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On page 303 in Peskin&Schroeder they give the vertex factor as

$$V = -ie\gamma^\mu \int d^4x$$

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while on page 304 they write

$$V_\times = -ie\gamma^\mu\int d^4x A_\mu(x).$$

enter image description here

Why are the diagrams pictorially different (one of the have the funny cross at the end of the photon line)? Why do the gauge field appear in $V_\times$ and not in $V$? Are $V$ and $V_\times$ identical?

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In the first case, the vertex is a vertex in the common sense (used to construct diagrams).

In the second case, the gauge field is not dynamic (in a path integral formulation, you do not integrate over), it is a background field that is fixed. In that case, we are interested on the effect of this non-dynamical field on the electron field. This is useful to study, for example, the probability to create electron-positron pairs from the vacuum when there is a (very) strong electric field (imposed by the outside, say, the experimentalists in the lab).

EDIT:

To be more technical on the second case (the non-dynamical field): let's have a look at the partition function $$Z[\tilde A]=\int D\psi e^{i S_0[\psi]+i S_A[\psi]}, $$ where $S_0$ is the standard free fermion action, and $$S_\tilde A[\psi]=-e\int d^4 x \tilde A_\mu\bar\psi\gamma^\mu \psi.$$ Notice that we do not integrate over $\tilde A_\mu$ in the functional integral. However, introducing $S_\tilde A[\psi]$ implies that a new vertex has to be used to compute the partition function, denoted by this wiggly line with crossed circle in the OP's question.

What is the point ? First, we see that $\tilde A_\mu$ couples to the fermions as a usual E&M field. Therefore, if the system we want to described is given by some fermions in a classical E&M field, we can modelize that by using this $\tilde A_\mu$ (the assumption here is that the effects of the fermions on the E&M is negligible). Second, by deriving $\ln Z$ with respect to $\tilde A_\mu$, we can compute the the correlation functions of the current. In this case, $\tilde A_\mu$ plays the role of a source term.

If the E&M field is dynamical, we have to integrate over and now $$Z=\int D\psi DA e^{iS[\psi,A]}$$ where $$S[\psi,A]=\int d^4x\left( \bar\psi \big(i\gamma^\mu(\partial_\mu+ie A_\mu)-m \big)\psi-\frac{1}{4}F_{\mu,\nu}F^{\mu,\nu}\right).$$ Now we integrate over $\psi$ and $A$ (with no ~) and the partition function does not depend on any sources. $A$ plays the role of a dynamical photon, with its own propagator and there is now the standard vertex interaction.

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  • $\begingroup$ Cool thanks for your answer. I've never really went through what a background/external field is in qft. $\endgroup$ – Your Majesty Mar 3 '14 at 21:01
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    $\begingroup$ @LoveLearning: It just means that the field is imposed (is not allowed to fluctuate). It can have two purposes: 1- it can describe a physical field, usually classical (for instance, a classical E&M field, the quantum fluctuations of which are assumed to be irrelevant); 2- it is used as a theoretical source, i.e. taking derivatives with respect to the sources allows to compute correlation functions. $\endgroup$ – Adam Mar 3 '14 at 22:24
  • $\begingroup$ That second paragraph I knew. It's just the physics that I've never read too much about. $\endgroup$ – Your Majesty Mar 3 '14 at 22:34
  • $\begingroup$ By not fluctuate and by imposed from the outside do you mean that it is a classical field and that all its derivatives are zero? Could you be more technical or give reference where this nonfluctuating imposed field is technically defined? Thanks. $\endgroup$ – Your Majesty Mar 13 '14 at 9:18
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    $\begingroup$ @LoveLearning: see my edit and tell me if it's technical enough now. $\endgroup$ – Adam Mar 13 '14 at 13:51

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