Talk:Selection rule

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some issues[edit]

First of all i think the notation in this page needs to be cleaned up, but i'm no good at it so someone should do it :-P. Second I have an issue with this statement:

"Thus there is no E0 (electric monopoles) or M0 (magnetic monopoles) radiation (the latter is forbidden because magnetic monopoles do not seem to exist)."

Would magnetic monopole radiation be allowed even if magnetic monopoles existed? I would think that the same reason there isn't electric monopole radiation (conserved current plus gauss' law) would also prevent magnetic monopole radiation. Achoo5000 19:29, 15 April 2006 (UTC)[reply]

n change in electric dipole radiation:

In considering a hydrogen atom in the 2p state, is a change in n allowed when considering dipole radiation? Does it allow for a transition to the 1s state?

"Thus there is no E0 (electric monopoles) ... radiation" Though this may be abusing the term "radiation" - cathode rays seem an appropriate candidate for E0 radiation, or simply electrons. They are electric monopoles and can be ejected in various reactions. 194.169.228.3 (talk) 05:45, 30 May 2009 (UTC)BChmura[reply]

The table of angular momentum selection rules is somewhat unclear as it doesn't distinguish between total (many-electron) L of the initial and final states and the l of the individual electron actually jumping. This means that students are confused by the apparent contradiction between rules (4) and (5) for LS coupling, which seem to suggest sometimes L can only change by +/-1 and other times by 0,+/-1 (of course the former applies to the orbital \Delta l of the jumping electron, e.g. configuration change, the latter to the \Delta L change between Terms). I would suggest 'L' throughout rule 4 is replaced by 'l'. Also, spin flips are forbidden in LS coupled E1 transitions, so 'If \Delta S=0' should be replaced by '\Delta S=0' in this case. — Preceding unsigned comment added by 81.141.93.180 (talk) 00:44, 24 February 2021 (UTC)[reply]

Something to add[edit]

If anyone could make one of those energy level diagrams for Hydrogen or something, would look nice. ArdClose (talk) 01:06, 27 March 2009 (UTC)[reply]

Good article[edit]

This strikes me as a very good article because of the way it first explains something in reasonably abstract language, then goes on to display the specific math in a practical way, along with a narrative that tries to explain the math so it makes sense. Just my opinion.

What is $\not \leftrightarrow$ ?[edit]

As a physics graduate student, I have never come across this notation. Since it is not standard in the usual QM textbooks (Griffiths, Sakurai, Merzbacher, Shankar, Schiff), perhaps it should either be taken out of this article or explicitly defined. The table on http://en.wikipedia.org/wiki/List_of_logic_symbols says it means "propositional logic", which is out of scope of an article on QM selection rules. — Preceding unsigned comment added by 128.143.102.247 (talk) 19:38, 5 January 2015 (UTC)[reply]

If it were a double arrow it would mean "not equivalent to" in mathematical logic. But the meaning here is different. I think it's being used informally to mean "the transition between these two states is forbidden". If the arrow were not crossed out, it would mean "the transition between these two states is permitted". I made this change. I have to admit I still don't understand the table. 178.38.105.134 (talk) 12:40, 27 February 2015 (UTC)[reply]

Doubtful formulation with "direct product"[edit]

The symmetry of the transition moment function is the direct product of the symmetries of its three components.

The link is to "direct product of groups". What are the groups here? A symmetry is not a group. Plus, I doubt that one should multiply together anything that is indexed by the spatial directions i=1,2,3, if that is what is meant by components There is something wrong with the formulation here. If someone can show me an example of what is meant, perhaps I can put it into words. 178.38.105.134 (talk) 12:21, 27 February 2015 (UTC)[reply]

I think it means that you "multiply" the parities together. You know how if you multiply an odd number by an even number, you always get an even number? And if you multiply an odd number by an odd number, you always get an odd number? I believe the meaning of the phrase in this context is that since there are three functions being multiplied together, you can simply "multiply" the parity of each function with the other two - like if you have three odd functions, you'll get a value of zero for the integral, but if you have two even functions and and odd function you'll get an odd function and the integral will be nonzero. 128.239.208.55 (talk) 01:13, 24 April 2015 (UTC)[reply]

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Where is the discussion on NMR?[edit]

Although selection rules is cited on a number of Wikipedia pages about topics in Nuclear magnetic resonance, this branch of spectroscopy has been omitted from the discussion. Can anyone add it? In the mean time I'll add NMR to the category list. Tachyon (talk) 19:59, 12 February 2019 (UTC)[reply]

Summary table, M_J=0->0 forbidden if delta J = 0[edit]

As can be seen from eg these lecture notes (page 6), $M_{J}=0\rightarrow M_{J}=0$ is forbidden if $\Delta J=0$ . This is not included in the summary table and as far as I can figure is not implied by it, for example I don't see how $2^{3}S_{1}(M_{J}=0)\rightarrow 2^{3}P_{1}(M_{J}=0)$ is shown to be forbidden from the table. In order: J=1 to 1 is allowed, M_J = 0 to 0 is allowed, the parity is reversed because it goes from L=0 to L=1 (this is the part I'm not 100% sure about), $\Delta L=1$ . I don't know if this rule applies to the higher order transitions. I've made an edit to add this, it might look better if this row is made 2 lines high and it's split over them but I started breaking things when I tried to do that. 149.248.103.106 (talk) 18:25, 9 February 2022 (UTC)[reply]