0
$\begingroup$

Place a sub-micron clump of crystal violet molecules in front of a multipixel detector. Raise the molecules to an electronically excited state with a beam of 590 nm light, illuminating from the side so you don't hit the detector with this beam.

Next, fire a collimated beam of 660 nm light at the molecules, which goes on to hit the detector at normal incidence. Qualitatively, how does the intensity pattern on the detector change due to the presence of the clump of electronically excited molecules? How will this pattern change as we vary the distance between the clump and the detector?

Background details:

Assume that the the illumination beam and the detector pixels are as ideal as possible; the detector pixels are as small as we care to make them and have negligible read noise, and the intensity of the beam is as uniform and stable as the shot-noise limit allows.

Crystal violet is an interesting chromophore. 590 nm light raises it to an electronically excited state which can fluoresce, but it can also relax to the electronic ground state via a nonradiative transition. Since the nonradiative relaxation is much faster than spontaneous emission, fluorescence from crystal violet molecules is nearly undetectable.

Min et. al showed that if you shoot crystal violet hard enough with 660 nm light, stimulated emission outcompetes nonradiative relaxation, and you can detect the stimulated emission as a small increase in the brightness of the stimulating beam.

I'm considering crystal violet molecules to eliminate the complication of spontaneous emission; they won't radiate on their own, the only light they'll emit is stimulated emission.

I'm considering a sub-micron clump of molecules to eliminate the complication of nonresonant scattering. Nonresonant scattering drops off extremely quickly with the radius of a spherical scatterer.

I'm considering a clump of molecules rather than a single molecule to ground the question in semiclassical mechanics (meaning QM for the molecules, but Maxwell's equations for the light). It's possible one cannot answer this question without reference to the photon model of light (meaning second quantization and Fock states, etc), but I'd be surprised if this is truly a quantum-optics phenomenon. I suspect the photon model of light and the quark model of the nucleus are equally (ir)relevant to the answer to this question.

| cite | improve this question | |
$\endgroup$
  • $\begingroup$ in my books stimulated emission IS a quantum mechanical phenomenon. "90 nm light raises it to an electronically excited state which can fluoresce," the language you use is quantum mechanics. contradicting the "irrelevant" in your last paragraph $\endgroup$ – anna v Dec 2 '14 at 19:03
  • $\begingroup$ You're confusing quantum field theory with quantum mechanics. If you genuinely need second quantization and Fock states to predict what the detector sees, go ahead and use it. $\endgroup$ – Andrew Dec 2 '14 at 21:16
  • $\begingroup$ ...and you're not the first to do so. Which implies the question could use some clarification. I'll think about how best to do so. $\endgroup$ – Andrew Dec 2 '14 at 21:17
  • $\begingroup$ The clarification I need is whether the 660nm has a level it can excite, or not. If not, nothing will happen other than the scattering of light off the molecules maybe changing a bit the kinetic energy/temperature. To see interference one needs coherence. If the second frequency can stimulate another level have a look at physics.stackexchange.com/questions/64707/… . $\endgroup$ – anna v Dec 3 '14 at 4:39
  • $\begingroup$ From the question: "...if you shoot crystal violet hard enough with 660 nm light, stimulated emission outcompetes nonradiative relaxation, and you can detect the stimulated emission..." $\endgroup$ – Andrew Dec 3 '14 at 6:49
1
$\begingroup$

This is a kind of optical amplifier -- see 1, 2.

Probably the main effect of the molecules on the beam will be scattering / refraction, just like if you put anything into of a beam of light. But a secondary effect, especially if the beam is much much larger than the clump of molecules, is that the profile of the beam will change a bit. The part of the beam that passes through the clump will become relatively brighter, while the rest of the beam won't change, loosely speaking.

To be more precise you need to look up gain guiding and play around with the equations. Sorry I cannot give a better description, I'm not sufficiently familiar with it myself.

But once again, I suspect that the gain guiding effect will be pretty hard to see compared to plain old scattering / refraction.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.