The superficial degree of divergence in Peskin and Schroeder

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Homework Help Overview

The discussion revolves around the superficial degree of divergence in quantum field theory, specifically referencing diagrams from Peskin and Schroeder's textbook. Participants are trying to understand the calculation of the degree of divergence, denoted as ##D##, and the interpretation of external lines in the context of the diagrams presented.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are examining the formula for ##D## and questioning the values assigned to ##N_{\gamma}## and ##N_e## in the context of specific diagrams. There is confusion regarding the interpretation of external fermion and photon lines, particularly how they contribute to the calculation of ##D##.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the definitions of external lines. Some have provided insights into the connections between photon and electron lines, but there is no explicit consensus on the correct values for ##N_{\gamma}## and ##N_e## in the diagrams.

Contextual Notes

Participants are working from a specific textbook reference and are constrained by the definitions provided therein. There is an emphasis on understanding the setup of the problem and the implications of the definitions used for the external lines.

MathematicalPhysicist
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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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MathematicalPhysicist said:

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##.
 
nrqed said:
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.
 
Is it because the photon lines are connected to the electron lines?
 
MathematicalPhysicist said:
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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