Quantum Optics IV Semi-classical description of matter-light interaction
fengxiaot Lv4
  2023-08-10 11:49:11 2023-08-10 11:49  

Semi-classical description of matter-light interaction means that we shall present the fundamental features of the interaction of an atom, which will be treated quantum mechanically, with a classical electromagnetic field, that is an electromagnetic field described by real electric and magnetic vectors obeying Maxwell’s equations.

Electrodynamics

Hamiltonian

It is known to all that the Maxwell equations are written

{∇⋅E=ρε0∇⋅B=0∇×E=−∂B∂t∇×B=μ0∂D∂t+μ0J\begin{cases} \nabla \cdot \boldsymbol{E} = \frac{\rho}{\varepsilon_0} \\ \nabla \cdot \boldsymbol{B} = 0 \\ \nabla \times \boldsymbol{E} = -\frac{\partial \boldsymbol{B}}{\partial t} \\ \nabla \times \boldsymbol{B} = \mu_0 \frac{\partial \boldsymbol{D}}{\partial t} + \mu_0 \boldsymbol{J} \end{cases}

which imply the existence of vector potential A(r,t)\boldsymbol{A}(\boldsymbol{r},t) and scalar potential U(r,t)U(\boldsymbol{r},t)

E=−∂A∂t−∇UB=∇×A\boldsymbol{E} = -\frac{\partial \boldsymbol{A}}{\partial t} - \nabla U \qquad \boldsymbol{B} = \nabla \times \boldsymbol{A}

The Lagrangian and Hamiltonian of a particle with mass mm and charge qq in the electromagnetic fields is

L=−mc21−β2+qAμdxμdt≈12mv2+qv⋅A−qU\mathcal{L} = -mc^2 \sqrt{1-\beta^2} +qA_\mu \frac{\mathrm{d}x^\mu}{\mathrm{d}t} \approx \frac{1}{2} m \boldsymbol{v}^2 + q \boldsymbol{v} \cdot \boldsymbol{A} - qU

H=12mv2+qU=12m(P−qA)2+qUH = \frac{1}{2}m\boldsymbol{v}^2 +qU = \frac{1}{2m} (\boldsymbol{P}-q\boldsymbol{A})^2 +qU

where P=−iℏ∇\boldsymbol{P} = - \mathrm{i}\hbar \nabla is the canonical momentum (in position space), which equals mv+qAm\boldsymbol{v} + q\boldsymbol{A} is the canonical momentum.

Long-wavelength approximation

In the atom–light interactions studied in quantum optics, the wavelength λ\lambda of the light is usually very large compared to atomic dimensions. Under these conditions, the amplitude of the external field is practically constant over the spatial extent of the atom and the vector potential A(r,t)\boldsymbol{A}(\boldsymbol{r},t) can be replaced by its value at the nucleus A(r0,t)\boldsymbol{A}(\boldsymbol{r}_0,t). This is the long-wavelength approximation.

H=12m[P−qA(r,t)]2+qU(r,t)≃12m[P−qA(r0,t)]2+qU(r0,t)H = \frac{1}{2m} [\boldsymbol{P}-q\boldsymbol{A}(\boldsymbol{r},t)]^2 +qU(\boldsymbol{r},t) \simeq \frac{1}{2m} [\boldsymbol{P}-q\boldsymbol{A}(\boldsymbol{r}_0 ,t)]^2 +qU(\boldsymbol{r}_0,t)

Transverse and longitudinal fields

Plane wave is the most basic mode of electromagnetic fields. It is always possible to decompose a complicated field into a superposition of plane waves with continuous k\boldsymbol{k} in free space or discrete kn\boldsymbol{k}_n in a cavity, using Fourier Transform.

For a plane wave, the curl and divergence of fields become

{ik⋅E=ρε0ik⋅B=0ik×E=−∂B∂tik×B=μ0∂D∂t+μ0J\begin{cases} \mathrm{i} \boldsymbol{k} \cdot \boldsymbol{E} = \frac{\rho}{\varepsilon_0} \\ \mathrm{i} \boldsymbol{k} \cdot \boldsymbol{B} = 0 \\ \mathrm{i} \boldsymbol{k} \times \boldsymbol{E} = -\frac{\partial \boldsymbol{B}}{\partial t} \\ \mathrm{i} \boldsymbol{k} \times \boldsymbol{B} = \mu_0 \frac{\partial \boldsymbol{D}}{\partial t} + \mu_0 \boldsymbol{J} \end{cases}

An electromagnetic field can be decomposed into longitudinal component E∥E_\parallel and transverse components E⊥E_{\perp}

