In electronic spectroscopy of molecules, why some quantum numbers are considered to be 'good quantum numbers'? For example, $n$ and $l$ are said to be not good quantum numbers while $j$ is considered to be a 'good quantum number'. What is the logic/idea behind these concepts?

  • $\begingroup$ Do you have any reference for the term? $\endgroup$ – Andreas H. Aug 20 '13 at 19:59

Good quantum number are associated with operators that commute with the Hamiltonian. They correspond to conseved quantities.

Overall angular momentum is conserved, but the portions of it due to orbital motion and due to spins are not themselves conserved. n is 'bad' in that there's no conserved physical quantity related to radial motion.

A decently good explanation is at http://mxp.physics.umn.edu/s09/projects/S09_OpticalPumping/atomicstructure.htm (A snapshot of the page can be found here)

  • $\begingroup$ I understood the good quantum numbers are those which commute with the Hamiltonian and hence conservative. But I didn't understand what is the importance or what additional information a "conserved quantum number" can give compared to one that is not conserved? What I think is; if the quantity (or quantum number) is conserved, it is invariant under space inversion. Means, the mirror image is symmetric. So, the mirror image reflects the properties of the original quantity. I think that is why this conservation is important. Am I right? Could you please correct me/comment? Many thanks! $\endgroup$ – albedo Aug 21 '13 at 13:38
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    $\begingroup$ @albedo Good quantum numbers are useful for labelling states. Recall that if you have $[H,A] = 0$ you can find simultaneous eigenstates of $H$ and $A$. That is, you have states with well defined energy and property $A$ (e.g. angular momentum). You couldn't do that if $[H,A] \neq 0$. $\endgroup$ – Goku Aug 22 '13 at 0:01
  • $\begingroup$ @Goku: Then why $l$ is a good quantum number in closed shell atomic systems? $l$ is still not conserved right? [Reference: Annu.Phys.Rev.Chem.2003,54,397-424; in page 401]. $\endgroup$ – albedo Aug 25 '13 at 8:12
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    $\begingroup$ @albedo I guess it is simply because in a closed shell $s=0$. Then we have $j=s+l=l$ conserved. $\endgroup$ – Goku Sep 13 '13 at 2:53
  • $\begingroup$ If $[H,A]\neq0$ then the $A$ (e.g. angular momentum) is not good quantum number. This means if you have a state with specific energy, and you measure the angular momentum, you will get different values. $\endgroup$ – xslittlegrass Oct 4 '13 at 15:32

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