I am having trouble reconciling two pieces of information.

Consider **supersymmetric QED**, i.e. a supersymmetric U(1) gauge theory with two chiral superfields of opposite charges, $h$ and $\hat{h}$.

The Kähler potential $K$ is:
$$
K = h^\dagger e^{2\,g\,q\,V} h + {\hat{h}}^\dagger e^{-2\,g\,q\,V} \hat{h} ,
$$
while the superpotential $W$ is the simplest possible:
$$
W = m\, h \, \hat{h}.
$$

**On the one hand**

Renormalized mass and fields are related to bare/original ones by $m_0 = Z_m m_r$, $h_0 = \sqrt{Z_h} h_r$, $\hat{h}_0 = \sqrt{Z_h} \hat{h}_r$. We also define $Z_m = 1 + \delta_m$, $Z_h = 1 + \delta_h$.

The SUSY non-renormalization theorems say that $W$ is not perturbatively renormalized, implying that
$$
Z_m Z_h = 1 \quad \Rightarrow \quad \delta_m = - \delta_h,
$$
at the one loop level.
If we then proceed to write the counterterm Feynman rule for the scalar propagator of $h$, we get:

i.e. the scalar propagator counterterm is proportional to $(p^2+m^2)$.

**On the other hand**

If one explicitly computes the divergent part of the $h$ self energy at one loop in dimensional regularization, one finds* that:
$$
i \Sigma_h (p^2) \bigg|_\textrm{div} = i \frac{g^2 q^2 }{(4\pi)^2} \frac{2}{\epsilon} \big(-4 m^2\big).
$$

i.e. the divergent part of the self-energy at one loop is proportional to just $m^2$.

However, I was expecting a divergent part proportional to $(p^2 + m^2)$, which is what can be cancelled by the aforementioned counterterm. Is this reasoning correct?

Any suggestions as to what might have gone wrong?

_{*In the literature one can find this result e.g. in arXiv:hep-ph/9907393, section 4.3, equation (150), by playing with the integrals. }

This post imported from StackExchange Physics at 2015-12-17 17:17 (UTC), posted by SE-user J-T