I'm slowly reading my way through Peskin and Schroeder. Near the end of section 7.1 they compare the mass shift of the electron from QFT to the classical value, both of which are divergent but in different ways.

The calculation from QFT gives:

$$\delta m = \frac{3\alpha}{4\pi}m_0\log\left(\frac{\Lambda^2}{m^2_0} \right)$$

Which diverges logarithmically as $\Lambda\to\infty$.

Versus the classical expression which diverges linearly, $\alpha\Lambda$.

The bit I don't understand is the argument they use after to explain why the divergence should be logarithmic:

"$\delta m$ must vanish when $m_0=0$. The mass shift must therefore be proportional to $m_0$, and so by dimensional analysis, it can depend only logarithmically on $\Lambda$".

(Taken from the paragraph immediately following the $\alpha\Lambda$ result.)

The understand the first part of this statement, its the dimensional analysis I don't get. Why is it logarithmic specifically and not some other dimensionless function of $\Lambda$?

  • $\begingroup$ I'd be interested in seeing a rigorous answer as to why it is the logarithm, perhaps using the $\log(xy) = \log(x)+\log(y)$ property. But in general it need not be I think-- it could be a polylogarithm or something, if the integral is complicated enough; as long as the function diverges at $\Lambda/ m_0 = 0$ and $\infty$. $\endgroup$
    – Dwagg
    Jun 24, 2019 at 17:36

3 Answers 3


From a dimensional analysis point of view: Assume you need linearity in $m_0$ (aka. $\delta m \propto m_0$) by the argument given by MadMax above. This means you have $ \delta m \sim m_0 f(\Lambda,m_0)$ for some function $f$. Since the LHS and RHS both must have dimensions of mass, this means $f$ needs to be dimensionless.

Yet it is a function of two quantities which have dimension so you can argue that $f$ needs to be a function of the dimensionless quantity $\frac{m_0}{\Lambda}$. So at this stage you argue that $f(\Lambda,m_0) = F\big(\frac{m_0}{\Lambda} \big)$ for some $F$.

It is true that in principle $F$ could be some really complicated function. However, the since the cutoff of the theory satisfies $m_0 \ll \Lambda$ (i.e. it is much bigger than all the scales in your problem), whatever the function $F$ is, you only need its series for $x\ll 1$.

So at this stage there are only a few options for how $F$ looks like. Either $F(x) \sim x^{n}$ for some $n$ or another option is $F(x) \sim \log(x)$. That's pretty much it (I think).

However, if you take $F(x) \sim x^n$ then you have the mass shift $\delta m \sim m_0 \left( \frac{m_0}{\Lambda} \right)^n$ which is linear in $m_0$ only when $n=0$, so this option is disqualified. You are left to conclude that $$ \delta m \propto m_0 \log\left( \frac{m_0}{\Lambda} \right) $$ up to some constants.


The logarithms in amplitudes are the major consequence of QFT and appear from integrals over 4-momentum. For the hard cutoff regularization integrals like: $$ \int_0^\Lambda \frac{d^4p}{(2 \pi)^4} \frac{1}{\left( p^2 + \Delta^2 \right)^n} $$ become $$ 2 \pi^2 i\int_0^\Lambda \frac{d |k|}{(2 \pi)^4} \frac{|k|^3}{\left( k^2 + \Delta^2 \right)^n} $$ after Wick rotation ($k_0 = i p_0$, $k^2 = k_0^2 + \vec{k}^2$) and using $d^4k = d\Omega_4 \ dk \ |k|^3$ and $\int d\Omega_4 = 2\pi^2$.

For $n > 2$ these quantities are convergent in the limit $\Lambda \rightarrow \infty$. For $n = 2$ the integral has logarithmic divergence.

For $n = 1$ the integral has solution with both quadratic and logarithmic terms: \begin{align} \frac{i}{16 \pi^2} \left( k^2 - \Delta^2 \log\left(k^2 + \Delta^2\right) \right) \Big\rvert^{k=\Lambda}_{k=0} = \frac{i}{16 \pi^2} \left( \Lambda^2 - \Delta^2 \log \frac{\Lambda^2 + \Delta^2}{\Delta^2} \right) \end{align}

For QED, however, it is not possible to have such integrals with $n = 1$, because there is no vertex with two photon and two electron edges. The simplest loop integral would have $n = 2$, and thus the amplitude can have only logarithmic divergence.


Fundamentally, it has to do with 't Hooft's technical naturalness argument: since the electron mass term breaks the chiral symmetry, the quantum correction to the electron mass has to be proportional to the weakly broken mass scale $m_0$, thus linear in $m_0$ (with logarithmic divergent factor).

A counter example is the quantum mass correction to a bosonic field, which is of the order $\Lambda^2$. Why? The root cause is that the bosonic mass does NOT break the chiral symmetry, thus it's NOT protected by 't Hooft's technical naturalness principle. The Higgs mass issue is exhibit A, hence the vexing hierarchy/naturalness problem that keeps physicists up at night.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.