# Light and atom interaction hamiltonian

Gold Member
I found this in a Phd thesis

consider a two level atom interacting with the electromagnetic field.
The atom is described by
##H_{at} = \hbar ω_0 J_z##
a monomode electric field is described by
##H_{em} = \hbar \omega (a^\dagger a + 1/2)##
We have ##E = E_0(a^\dagger + a)## and the dipolar moment is ##D = d_o(J_+ - J_-)##
We have then ##H_{int} = -DE##

So this interference hamitonian contains four terms
a) ##a J_+ and a^\dagger J_-##
but also
b) ##a J_- and a^\dagger J_+##
the author writes later a formula without the b) terms.
How can we make them disappear (mathematically)?

Mentor
Physically, they are zero because they correspond to transitions to non-existing states (as we just have two states). There is also some more mathematical argument that I forgot, but they are really zero.

Gold Member
I have no doubt about it. But as he begins with a hamiltonian with the b) terms i think that there is a mathematicall trick to erase them.

Take (g,n). Does his hamiltonian permit to get (e,n+1) which exists?

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Gold Member
Has an interference hamiltonian to commute with the free hamiltonian?

I have no doubt about it. But as he begins with a hamiltonian with the b) terms i think that there is a mathematicall trick to erase them.

Take (g,n). Does his hamiltonian permit to get (e,n+1) which exists?

If I understand your notation, the terms in (b) either lower the state and absorb a photon (which cannot happen) or raise the state and emit a photon (which also cannot happen). As for the above, the interaction Hamiltonian connects (g,n) with (e, n-1).

Gold Member
My question is about the DE hamiltonian (D is the dipole,E is the electric field)
How is it quantized?
The answer may be in the fact that DE is a scalar product and that E is transverse so it gives 2 components?

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