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Aug 12, 2015 at 13:56 answer added MikeV timeline score: 2
Apr 2, 2015 at 22:53 comment added HDE 226868 @UserAnonymous Already did, when he posted it. But merging your accounts will mean that you can deal with something that comes up cross-site. All that would happen would be that you can access both accounts, and they'd have the same name. And you'll soon get those privileges, because you're close to getting the association bonus.
Apr 2, 2015 at 18:56 comment added User Anonymous @HDE226868 - I think my interests are better suited to this site than Astronomy. I won't even even have any good question to contribute there. Besides, if I create another UserAnonymous account there, I won't have upvoting and commenting privileges, and asking fake questions just to get enough reputation to be able to do that doesn't make sense to me. Could you please upvote that answer and thank Rob on my behalf? Or post a link to this question here, in a comment? Thanks.
Apr 2, 2015 at 18:51 comment added HDE 226868 @UserAnonymous Huh, weird. You should be able to contact SE and get your accounts merged, if you want.
Apr 2, 2015 at 18:48 comment added User Anonymous Hi @HDE226868. That was me only, while I was still unregistered. I landed up on that site first, saw some ques which looked interesting, posted the question, didn't get an answer, and was disappointed. (I now know answers don't come up immediately, but back then I thought they do.) A little later, I landed up here and thought the same is more likely to be answered here. So, I signed up, and while I never got an "answer", dmckee answered it with his comments. Somehow, in the middle of all of this, I never checked at how it fared over there. It got an answer after 9 days!
Apr 1, 2015 at 18:49 comment added HDE 226868 Related question on Astronomy: astronomy.stackexchange.com/q/6257
Sep 24, 2014 at 11:18 comment added User Anonymous @dmckee - Saw your comment today. Thank you. If you post this as an answer, I can accept and up vote.
Sep 6, 2014 at 15:41 comment added dmckee --- ex-moderator kitten "Wave" doesn't have any importance. It's not a marker for something that you have to do or remember of calculate. It means the same thing as "state of the system with angular momentum <l == 1>" only it's shorter. You could say "l-shell", too, but it would be unusual.
Sep 6, 2014 at 5:44 comment added User Anonymous @dmckee - Thanks, so now we agree that there has to be a scattering context (rest of it was fine right from the start). But that's precisely what I don't get - if it is a particle with $l=1$, produced in a scattering experiment (like all are), why does it have to INHERIT the ''wave'' label from that context?
Sep 5, 2014 at 19:12 comment added dmckee --- ex-moderator kitten I don't know from personal experience or from primary testimony, but I assume the "wave" usage was popularized in the scattering domain. It's just that the existence of angular momentum quantum numbers that have to be described applies to all states, bound or otherwise.
Sep 5, 2014 at 9:20 comment added User Anonymous @dmckee - Thanks. That makes sense. But how did that ''habit'' come into existence at all. What is so wavy about any particle state in the first place? That was the reason why I refered to scattering theory here - at least there are waves over there!
Sep 4, 2014 at 16:54 history edited David Z CC BY-SA 3.0
edited title
Sep 4, 2014 at 14:59 comment added dmckee --- ex-moderator kitten Recall that the s, p, d, f naming scheme originally came from atomic spectroscopy not from scattering theory. It applies just fine to bound systems such as charmonium without any need to talk about wave functions. The state has such-and-so angular momentum. The appellation "p-wave" is just habit.
Sep 4, 2014 at 14:39 comment added User Anonymous Kyle Kanos - I'm not too sure about that. In non-relativistic treatments, one does talk about the wave function, but in general in particle Physics, single particle wavefunctions are problematic aren't they? (e.g. in the simple Klein Gordon, or Dirac equation solutions etc.) So, wave standing for wavefunction does not look right to me, especially since notation would be general and not specific to NR. But nevertheless, thanks for your participation in this. You certainly improved my question.
Sep 4, 2014 at 14:34 comment added Kyle Kanos An off-the-cuff guess: it's the wave function for the P-state. I'm not much of a particle physicist, so I'll have to leave an actual answer to someone else.
Sep 4, 2014 at 14:29 comment added User Anonymous Kyle Kanos - You are right. Even the wiki article says it is the spectroscopic notation. But I am not understanding the meaning of wave, I get the P part. This wave notation is very common while refering to these states, most people call it P wave charmonia. But I have not come across any occasion where it is explained what the meaning of this ''wave'' is. Why not just charmonia P state?
Sep 4, 2014 at 14:29 history tweeted twitter.com/#!/StackPhysics/status/507535821726298113
Sep 4, 2014 at 14:25 comment added Kyle Kanos I see no reason to think that it is not spectroscopic notation they are using.
Sep 4, 2014 at 14:20 comment added User Anonymous Kyle Kanos - thanks for the edit and yes, they are talking about charmonium. I have read that wiki article, but that doesn't answer the wave part. Thanks anyways.
Sep 4, 2014 at 14:17 comment added Kyle Kanos They are talking about charmonium which is the meson $c\bar{c}$, see this Wikipedia article on the J/$\psi$ meson (this is mentioned in the opening paragraph of the linked paper as well).
Sep 4, 2014 at 14:15 history edited Kyle Kanos CC BY-SA 3.0
minor formatting, added tag, fixed title
Sep 4, 2014 at 14:15 review First posts
Sep 4, 2014 at 14:18
Sep 4, 2014 at 14:13 history asked User Anonymous CC BY-SA 3.0