# Dimensional analysis of the fermion mass renormalization

1. Mar 29, 2013

### phypar

In the textbook, usually the fermion mass renormalization is introduced as follows: the mass shift $\delta m$ must vanish when $m_0=0$. The mass shift must therefore be proportional to $m_0$. By dimensional analysis, it can only depend logarithmically on $\Lambda$ (the ultraviolet cutoff): $\delta m \sim \ln \Lambda$.

Here I don't understand how the "dimensional analysis" is performed, can somebody explain it in detail? Thanks in advance.

2. Mar 29, 2013

### The_Duck

Take a look at e.g. the diagram (a) here, in QED: http://ej.iop.org/images/1751-8121/45/38/383001/Full/jpa374448f4_online.jpg

This diagram will correct the mass of the electron. Schematically, this diagram has a value that looks like

$\int d^4 p \frac{\gamma^\mu p_\mu + m_e}{p^2 + m^2}\frac{1}{p^2}$

The integrand has dimension 1/[mass]^3, so the diagram has a whole has dimension [mass]^1.

Now, the integral superficially seems to be linearly divergent, like $\int d^4 p (1/p^3)$. In that case the result would be proportional to $\Lambda$ times a dimensionless constant (since it must have dimensions of [mass]^1). However, looking a little closer we see that the term involving $\gamma^\mu p_\mu$ is odd in $p$ and therefore cancels out, leaving

$m_e \int d^4 p \frac{1}{p^2 + m^2}\frac{1}{p^2}$

Now we can see that the integral is actually logarithmically divergent, so the diagram is proportional to $m_e \ln \Lambda$. Here $m_e$ has units of [mass]^1 and $\ln \Lambda$ is dimensionless, so the diagram has units of [mass]^1, as required.

The point of the symmetry+dimensional analysis argument is to argue that this same thing will happen generally. The diagram has to have units of [mass]^1, and must be proportional to $m_e$. So whatever the diagram evaluates to, it has to look like $m_e$ times something dimensionless. The only dimensionless thing you can make out of $\Lambda$ is $\ln \Lambda$, so at worst diagrams like this can only diverge like $\ln \Lambda$ and not, say, $\Lambda^2$.

3. Mar 29, 2013

### phypar

Thanks a lot, you gave a very clear explanation. Now I understand it fully.