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How to prove that Feynman propagator are equivalent in Coulomb gauge and $R_\zeta$ gauge? (Be more specific, they are same when they contract with external current)

In $R_\zeta$ gauge, the propagator takes the form $$ D_{F\mu\nu}=-\frac{1}{k^2+i\epsilon}\left[g_{\mu\nu}-(1-\zeta)\frac{k_{\mu}k_\nu}{k^2}\right] $$

When $\zeta=1$ it's the Feynman gauge, $D_{F\mu\nu}=-\frac{g_{\mu\nu} }{k^2+i\epsilon} $. It's the usual form we use in Feynman rules. Because the $A_{\mu}$ is always coupled to the conserved current,i.e. $\partial_\mu J^\mu(x)=0$ so $k_\mu J^{\mu}(k)=0$. So it proves that propagator in $R_{\zeta}$ gauge is equivalent to the Feynman gauge.

In Coulomb gauge, the propagator takes the form: $$ \begin{aligned} D^C_{F\mu\nu}&=\frac{1}{k^2+i\epsilon}\left(\sum_{i=1,2}\epsilon_\mu(k,i)\epsilon_\nu(k,i)\right)\\ &= -\frac{g_{\mu\nu} }{k^2+i\epsilon} -\frac{n_{\mu}n_{\nu}}{(k\cdot n)^2-k^2}-\frac{1} {k^2+i\epsilon} \frac{k_\mu k_\nu-(k_\mu n_\nu+k_\nu n_\mu )(k.n)}{(k\cdot n)^2-k^2} \end{aligned} $$

Using the same argument of above, the 3rd term has no contribution. But the second term is an instantaneous Coulomb interaction. If we choose $n^{\mu}=(1,0,0,0)$, the second the term is $$ \frac{\delta_{\mu,0}\delta_{\nu,0}}{|\mathbf{k}|^2} $$

In coordinate space, this term is $$ \delta_{\mu,0}\delta_{\nu,0}\frac{\delta(x_0-y_0)}{4\pi|\mathbf{x}-\mathbf{y}|} $$

I can't see why it's zero when it couples with external current.

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  • $\begingroup$ This term will not vanish. It's obvious why this is called Coulomb gauge, right? ;) in other gauges you have to work a bit harder to extract this term. Try using an explicit expression for $k=(E, p,0,0)$ and see how it works in Feynman gauge. $\endgroup$ – rwold Apr 21 '17 at 0:56
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    $\begingroup$ @rwold If it's not vanishing, how to prove Coulomb gauge is equivalent to Feynman gauge? $\endgroup$ – user153663 Apr 21 '17 at 0:59
  • $\begingroup$ It has to vanish on the physical (transverse) polarizations, right? I am not sure what you mean by equivalence, but for what I know as long as they have the same S-matrix they can be considered equivalent. $\endgroup$ – Prof. Legolasov Apr 21 '17 at 2:30
  • $\begingroup$ @SolenodonParadoxus I want to prove the instantaneous Coulomb interaction vanishes when instantaneous Coulomb interaction contract with externel source. But it seems impossible. $\endgroup$ – user153663 Apr 21 '17 at 2:33
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The term $$ \delta_{\mu0}\delta_{\nu0}\frac{\delta(x_0-y_0)}{4\pi|\boldsymbol{x}-\boldsymbol{y}|} $$ does not vanish on its own. Rather, it is cancelled by the instantaneous (non-local) Coulomb term $$ \begin{aligned} \mathcal L_\mathrm{Coul.}&=\frac12\int\mathrm d\boldsymbol y\ \frac{1}{4\pi|\boldsymbol x-\boldsymbol y|}J^0(x^0,\boldsymbol x)J^0(x^0,\boldsymbol y)\\ &=\frac12\int\mathrm dy\ \delta_{\mu0}\delta_{\nu0}\frac{\delta(x^0-y^0)}{4\pi|\boldsymbol x-\boldsymbol y|}J^\mu(x)J^\nu(y) \end{aligned} $$

For more details, see Srednicki, chapters 55 and 56.

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  • $\begingroup$ See also Bjorken, Drell, chapter 17.9. $\endgroup$ – AccidentalFourierTransform Apr 21 '17 at 8:43
  • $\begingroup$ Oh~I see . In Coulomb gauge with external source, $A^0$ is decided by external source. So there will be a nonlocal term in original Lagrangian. Thanks. $\endgroup$ – user153663 Apr 21 '17 at 14:41
  • $\begingroup$ @fff123123 exactly :-) I'm glad I could help. $\endgroup$ – AccidentalFourierTransform Apr 22 '17 at 7:51

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