# QED Vertex Factor/Rule

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

$$V = -ie\gamma^\mu \int d^4x$$ while on page 304 they write

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

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.

• Cool thanks for your answer. I've never really went through what a background/external field is in qft. – Your Majesty Mar 3 '14 at 21:01
• @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. – Adam Mar 3 '14 at 22:24
• That second paragraph I knew. It's just the physics that I've never read too much about. – Your Majesty Mar 3 '14 at 22:34
• 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. – Your Majesty Mar 13 '14 at 9:18
• @LoveLearning: see my edit and tell me if it's technical enough now. – Adam Mar 13 '14 at 13:51