Atom-Light Interaction: Understanding d.E vs p.A Hamiltonian

In summary, Cohen-Tannoudji 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). For calculations using the electric dipole Hamiltonian, one needs to first carry out the transformation on the initial state. 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
  • #1
zxontt
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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?
 
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  • #2
I think that in lowest order in the field strength you can replace T by 1.
 
  • #3
DrDu said:
I think that in lowest order in the field strength you can replace T by 1.
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 vacuum 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
Sounds reasonable.
 
  • #5
DrDu said:
Sounds reasonable.
Thanks a lot!
 

1. What is the d.E vs p.A Hamiltonian in atom-light interaction?

The d.E vs p.A Hamiltonian is a mathematical representation of the interaction between atoms and light. It takes into account the energy (d.E) and momentum (p.A) of the atoms and light particles, and describes how they exchange energy and momentum during the interaction.

2. How does the d.E vs p.A Hamiltonian help us understand atom-light interaction?

The d.E vs p.A Hamiltonian allows us to make quantitative predictions about the behavior of atoms and light during their interaction. It helps us understand how the energy and momentum of the particles are affected by the interaction and how this affects the overall behavior of the system.

3. How is the d.E vs p.A Hamiltonian derived?

The d.E vs p.A Hamiltonian is derived from the fundamental principles of quantum mechanics. It takes into account the properties of atoms and light and their interactions, and uses mathematical equations to describe their behavior.

4. What factors can affect the d.E vs p.A Hamiltonian?

The d.E vs p.A Hamiltonian can be affected by a variety of factors, including the intensity and frequency of the light, the properties of the atoms (such as their energy levels), and the distance between the atoms and the light source.

5. How is the d.E vs p.A Hamiltonian used in practical applications?

The d.E vs p.A Hamiltonian is used in a wide range of practical applications, such as in the development of new technologies like lasers and quantum computing. It is also used in research to study the behavior of atoms and light in various environments, and to make predictions about their interactions in different scenarios.

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