Stupid question on anomalous magnetic moment of the electron

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SUMMARY

The discussion focuses on the calculation of the anomalous magnetic moment of the electron, specifically addressing the contributions from the vertex corrections denoted as \(\Lambda^{(1)}\) and \(\Lambda^{(2)}\). Participants clarify that while \(\Lambda^{(1)}\) contains both divergent and finite terms, it ultimately does not contribute to the magnetic moment due to its behavior as \(q\) approaches zero. The relevant literature suggested includes "An Introduction to Quantum Field Theory" by Peskin & Schroeder, which explains the significance of the \(\Lambda^{(2)}\) term in this context.

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  • Understanding of quantum field theory concepts
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  • Knowledge of the notation used in Ryder's and Peskin & Schroeder's texts
  • Basic grasp of form factors and their role in particle physics
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  • Study the vertex corrections in quantum electrodynamics as outlined in Ryder's work
  • Examine the relevant sections in "An Introduction to Quantum Field Theory" by Peskin & Schroeder
  • Learn about the implications of form factors on particle interactions
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paolorossi
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I don't understand one think: when we calculated the correction of the magnetic moment of the electron, we consider only the finite terms of the vertex correction that non depend on the parameter μ. Using Ryder notation , we consider only

\bar{u}(p')(γμ+\Lambdaμ(2))u(p)

but there is also another finite contribution in \Lambdaμ(1)
\Lambdaμ(1) = divergent + finite

, why we don't consider that? some book to view this?
 
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paolorossi said:
, why we don't consider that? some book to view this?
Try Peskin & Schroeder. If I remember correctly, they show how only the ##\Lambda^{(2)}## term you mentioned is relevant to the anomalous magnetic moment, because the ##\Lambda^{(1)}## term goes to 1 as ##q\to 0##. (I'm guessing here that Ryder's ##\Lambda## corresponds to P&S's ##F##.)
 
yes,there are two form factors .The first of which becomes 1 in zero momentum transfer case.However I have not read ryder,so may be it is something different.I will reply after taking a look somewhere else.
 
yes,there are two form factors .The first of which becomes 1 in zero momentum transfer case.However I have not read ryder,so may be it is something different.I will reply after taking a look somewhere else.
thanks . however the \Lambdaμ is the vertex correction... when the ryder regularize it, whit the dimensional method, he writes it as the sum of two terms: \Lambda(2)μ that is convergent, and \Lambda(1)μ that contains the divergent term plus a finite term. Then he calculates the quantity
\bar{u}(p')(\gammaμ+\Lambdaμ(2))u(p)
whit q=p'-p (q is the momentum associated to the external photon ) different fo zero. He find the correction:
g/2=1+α/2π

but why he doesn't consider the finite term in \Lambda(1)μ ?
 
strangerep said:
Try Peskin & Schroeder. If I remember correctly, they show how only the ##\Lambda^{(2)}## term you mentioned is relevant to the anomalous magnetic moment, because the ##\Lambda^{(1)}## term goes to 1 as ##q\to 0##. (I'm guessing here that Ryder's ##\Lambda## corresponds to P&S's ##F##.)

thanks now I try to read something
 
ok now I understand. I calculate Λ(1)μ : the finite term in it is proportional to the matrix γμ, so it doesn't give any contribution to the magentic moment of the electron ( it contributes only to the form factor F1 )

certain that the ryder could write instead of "finite" in the expression of Λ(1)μ the term itself, or say something about ...
 
they are just the form factors.By taking the non relativistic limit you can interpret F1(0) as the unit of charge(zero momentum transfer case it is 1) and other factor F2(0) which using gordon identity can be interpreted as electron having a magnetic moment (1+F2(0)) which interacts with magnetic field .
 
they are just the form factors
... I think it's more correct to say that they are connected to form factors, in fact, using the identity of gordon we find that only a part of the finite term \Lambda(2)μ contributes to the form factor F2 , and so to the magnetic moment of the electron... anyway thanks to all for your answers, bye
 

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