The superficial degree of divergence in Peskin and Schroeder

  • #1
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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.
Peskin.png

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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Answers and Replies

  • #2
nrqed
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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.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##.
 
  • #3
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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.
 
  • #4
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Is it because the photon lines are connected to the electron lines?
 
  • #5
nrqed
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Is it because the photon lines are connected to the electron lines?

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

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