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Recently I read about something called photon up-conversion, in which a lower frequency of light is absorbed by a material that then proceeds to emit a higher frequency of light. When the mechanism for up-conversion is two-photon absorption, a molecule absorbs two lower-frequency photons, emits a higher-frequency photon, and then returns to it's initial state.

How would the change in entropy be calculated in a situation like this?

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    $\begingroup$ Why should it violate the second law? $\endgroup$ – Jon Custer Jul 7 '17 at 17:55
  • $\begingroup$ Because it's effectively concentrating energy $\endgroup$ – J. Antonio Perez Jul 7 '17 at 17:56
  • $\begingroup$ I don't think the second law means what you think it means. Multiphoton absorption, with or without using intermediate states, is certainly an allowed process. Consider process A, where a three-state system allows you to absorb one photon (going from ground to 1st excited state), and then another (going from 1st to 2nd excited state). Do you think that violates the second law? And if so, why? $\endgroup$ – Jon Custer Jul 7 '17 at 18:11
  • $\begingroup$ @JorgePerez so does a magnifying glass. One must be very careful with simplified statements about what the second law does and does not allow. $\endgroup$ – Rococo Jul 7 '17 at 18:13
  • $\begingroup$ More specifically, double photon absorption seems to make a reduction in the number of quantum micro states more probable than not. Under the second law of thermodynamics, the expected change in the number of possible quantum states is positive (an increase in the number of possible states), but a system in which double photon absorption is taking place seems like it'd be expected to slowly decrease the number of possible quantum states $\endgroup$ – J. Antonio Perez Jul 7 '17 at 18:42
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The intuition you're using is that entropy should be linear in particle number, since each particle contributes some amount of entropy on its own, so that reactions that decrease the particle number should decrease the energy.

This isn't quite accurate for a number of reasons, but the most important one here is that the entropy isn't linear in the number of particles, it's sublinear, since you have to divide the partition function by $N!$ because the particles are identical. So the first high-energy photon that comes out of this reaction can contribute a lot more to the entropy than the two low-energy photons did. (An additional factor is that each high-energy photon already contributes more entropy because it has more possible momentum states.)

As more and more high-energy photons are produced, the entropy gain of the reaction decreases. Eventually, when the maximum entropy is reached, the reaction will run backwards at the same rate it runs forwards.

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  • $\begingroup$ Actually, now that I think about this I am not completely sure how upconversion is unitary (this might be another version of the OP's question). Naively, I might think that the two-photon states of $|k_1 \omega_1, k_2\omega_2\rangle$ and $|(k_1+\delta k) \omega_1, (k_2-\delta k) \omega_2\rangle$ would both upconvert to the same outgoing photon $|(k_1+k_2) (\omega_1+\omega_2) \rangle$, but clearly that isn't correct. Maybe the phase-matching conditions make this work out? $\endgroup$ – Rococo Jul 7 '17 at 20:47
  • $\begingroup$ @Rococo Hmm, that's confusing. Are you sure the outgoing photon has to have momentum $k_1 + k_2$? I would have thought it could come out in a range of directions. (Plus, if it were always $k_1 + k_2$ it wouldn't be on-shell.) $\endgroup$ – knzhou Jul 7 '17 at 21:07
  • $\begingroup$ For photon downconversion it is true that energy and momentum must both be conserved by the process, and therefore the downconverted photons can only be emitted at frequencies and angles that satisfy these contraints. I would imagine that the same is true for upconversion, but I do not know for a fact. If the processes are just time-reversed pairs, then the resolution is simply that only certain angles of incident photons allow upconversion in a given medium. $\endgroup$ – Rococo Jul 8 '17 at 1:49
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Two people go into your room, each with a cup of water. You pour the two cups together and walk out with a pint. All the water is conserved. Nothing was created from thin air.

Two little photons go in, on big photon comes out is also not inherently wrong, as long as the energies of the two incoming photons adds up to the energy of the exiting photon. This assumes the material is in the same energy state before and after, which appears to be the case you are asking about.

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    $\begingroup$ I know energy is conserved; I want to understand why the entropy increases in this situation. $\endgroup$ – J. Antonio Perez Jul 7 '17 at 18:54
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I believe this question suffers from trying to explain why a microscopic process does or doesn’t occur, with reference to the second law. Perhaps it is best to consider entropy as an emergent property that cannot be understood when considering only one (or two/three) photons and a single atom. The same issue would ensue if we followed two molecules of a gas in a mixture that have a collision and “concentrate” at one end of a container. You could argue that they also break the second law. Entropy would only decrease here if there was a bulk concentration that occurred spontaneously. I would ask if there was net energy dispersal that accompanied the”concentration” of the two photon absorption and single emission.

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