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Given the following action $$ \mathcal{S}=\int{d^4x\;d^2\theta\;d^2\bar{\theta}\left(\bar{Q}_+e^{2V}Q_++\bar{Q}_-e^{-2V}Q_--2\xi V\right)}+\int{d^4x\;d^2\theta\left(mQ_-Q_++\frac{\tau}{16\pi i}W^\alpha W_\alpha\right)}+h.c. $$

Where $\bar{Q}$ is the hermitian conjugated $Q$, $\tau = \frac{\theta}{2\pi}+i\frac{4\pi}{g^2}$ the gauge coupling, $\xi$ the Fayet-Iliopoulos parameter, $W_\alpha=-\frac{1}{4}\bar{D}^2D_\alpha V$ is the gaugino superfield and $V$ the abelian vector superfield.

How can I show that the action is invariant under $$ Q_+\rightarrow e^{i\Lambda}Q_+ \;\\ Q_-\rightarrow e^{-i\Lambda}Q_- \;\\ V\rightarrow V-\frac{i}{2}\Lambda +\frac{i}{2}\bar{\Lambda} $$

When I try the transformation on the first term I get the following: $$ \bar{Q}_+e^{2V}Q_+\rightarrow \bar{Q}_+e^{2V}Q_+ e^{Im(\Lambda)} $$ I have a similar problem with the next term

Edit:

$Q_+$ is a chiral superfield with charge 1

$Q_-$ is a chiral superfield with charge -1

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1 Answer 1

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Let's check invariant of first term. Using transformation laws and fact, that transformation are abelian:

$$Q_+\rightarrow e^{i\Lambda}Q_+ \Rightarrow \bar{Q}_+\rightarrow \bar{Q}_+ e^{-i\bar{\Lambda}}$$

$$ V\rightarrow V-\frac{i}{2}\Lambda +\frac{i}{2}\bar{\Lambda} \Rightarrow e^{2V}\rightarrow e^{i\bar{\Lambda}}e^{2V} e^{-i\Lambda} $$

We obtain:

$$ \bar{Q}_+ e^{2V} Q_+ \to \bar{Q}_+ e^{-i\bar{\Lambda}} e^{i\bar{\Lambda}}e^{2V} e^{-i\Lambda} e^{i\Lambda}Q_+ = \bar{Q}_+ e^{2V} Q_+ $$

For gauge multiplet term:

$$ W_\alpha=-\frac{1}{4}\bar{D}^2D_\alpha V \to -\frac{1}{4}\bar{D}^2D_\alpha V -\frac{i}{8}\bar{D}^2D_\alpha\Lambda +\frac{i}{8}\bar{D}^2D_\alpha\bar{\Lambda} $$

Now using that $\Lambda$ is chiral, $\bar{\Lambda}$ antichiral and $[\bar{D}^2,D_\alpha]=0$: $$ D_\alpha\bar{\Lambda} =0 $$ $$ \bar{D}^2D_\alpha\Lambda = D_\alpha\bar{D}^2\Lambda = 0 $$

So $W_\alpha$ is gauge invariant quantity.

Fayet-Iliopoulos term:

$$ \int d^4x \;d^2\theta d^2 \bar{\theta}\; \xi V\rightarrow \int d^4x \;d^2\theta d^2 \bar{\theta}\; \xi \left(V-\frac{i}{2}\Lambda +\frac{i}{2}\bar{\Lambda}\right) $$

Using that $\int d\theta_\alpha = D_\alpha$ and $\int d\theta_{\dot{\alpha}} = \bar{D}_{\dot{\alpha}}$ (up to the boundary term) and chirality properties, we immediately see, that FI term is gauge invariant.

The same idea works with other terms.

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  • $\begingroup$ Can you show this for the $W^\alpha W_\alpha$ term? $\endgroup$ Apr 25, 2020 at 23:21
  • $\begingroup$ @fielder, I updated answer $\endgroup$
    – Nikita
    Apr 25, 2020 at 23:37
  • $\begingroup$ thank you very much! Just a final part, how do you show that for the Fayet-Iliopoulos term ? $\endgroup$ Apr 26, 2020 at 0:22
  • $\begingroup$ @fielder, I updated answer $\endgroup$
    – Nikita
    Apr 26, 2020 at 0:36
  • $\begingroup$ @fielder, is my answer clear to you? $\endgroup$
    – Nikita
    Apr 26, 2020 at 18:55

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