Clarification of a passage from the Schwartz QFT text

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In summary, Schwartz discusses the concept of renormalization in QFT and how it applies to the ##\phi^4## theory. He explains that while the scattering matrix for this theory is infinite, the difference between two different scales is finite. However, it is impossible to extract the bare coupling constant, ##\lambda##, from the scattering process as it is also affected by a ##\lambda^2## correction. To determine a finite value for ##\lambda##, one must use the elastic scattering cross section and renormalize the coupling constant at a specific scale, which then leads to the running of the renormalized coupling constant, wavefunction normalization constants, and masses governed by the renormalization-group equation.
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HomogenousCow
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Hi, would like some clarification on this passage from Matthew D. Schwartz's QFT text regarding renormalization on page 297, 15.4.1 Renormalization of ##\lambda##:

"First of all, notice that, while ##M(s)## (the ##\phi\phi\rightarrow\phi\phi## scattering matrix for ##\phi^4## theory) is infinite, the difference between ##M(s_1)## and ##M(s_2)## at two different scales is finite:$$M(s_1)-M(s_2)=\frac{\lambda^2}{32\pi^2}\ln(\frac{s_2}{s_1}).$$ Should we also expect that ##M(s)## itself be finite? After all, ##M^2## is supposed to be a physical cross section.
To answer this, let us think more about ##\lambda##. It should be characterizing the strength of the ##phi^4## interaction. So to measure ##\lambda## we would simply measure the cross section for ##\phi\phi\rightarrow\phi\phi## scattering, or equivalently, ##M##. But this matrix element is not just proportional to ##\lambda## but also has the ##\lambda^2## correction above. Thus, it is impossssible to simply extract ##\lambda## from this scattering process."

I'm a bit confused by the part in bold, can't we determine ##\lambda## from the difference in scattering cross-sections that he just wrote down?
 
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I'm not sure what Schwartz is after. I've the book not here, but I guess what he means is the following: The ##\lambda## is the bare coupling and is infinite and unobservable. However, you can use the elastic scattering cross section to define a finite renormalized value for ##\lambda## at some given scale ##s_R## by measuring the cross section. Then you express everything in terms of renormalized quantities and everything is finite. However, the renormalized coupling constants (and wave-function normalization as well as masses) have to be changed, when the renormalization scale ##s_R## is changed, such that the cross section stays unchanged. That demand leads to the running of the renormalized coupling constant, wavefunction normalization constants, and masses governed by the socalled renormalization-group equation. For my treatment of renormalization, see

https://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf
 

1. What is the purpose of clarifying a passage from the Schwartz QFT text?

The purpose of clarifying a passage from the Schwartz QFT text is to ensure a clear understanding of the content and concepts presented in the text. This can help to prevent misinterpretation and promote accurate comprehension.

2. How do you determine which passages need clarification in the Schwartz QFT text?

Determining which passages need clarification in the Schwartz QFT text involves carefully reading and analyzing the text, as well as considering any questions or confusion that may arise from the reader. It may also be helpful to consult with other experts in the field.

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4. Is it necessary to clarify every passage in the Schwartz QFT text?

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5. Can clarification of a passage from the Schwartz QFT text change the intended meaning?

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