# Light and atom interaction hamiltonian

#### naima

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)?

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#### mfb

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.

#### naima

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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#### naima

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

#### Andy Resnick

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).

#### naima

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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#### naima

Gold Member
Thank you for the link. It says that two of the four terms can be omitted according to the Rotating Frame Approximation (RWA)
So in a full correct calculus we would have a small term corresponding to an electron emitting a photon and getting a higher energy!

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