Fine-structure constant, coupling strength and their experimental manifestation in the field of light-matter interaction

Question

I am currently studying the quantum Rabi model, whose Hamiltonian reads $$ H = \frac{\Omega}{2} \sigma_z + \hbar \omega a^\dagger a + g (a^\dagger+a) \sigma_x. $$ In mainstream papers, this is the Hamiltonian we work with. However, I am asked the question, what the coupling strength $g$ actually is? Clearly, it characterizes the coupling strength between a two-level atom and a bosonic field, so I believe it must be dependent on the fine-structure constant $\alpha$. But I don't know the exact functional expression in terms of $\alpha$.

In addition, we all talk about the experimental implementation of this model, how will $g$ and $\alpha$ manifest in experiments?

Since I study this model from a theoretical perspective, any recommendations of papers or books would be appreciated. Thanks!

Answer 1

This model is widely used to study a single mode of an electric field (e.g. in a cavity) inducing oscillations above the ground state of an atom via the Stark effect. This has the interaction Hamiltonian \begin{align} H_{\text{int}} = -\vec{p} \cdot \vec{E} = -E p_z (a + a^\dagger) \end{align} where we have taken the field to point in the $z$ direction without loss of generality. If we now assume that the atom is Hydrogen, $p_z = -ez = -er \cos \theta$ because the proton and electron are equal and opposite point charges separated by $z$. The ground state of Hydrogen is obvious, but for the excited state, we will need to choose one with $\ell = 1$, rather than $\ell = 0$ for oscillations with this Hamiltonian to be possible. Looking up the wavefunctions, the off-diagonal element of this matrix that we care about is \begin{align} \left < \psi_{1,0,0} | H_{\text{int}} | \psi_{2,1,0} \right > &= eE(a + a^\dagger) \int_0^{2\pi} \int_0^\pi \int_0^\infty \psi_{1,0,0}(r,\theta,\phi) r\cos\theta \psi_{2,1,0}(r,\theta,\phi) r^2 \sin \theta dr d\theta d\phi \\ &= eE(a + a^\dagger) \frac{2\pi}{\sqrt{6} a_0^4} \int_0^\pi \int_0^\infty r^4 e^{-\frac{3r}{2a_0}} \frac{1}{4} \frac{\sqrt{3}}{\pi} \cos^2 \theta \sin \theta dr d\theta \\ &= eE(a + a^\dagger) \frac{128 \sqrt{2}}{243} a_0. \end{align} So in this idealized situation, you get $g$ by multiplying the Bohr radius, applied field and elementary charge. These can be re-expressed in terms of the fine structure constant if desired.