1
$\begingroup$

In Peskin's QFT textbook, when discussing the superficial divergence of loops in QED, the book says (page 317):

"To analyse the photon one-point function,note that the external photon must be attached to a QED vertex. Neglecting the external photon propagator,this amplitude is therefore

$-i e \int d^4x e^{-iqx} \langle \Omega| T j_{\mu}(x)|\Omega \rangle $

where $j_{\mu}=\bar{\psi} \gamma^{\mu} \psi$ is the electromagnetic current operator. "

But I cannot figure out why amplitude of this diagram can be represented by the expectation value of electricity current. I want to know how to get the formula through Feynman Rules.The formula coped from Peskin's book

$\endgroup$
1
$\begingroup$

The result given by P&S goes beyond perturbation theory (and Feynman diagrammatics), and is exact. To recover this using Feynman rules, you would have to resum all terms of the perturbative expansion. But you might want to show that it is indeed true at some given order in $e$.

The best way to see that this is an exact result is to use the path integral formalism. The partition function of QED with an external source for the photon field is (up to gauge fixing terms that are irrelevant here) $$ Z[J]=\int DA D\bar\psi D\psi \exp\left(-S_f[\bar\psi,\psi]-S_p[A]-S_i[\bar\psi,\psi,A]+\int_x J_\mu(x)A^\mu(x)\right)\ , $$ where $S_f[\bar\psi,\psi]$ is the action of the fermions (without the photon field), $S_p[A]$ that of the photons, $S_i[\bar\psi,\psi,A]=ie\int_x j_\mu(x)A^\mu(x)$ is the interaction action. The last term is the source term. Integrating out the Gaussian photon field, we obtain, $$ Z[J]\propto\int D\bar\psi D\psi \exp\left(-S_f[\bar\psi,\psi]+\frac12\int_{x,y} \left(J_\mu(x)-iej_\mu(x)\right)G^{\mu\nu}(x,y)\left(J_\nu(y)-iej_\nu(y)\right)\right)\ , $$ where $G^{\mu\nu}(x,y)$ is the bare photon propagator. We thus find the average value of the EM field from $$ \langle A_\mu(x)\rangle=\frac{\delta \ln Z}{\delta J_\mu(x)}\bigg|_{J=0} \ , $$ which gives $$ \langle A_\mu(x)\rangle= -ie\int_y G^{\mu\nu}(x,y)\langle j_\nu(y) \rangle\ , $$ which corresponds to the solution of Maxwell equations for a source $\langle j_\nu(y)\rangle$, which is computed from the QED problem, since $$ \langle j_\nu(y) \rangle=\frac{1}{Z[0]}\int DA D\bar\psi D\psi\;j_\nu(y) \exp\left(-S_f[\bar\psi,\psi]-S_p[A]-S_i[\bar\psi,\psi,A]\right)\ . $$

$\endgroup$
  • $\begingroup$ If the result is exact, but how should I prove it. I can hardly see the connection between the two sides of the equation. $\endgroup$ – Eric Yang Mar 28 '16 at 15:13
  • 1
    $\begingroup$ @EricYang: might be clearer with my edit. The only left out is the photon propagator, which is only the bare one in my answer... $\endgroup$ – Adam Mar 28 '16 at 15:51
  • $\begingroup$ Thank you so much for your answer. Now I know how to get this formula. $\endgroup$ – Eric Yang Mar 29 '16 at 15:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.