if I apply, say, $S_+$ on this wavefunction (assuming I am not applying it to the top of the ladder), where this is the raising operator in the molecule frame. In principle this should give: $|e,\Lambda,\Omega,S>|J,\Sigma+1,\Lambda,\Omega,M>$. However initially I had, $\Omega= \Sigma + \Lambda$, but now I would have $\Omega= \Sigma + 1 + \Lambda$, which doesn't make sense. Given that by definition $\Omega= \Sigma + \Lambda$, I would expect that applying $S_+$ would also turn $\Omega$ to $\Omega+1$, but in many books I see that the $S_+$ operator is usually multiplied by $J_-$, which is the rising operator for $\Omega$ (it is "-" because I work in the molecule frame and the signs are inverted). But if that is the case, my previous argument doesn't hold anymore. So my questions is, if I apply $S_+$, do I need to apply $J_-$, too, such that the $\Omega$ is consistent with $\Omega= \Sigma + \Lambda$? And if so, why isn't $\Omega$ automatically changed by 1 when I apply $S_+$? Does it even make sense mathematically to apply $S_+$ alone at all given this? Thank you!

 

[eljose79]

Hello! Thank you for your question. The confusion here stems from the fact that in the molecular frame, the quantum numbers Ω and Σ are not independent, but are related through the equation Ω = Σ + Λ. This means that when you apply the raising operator S+ to the wavefunction, you are not only changing the value of Σ, but also the value of Ω. This is why you see the Ω+1 term in the resulting wavefunction.

In order to keep the Ω=Σ+Λ relation consistent, you would need to apply the lowering operator J- as well. This is because the J- operator only affects the Ω quantum number, and does not change the values of Λ or Σ. By applying both the S+ and J- operators, you are effectively changing the Σ value while keeping the Ω=Σ+Λ relation intact.

To answer your question about why the Ω quantum number is not automatically changed by 1 when you apply S+, it is because the S+ operator is not specifically designed to change the value of Ω. Its main purpose is to change the value of Σ, and the change in Ω is a consequence of this.

In summary, it does make sense mathematically to apply S+ alone, but in order to maintain the consistency of the Ω=Σ+Λ relation, you would need to also apply J-. I hope this helps clarify your doubts. Thank you for your interest in quantum mechanics!