- #1
lfqm
- 22
- 0
Hey guys, I've recently read about the Tavis-Cummings and Dicke models and I got a little bit confused about them. They are suppoused to model N identical atoms interacting with a one-mode EM field, however the atomic operators are defined in the basis (for the case of two atoms):
[tex]\left\{{|e_{1},e_{2}>, |e_{1},g_{2}>, |g_{1},e_{2}>, |g_{1},g_{2}>}\right\}[/tex]
which obviously makes a distinction between the atoms.
Then it gets even more confusing, as they start working in a spin basis [tex]\left\{{|j,m>}\right\}[/tex] which makes the atoms identical for the case j=N/2... I don't even undestand why they fix j=N/2
Concretely, my question is: What is the basis of the hilbert space the Dicke hamiltonian is acting on (the atomic part)?
The [tex]2^N[/tex] elements basis (distinguishable atoms), the [tex]\displaystyle\sum_{j=0}^{N/2}(2j+1)[/tex] elements basis (considering all spin values) or the [tex]N+1[/tex] elements basis (fixing j=N/2).
And what is the form of the atomic operators in this basis?
Thanks
[tex]\left\{{|e_{1},e_{2}>, |e_{1},g_{2}>, |g_{1},e_{2}>, |g_{1},g_{2}>}\right\}[/tex]
which obviously makes a distinction between the atoms.
Then it gets even more confusing, as they start working in a spin basis [tex]\left\{{|j,m>}\right\}[/tex] which makes the atoms identical for the case j=N/2... I don't even undestand why they fix j=N/2
Concretely, my question is: What is the basis of the hilbert space the Dicke hamiltonian is acting on (the atomic part)?
The [tex]2^N[/tex] elements basis (distinguishable atoms), the [tex]\displaystyle\sum_{j=0}^{N/2}(2j+1)[/tex] elements basis (considering all spin values) or the [tex]N+1[/tex] elements basis (fixing j=N/2).
And what is the form of the atomic operators in this basis?
Thanks