For the free photon-field:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]Z[J]=exp(\frac{i}{2} \int d^4x \int d^4y J^\mu(x)\Delta_{\mu \nu}} (x-y) J^\nu(y)) [/tex]

where [tex]\Delta_{\mu \nu}}=g_{\mu \nu}/(k^2-i\epsilon) [/tex] is the free-photon propagator. This leads to Coloumb's law for the electrostatic energy between two charges at [tex]x_1 [/tex] and [tex]x_2 [/tex] when you plug in [tex]J(x)=e_1\delta(\vec{x}-\vec{x_1})+e_2\delta(\vec{x}-\vec{x_2}) [/tex].

What happens if you use the full-photon propagator [tex]\Delta_{\mu \nu}}=g_{\mu \nu}/(k^2-\Pi(k^2)-i\epsilon) [/tex] instead? Since [tex]\Pi(k^2)[/tex] is of the order of the fine-structure constant when the fine-structure constant is small, wouldn't you get Coloumb's law plus a small correction?

Or is this cheating, and you need to include the full interaction Lagrangian in Z[J]:

[tex]Z[J]=exp(i \int d^4t \mathcal L_I(\frac{1}{i}\frac{\delta}{\delta J(t})) exp(\frac{i}{2} \int d^4x \int d^4y J^\mu(x)\Delta_{\mu \nu}} (x-y) J^\nu(y)) [/tex]

to get the correction to Coloumb's law? Or is this all bad anyways because in the end, we will treat the surviving source functions, J(x), the electrons, as classical electrons, stationary point particles with delta functions? The correct fully quantum-mechanical way would be to just amputate the sources and plug in the external momenta instead of plugging in classical sources?

Because it seems to me that the only difference between non-Abelian gauge theories and Abelian ones are self-interactions of the boson field. If we can treat the self-interaction of the gluons as corrections to the propagator, then why can't we calculate perturbatively the gluon force?

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# Quantum corrections to force laws

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