# Does Feynman's derivation of Maxwell's equations have a physical interpretation?

There are so many times that something leaves you stumped. I was recently reading the paper "Feynman's derivation of Maxwell's equations and extra dimensions" and the derivation of the Maxwell's equations from just Newton's second law and the quantum mechanical commutation relations really intrigued me. They only derived the Bianchi set, yet with slight tweakings with relativity, the other two can be derived.

Awesome as it is, does this even have a physical interpretation? How is it possible to mix together classical and quantum equations for a single particle, which aren't even compatible, and produce a description of the electromagnetic field?

• beautiful paper – lurscher Mar 12 '18 at 11:56
• it seems to be missing the connection between the algebraical fields and the sources. The actual Maxwell theory asserts that these fields arise from sources, it is not clear in the paper what role is there for sources – lurscher Mar 12 '18 at 11:58
• Well, you can obviously interpret the fields as quantum fields giving rise to the field sources. We obviously need an extension, but is the form that he derives just a bitter coincidence? – Yuzuriha Inori Mar 12 '18 at 12:01
• The physical meaning of the Feynman's proof is that the coupling to external fields is dictated by the Poisson geometry of the configuration space. The solutions of the Feynman problem for configuration spaces of interest in physics are the minimal coupling interactions: Lorentz force, its non-Abelian generalizations, gravity and Kaluza-Klein. The Feynman's proof can be performed completely within classical mechanics where the commutations are given by means of Poisson brackets. It is a profound result and completely nonintuitive in spite of its apparently simple derivation. – David Bar Moshe Mar 13 '18 at 13:04
• cont. The Feynman's proof is a special case of invese problems in the calculus of variation. Please see carinena, Ibort, Marmo and Stern cds.cern.ch/record/283762/files/SCAN-9506196.pdf – David Bar Moshe Mar 13 '18 at 13:07

Feynman's derivation is wonderful, and I want to sketch why we would expect it to work, and what implicit assumptions it's really making. The real issue is that by switching back and forth between quantum and classical notation, Feynman sneaks in physical assumptions that are sufficiently restrictive to determine Maxwell's equations uniquely.

To show this, I'll give a similar proof in fully classical, relativistic notation. By locality, we expect the force on a particle at position $x^\mu$ with momentum $p^\mu$ depends solely on $p^\mu$ and $F(x^\mu$). (This is Eq. 1 in the paper.) Then the most general possible expression for the relativistic four-force is $$\frac{d p^\mu}{d\tau}= F_1^\mu(x^\mu) + F_2^{\mu\nu}(x^\mu)\, p_\nu + F_3^{\mu\nu\rho}(x^\mu)\, p_\nu p_\rho + \ldots$$ where we have an infinite series of $F_i$ tensors representing the field $F$. (Of course, we already implicitly used rotational invariance to get this.) I'll suppress the $x^\mu$ argument to save space.

It's clear that we need more physical assumptions at this point since the $F_i$ are much too general. The next step is to assume that the Lagrangian $L(x^\mu, \dot{x}^\mu, t)$ is quadratic in velocity. Differentiating, this implies that the force must be at most linear in momentum, so we have $$\frac{d p^\mu}{d\tau}= F_1^\mu + F_2^{\mu\nu}\, p_\nu.$$ This is a rather strong assumption, so how did Feynman slip it in? It's in equation 2, $$[x_i, v_j] = i \frac{\hbar}{m} \delta_{ij}.$$ Now, to go from classical Hamiltonian mechanics to quantum mechanics, we perform Dirac's prescription of replacing Poisson brackets with commutators, which yields the canonical commutation relations $[x_i, p_j] = i \hbar \delta_{ij}$ where $x_i$ and $p_i$ are classically canonically conjugate. Thus, Feynman's Eq. 2 implicitly uses the innocuous-looking equation $$\mathbf{p} = m \mathbf{v}.$$ However, since the momentum is defined as $$p \equiv \frac{\partial L}{\partial \dot{x}}$$ this is really a statement that the Lagrangian is quadratic in velocity, so the force is at most linear in velocity. Thus we get a strong mathematical constraint by using a familiar, intuitive physical result.

