Submitted by TheGandPTurtle t3_111g7s9 in askscience
Weed_O_Whirler t1_j8gn3wr wrote
Reply to comment by dack42 in Light traveling through a medium that slows it. Does the same photon emerge? by TheGandPTurtle
While it's true if a single photon is absorbed by a single atom, you cannot predict which direction the photon will be re-emitted, so how is this different? Well, the process is not so different from light "reflecting" off a mirror- while we say light reflects, a mirror is also an absorption and re-emission situation, and obviously those don't go in a random direction.
The answer comes down to conservation of momentum and interference. Since the incoming photon has momentum, there is a higher probability that the photon will be emitted in the same direction to conserve momentum. Obviously, not all of the photons are emitted in that direction, but due to the probability there will be constructive interference in the same direction and destructive in all other directions. In general, things like Snell's Law of Refraction, and angle of incidence equalling angle of reflection occur with large numbers of photons, and they do not describe what happens when a single photon is absorbed.
MasterPatricko t1_j8ib8hl wrote
> a mirror is also an absorption and re-emission situation
I already addressed the use of "absorption and re-emission" in a previous reply to you, so I won't repeat that -- but for anyone reading, this is another case where you must not confuse "classical particle absorption and re-emission" (photon comes in, is absorbed, is gone for a moment, is emitted) with the collective scattering of probabilities/wavefunctions that we calculate in QM.
The law of reflection does not (cannot) come from a classical particle scattering model. It can be derived from either a classical wave model or a properly quantum mechanical model (wavefunctions are waves after all).
> The answer comes down to conservation of momentum and interference. Since the incoming photon has momentum, there is a higher probability that the photon will be emitted in the same direction to conserve momentum.
We've already discussed why you should not imagine the photon being completely absorbed by an atom when trying to derive the laws of optics. But let me address the momentum part of this answer (interference part is fine) as well because it's also misleading. Why should an atom prefer to return exactly to its previous state of "zero momentum"? Answer: It doesn't. Further: First of all the momentum we are talking about is tiny when a photon is absorbed by an atom, so even the thermal motion of the atoms is already going to wash out any momentum change. Second the atom is free to exchange momentum with its neighbours via phonons, so there's no reason why it has to return to zero by itself. Third, we measure single photon/single atom absorption and re-emission behaviour experimentally and it truly is isotropic (well, kinda), while your answer effectively claims it is not (contradicting your very first statement). (Incidentally, when 1 and 2 above are not true any more, like for a very cold gas, we can do fun stuff like laser Doppler cooling.)
The real answer is that we're not dealing with classical absorption/emission of photons by one atom, we're dealing with a collective behaviour and scattered probabilities/partial waves (see previous linked discussion, etc.) -- even in the case of few photons. So though all the scattered components are random in direction, we add them coherently (interference, as you said). The newly constructed wave/wavefunction has a clear direction of travel obeying the law of reflection.
WagonWheelsRX8 t1_j8ipiy5 wrote
I've been following and trying to wrap my head around these concepts as well. Did not realize something we see every day (light passing through glass) would be such a deep topic.
Based on your description, my understanding is that if you had a laser, a piece of glass (transparent material with index of refraction) and a detector placed in a dark box in a vacuum, and
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you emitted a single photon, there is a high probability that photon would be detected by the detector.
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you emitted many photons, the photon detector on the other side of the transparent material would not detect the same quantity of photons emitted, because there is a probability that some of them would be reflected back at the emitter. However, because they are all part of a wavefront, none of them would be emitted out the sides of the transparent material?
MasterPatricko t1_j8j3fc4 wrote
I think you're on the right track, mostly. I can try to add some detail/clarification.
> 1) you emitted a single photon, there is a high probability that photon would be detected by the detector.
Yeah, more or less -- the probability depends on the coefficient of reflection of the surface of the glass. It could be high, or low, depending on the angle and refractive index of the glass, and will match the classical calculation of the relative amplitudes of transmitted and reflected waves (Fresnel equations), because fundamentally, the photon wavefunction is obeying pretty much the same wave propagation laws. Note you only get this behaviour if the photon wavefunction spatial extent (typically many photon wavelengths) means that it has the possibility to interact with many identical atoms which are spaced in a uniform way (relative to the wavelength of the photon). This is mostly true for optical wavelengths hitting normal materials (typical atom spacing 0.1nm, optical wavelength 500nm, any random variation is too small to be "seen" by a photon) but is not necessarily true at shorter wavelengths like X-rays, or if your "glass" is only a few atoms big. Those scenarios are more complicated and you don't always end up with the "normal" ray optics rules after summing those probabilities.
The only real difference to the classical wave picture is in the moment of detection -- instead of measuring a classical wave, with some part reflected and some transmitted, we are set up for a discrete, quantum mechanical interaction in the detector. A 1 or a 0. This collapses* the wavefunction according to the probabilities mentioned before, to either interact with the detector (and so we say the photon was transmitted and then absorbed in the detector) or not (and here we can say the photon was reflected).
> 2) ... However, because they are all part of a wavefront, none of them would be emitted out the sides of the transparent material?
You can hopefully see how the previous explanation scales up to emitting many photons, the numbers of photons detected will match the probability calculations. But to address your last sentence -- in this toy example, I assumed that there was no internal absorption or scattering inside the glass. In such a case, the original+scattered wavefunctions sum up such that the only possibilities are transmission straight through or reflection right at the interface, and there is no chance for a "random" photon to emerge out of the side of the glass at some angle. This isn't to do with photons being "part of a wavefront", unless I misunderstand you. More fundamentally, photons are waves; or at least, they travel in the same ways that classical waves do, and waves moving through a uniform medium don't just randomly scatter.
Now if we make things a bit more realistic -- the glass is not going to be a perfect crystal, and the uniform background of atoms assumption is not going to be exactly true. There will now be a small possibility of the wavefunction scattering differently off some imperfection in the atomic structure. Because it's associated with an imperfection, this part of the wavefunction won't be cancelled out by all the neighbouring scattered amplitudes, and you will end up with a real (but small) probability to have a photon emerging out of the side of the glass in some random direction, in addition to the main probabilities of transmission or reflection at the surfaces.
* "The measurement problem", or what exactly wave-function collapse really means physically, is a very thorny issue to which there isn't a good answer. But we know that mathematically, it works.
WagonWheelsRX8 t1_j8je7x8 wrote
Very interesting, thanks this is helpful! Yes, I suppose in the real world there are a lot of additional factors that need to be considered as well (such as the actual uniformity of the glass, etc.)
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