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Quantization axis |
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| Jul14-12, 07:29 AM | #1 |
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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. |
| Jul14-12, 07:42 AM | #2 |
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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 |
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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 |
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Quantization axis |
| Jul17-12, 08:08 AM | #5 |
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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 |
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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 |
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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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