I am reading Cohen-Tannoudji's Atom photon interactions (2004 version), in the Appendix he explains that for atom-light interaction, the electric dipole Hamiltonian (d.E form) is got from the original, "physical" (in line with his language) p.A form Hamiltonian by a time-independent unitary transformation (eq. (72) of the Appendix at page 635).(adsbygoogle = window.adsbygoogle || []).push({});

Basically, ##H_{d.E}=TH_{p.A}T^{\dagger}##, where ##H_{d.E}## is the Hamiltonian in electric dipole form, while ##H_{p.A}## is the p.A form Hamiltonian, and ##T=\exp[ -\frac{i}{\hbar} \mathbf{d} \cdot \mathbf{A}_{\perp}(0) ]=\exp\{ \sum_j (\lambda_j ^* a_j- \lambda_j a_j^{\dagger}) \}## where ##\lambda_j = \frac{i}{\sqrt{2 \epsilon_0 \hbar \omega_j L^3}}\mathbf{\epsilon}_j \cdot \mathbf{d}##, is the unitary transformation.

Then isn't it that if one wants to make calculations using the electric dipole Hamiltonian, one needs to first carry out the transformation on the initial state? That is, the initial state should be ##T \left | \phi_{ini} \right \rangle ## but not simply ##\left | \phi_{ini} \right \rangle##? where ##\left | \phi_{ini} \right \rangle## is the ("physical") initial state under the p.A representation.

However when he calculates the photondetection signal in Complement ##\textrm{A}_{\textrm{II}}## (more specifically see eq. (22) and texts around it in this Complement), there seems no indication that this transformation is carried out. Also, according to my experience, usually when using the d.E form one does not carry out any such transformation on the state. Do one need to or not need to carry out such transformation on state when using the d.E form Hamiltonian?

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# A To use the the d.E form of the Hamiltonian, does the state need to be transformed?

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