Construction of Hamiltonian for interacting optical fields

[Api]

Hi. I am going to start my MSc in a couple of months majoring in nonlinear quantum optics. I have a good basic in quantum mechanics, but have never looked at quantum optics before. My topic will be to investigate quantum properties of nonlinear optical coupler but i have problem with the construction of the Hamiltonian for interacting optical fields. For starting, i start with two mode nonlinear optical coupler

H= hω(a1b1+a2b2)+hωg(a1^2 b1^2+a2^2 b2^2) + hk(a1b2+b1a2)

h=planck constant (h bar)
ω=angular frequency
an= creation operator associated with the optical mode
bn=annihilation operator associated with the optical mode
g=nonlinear susceptibility
k= coupling constant

I will really appreciate it if anyone can explain the physical meaning of this Hamiltonian and how do we construct such equation to describe the system.

 

[strangerep]

Hi. I am going to start my MSc in a couple of months majoring in nonlinear quantum optics. I have a good basic in quantum mechanics, but have never looked at quantum optics before. My topic will be to investigate quantum properties of nonlinear optical coupler [...]

If you never looked at quantum optics, then definitely get a copy of Mandel & Wolf, "Optical Coherence & Quantum Optics" asap. Some have described it as a "bible" of this subject. (I'm no expert on it, but I can say that it's a surprisingly interesting subject after one gets into it a bit.)

In a 30-sec search, I didn't find the exact Hamiltonian that you wrote, but something similar appears in eq(22.3-1) of M&W which deals with harmonic generation in nonlinear media.
$$H= \sum_{i=1}^2 \hbar \omega(n_i + 1/2)<br /> ~+~ \hbar g \left(a^\dagger_2 a_1^2 + {a^\dagger_1}^2 a_2 \right)$$
The preceding section 22.2 (Energy of the field in a dielectric) gives some of the motivation for the form of this Hamiltonian, but note that this is in ch22 and there's lots of stuff earlier in the book that you'd need to make sure you understand before diving into the above.

 

[Api]

ok.thanks for the reply