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Quantization axis

 
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Jul14-12, 07:29 AM   #1
 

Quantization axis


Hi

I am looking at a bunch of atoms in a homogeneous magnetic field, irridiated by a monochromatic EM wave. I am trying to figure out how to intensity pattern of the emitted light by the atoms looks.

Case 1) I have attached a picture of the situation called "case_1.jpg". It is very clear that only π-transitions are being driven, i.e. Δm=0 transitions.

Case 2) I have attached a picture of the situation again. The quantization axis points along the magnetic field, but the polarization is orthogonal to it. So somehow I need to decompose the polarization into something in the same plane as the B-field. How can I do that?

I would be very happy to recieve some feedback.

Best,
Niles.
Attached Thumbnails
case_1.jpg   case_2.jpg  
 
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Jul14-12, 07:42 AM   #2
 
Ah, ok. I think I figured it out entirely by myself. I can of course always decompose it into circularly polarized light along k. So they will drive the Δm=+1 and Δm=-1 transition. But then what happens when B is perpendicular to both k and E? Then my "trick" doesn't work anymore.



Best,
Niles.
 
Jul14-12, 04:58 PM   #3
 
Quote by Niles View Post
Ah, ok. I think I figured it out entirely by myself. I can of course always decompose it into circularly polarized light along k. So they will drive the Δm=+1 and Δm=-1 transition. But then what happens when B is perpendicular to both k and E? Then my "trick" doesn't work anymore.



Best,
Niles.
Maybe that is a forbidden transition.
Or maybe you better look at quadropole moments, or magnetic dipole moments.
I thought that the m=0 transition is dipole forbidden, anyway.
 
Jul14-12, 05:03 PM   #4
 

Quantization axis


Quote by Darwin123 View Post
Maybe that is a forbidden transition.
Or maybe you better look at quadropole moments, or magnetic dipole moments.
I thought that the m=0 transition is dipole forbidden, anyway.
I'm pretty sure having E perp. to k perp. to B will still yield a signal. I just don't see how I can ever decompose E into something along B, but I know it is possible.
 
Jul17-12, 08:08 AM   #5
 
You can still decompose the linear polarization into two circular ones. What changes wrt case_1 is the relative phase between the two circular waves.

Therefore you should get the same spectrum as in case_1, i.e. delta-m=0.
 
Jul18-12, 01:30 PM   #6
 
Hi

Thanks for replying! However I have to disagree. So B is perp. to k, which is perp. to E: If the electron is oscillating circularly along B, then looking "edge on", it looks linear. And it is exactly this motion that the E-field excites. So the transitions being driven are delta-m = +/- 1.

Does this sound reasonable to you?
 
Jul21-12, 02:40 PM   #7
 
When I find a moment I will work this out in detail.

You can write the dipole operator ε.r as Ʃ_m |r| ε_m Y_1,m
where m=-1,0,1 and Y_1,m is a spherical harmonic.

The matrix element then reduces to an amplitude prefactor and some
Clebsch-Gordans. If you know the initial and final angular momenta
this is easy to write down exactly.
 
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