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### atomic transitions

Posted: **Sat Mar 27, 2010 5:21 pm**

by **awonen**

Do photons originating from electronic transitions have random polarizations? Is the polarization somehow coupled to the spin state of the decaying electron? Is there any way to distinguish the spin state of the excited state of the decaying electrons by observing the emitted photon?

### Re: atomic transitions

Posted: **Sun Mar 28, 2010 12:00 pm**

by **grae313**

awonen wrote:Do photons originating from electronic transitions have random polarizations? Is the polarization somehow coupled to the spin state of the decaying electron? Is there any way to distinguish the spin state of the excited state of the decaying electrons by observing the emitted photon?

I think it depends on whether the transition occurs in the presence of a magnetic field. Without a magnetic field the different m levels are degenerate and oriented randomly, and transitions in this case will be randomly polarized. The presence of a magnetic field lifts the space isotropy and separates the different m levels. When this is the case, transitions for which delta m is zero (pi transition) emit linearly polarized radiation and when delta m is plus/minus 1 (sigma transition) the radiation is circularly polarized along the spin axis and linearly polarized along the tangent plane.

You can imagine an electron undergoing a pi transition as oscillating back and forth in the z-axis (defined by the external B-field), then the radiation field is just like an oscillating dipole. You can imagine an electron undergoing a sigma transition as circulating around in the xy-plane, so from above or below you see circularly polarized light, and looking at it in profile from the xy-plane it looks like it's oscillating in the plane so you get linearly polarized light.

### Re: atomic transitions

Posted: **Sun Mar 28, 2010 2:05 pm**

by **mobytish**

You may also want to look at little bit at how a laser works. Using polarized light to stimulate emission of photons does help to increase the likelihood of a given polarization, but even that won't result in 100% polarization of all of the emitted photons. And adding lots of polarizing optics can help, with the highest quality optics increasing the polarization to at most 99.97% (I think the number may be higher, but there's still a bit that's not perfectly polarized). Sorry if this is a little tangent to your question...

### Re: atomic transitions

Posted: **Sun Mar 28, 2010 10:03 pm**

by **grae313**

grae313 wrote:I think it depends on whether the transition occurs in the presence of a magnetic field. Without a magnetic field the different m levels are degenerate and oriented randomly, and transitions in this case will be randomly polarized. The presence of a magnetic field lifts the space isotropy and separates the different m levels. When this is the case, transitions for which delta m is zero (pi transition) emit linearly polarized radiation and when delta m is plus/minus 1 (sigma transition) the radiation is circularly polarized along the spin axis and linearly polarized along the tangent plane.

You can imagine an electron undergoing a pi transition as oscillating back and forth in the z-axis (defined by the external B-field), then the radiation field is just like an oscillating dipole. You can imagine an electron undergoing a sigma transition as circulating around in the xy-plane, so from above or below you see circularly polarized light, and looking at it in profile from the xy-plane it looks like it's oscillating in the plane so you get linearly polarized light.

Oh, and also it's helpful to realize that circularly polarized light carries angular momentum, and then the polarization rules for transitions make more sense. For circularly polarized light to carry away some angular momentum then the angular momentum of the atom must have changed in order for it to be conserved, hence delta m = plus/minus 1, whereas linearly polarized light does not have angular momentum associated with it so the angular momentum change of the atom must also be zero, hence delta m = 0.