# The superficial degree of divergence in Peskin and Schroeder

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.

## 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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• Peskin.png
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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##.

Gold Member
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.

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

nrqed