# In charge renormalization equation, $e=e_{0}^{2}\left[1-e_{0}^{2}A\right]$, how can an infinite $e_{0}$ and $A$ give finite $e$ in any limit?

In Griffiths elementary particle book (chapter 7, 'Quantum electrodynamics', equation 7.188), one gets the following equation for the vacuum polarization calculated to one loop correction.

$$\frac{e_{0}^{2}}{q^{2}}\left(1-\frac{e_{0}^{2}}{12\pi^{2}}\left[\ln\left(\frac{M^{2}}{m^{2}}\right)-f\left(\frac{-q^{2}}{m^{2}c^{2}}\right)\right]\right)$$

Here, $e_{0}$ is the bare charge of the electron, $m$ is its mass, $M$ is a cutoff parameter, $q$ is the momentum related to the specific Feynman diagram. Also, $f(q)$ is a finite function of $q$.

Now, the renormalization procedure at this point is to absorb the divergent $\ln\left(\frac{M^{2}}{m^{2}}\right)$ term by defining the physical charge $e$ of the electron as:

$$e^{2}=e_{0}^{2}\left(1-\frac{e_{0}^{2}}{12\pi^{2}}\ln\left(\frac{M^{2}}{m^{2}}\right)\right)\tag{1}$$

Here, it is usually stated that although $\lim_{M \to \infty}e_{0}=\infty$ and $\lim_{M \to \infty} \ln \left(\frac{M^{2}}{m^{2}}\right)=\infty$, $e$ converges to a finite value that can be measured in laboratory.

From my understanding of the procedure of limits, I don't understand how an infinite $e_{0}$ and $\ln\left(\frac{M^{2}}{m^{2}}\right)$ can give rise to a finite $e$ in equation $(1)$.

My argument is as following:

If $\lim_{M \to \infty} e_{0}^{2}\left(1-\frac{e_{0}^{2}}{12\pi^{2}}\ln\left(\frac{M^{2}}{m^{2}}\right)\right)$ is finite, and if $\lim_{M \to \infty}e_0^{2} =\infty$ then,

$$\lim_{M \to \infty}\left(1-\frac{e_{0}^{2}}{12\pi^{2}}\ln\left(\frac{M^{2}}{m^{2}}\right)\right)=0$$

$$\implies \lim_{M \to \infty}\left(\frac{e_{0}^{2}}{12\pi^{2}}\ln\left(\frac{M^{2}}{m^{2}}\right)\right)=1$$

$$\implies \lim_{M \to \infty} \ln \left(\frac{M^2}{m^2}\right)= \lim_{M \to \infty}\left(\frac{12\pi^2}{e_{0}^{2}}\right)$$

$$=0$$

But we already know that $\lim_{M \to \infty} \ln \left(\frac{M^2}{m^2}\right)=\infty$.

Hence, the bare charge $e_0$ of the electron cannot be infinite.

• Since it is $e_0(M)$, is it true that $$\frac{\lim_{M \to \infty}\left(\frac{e_{0}^{2}}{12\pi^{2}}\ln\left(\frac{M^{2}}{m^{2}}\right)\right)}{\frac{e_{0}^{2}}{12\pi^{2}}} = \frac{12\pi^{2}}{e_{0}^{2}}$$ ? – Hal Hollis Sep 20 '18 at 16:51
• @halhollis I should have introduced limit on the rhs as well. I have edited the question, although i am not sure if this is what you are asking. – Prem kumar Sep 20 '18 at 17:42

• You're working in "bare" perturbation theory, performing a "Taylor expansion" in the formally infinite $e_0$. So it should be no surprise that you get nonsensical results if you cut off the series.
• Here is a simple analogy: consider the function $e^{-x}$ which limits to zero as $x \to \infty$. Naively taking the first two terms of the Taylor series gives $$e^{-x} "\approx" 1 - x$$ which certainly does not limit to zero as $x \to \infty$. That doesn't mean that $e^{-x}$ can't.
• If you insist that your theory is well-defined at all energies, then the conclusion that you cannot make $e$ finite is correct, even though your argument is wrong. This is the Landau pole problem of QED. The only way to avoid the coupling blowing up at some absurdly high energy scale (far above the Planck scale) is to have it be identically zero.
• In the modern effective field theory standpoint, QED is, at best, only valid up to the Planck scale. Therefore, the coefficient $\log(M^2/m^2)$ is not infinite, but rather on the order of $10$ to $100$. Bare perturbation theory works. The quantity $e_0$ is larger than $e$, but still well under $1$.