E=E∥ek+E⊥,1εk,1+E⊥,2εk,2\boldsymbol{E} = E_\parallel \boldsymbol{e}_\boldsymbol{k} + E_{\perp,1} \boldsymbol{\varepsilon}_{\boldsymbol{k},1} + E_{\perp,2} \boldsymbol{\varepsilon}_{\boldsymbol{k},2}

where ek\boldsymbol{e}_\boldsymbol{k} is a unit vector in the direction k\boldsymbol{k} . As a corollary, the Maxwell equations can be simplified into

{ik⋅E∥=ρε0ik⋅B∥=0ik×E⊥=−∂B⊥∂tik×B⊥=μ0∂D⊥∂t+μ0J⊥\begin{cases} \mathrm{i} \boldsymbol{k} \cdot \boldsymbol{E}_\parallel = \frac{\rho}{\varepsilon_0} \\ \mathrm{i} \boldsymbol{k} \cdot \boldsymbol{B}_\parallel = 0 \\ \mathrm{i} \boldsymbol{k} \times \boldsymbol{E}_\perp = -\frac{\partial \boldsymbol{B}_\perp}{\partial t} \\ \mathrm{i} \boldsymbol{k} \times \boldsymbol{B}_\perp = \mu_0 \frac{\partial \boldsymbol{D}_\perp}{\partial t} + \mu_0 \boldsymbol{J}_\perp \end{cases}

Coulomb gauge

Coulomb gauge requires ∇⋅A=0\nabla \cdot \boldsymbol{A} = 0 . Here is a typical example of electromagnetic fields that satisfy Coulomb gauge.

E=E0cos⁡(ωt−k⋅r)B=k×E0ωcos⁡(ωt−k⋅r)E0⋅k=0\begin{gathered} \boldsymbol{E}=\boldsymbol{E}_0 \cos (\omega t-\boldsymbol{k} \cdot \boldsymbol{r}) \\ \boldsymbol{B}=\frac{\boldsymbol{k} \times \boldsymbol{E}_0}{\omega} \cos (\omega t-\boldsymbol{k} \cdot \boldsymbol{r}) \\ \boldsymbol{E}_0 \cdot \boldsymbol{k}=0 \end{gathered}

The corresponding vector potential and scalar potential are

A⊥=−E0ωsin⁡(ωt−k⋅r)U(r,t)=0\boldsymbol{A}_\perp = -\frac{\boldsymbol{E}_0}{\omega} \sin (\omega t - \boldsymbol{k}\cdot \boldsymbol{r}) \qquad U(\boldsymbol{r},t)=0

Since A\boldsymbol{A} is purely transverse, which means A=A⊥\boldsymbol{A} = \boldsymbol{A}_\perp, the Coulomb gauge has been automatically satisfied.

Göppert-Mayer gauge

Perform a gauge transformation

{A′(r,t)=A(r,t)+∇fU′(r,t)=U(r,t)−∂f/∂t\begin{cases} \boldsymbol{A}^\prime(\boldsymbol{r},t) = \boldsymbol{A}(\boldsymbol{r},t) + \nabla f \\ U^\prime(\boldsymbol{r},t) = U(\boldsymbol{r},t) - \partial f/\partial t \end{cases}

in which f(r,t)=−(r−r0)⋅A⊥(r0,t)f(\boldsymbol{r},t) = -(\boldsymbol{r}-\boldsymbol{r}_0)\cdot \boldsymbol{A}_\perp(\boldsymbol{r}_0,t) . Introducing the electric dipole operator of the atom

D=q(r−r0)\boldsymbol{D} = q(\boldsymbol{r}-\boldsymbol{r}_0)

The Hamiltonian under Göppert-Mayer gauge will become

H=12m[P−qA′(r,t)]2−D⋅E(r,t)H = \frac{1}{2m} [\boldsymbol{P}-q\boldsymbol{A}^\prime(\boldsymbol{r},t)]^2 - \boldsymbol{D} \cdot \boldsymbol{E}(\boldsymbol{r},t)

Make the long-wavelength approximation which, as before, enables us to replace the potentials associated with the applied field with their values evaluated at the atomic nucleus. We therefore replace A′(r,t)\boldsymbol{A}^\prime(\boldsymbol{r},t) by A′(r0,t)\boldsymbol{A}^\prime(\boldsymbol{r}_0,t). But notice A′(r0,t)=0\boldsymbol{A}^\prime(\boldsymbol{r}_0,t) = 0 , so the Hamiltonian finally becomes

H=P22m−D⋅E(r,t)H = \frac{\boldsymbol{P}^2}{2m} - \boldsymbol{D} \cdot \boldsymbol{E}(\boldsymbol{r},t)