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A vector superfield is defined by postulating an invariance under a linear transformation in the space of vector superfields:

$V \longrightarrow V + i\Lambda - i\Lambda^{\dagger}$

where $i\Lambda - i\Lambda^{\dagger}$ is a vector superfield.

My question, however, is concerning the mass dimension of the superfields involved. We know that the vector superfield V has a zero mass dimension, whereas the chiral superfield $\Lambda$ has a mass dimension of 1 (The combination $i\Lambda - i\Lambda^{\dagger}$as stated is a vector superfield.)

So how is it possible to define this supergauge transformation, where we have added a mass dimensional quantity to one which has no mass dimension?

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    $\begingroup$ $\Lambda$ has mass dimension 0 $\endgroup$
    – Elliot Schneider
    Commented Jul 6, 2016 at 15:46
  • $\begingroup$ @user81003 I don't quite agree with you here. A chiral superfield has a mass dimension 1, since the chiral supefield consists of a term like $\Phi(y) \supset \sqrt{2} \theta \xi(y) $, where the $\xi(y)$ is the fermionic component of the chiral superfield, and hence has a mass dimension of 3/2, whereas the mass dimension of the Grassmann variable $\theta $ is -1/2, giving the chiral superfield a mass dimension of 1. $\endgroup$
    – RM2401
    Commented Jul 8, 2016 at 15:16
  • $\begingroup$ For example, a superfield $\Phi$ coupled to the gauge theory transforms as $\Phi \to e^{iq\Lambda} \Phi$ in the usual convention, so $\Lambda$ must be dimensionless. $\endgroup$
    – Elliot Schneider
    Commented Jul 8, 2016 at 18:46
  • $\begingroup$ Yes, that is itself a part of the conundrum. Your comment and my previous comment both seem to be true, but can't be. I, obviously, see that $\Lambda$ must be mass dimension zero, but from the argument I mentioned, it must have a mass dimension of 1. I probably should have framed the question better. $\endgroup$
    – RM2401
    Commented Jul 8, 2016 at 20:51

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You assume canonical mass dimensions for $V$ and $\Lambda$. But you can in general choose non-canonical ones. So $\Lambda$ here can have $0$ mass dimension. If you insist on it having mass dimension 1, you can compensate it by a mass parameter of your theory.

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