In standard input-output theory, we model the interaction between light and a two level system by the following hamiltonian which assumes rotating wave approximation:
The interaction between a classical electric field, polarized along a single axis: $x$, and a dipole is the following:
$$H=-\frac{\hbar \omega_0}{2} \sigma_z + \epsilon(e^{j \phi} \sigma + e^{-j\phi} \sigma^{\dagger})E_x(t)$$
When ones apply the rotating wave approximation we have:
$$H=-\frac{\hbar \omega_0}{2} \sigma_z + \sum_{n=-\infty}^{+\infty} \epsilon (E^*(k_n) e^{j(\phi+\omega_n t)} \sigma + E(k_n)e^{-j(\phi+\omega_n t)} \sigma^{\dagger})$$
And, to make easy "comparisons" with the quantized version of the field, we can rewrite:
$$E(k_n)=\sqrt{\frac{\hbar \omega_n}{2 \epsilon_0 V}} \alpha(k_n)$$
$$H=-\frac{\hbar \omega_0}{2} \sigma_z +\sum_n \epsilon \sqrt{\frac{\hbar \omega_n}{2 \epsilon_0 V}} (e^{j \phi} \alpha^{*}(k_n) \sigma + e^{-j \phi} \alpha(k_n) \sigma^{\dagger})$$
In the quantized version (E.M field quantized), we just have to replace $\alpha \rightarrow a^{\dagger}$ which leads to:
$$H=-\frac{\hbar \omega_0}{2} \sigma_z + \sum_{n=-\infty}^{+\infty} \hbar \omega_n \widehat{a}^{\dagger}(k_n)\widehat{a}(k_n)+\sum_n \hbar g(\omega_n) (e^{j \phi} \widehat{a}^{\dagger}(k_n) \sigma + e^{-j \phi} \widehat{a}(k_n)\sigma^{\dagger})$$
$$\hbar g(\omega_n)=\epsilon \sqrt{\frac{\hbar \omega_n}{2 \epsilon_0 V}}$$
At this point, if we look at the classical Hamiltonian, because the coupling is flat in frequency between the field and the dipole, there are no memory effect. It can be seen on the first equation: if I replace $E_x(t)$ by its Fourier serie each fourier coefficient of the field will be coupled by a constant term to the TLS operators (the interaction is independant of $\omega$).
More directly, if I cut the electric field at time $t_0$, the interaction is instantaneously cut. There is no latence (it is another way of saying the coupling is flat in frequency).
In the quantum regime it is thus the case as well. The fact the constant $g(\omega_n)$ depends on $\omega_n$ is simply a matter of rewriting that can be seen on the classical level in the derivations done here. (It is basically when the change of variable $E \rightarrow \alpha$ is done).
Now, in input output theory, we usually approximate: $g(\omega_n) \approx g$, i.e the coupling constant becomes frequency-independant. This is called Markov approximation.
My questions are the following.
- Why is the approximation $g(\omega_n) \approx g$ valid ? The context of what I want to understand is basically how to drive a two level system to induce single qubit gates under the R.W.A. Here, Electric dipole interaction: flat coupling semi-classically, non flat with E.M field quantized: why ? What does that mean? it is said the it is implied by the R.W.A but I am not sure to exactly see why.
- Why is it called Markov approximation ? Indeed on the classical level we can directly see that there is no memory effect seen by the system, the coupling being flat in frequency. For me, stating $g(\omega_n) \approx g$, if we go back to the classical Hamiltonian it will induce a frequency dependent coupling between the field and the dipole and will thus induce memory effect. It confuses me.
[edit]: I think I understand somehow the first point, but I would like to check it.
In the context of performing Rabi oscillations with this field, we will apply a pulse of duration $\Delta T$, where $\Delta T \sim \frac{1}{\Omega}$ ($\Omega$ being the classical Rabi frequency induced). For example to perform a $\pi$ pulse, it requires a pulse length = time of interaction being $\Delta T = \frac{\pi}{\Omega}$.
The width of the modes composing the pulse is $\Delta \omega \sim \frac{1}{\Delta T} = \Omega$.
Finally, as the RWA is valid if $\Omega \ll \omega_0$, we have: $RWA \Rightarrow \frac{\Delta \omega}{\omega_0} \ll 1$: the range of populated frequencies of the drive is a narrow band around $\omega_0$. Thus, we can say $g(\omega_n) \approx g(\omega_0)$.
Does that make sense for you ?
