Coupling with Light

We now bring light into the picture. In most cases, we can treat the light classically as a electric field \(E(t)\). The total Hamiltonian is then $$\hat{H}(t) = \hat{H}_ {0} + \hat{V}_ {\mathrm{ext}}(t)$$ with the time-independent system Hamiltonian \(\hat{H}_ {0}\) and the time-dependent external potential operator \(\hat{V}_ {\mathrm{ext}}(t)\).

Applying dipole approximation, the external potential of an electric field can be expressed as $$V_{\mathrm{ext}}(t) = \mu_{fi} E(t)$$ with the dipole coupling strength between initial and final states \(\mu_{fi}\). For a two-state system, this operator can thus be represented by the matrix $$ \mathbf{V_{\mathrm{ext}}(t)} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \mu E(t) $$

The split operator formalism with time-dependent potential looks like:

$$ |\psi(t+\Delta t)\rangle = \exp\left(-\frac{\mathrm{i}}{\hbar}\frac{\hat{V}(t + \Delta t)}{2}\Delta t\right) \exp\left(-\frac{\mathrm{i}}{\hbar}\hat{T}\Delta t\right) \exp\left(-\frac{\mathrm{i}}{\hbar}\frac{\hat{V} (t)}{2}\Delta t\right) |\psi(t)\rangle $$

Note that the first factor contains potential energy at \(t + \Delta t\) and the third factor potential energy at \(t\). This asymmetry arises from the derivation and decreases propagation error in \(\Delta t\). One can also do it symmetrically, but at a cost of higher propagation error.