Can someone help me to expand the superpotential for $\mathcal{N}=2$ supersymmetry?

I want to expand the term in the superpotential for supersymmetry $\mathcal{N}=2$. I'm interested in just the hypermultiplet part, so just the coupling \begin{equation} \int d^2\theta \sqrt{2}gH_1\Phi H_2, \end{equation} where the $H_1=(H_1,\psi_{1|\alpha},F_1)$ and $\bar{H}_2=(\bar{H}_2,\bar{\psi}_{2}^{\dot\alpha},\bar{F}_2)$ are the two chiral fields belonging to the hypermultiplets and $\Phi=(\phi,\psi_\alpha,F)$ is the chiral field of the vector multiplet. My conventions for the calculation are: \begin{align} \Phi&=\phi+\sqrt{2}\theta\psi+i\theta\sigma^\mu\bar\theta\partial_\mu\phi-\theta\theta F-\frac{i}{\sqrt{2}}\theta\theta\partial_\mu\psi\sigma^\mu\bar\theta-\frac{1}{4}\theta\theta\bar\theta\bar\theta \square \phi\\ H_1&=H_1+\sqrt{2}\theta\psi_1+i\theta\sigma^\mu\bar\theta\partial_\mu H_1-\theta\theta F_1-\frac{i}{\sqrt{2}}\theta\theta\partial_\mu\psi_1\sigma^\mu\bar\theta-\frac{1}{4}\theta\theta\bar\theta\bar\theta \square H_1\\ H_2&=H_2+\sqrt{2}\theta\psi_2+i\theta\sigma^\mu\bar\theta\partial_\mu H_2-\theta\theta F_2-\frac{i}{\sqrt{2}}\theta\theta\partial_\mu\psi_2\sigma^\mu\bar\theta-\frac{1}{4}\theta\theta\bar\theta\bar\theta \square H_2\\ \end{align} I would say also that $H_1$ and $\bar{H}_2$ transform in the same representation of the internal symmetries, while $H_2$ in the complex conjugate. I suppose a non-abelian gauge theory of group $G$ where $\Phi$ transforms in the adjoint as \begin{equation} \Phi=\Phi^a T^a, \end{equation} where $a=1,\ldots,\dim G$. In this way $H_1$ and $H_2$ are respectively in the fundamental and anti-fundamental of the gauge group. I would say: \begin{equation} H_{1|i}\qquad H_{2}^i \end{equation} Is that correct? The index $i$ goes from $i=1,\ldots \dim R$, i.e. the dimension of the representation, such that the matrix is \begin{equation} T^a=(T^a)^i_j. \end{equation} Is this right? I have already done some calculation in the variable

\begin{equation} y^\mu=x^\mu-i(\theta\sigma^\mu\bar\theta) \end{equation}

and I think, after the calculation, I have to take just the $\theta\theta$ component since the integration kills all the others, right? Can someone check that the final result is

\begin{equation} H_{1|i}F^a(T^a)^i_jH_2^j-\psi_{1|i}\psi^a(T^a)^i_jH_2^j +\phi^a(T^a)^i_jF_{1|i}H_2^j-H_{1|i}\psi^a(T^a)^i_j\psi_2^j-\phi^a(T^a)^i_j\psi_{1|i}\psi_2^j+H_{1|i}\phi^a(T^a)^i_j F_2^j +\text{h.c.} \end{equation}

In the end I would like that someone explain to me why some notes said that substituing in the Lagrangian density the equation of motion for the auxiliary $F$ fields, it is possible to reduce the scalar potential to just $D-$terms. I think that this is not right, since if I substitute the terms from the equation of motion for $F$, I will get some $F-$terms that are the relation of traceless of the mesonic mass $H_{1|i}H_2^j$, and others things.

Is there some notes where they do the calculation explicitly for a generic gauge group in such a way I can check their result? Is it true that I can reduce the scalar potential to just $D-$terms?