The next physical assumption is that the force does not change the mass of the particle. Feynman does this implicitly when moving from Eq. 2 to Eq. 4 by not including a $dm/dt$ term. On the other hand, since $p^\mu p_\mu = m^2$, in our notation $dm/dt = 0$ is equivalent to the nontrivial constraint $$0 = p_\mu \frac{dp^\mu}{d\tau} = F_1^\mu p_\mu + F_2^{\mu\nu} p_\mu p_\nu.$$ For this to always hold, we need $F_1 = 0$ and $F_2$ (hereafter called $F$) to be an antisymmetric tensor and hence a rank two differential form. We've now recovered the Lorentz force law $$\frac{d p^\mu}{d\tau} = F^{\mu\nu} p_\nu.$$

Our next task is to restore Maxwell's equations. That seems impossible because we don't know anything about the field's dynamics, but again the simplicity of the Hamiltonian helps. Since it is at most quadratic in momentum, the most general form is $$H = \frac{p^2}{2m} + \mathbf{A}_1 \cdot \mathbf{p} + A_2.$$ Collecting $\mathbf{A}_1$ and $A_2$ into a four-vector $A^\mu$, Hamilton's equations are $$\frac{dp^\mu}{d\tau} = (dA)^{\mu\nu} p_\nu$$ where $d$ is the exterior derivative. That is, the simplicity of the Hamiltonian forces the field $F$ to be described in terms of a potential, $F = dA$. Since $d^2 = 0$ we conclude $$dF = 0$$ which contains two of Maxwell's equations, specifically Gauss's law for magnetism and Faraday's law. So far we haven't actually used relativity, just worked in relativistic notation, and indeed this is where our derivation and Feynman's run out of steam. To get the other two equations, we need relativity proper.

The basic conclusion is that Feynman's derivation is great, but not completely mysterious. In particular, it isn't really mixing classical and quantum mechanics at all -- the quantum equations that Feynman uses are equivalent to classical ones derived from Hamilton's equations, because he is using the Dirac quantization procedure, so the only real purpose of the quantum mechanics is to slip in $\mathbf{p} = m \mathbf{v}$, and by extension, the fact that the Hamiltonian is very simple, i.e. quadratic in $\mathbf{p}$. The other assumptions are locality and mass conservation.

It's not surprising that electromagnetism pops out almost 'for free', because the space of possible theories really is quite constrained. In the more general framework of quantum field theory, we can get Maxwell's equations by assuming locality, parity symmetry, Lorentz invariance, and that there exists a long-range force mediated by a spin 1 particle, as explained elsewhere on this site. This has consequences for classical physics, because the only classical physics we can observe are those quantum fields which have a sensible classical limit.

• @YuzurihaInori What I try to emphasize in the answer is that it's not mathematical trickery. There is a lot of trickery but it's all physical; Feynman is sneaking in experimentally well-tested assumptions about the physical world in between the lines. – knzhou Mar 12 '18 at 12:40
• If we use relativity from the very beginning, can we expect Einstein to pop out instead of Maxwell (or rather, along with Maxwell) ? – Yuzuriha Inori Mar 12 '18 at 12:44
• @YuzurihaInori Also, yeah, it's quite interesting that we can get such strong constraints on fields just by how they act on particles. On the other hand, that's not actually that surprising because that is the only way we've ever gotten information about fields. Nobody has ever measured an electric field -- only how it acts on particles. – knzhou Mar 12 '18 at 12:44
• @YuzurihaInori Yes! See Feynman's "Lectures on Gravitation" where he generalizes what I said in my last paragraph to get Einstein's equations for free. – knzhou Mar 12 '18 at 12:44
• @YuzurihaInori Look here! – knzhou Mar 12 '18 at 12:46