GREPhysics.NET: GR8677 #92

GR8677 #92

Problem

GREPhysics.NET Official Solution

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

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.

lattes
2008-08-05 22:40:54

Good observation!

Poop Loops
2008-11-05 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.

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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.

yosun
2005-11-10 11:33:18

if $\Delta l =\pm 1$ and $\Delta s$ is unrestricted, then shouldn't $\Delta j$ also be unrestricted?

keflavich
2005-11-10 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.

yosun
2005-11-10 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

keflavich
2005-11-10 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 $\Delta m_l = -1$ ?

keflavich
2005-11-10 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.

mhas035
2007-04-12 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

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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.

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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.

yosun
2005-11-05 23:09:46

Thanks snimi1 for the correction!

athithi
2008-04-10 22:50:20

i am wondering about the option A ..how n can take negative values ?

thebigshow500
2008-10-14 10:56:46

Same question on choice (A)...the subject GRE test is coming soon, anyone care to explain?

kobayashi_maru
2008-10-15 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.

thebigshow500
2008-10-15 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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