Ok, I have found all the Lagrangian:

\begin{equation} \mathcal{L}= \text{Tr}\left[-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-i\lambda \sigma^\mu D_\mu \bar\lambda +\frac{1}{2}D^2\right]+\frac{\theta_{YM}}{32\pi^2}g^2 \text{Tr}\left[F_{\mu\nu}\tilde{F}^{\mu\nu}\right]+\text{Tr}\left[(\overline{D_\mu\phi})D^\mu\phi-i\psi\sigma^\mu D_\mu\bar\psi+\bar{F}F+i\sqrt{2}g\bar\phi\{\lambda,\psi\}-i\sqrt{2}\{\bar\psi,\bar\lambda\}\phi +gD[\phi,\bar\phi]\right]+(\overline{D_\mu H_1})^i(D^\mu H_1)_i-i\psi_{1|i}(D_\mu)^i_j \bar\psi_1^j+\bar{F}_1^iF_{1|i}+i\sqrt{2}g\bar{H}_1^i\lambda^a (T^a)_i^j \psi_{1|j}-i\sqrt{2}g\bar\psi_1^i\bar{\lambda}^a(T^a)_i^jH_{1|j}+g\bar{H}_1^iD^a(T^a)^j_iH_{1|j}+(\overline{D_\mu H_2})_i(D^\mu H_2)^i-i\psi_{2}^i(D_\mu)_i^j \bar\psi_{2|j}+\bar{F}_{2|i}F_{2}^i-ig\sqrt{2}\bar{H}_{2|i}\lambda^a (T^a)^i_j \psi_{2}^k+i\sqrt{2}g\bar\psi_{2|i}\bar{\lambda}^a(T^a)^i_jH_{2}^j-g\bar{H}_{2|i}D^a(T^a)^i_jH_{2}^j+\sqrt{2}g\left(H_{1|i}\phi^a(T^a)^i_j F_2^j-H_{1|i}\psi^a(T^a)^i_j\psi_2^j+H_{1|j}F^a(T^a)^j_iH_2^i-\psi_{1|j}\psi^a(T^a)^j_iH_2^i+F_{1|i}\phi^a(T^a)^i_jH_2^j-\psi_{1|i}\phi^a(T^a)^i_j\psi_2^j\right)+\sqrt{2}g\left(\bar{H}_{1}^i\bar\phi^a(T^a)_i^j \bar{F}_{2|j}-\bar{H}_{1}^i\bar\psi^a(T^a)_i^j\bar\psi_{2|j}+\bar{H}_{1}^j\bar{F}^a(T^a)_j^i\bar{H}_{2|i}-\bar\psi_{1}^j\bar\psi^a(T^a)_j^i\bar{H}_{2|i}+\bar{F}_{1}^i\bar\phi^a(T^a)_i^j\bar{H}_{2|j}-\psi_{1}^i\bar\phi^a(T^a)_i^j\bar\psi_{2|j}\right) \end{equation}

The equations of motion for the auxiliary fields are

\begin{equation} \begin{aligned} D^a&=-g[\phi,\bar\phi]^a-g\bar{H}_1^i(T^a)^j_iH_{1|j}+g\bar{H}_{2|i}(T^a)^i_jH_{2}^j\\ \bar{F}^a&=-\sqrt{2}gH_{1|j}(T^a)^j_iH^i_2\\ \bar{F}_{1}^i&=-\sqrt{2}g\phi^a(T^a)^i_jH_2^j\\ \bar{F}_{2|j}&=-\sqrt{2}gH_{1|i}\phi^a(T^a)^i_j \end{aligned} \end{equation}

So that the scalar potential is

\begin{equation} V=\sum_i \bar{F}_iF_i+\frac{1}{2}D^2=2g^2H_{1|j}(T^a)^j_iH^i_2\bar{H}_{1}^n(T^a)_n^m\bar{H}_{2|m}+2g^2\phi^a(T^a)^i_nH_2^n\bar\phi^b(T^b)^m_i\bar{H}_{2|m}+2g^2H_{1|i}\phi^a(T^a)^i_jH_{1}^m\bar\phi^a(T^a)^j_m+\frac{1}{2}\left(g[\phi,\bar\phi]^a-g(\bar{H}_1^i(T^a)^j_iH_{1|j}-\bar{H}_{2|i}(T^a)^i_jH_{2}^j)\right)^2 \end{equation}

Can someone help me to rewrite it in a more convenient way? I see that there is some factor repeated if we expand the squares, but I cannot manage to rewrite it better. How can I rewrite it so that is clearer the division in Coulomb and Higgs branches?

• Have you looked at standard textbooks like Terning's Supersymmetry, or Shifman's Advanced QFT ? – Antoine Apr 10 '17 at 8:07
• @user40085 No, there is something on the Weinberg Volume 3 but it is not useful for the part of the question I have edited now. I am more interested in the scalar potential written in such a way that the F-term and the D-term gives immediate the Higgs branch and the Coulomb branch of the theory. Can you from the potential that I have written say to me how to rewrite it better? – Alessandro Mininno Apr 10 '17 at 10:14