Basic reminders of dipole-light interaction
In the dipolar approximation, the electric-field dipole interaction is written as:
$$H=-\mathbf{d}.\mathbf{E}(t) = -q\mathbf{r}.\mathbf{E}(t)$$ $\mathbf{r}$ being the distance between the two charge that make the dipole.
Semi-classical modelling of quantum dipole interacting with light
Now, I start to do quantum physics and I assume that the dipole is a two level system. I thus have:
$$ \mathbf{r} = \widehat{\mathbf{r}}=\langle 0|\widehat{\mathbf{r}}|1\rangle |0\rangle \langle 1|+\langle 1|\widehat{\mathbf{r}}|0\rangle |1\rangle \langle 0|$$
Indeed, as I work with a TLS, I can decompose its operator on a qubit basis, and the diagonal elements give $0$ by symmetry because the position operator is odd, and would be integrated by an even function (square modulus of $|0\rangle$ or $|1\rangle$).
A particular case that I will focus on now is the following:
$$q \widehat{r}=\epsilon \sigma_x$$
$$H=-q\widehat{\mathbf{r}}.\mathbf{E}(t)=\epsilon \widehat{\sigma}_x E_x(t)$$
This is the semi-classical modelling of light-matter interaction for the matter being a dipole. As we see, the coupling is flat in frequency (I can expand $E_x(t)=\int d \omega E_x(\omega)e^{j \omega t}$ to see it).
Full quantum modelling
To go further, we can quantize the electric field and write down the hamiltonian. Using periodic boundary conditions, focusing on $x=0$ where the dipole is, the electric field has the form (I assume a single polarization to simplify)
$$\widehat{E_x}=\sum_k \sqrt{\frac{\hbar \omega_k}{2 \epsilon_0 V}} \left( \widehat{a}(\omega_k) e^{-j\omega_t} + \widehat{a}^{\dagger}(\omega_k) e^{j\omega_t} \right)$$
Now, the Hamiltonian takes the form:
$$H=-q\widehat{\mathbf{r}}.\mathbf{E}(t)=\widehat{\sigma}_x \sum_k \epsilon \sqrt{\frac{\hbar \omega_k}{2 \epsilon_0 V}} \left( \widehat{a}(\omega_k) e^{-j\omega_t} + \widehat{a}^{\dagger}(\omega_k) e^{j\omega_t} \right)$$
My question
And here, we see that the coupling is no longer flat in frequency, in the opposite of the classical case.
This disturbs me a lot. A flat coupling has some physical interpretations. Basically in the classical modelling if I send a pulse on the TLS, once the pulse is passed, the qubit no longer evolves as a consequence of flat frequency coupling $\Leftrightarrow$ instantaneous time response (no memory effect).
In the quantum regime, the description should be more accurate physically but we should still have this no memory effect.
I don't understand.
