# How do you go from quantum electrodynamics to Maxwell's equations?

I've read and heard that quantum electrodynamics is more fundamental than maxwells equations. How do you go from quantum electrodynamics to Maxwell's equations?

• The first, very oversimplified, answer would be "taking the limit $\hbar\to 0$". I will transform it in a more detailed answer as soon as I have time. – yuggib Aug 29 '14 at 7:55
• – leongz Aug 29 '14 at 20:30

Disclaimer: This is answer is given from a mathematical physics point of view, and it is a little bit technical. Any comment or additional answer from other points of view is welcome.

The classical limit of quantum theories and quantum field theories is not straightforward. It is now a very active research topic in mathematical physics and analysis.

The idea is simple: by its own construction, quantum mechanics should reduce to classical mechanics in the limit $\hslash\to 0$. I don't think it is necessary to go into details, however for QM this procedure is now well understood and rigorous form a mathematical standpoint.

For QFTs, such as QED, the situation is similar, although more complicated, and it can be mathematically handled only in few situations. Although it has not been proved yet, I think it is possible to prove convergence to classical dynamics for a (simple) model of QED, describing rigid charges interacting with the quantized EM field.

The Hilbert space is $\mathscr{H}=L^2(\mathbb{R}^{3})\otimes \Gamma_s(\mathbb{C}^2\otimes L^2(\mathbb{R}^3))$ ($\Gamma_s$ is the symmetric Fock space). The Hamiltonian describes an extended charge (with charge/mass ratio $1$) coupled with a quantized EM field in the Coulomb gauge: \begin{equation*} \hat{H}=(\hat{p} - c^{-1} \hat{A}(\hat{x}))^2+\sum_{\lambda=1,2}\hslash\int dk\;\omega(k)a^*(k,\lambda)a(k,\lambda)\; , \end{equation*} where $\hat{p}=-i\sqrt{\hslash}\nabla$ and $\hat{x}=i\sqrt{\hslash}x$ are the momentum and position operators of the particle, $a^{\#}(k, \lambda)$ are the annihilation/creation operators of the EM field (in the two polarizations) and $\hat{A}(x)$ is the quantized vector potential \begin{equation*} \hat{A}(x)=\sum_{\lambda=1,2}\int \frac{dk}{(2\pi)^{-3/2}}\;c\sqrt{\hslash/2\lvert k\rvert}\;e_\lambda(k)\chi(k)(a(k,\lambda)e^{ik\cdot x}+a^*(k,\lambda)e^{-ik\cdot x})\; ; \end{equation*} with $e_\lambda(k)$ orthonormal vectors such that $k\cdot e_\lambda(k)=0$ (they implement the Coulomb gauge) and $\chi$ is the Fourier transform of the charge distribution of the particle. The magnetic field operator is $\hat{B}(x)=\nabla\times \hat{A}(x)$ and the (perpendicular) electric field is $$\hat{E}(x)=\sum_{\lambda=1,2}\int \frac{dk}{(2\pi)^{-3/2}}\;\sqrt{\hslash\lvert k\rvert/2}\;e_\lambda(k)\chi(k)i(a(k,\lambda)e^{ik\cdot x}-a^*(k,\lambda)e^{-ik\cdot x})\;$$

$\hat{H}$ is a self adjoint operator on $\mathscr{H}$, if $\chi(k)/\sqrt{\lvert k\rvert}\in L^2(\mathbb{R}^3)$, so there is a well defined quantum dynamics $U(t)=e^{-it\hat{H}/\hslash}$. Consider now the $\hslash$-dependent coherent states \begin{equation*} \lvert C_\hslash(\xi,\pi,\alpha_1,\alpha_2)\rangle=\exp\Bigl(i\hslash^{-1/2}(\pi\cdot x+i\xi \cdot\nabla)\Bigr)\otimes\exp\Bigl(\hslash^{-1/2}\sum_{\lambda=1,2}(a^*_\lambda(\alpha_\lambda)-a_\lambda(\bar{\alpha}_\lambda))\Bigr)\Omega\; , \end{equation*} where $\Omega=\Omega_1\otimes\Omega_2$ with $\Omega_1\in C_0^\infty(\mathbb{R}^3)$ (or in general regular enough, and with norm one) and $\Omega_2$ the Fock space vacuum.

