Faddeev-Popov trick in QED Peskin and Schroeder

Question

On page 297 of Peskin and Schroeder, the book obtains the propogator

$$\tag{9.58} \tilde{D}_F^{\mu\nu}(k)=\frac{-i}{k^2+i\epsilon}\bigg(g^{\mu\nu}-(1-\xi)\frac{k^\mu k^\nu}{k^2}\bigg).$$

The book then says"

the Faddeev-Popov procedure guarantees that the value of any correlation function of gauge invariant operators computed from Feynman diagrams will be independent of the value of $\xi$ used in the calculation. In the case of QED, it is not difficult to prove this $\xi$ independence directly. Notice in equation 9.58 that $\xi$ multiplies a term in the photon propagator proportional to $k^\mu k^\nu$. According to the Ward Takahashi identity the replacement in a Green's function of any photon propagator by $k^\mu k^\nu$ yields zero, except for terms involving external off shell fermions. These terms are equal and opposite for particle and antiparticle and vanish when the fermions are grouped into gauge invariant combinations."

What exactly does Peskin and Schroeder mean by "These terms are equal and opposite for particle and antiparticle and vanish when the fermions are grouped into gauge invariant combinations"? What would be an example of this?

Answer 1

By gauge invariant combinations, he is referring to interaction vertices allowed by the standard model. The standard model Lagrangian is gauge invariant, so each interaction term—which corresponds to a legal interaction vertex in a Feynman diagram—must also be gauge invariant. For example, you can have an electron, positron, and photon vertex, but not a vertex where an electron is connected to two positrons. I believe he means that if you are writing down the matrix element for a legal, gauge invariant Feynman diagram, then you will obtain the desired cancellation of terms.