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GR8677 #92
Problem
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Quantum Mechanics$\Rightarrow$}Perturbation Theory

One can derive the selection rules by applying the electric dipole approximation in time-dependent perturbation theory. The results are the following: $\Delta m = 0, \pm 1$, $\delta l = \pm 1$. Choices (B) and (C) are (exactly this, thus) immediately out. There is no selection rule for spin, and thus choice (D) is it. (The correction is due to user snim1.)

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Comments
kevintah
2015-09-09 09:05:11
This site has more information including the full derivation and results. http://quantummechanics.ucsd.edu/ph130a/130_notes/node422.html\r\n There is no selection rule for spin as mentioned\r\n
 sam162015-10-20 14:12:33 page not found!
 NoPhysicist22017-03-19 11:36:32 http://quantummechanics.ucsd.edu/ph130a/130_notes/node422.html
flyboy621
2010-11-15 19:54:48
$\Delta s$ is always 0, and $\Delta l$ is always +/- 1. Therefore $\Delta j$ is also +/- 1. Thus (B) and (E) are compatible and (D) is not.
dvarjas
2010-10-06 07:52:54
My oppinion after much thougth about why D is incorrect and E correct: If there are two electrons on the same orbit, their spatial wave functions are symmetric, therefore they form a spin singlet with s=0. It could happen, that an unpaired electron with s=1/2 jumps to a half filled orbit with another electron with s=1/2, making a transition from s=1/2+1/2=1 to s=0, or the reverse process could also happen. But the total angular momentum j is conserved for the whole system, photon takes away j=1, so $\Delta j=\pm 1$ as in (E). But $\Delta l=\pm 1$ and $\Delta s$ cannot be greater than 1 for a single electron process, $\Delta s=0$ is needed for conservation of total angular momentum. So the only process allowed is that an unpaired electron jumps to an empty orbit.
 gman2010-11-11 23:11:59 How about this... When the electron transitions, $\Delta l = \pm 1$. Since it's an electron, and it doesn't change spin, $\Delta s=0$. So $\Delta j=\pm 1$. Now the photon has spin 1, and has no orbital angular momentum (right? what's it.. orbiting?) So $l=0$, $s=\pm 1$, and thus $\pm j=1$. All consistent with the above selection rules. The orbital angular momentum lost becomes spin angular momentum. Thus, $\Delta l = \pm 1$ $\Delta s=0$ $\Rightarrow \Delta j= \pm 1$
hanmas
2008-04-11 20:48:21
s always equals to 1/2 for electron, so there is no such thing as $\Delta s$ = -1.
 lattes2008-08-05 22:40:54 Good observation!
 Poop Loops2008-11-05 21:43:06 But can't you go +1/2 or -1/2? I thought that's what it meant. Either way, it still doesn't work so that's the answer.
 wittensdog2009-07-28 08:17:19 Indeed, s is the magnitude of spin, and m_s would be the value of it's projection along a given axis, often denoted z by convention. It's the same as how l is the magnitude of orbital angular momentum, and m_l is it's projection along an axis. Since in the answers they did make note of l vs. m_l (and not simply just m), I believe they meant to distinguish between s and m_s. While m_s can change for an electron, between -1/2 and 1/2, the MAGNITUDE of spin for any particle is fixed, so having any change in s is nonsensical. I actually didn't even notice this myself, and was originally thinking it meant m_s. Good observation!
keflavich
2005-11-10 11:04:25
Since $\Delta l =\pm 1$, and $\Delta s$ is unrestrictied (i.e. it can equal zero), E is compatible with electric dipole selection rules. This whole question is about photon emission: photons are emitted when electrons change energy levels, and it\'s the change in energy level\'s that\'s being restricted.
 yosun2005-11-10 11:33:18 if $\Delta l= \pm 1$ and $\Delta s$ is unrestricted, then shouldn't $\Delta j$ also be unrestricted?
 keflavich2005-11-10 12:05:46 Maybe my logic was wrong. In retrospect, I think any $\Delta s$ restriction should be incompatible with electric dipole emission since changes in the spin are related to hyperfine splitting. If you were to consider only electric dipole emission by excited states of atoms, I think you must assume that spin doesn't change and therefore $\Delta j = \Delta l$. Any $\Delta s$ would not involve or be involved in a transition between excitation states of an atom. I think it would require two emissions - one hyperfine, one electric dipole - to change both $l$ and $s$. Anyway, $\Delta j = \pm 1$ certainly wouldn't violate any selection rules, but $\Delta s = \pm 1$ would prevent any transition that did NOT change the spin of an electron, which is not a condition of electric dipole emission.
 yosun2005-11-10 12:11:05 on the issue of $\Delta j = \Delta l$ ... consider transition B of GR0177.84 at http://grephysics.net/disp.php?yload=prob&serial=3&prob=84
 keflavich2005-11-10 17:48:31 Right, valid point (though maybe you're refering to C). Instead of arguing more, I decided to look it up... http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydazi.html /> $\Delta j = \pm 1$ is a valid restriction for some case (though I don't really understand what a j=0 state would be since $j=l \pm s$), there is no spin restriction. As it turns out, hyperphysics claims that in photon emission the electron spin quantum number doesn't change, so perhaps the C transition (in 0177:82) is allowed because $\Delta m_l = -1$?
 keflavich2005-11-10 17:53:07 Ignore my comments, just check out the website. I've contradicted myself enough on this problem that my answers are probably unreliable.
 mhas0352007-04-12 23:10:09 The electric dipole operator is not dependant on the spin of the electron so for electirc dipole transitions, transitions between different spin states will give a probability of zero
Buli
2005-11-08 21:14:27
If there is no selection rule for spin, then there would be no for j too. Recall that j=l+s. I was wondering this is because the photon emission obeys the conservation of spin momentum so that delta s is zero. Can anybody tell your opinion? Thanks.
 mpdude82012-04-19 16:56:25 You said it yourself: j = l +s, there IS a selection rule for l (it must change by 1) but no selection rule for s, thus j must change with l. If there was a selection rule for s, you may be able to have the change in l and s cancel perfectly.
snimi1
2005-11-05 00:54:17
The official answer is D, which I tend to agree because there is no spin related selection rule.
 yosun2005-11-05 23:09:46 Thanks snimi1 for the correction!
 athithi2008-04-10 22:50:20 i am wondering about the option A ..how n can take negative values ?
 thebigshow5002008-10-14 10:56:46 Same question on choice (A)...the subject GRE test is coming soon, anyone care to explain?
 kobayashi_maru2008-10-15 07:31:37 The problem specifies emission of photons by excited states. For an electron to emit a photon, it has to drop from a higher excited state to a lower one (or ground state). Thus, $\Delta n$ is negative.
 thebigshow5002008-10-15 18:45:29 I am an idiot. I thought $\Delta n$ has to be a positive value. But it should be $\Delta n = n_f-n_i$ = negative value. $n_f$ = lower state $n_i$ = higher state Thanks a lot!

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