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.


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:


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,

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

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

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


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.

  • $\begingroup$ Since it is $e_0(M)$, is it true that $$\frac{\lim_{M \to \infty}\left(\frac{e_{0}^{2}}{12\pi^{2}}\ln\left(\frac{M^{2}}{m^{2}}\right)\right)}{\frac{e_{0}^{2}}{12\pi^{2}}} = \frac{12\pi^{2}}{e_{0}^{2}}$$ ? $\endgroup$
    – Hal Hollis
    Commented Sep 20, 2018 at 16:51
  • $\begingroup$ @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. $\endgroup$
    – Prem
    Commented Sep 20, 2018 at 17:42

1 Answer 1


Griffiths' argument is right in spirit but a bit too naive here.

  • 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 switch to the better-defined "renormalized" perturbation theory, you'll have a Taylor series in a finite variable, and there is no problem at all.
  • 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 $1$.

In general it's clearest to think of everything as an effective field theory, in which case many of the conceptual problems with renormalization disappear. "Logarithmic divergences" really just turn into moderate running of the coupling constant before you hit the Planck scale. "Quadratic divergences" simply denote sensitive dependence on UV physics. I talk about this in more detail here.


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