# A I have some problems with units, can someone help me?

1. Aug 12, 2016

### amelos

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

2. Aug 12, 2016

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.

Last edited: Aug 12, 2016
3. Aug 12, 2016

And a follow-on to post #2: I think Fermi's Golden Rule or a similar relation may be used to get from the Hamiltonian given by your textbook to the form of the intensity or power equation result that is given in Boyd's book. Just maybe the two are completely consistent, but it would be neat to be able to show that if that is the case.

4. Aug 13, 2016

### amelos

Hey Charles,

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

5. Aug 13, 2016

The subject is difficult enough (a detailed quantum mechanical calculation in non-linear optics) that if you are really interested in it, I would recommend trying to find the same calculation in another textbook to compare the results. I am thinking there should perhaps also be an $a_3^{+}$ creation operator in the Hamiltonian. In any case I got a number of the terms in his $\Gamma$ to agree with Boyd's result by using Fermi's golden rule, but so far I don't have complete agreement.

6. Aug 13, 2016

### Jilang

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

7. Aug 13, 2016

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.