The superficial degree of divergence in Peskin and Schroeder

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Homework Statement

I have in the picture attached a screenshot from Peskin's textbook.

I don't understand how did they get that for the two last diagrams that $D=-2$.
The question is from pages 316-317 of Peskin's textbook.

Homework Equations

$$D=4-N_{\gamma}-3/2N_e$$
where $N_e$=number of external electron lines;
$N_\gamma$=number of external photon lines.

The Attempt at a Solution

It seems that for the two last diagrams, we have: $N_{\gamma}=N_e=2$, which means that $D=4-2-2*3/2=2-3=-1$, why is it that $D=-2$?

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nrqed
Homework Helper
Gold Member

Homework Statement

I have in the picture attached a screenshot from Peskin's textbook.

I don't understand how did they get that for the two last diagrams that $D=-2$.
The question is from pages 316-317 of Peskin's textbook.View attachment 226716

Homework Equations

$$D=4-N_{\gamma}-3/2N_e$$
where $N_e$=number of external electron lines;
$N_\gamma$=number of external photon lines.

The Attempt at a Solution

It seems that for the two last diagrams, we have: $N_{\gamma}=N_e=2$, which means that $D=4-2-2*3/2=2-3=-1$, why is it that $D=-2$?
When they say that $N_e$ is the number of external fermion lines, it means the number of external incoming fermion lines plus th number of outgoing fermion lines. In the last two diagrams, $N_e=4$.

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When they say that $N_e$ is the number of external fermion lines, it means the number of external incoming fermion lines plus th number of outgoing fermion lines. In the last two diagrams, $N_e=4$.
So $N_\gamma=0$, why is that? I see two external photon lines.

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Is it because the photon lines are connected to the electron lines?

nrqed
Yes. Note that in the first diagrams you had to use $N_e=2$