Stokes, Adam
Ω = ω<sub>0</sub> and δ<sub><em>l</em></sub> = 0
<p><strong>Figure 2.</strong> Ω = ω<sub>0</sub> and δ<sub><em>l</em></sub> = 0. In (a) and (b) the lineshapes associated with the Coulomb gauge, Poincaré gauge and the symmetric representation are plotted, each lineshape includes the laser contribution. In (a) Γ = ω<sub>0</sub>/10 and in (b) Γ = ω<sub>0</sub>/100. In (c) and (d) the lineshape in the symmetric representation including the laser contribution is compared to the bare Lorentzian curve (\Gamma /2\pi )/(\delta _k^2+\Gamma ^2/4), and to the Lorentzian including the laser contribution. In (c) Γ = ω<sub>0</sub>/10 and in (d) Γ = ω<sub>0</sub>/100.</p> <p><strong>Abstract</strong></p> <p>We use a general formulation of non-relativistic quantum electrodynamics in which the gauge freedom is carried by the arbitrary transverse component of the Green's function for the divergence operator to calculate the natural lineshape of spontaneous emission, thus discerning the full dependence of the result on the choice of gauge. We also use a representation of the Hamiltonian in which the virtual field associated with the atomic ground state is explicitly absent. We consider two processes by which the atom is excited; the first is resonant absorption of incident radiation with a sharp line. This treatment is then adapted to derive a resonance fluorescence rate associated with the Lamb line in atomic hydrogen. Second we consider the atom's excitation due to irradiation with a laser pulse treated semi-classically. An experiment could be used to reveal which of the calculated lineshape distributions is closest to the measured one. This would provide an answer to a question of fundamental importance; how does one best describe atom–radiation interactions with the canonical formalism?</p>
resonance fluorescence rate;laser contribution;lorentzian;lineshape;representation
2013-06-27
https://iop.figshare.com/articles/__sub_0_sub_and_sub_em_l_em_sub_0/1012259

10.6084/m9.figshare.1012259.v1