Physics Help Forum To use the d.E Hamiltonian, need state be transformed?

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 Feb 15th 2017, 07:34 PM #1 Junior Member   Join Date: Feb 2017 Posts: 1 To use the electric dipole (d.E) Hamiltonian, need state be transformed? 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). 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 $T$ 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? Am I missing something? Last edited by zxontt; Feb 16th 2017 at 04:00 AM.
Feb 16th 2017, 07:06 AM   #2

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 Originally Posted by zxontt 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). 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 $T$ 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? Am I missing something?
Am I right in assuming that you are using the Hamiltonian representation? In this representation the state kets are independent of the transformation and the operators contain the dynamical information for the system.

-Dan
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