Atomic Transitions - Dipole Approximation

Question

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.

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.