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Can someone help me to calculate the unit of gamma from the eq. (2.3). From Eq. (2.2) I assume, it must be [s^-1], but I can not produce it.

Thanks

Amelos

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Can someone help me to calculate the unit of gamma from the eq. (2.3). From Eq. (2.2) I assume, it must be [s^-1], but I can not produce it.

Thanks

Amelos

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A couple of questions and comments is what is "C"? and it is kind of surprising to see an ## \epsilon_o ## to the 3rd power in the denominator without an ## e^6 ## where ## e ## is the elementary charge in the numerator. Otherwise a ## \mu_o ## to the 3rd power should be present since speed of light ## c=1/\sqrt{\mu_o \epsilon_o} ##. And I assume ## n_1 ##, ## n_2 ##, and ## n_3 ## are dimensionless=either indices of refraction or photon numbers...editing... additional comment is it resembles something found in the early chapters of Boyd's Non-Linear Optics book, but the exponents on the terms including the sinc function are different.(Boyd's book has ## sinc^2 (\Delta k l/2) ##). editing... Additional item=when comparing to Boyd's result, I think perhaps this even needs a ## \Gamma^2 ## and the operator terms may need to be ## (a_1^{+}a_1^{-})( a_2^{+}a_2^{-}) ## (number operators), but perhaps somehow a second order term arises out of this Hamiltonian,(from perturbation theory, etc.), that makes the results agree. ... additional edit: It appears the author is doing sum-frequency generation which is covered in pages 62-66 of Boyd's book using coupled-wave equations. The question is, does the author get the Hamiltonian correct and get one that is consistent with Boyd's book? I am on a learning curve as well with this material, but at least this might help you to figure out what the author of your textbook is doing.

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Thank you very much for your reply.

C: light of speed

n1,n2,n3 are unitless refractive indices

deff : V/m

I3: intensity of pump (assume W/m^2)

I am not that familiar with quantum mechanics. I thought maybe he is not doing the calculation in SI units.

Thanks

Halil

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I have been looking at this. I just can't get the amps ( or coulombs) to cancel.

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This one is similar enough to Boyd's result that much of the difference could be the use of SI here vs. Boyd's c.g.s. units. Fermi's Golden Rule seems to be the way to proceed with this. One item that is puzzling in the above though is the parameter that is being solved for is ## I_3 ## and you get an expression for ## I_3 ## that contains ## \Gamma^2 ##. The ## I_3 ## term in the expression for ## \Gamma^2 ## is one item that appears to be in error. Most of the other terms are in agreement with Boyd's result. Boyd also has an ## L^2 ## in his expression for ## I_3 ## (and ## \Gamma^2 ##), and Boyd gets an ## \omega_3^2 ## where I can only account for a first power of ## \omega_3 ## by multiplying the number of transitions (Fermi's Golden Rule) by the energy ## \hbar \omega_3 ##. The ## \epsilon_o^3 ## and differences in the other constant factors could largely be due to the choice of units, but the extra ## I_3 ## mentioned above and the missing ## L^2 ## can not be explained by the choice of units. As previously mentioned, I think the Hamiltonian may need to include some kind of ## a_3^{+} ## operator. In any case, the result is close to being correct, but it doesn't appear to be 100% correct. It would make sense that this ## a_3^{+} ## operator would be weighted by the ## a_1^{-}a_2^{-} ## factors so that the Hamiltonian should contain (but doesn't by the textbook of post #1) a term of the form ## a_1^{-}a_2^{-}a_3^{+} ##. And then of course, there is the reverse transition with a term of the form ## a_1^{+}a_2^{+}a_3^{-} ##. editing... and in a google of the literature of non-linear optics, the tri-linear Hamiltonian was found in a number of the papers on the subject.

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