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

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1. Feb 15, 2017

### 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 $T=\exp[ -\frac{i}{\hbar} \mathbf{d} \cdot \mathbf{A}_{\perp}(0) ]=\exp\{ \sum_j (\lambda_j ^* a_j- \lambda_j a_j^{\dagger}) \}$ 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?

Last edited: Feb 15, 2017
2. Feb 16, 2017

### DrDu

I think that in lowest order in the field strength you can replace T by 1.

3. Feb 17, 2017

### zxontt

I did a very crude estimation to convince myself of $d\cdot |A|/\hbar \ll 1$ and so that $T \approx 1$:

I took the dipole moment to be that of a typical atom, approximately $2 e a_0 \approx 1.6 \times 10^{-29} C \cdot m$ (where $e$ is the elementary charge and $a_0$ the bohr radius), and though I am intending this formalism for single optical photon level calculation, I took a quite intense laser intensity for safety, an intensity of $I= 50 W/ \textrm{cm}^2$, and $\omega = 0.4 \times 10^{15} \times 2 \pi \, \textrm{Hz}$. Then using the formulas $I= \frac{c \epsilon_0}{2} |E|^2$, $|A| \approx |E|/ \omega$ where $c$ is the speed of light and $\epsilon_0$ the vaccum permittivity, I end up with the dimensionless number $d\cdot |A|/\hbar \approx 1.2 \times 10^{-6}$, indeed it is quite small and it should be quite plausible to replace $T$ by 1.

Is this estimation valid?

4. Feb 17, 2017

### DrDu

Sounds reasonable.

5. Feb 17, 2017

### zxontt

Thanks a lot!