$\require{cancel}$In Peskin & Schroeder chapter 10 page 332 we have the renormalization condition $$\left. \Sigma (\cancel{p})\right|_{\cancel{p}=m} ~=~ 0. \tag{10.40} $$

How is it possible to set the sum of traceless matrices equal to a diagonal matrix $m$? How do we interpret this?


1 Answer 1


$\cancel{p} = p^\mu\gamma_\mu = m$ seems to be a typo. It should be $p^2 = p^\mu p_\mu = m^2$ in stead.

Fermion propagator is $$ G = \frac{i}{\cancel{p}-m_0 - \Sigma(\cancel{p})+i\epsilon} = \frac{1}{1-b(p^2)}\frac{i}{\cancel{p}-\frac{m_0 + a(p^2)}{1-b(p^2)} + i\epsilon}, $$ where $$ \Sigma(p) = a(p^2) + b(p^2)\cancel{p}. $$ The propagator has a pole at $$ m_p = \frac{m_0 + a(m_p^2)}{1-b(m_p^2)}, $$ where $m_0$ is bare mass (infinite) and $m_p$ is the physical mass (finite).

One may rearrange the above Fermion propagator via introducing modified self energy $\hat{\Sigma}(\cancel{p})$ (alternatively and often obfuscating to the neophytes, via introducing counterterms as the preferred route in most text books) so that $$ G = \frac{iZ}{\cancel{p}-m_p - \hat{\Sigma}(\cancel{p})+i\epsilon}, $$ where $$ Z = \frac{1}{1-b(m_p^2)}, $$ and $$ Z^{-1}\hat{\Sigma}(\cancel{p}) = [a(p^2)-a(m_p^2)] + [b(p^2)-b(m_p^2)]\cancel{p}. $$ Note that the difference $a(p^2)-a(m_p^2)$ is finite, even though $a(p^2)$ and $a(m_p^2)$ are individually infinite. Explicit regularization schemes (such as the widely used dimensional regularization) are not necessary at all! if we follow the regime of sticking with finite differences (i.e. $a(p^2)-a(m_p^2)$)and measurable quantities (i.e. $m_p$).

The on-mass-shell renormalization conditions are just trivial identities stemming from the above definition of modified self energy $\hat{\Sigma}(\cancel{p})$, $$ \hat{\Sigma}(\cancel{p})|_{p^2 = m_p^2} = 0, $$ $$ \hat{\Sigma}'(\cancel{p})|_{p^2 = m_p^2} = 0. $$

Of cause, we can redo the above excise at a different mass scale $\mu$ (the renormalization scale) other than $m_p$. But it won't change the physical picture. The renormalization group technic (via sliding the renormalization scale $\mu$) is also dispensable (at least in the context of high energy physics), since simply resuming the geometric series can achieve the same.

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    $\begingroup$ $p^\mu\gamma_\mu=m$ is not a typo. It is a very standard way to renormalise fermionic correlation functions. One regards $\not p$ as a formal complex number, and evaluates the residue at $\not p=m$. $\endgroup$ Jun 19, 2018 at 15:34
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    $\begingroup$ As stated by Yoni, $\not{p} = p^\mu\gamma_\mu$ is the sum of traceless matrices. It can't be a complex number. $\endgroup$
    – MadMax
    Jun 19, 2018 at 15:57
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    $\begingroup$ Sure, a matrix is not a complex number. As I said, it is a rather standard trick. It is a formal complex number. Pick any book where fermionic correlation functions are discussed; the normalisation conditions are $\Sigma(\not p=m)=0$ and $\frac{\mathrm d\Sigma}{\mathrm d\not p}(\not p=m)=0$. $\endgroup$ Jun 19, 2018 at 16:00
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    $\begingroup$ No, by $\not p=m$ I mean $\not p=m$. Have a look at Bjorken&Drell, eq. 19.27, or Itzykson&Zuber, eq. 7.32, or Mandl&Shaw, eq. 9.24, or Peskin&Schroëder, eq. 10.40, or Schwartz, eqs. 19.60-19.64, or Srednicki, eq. 62.27, or Ticciati, eq. 10.13.14, or Weinberg, eq. 11.4.9, or, as I said before, any book where fermionic correlation functions are discussed. $\endgroup$ Jun 19, 2018 at 16:46
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    $\begingroup$ Well in that case, I strongly suggest correcting the typo in these books. $\endgroup$
    – MadMax
    Jun 19, 2018 at 16:53

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