Light and atom interaction hamiltonian

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  • #1
naima
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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)?
 

Answers and Replies

  • #2
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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.
 
  • #3
naima
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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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  • #4
naima
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Has an interference hamiltonian to commute with the free hamiltonian?
 
  • #5
Andy Resnick
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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).
 
  • #6
naima
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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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  • #8
naima
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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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