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For the abelian QED theory, $$[D_\mu, D_\nu]=ieQF_{\mu\nu}$$ where $D_\mu=\partial_\mu+ieQA_\mu$ is the gauge covariant derivative in QED and $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$ is the electromagnetic field strength tensor.

For the nonabelian $SU(N)$ gauge theory, $$[D_\mu, D_\nu]=igT^a G^a_{\mu\nu}$$ where $D_\mu=\partial_\mu+igT^a G^a_\mu, G_{\mu\nu}^a=\partial_\mu G^a_\nu-\partial_\nu G^a_\mu-gf_{abc}G^b_\mu G_\nu^c$ are the corresponding field strength tensors.

Though I can derive these results quite easily, I wonder what is the physical content of these two equations. Here a similar question was asked but the answer involves advanced mathematical concepts and notations such as wedge product, exterior derivatives, holonomy etc. Can someone dumb down the notation and state the significance of these relations? Thanks in advance!

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It may be worth looking at this formula as the infinitesimal limit of an exponentiated relation. We have

$$ \exp(- a^\mu D_\mu) \exp(- b^\nu D_\nu) \exp( a^\mu D_\mu) \exp( b^\nu D_\nu) \approx 1 + a^\mu b^\nu [D_\mu, D_\nu] + \dots $$

Just like $\exp( a^\mu \partial_\mu)$ is an operator of translation in a flat space, $\exp( a^\mu D_\mu)$ is an operator of parallel transport, basically translation in a space with nontrivial connection/geometry. The product of these four exponents is a transport along a parallelogram. Therefore we can conclude that the field strength tensors $F_{\mu\nu}/G^a_{\mu\nu}$ are measure of how much parallel transport around a small closed curve changes the object transported.

This also show that field strength tensors are curvatures of a certain (non-metric) connection. They are the gauge field theory equivalent of Riemann tensor from general relativity.

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To illustrate the physics behind the relation, let's take a physical example of a charged particle moving in the $x$-$y$ plane, in the presence of a magnetic field $B$ pointing in the $z$ direction. We want to interpret the relation $$ [D_x,D_y] = i q F_{xy} = i q B \,, $$ as an operator acting on the wavefunction $\psi$ of a particle of charge $q$. Now, $\psi$ has an unphysical phase, because it changes under gauge transformations: $$ A_\mu \longrightarrow A_\mu +\partial_\mu\lambda,\qquad \psi \longrightarrow e^{-iq\lambda}\psi $$ However, relative phases are important for interference effects. The covariant derivative gives a gauge invariant way to say that the phase of the wavefunction is "constant", $D_\mu \psi=0$.

Now, let's say the particle has two possible paths to get from the origin $x=y=0$ to a point $x=y=\epsilon$: either it moves first in the $x$ direction to $x=\epsilon$, $y=0$ and then in the $y$ direction, or the other way round. The commutator $\epsilon^2[D_x,D_y]\psi$ compares how the magnetic field affects the wavefunction differently for the two paths: we get $i q B \epsilon^2 \psi$, which tells us that there's a small relative phase $e^{i q B \epsilon^2}$.

The relative phase can lead to interference effects from a superposition of the two paths, which is the Aharonov-Bohm effect. For a finite path, the relative phase will be $q$ times the total magnetic flux $\int\! B$ between the paths. More covariantly, this can also be expressed as a surface integral of $F_{\mu\nu}$, or using Stokes' theorem, a line integral $\oint A_\mu dx^\mu$ following the two paths in opposite directions. While $A_\mu$ is not gauge invariant, this integral along a closed path (a `Wilson loop') is. The commutator relation is therefore a version of Stokes' theorem, relating an integral round a loop to a flux through it, for infinitesimal loops.

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The physical meaning is that a photon does not interact with other photons (abelian group) while gluons interact with each other and that is why in QCD you have to consider a non abelian group (SU(3)).

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  • $\begingroup$ How does that follow from the commutator relations? $\endgroup$ – mithusengupta123 Feb 2 at 14:07
  • $\begingroup$ $G^{j}_{μν}$ has an extra term. The mathematical reason for that extra term is that $SU(N)$ is a non abelian group and the physical reason is that gluons "carry" color charge and interact with its other while a photon does not "carry" any charge and does not interact with other photons. $\endgroup$ – ApolloRa Feb 2 at 14:40
  • $\begingroup$ That does not require you to know the commutator. Does it? $\endgroup$ – mithusengupta123 Feb 2 at 14:51
  • $\begingroup$ The absence of that extra term would mean that there are no interactions between gluons. $\endgroup$ – ApolloRa Feb 2 at 14:58

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