Gauge invariance of superpotential

[RedX]

The superpotential is basically a product of left chiral superfields, taking the $$\theta \theta$$ component.

However, under a supergauge transformation, the left chiral superfields change, and the superpotential does not seem to be supergauge invariant.

In fact, under supergauge transformation, the left chiral superfield no longer seems to be a left chiral superfield.

Am I doing something wrong?

 

[Avodyne]

A chiral superfield that is in a nontrivial rep of the gauge group will change under a supergauge transformation, but it remains a chiral superfield. The superpotential must be constructed to be supergauge invariant.

You'll need to provide more details if you don't agree.

 

[RedX]

I agree that under a supergauge transformation, a chiral superfield remains a chiral superfield - I messed up on that one.

But take as a superpotential the $$\theta \theta$$ component of a chiral superfield {in Srednicki's book, equation (95.29), whose component would be F(x)}. A generic supergauge transformation is the exponential of i times another generic chiral superfield {equation 95.51 of Srednicki). Clearly the $$\theta \theta$$ term of the new field will be different. In the examples of Srednicki, for instance, the B(x) of 95.51 will multiply the F(x) of 95.29 to get a new $$\theta \theta$$ term. Therefore the Lagrangian is not invariant under supergauge transformation.

 

[Avodyne]

The superpotential must be a gauge invariant product of fields, just like, in non-supersymmetric theories, the scalar potential and Yukawa couplings must be gauge invariant. Terms that are not gauge invariant are not allowed. Take a look at the next section of Srednicki, where he writes down the gauge-invariant superpotential for the supersymmetric version of the Standard Model.

 

[RedX]

There was a question on the high-energy physics board about a calculation in the Wess-Zumino model, which Srednicki defines as eqn. (95.47). It just bothered me that the super-potential in the Wess-Zumino model doesn't look super-gauge invariant, but I'm just starting the next chapter on the supersymmetric Standard Model so maybe they modify it.

But chapter 95 was good as I at least got to see the super-partner of the electron (the A field) and the superpartner of the photon (the $$\lambda$$ field), although Srednicki strangely never uses the term super-partner but instead calls the $$\lambda$$ field the gaugino.

 

[malawi_glenn]

But chapter 95 was good as I at least got to see the super-partner of the electron (the A field) and the superpartner of the photon (the $$\lambda$$ field), although Srednicki strangely never uses the term super-partner but instead calls the $$\lambda$$ field the gaugino.

That it is also standard, also the photon does not really have a superpartner in that sense, you have to look to the superpartners of the SU(2)_L x U(1)_Y gauge group, they are then called Winos and Binos, which you, as in the EW SM, combine to get the corresponding Zino and Photino.

 

[Avodyne]

There was a question on the high-energy physics board about a calculation in the Wess-Zumino model, which Srednicki defines as eqn. (95.47). It just bothered me that the super-potential in the Wess-Zumino model doesn't look super-gauge invariant

It's not! The Wess-Zumino model does not have a gauge symmetry, or even a global symmetry that could be gauged.