Timeline for Why doesn't photon up-conversion violate the second law of thermodynamics?
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4 events
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| Jul 8, 2017 at 1:49 | comment | added | Rococo | 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. | |
| Jul 7, 2017 at 21:07 | comment | added | knzhou | @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.) | |
| Jul 7, 2017 at 20:47 | comment | added | Rococo | 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? | |
| Jul 7, 2017 at 19:33 | history | answered | knzhou | CC BY-SA 3.0 |