What it should be at least possible to prove is that ($\alpha_\lambda$ is the classical correspondent of $\sqrt{\hslash}a_\lambda$, and it appears inside $E(t,x)$ and $B(t,x)$ below): \begin{gather*} \lim_{\hslash\to 0}\langle C_\hslash(\xi,\pi,\alpha_1,\alpha_2),U^*(t)\hat{p}U(t)C_\hslash(\xi,\pi,\alpha_1,\alpha_2)\rangle=\pi(t)\\ \lim_{\hslash\to 0}\langle C_\hslash(\xi,\pi,\alpha_1,\alpha_2),U^*(t)\hat{x}U(t)C_\hslash(\xi,\pi,\alpha_1,\alpha_2)\rangle=\xi(t)\\ \lim_{\hslash\to 0}\langle C_\hslash(\xi,\pi,\alpha_1,\alpha_2),U^*(t)\hat{E}(x)U(t)C_\hslash(\xi,\pi,\alpha_1,\alpha_2)\rangle=E(t,x)\\ \lim_{\hslash\to 0}\langle C_\hslash(\xi,\pi,\alpha_1,\alpha_2),U^*(t)\hat{B}(x)U(t)C_\hslash(\xi,\pi,\alpha_1,\alpha_2)\rangle=B(t,x)\; ; \end{gather*} where $(\pi(t),\xi(t),E(t,x),B(t,x))$ is the solution of the classical equation of motion of a rigid charge coupled to the electromagnetic field: \begin{equation*} % \left\{ \begin{aligned} &\left\{\begin{aligned} \partial_t &B + \nabla\times E=0\\ \partial_t &E - \nabla\times B=-j \end{aligned}\right. \mspace{20mu} \left\{\begin{aligned} \nabla\cdot &E=\rho\\ \nabla\cdot &B=0 \end{aligned}\right.\\ &\left\{\begin{aligned} \dot{\xi}&= 2\pi\\ \dot{\pi}&= \frac{1}{2}[(\check{\chi}*E)(\xi)+2\pi\times(\check{\chi}*B)(\xi)] \end{aligned}\right. \end{aligned} % \right. \end{equation*} with $j=2\pi\check{\chi}(\xi-x)$, and $\rho=\check{\chi}(\xi-x)$ (charge density and current).

To sum up: the time evolved quantum observables averaged over $\hslash$-dependent coherent states converge in the limit $\hslash\to 0$ to the corresponding classical quantities, evolved by the classical dynamics.

Hoping this is not too technical, this picture gives a precise idea of the correspondence between the classical and quantum dynamics for an EM field coupled to a charge with extended distribution (point charges cannot be treated mathematically on a completely rigorous level both classically and quantum mechanically).

• Have a look at this blog entry motls.blogspot.com/2011/11/… on how classical fields emerge from the photon – anna v Sep 3 '14 at 16:31
• @annav Have a look at the (mathematical physics) literature (at the very least the classics, i.e. this and this) to see how classical variables emerge from quantum observables with a proper level of rigor. – yuggib Sep 3 '14 at 18:49
• @annav Thanks to you. The system considered there is a many body theory for bosons, however the technique can be extended to situations where the number of particles is not conserved. The modern literature on the subject usually deals with mean field limit (because it is, mathematicaly at least, almost equivalent) and it is very extended. If you are interested I can provide some additional links. – yuggib Sep 3 '14 at 19:14
• Nice answer! Thanks. Is the coherent state $\vert C_\hbar\rangle$ above the same as the coherent state conventionally used in quantum optics or QED, namely the infinite-mode version of harmonic oscillator's coherent state? Are the $\hat{x},\hat{p}$ in the first two limits the quadrature operators in quantum optics? – xiaohuamao Oct 19 '14 at 5:45
• @huotuichang The coherent state above is the tensor product of two coherent states: an harmonic oscillator coherent state (for the particle), and an infinite-mode one for the EM field (to use your terminology ;-) ). Concerning the second question: well, in principle you can see them like that (formally), but here the physical meaning is really as the position and momentum of a particle in the EM field. I don't know if there is a meaningful model in quantum optics where one mode of radiation (described by $\hat{x}$ and $\hat{p}$) is coupled with another radiation in the way described above... – yuggib Oct 19 '14 at 9:35