According to "Introductory Lectures on Quantum Field Theory", by L. Álvarez-Gaumé and M. A. Vázquez-Mozo, (P. 84) the dependence of the bare charge $e_{0}(\Lambda)$ on the cutoff $\Lambda$ is determined by the identity

$$ e(\mu)^{2} = e_{0}(\Lambda)^{2}\left( 1+\frac{e_{0}(\Lambda)^{2}}{12\pi^{2}}\log \left( \frac{\mu^{2}}{\Lambda^{2}}\right)\right).\tag{8.13}$$

Taking into account that we are working in perturbation theory in $e(\mu)^{2}$, we can express the bare charge $e_{0}(\Lambda)^{2}$ in terms of $e(\mu)^{2}$ as

$$e(\Lambda)^{2} = e(\mu)^{2}\left( 1+\frac{e(\mu)^{2}}{12\pi^{2}}\log \left( \frac{\mu^{2}}{\Lambda^{2}}\right)\right) + O(e(\mu)^{ 6}).\tag{8.14}$$

How does one obtain the second equation? I tried solving for the bare charge in the first one and then replacing it twice in the result, but I end up with an extra minus sign.

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    $\begingroup$ I also seem to get a minus sign :( $\endgroup$ Jun 21, 2022 at 19:38
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    $\begingroup$ Indeed, the scale of the correction coupling, the term with 12 in the denominator, should agree with the denominator of the log argument... This is what allows the elimination of Λ and the derivation of (8.15). Consequently, (8.14) has the wrong sign, as you suspect. $\endgroup$ Jun 22, 2022 at 1:25

1 Answer 1


Your second equation has the sign wrong (obviously; read on). It is probably a typo. It should be, instead,
$$e(\Lambda)^{2} = e(\mu)^{2}\left( 1+\frac{e(\mu)^{2}}{12\pi^{2}}\log \left( \frac{\Lambda^{2}}{\mu^{2}}\right)\right) + O(e(\mu)^{ 6}),\tag{8.14}$$ which results from inverting $$ e(\mu)^{2} = e (\Lambda)^{2}\left( 1+\frac{e (\Lambda)^{2}}{12\pi^{2}}\log \left( \frac{\mu^{2}}{\Lambda^{2}}\right)\right),\tag{8.13}$$ since $$ { e(\Lambda)^{2}\over e(\mu)^{2}}= { 1\over 1+\frac{e (\Lambda)^{2}}{12\pi^{2}}\log \left( \frac{\mu^{2}}{\Lambda^{2}}\right ) }\\ = 1-\frac{e (\Lambda)^{2}}{12\pi^{2}}\log \left( \frac{\mu^{2}}{\Lambda^{2}}\right ) +... \\ = 1+\frac{e (\mu)^{2}}{12\pi^{2}}\log \left( \frac{\Lambda^{2}}{\mu^{2}}\right ) +... $$

It is this "self-similar scale" feature (the term with 12 in the denominator agreement with the denominator of the log argument) that makes elimination of $\log \Lambda^2$ possible to yield the epochal 1954 Gell-Mann—Low RG equation, $${e(\mu)^{2}\over e(\mu_0 )^{2} } = 1+\frac{e(\mu_0)^{2}}{12\pi^{2}}\log \left( \frac{\mu^{2}}{\mu_0^{2}}\right) .\tag{8.15}$$ $e(\mu)$ is an increasing function of μ.

(Appendix B, with suitable gratitude expressed to T D Lee by the authors: a major moment in 20th century intellectual history. Feynman has said it weighed heavily in his hiring MGM at Caltech.

Geeky: In WP, consider $~ G (x) =\exp( 1/x)$ as your Wegner scaling function and $d= -1/6\pi^2$.)


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