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I'm self studying QFT from Peskin and Schroeder. In chapter 19 of this book, page 653 (Perturbation theory anomalies) the expectation value of the induced current is calculated. I'm confused with the assertion that the corresponding feynman graph of this expected value is

enter image description here

$\uparrow$ Figure 19.1.

The expression given for the expected value is $$ \int \mathrm d^2 x e^{iqx} \langle j(x)\rangle = \frac{i}{e} i \Pi^{\mu \nu}(q)A_{\nu}(q).\tag{19.15}$$

I don't understand 2 things, first how is the Feynman graph related to the expected value of the current and how is this Feynman graph related to the right part of the equality, I understand that $\Pi^{\mu \nu}$ is the loop part, but the $A_{\nu}$ confuses me, how is it related to the external photon leg?

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The calculation you are referring to is for the current induced by a fixed background electromagnetic field in two-dimensional massless QED.

Having a background field means quantising the usual fermionic action $$ \bar{\psi}(iD\!\!\!/)\psi $$

with $D_\mu=\partial_\mu + i e A_\mu(x)+i e B_\mu(x)$, where $A_{\mu}(x)$ is a fixed, classical gauge field (external background field), while $B_\mu$ is the usual photon field. It might be a bit confusing since $A_\mu$ normally indicates the photon field, but I am using this notation to be consistent with your question and that particular section of PS.

Doing so gives you the usual Feynman rules for QED, plus the additional (two-fermion) vertex

External field vertex

with value (in Fourier space) $-i e \gamma^\mu A_\mu(q)$ (see e.g. page 304 of PS). You can then use these rules to compute the expected value of $j^\mu=\bar{\psi}\gamma^\mu\psi$.

Without an external field the expected value of the current would be zero, as it is not possible to build diagrams due to momentum conservation. Having an external field means that you can insert external legs pinned to the $\otimes$ (and corresponding factors of $A_\mu$).

The figure you posted is the lowest order diagram contributing to such an expansion. Note that it corresponds to the first correction to the photon propagator, with both legs amputated, and one of them replaced by the background field leg. The black dot on the left corresponds to $\bar{\psi}\gamma^\mu\psi$ and not to a fermion-photon vertex, introducing a factor of $1/(-i e)$ wrt the usual self energy $i\Pi^{\mu\nu}$.

If you think about it, it should be clear that all the diagrams contributing to $\langle j^\mu\rangle$ with one external $A_\nu$ leg correspond to those appearing in the expansion for the photon propagator.

The sum of those (amputated) diagrams is exactly the definition of $i\Pi^{\mu\nu}$. So in principle the diagram in the figure only gives the first non-vanishing contribution to $\langle j^\mu \rangle$. However as they say in the book, for 2d massless QED Schwinger showed that the one-loop calculation of the photon propagator is exact, so the one-loop diagram is all you need also in this case.

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  • $\begingroup$ "Without an external field the expected value of the current would be zero, as it is not possible to build diagrams due to momentum conservation." - Is there some momentum income expected from the left side into the $\psi \gamma \psi$-vertex? Otherwise the momentum would be conserved through the loop alone. Why do we need a loop anyway? $\endgroup$ – Statics Sep 16 '16 at 14:33

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