# The Beta Function and the Bare Charge

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.

• I also seem to get a minus sign :( Jun 21, 2022 at 19:38
• 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. Jun 22, 2022 at 1:25

$$e(\Lambda)^{2} = e(\mu)^{2}\left( 1+\frac{e(\mu)^{2}}{12\pi^{2}}\log \left( \frac{\Lambda^{2}}{\mu^{2}}\right)\right) + O(e(\mu)^{ 6}),\tag{8.14}$$ 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.
Geeky: In WP, consider $$~ G (x) =\exp( 1/x)$$ as your Wegner scaling function and $$d= -1/6\pi^2$$.)