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my question concerns the interaction of light and matter in a semi-classical approach. (Quantized Atoms, Classical Fields)

In the Coulomb gauge (div A = 0 , $\phi$=0) we have $E = -\frac{1}{c}\partial_{t}A$ and $B = rot(A)$.

Let's now consider a system of N charged particles (all denoted by an index k).

Let $H_{0}$ be the unperturbed time-independent Hamiltonian and suppose that the eigenvalue problem corresponding to this Hamiltoninan is solved.

Let $H_{1}=-\sum\limits_{k=0}^{N} \frac{e_{k}}{m_{k} c} A(x_{k},t) \cdot p_{k} + \sum\limits_{k=0}^{N} \frac{e_{k}^2}{2 m_{k} c^2} A(x_{k},t)^{2}$ be the time-dependent pertubation.

The atom is now placed at the origin. The wavelength of the considered light is large compared to the diameter of the atom. Hence we have approximately $A_(x_{k},t)\psi(x_{1},...,x_{n}) \approx A_(0,t)\psi(x_{1},...,x_{n})$. (Note that the atom is placed at the origin).

Now the second term in the perturbed Hamiltonian $H_{1}(t)$ (proportional to A^{2}) is of the form $\lambda \cdot f(t)$ for some real number $\lambda$ and some scalar time-dependent function f(t) using this approximation.

Nowit is said that this term only causes a phase factor in the wave function $\phi_{I}(t)$ viewed in the interaaction picture. I don't really understand why that is the case. A time -depnedent Hamiltonian does not trigger in genereal a propagator of the form exp( i*...) .

Can someone help me with this question??

Thanks in advance for the responses.

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I think I just figured out a (rather trivial) explanation. It's because we're basically looking for a solution of a ODE of the form f'(t) = i * c(t) * f(t) where c(t) is a real-valued function of t. –  Stan Jul 17 '11 at 21:17
Is this correct? –  Stan Jul 17 '11 at 21:18
The $A^2$ term is usually left out since it is much smaller than the other one. It is however needed if you want to consider an acceleration of the electron. I have never heard of this "only phasefactor" argumentation and it seems unnatural to me. Greets –  Robert Filter Jul 17 '11 at 21:53
A minor comment to the question (v1): The condition $\phi=0$ (known as the temporal gauge) is not part of the Coulomb gauge fixing condition. –  Qmechanic Jul 18 '11 at 9:42
just a warning: there is a subtle issue of correct gauges choice both for the field and the wave-functions which leads to what is known to "d.E" versus "p.A" form of the dipole approximation. Scully and Zubriary in Quantum Optics has an explanation of where the problem is and Cohen-Tannoudji has the most systemat resolution in among quantum optics textbooks that I know –  Slaviks Aug 17 '11 at 19:03

1 Answer 1

If you use the approximation $A(x_k,t) \sim A(0,t)$ then the $A^2$ part of the Hamiltonian is just $f(t) \widehat{\mathbf 1}$ and so is quite uninteresting. This is because the identity operator $\widehat{\mathbf 1}$ commutes with everything and so does the associated evolution operator $$U(t', t'') = \exp \left(-{i \over \hbar} \int_{t'}^{t''} f(t) {\rm d} t \right) \widehat{\mathbf 1}.$$ This is the complex phase you were looking for.

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