MSSM (Minimal Supersymmetric Standard Model)

I have to write the action in superfield for SM with the Higgs, but I have problems with the mass term.

This is what I have (I don't know if it is right):

$$S_{1}=Tr\left(W^{\alpha}W_{\alpha}\right)\vert_{\theta\theta}+Tr\left(\bar{W}_{\dot\alpha}\bar{W}^{\dot\alpha}\right)\vert_{\bar\theta\bar\theta}+Tr\left(V^{\alpha}V_{\alpha}\right)\vert_{\theta\theta}+Tr\left(\bar{V}_{\dot\alpha}\bar{V}^{\dot\alpha}\right)\vert_{\bar\theta\bar\theta}+U^{\alpha}U_{\alpha}\vert_{\theta\theta}+\bar{U}_{\dot\alpha}\bar{U}^{\dot\alpha}\vert_{\bar\theta\bar\theta}.$$

$$S_{1}$$ is the part of the Lagrangian concerns with the kinetic term of the gauge field. In this case we have 8 gauge fields, gluons ($$SU(3)$$) in $$W_{\alpha}=W_{\alpha}^{a}(t^{a})_{ij}$$, $$a=1,2,...,8.$$ Analogously for $$V_{\alpha}$$ where we have 3 gauge fields related with weak interaction ($$SU(2)$$) and a gauge field for the $$U(1)$$ algebra.
Now we want to introduce matter fields.

$$\bullet$$(3,2,1/6) left handed quark doublet (A)
$$\bullet$$($$\bar{3}$$,1,-2/3) right handed up-type anti-quark (B)
$$\bullet$$($$\bar{3}$$,1,1/3) right handed down-type anti-quark (C)
$$\bullet$$(1,2,-1/2) left handed lepton doublet (D)
$$\bullet$$(1,1,1) right handed anti-lepton (E)
$$\bullet$$(1,2,1/2) Higgs ($$H_{u}$$)
$$\bullet$$(1,2,-1/2) Higgs ($$H_{d}$$)

The action related to the matter, Here $$N,M=1,2,3$$ and $$I,J=1,2$$.
$$\bullet$$ $$S_{A}=\bar{A}^{NI}\left(e^{V_{W}\oplus V_{V}}\right)_{NI}^{MJ}A_{MJ}\vert_{\theta\theta\bar{\theta}\bar{\theta}}=\bar{A}^{NI}\left[\left(e^{V_{W}}\right)_{N}^{M}\otimes\left(e^{V_{V}}\right)_{I}^{J}\right]A_{MJ}\vert_{\theta\theta\bar{\theta}\bar{\theta}}$$

$$\bullet$$ $$S_{B}=B^{N}\left(e^{-V_{W}}\right)_{N}^{M}\bar{B}_{M}\vert_{\theta\theta\bar{\theta}\bar{\theta}}$$

$$\bulletS_{C}=C^{N}\left(e^{-V_{W}}\right)_{N}^{M}\bar{C}_{M}\vert_{\theta\theta\bar{\theta}\bar{\theta}}$$

$$\bulletS_{D}=\bar{D}^{I}\left(e^{V_{V}}\right)_{I}^{J}D_{J}\vert_{\theta\theta\bar{\theta}\bar{\theta}}$$

$$\bulletS_{E}=E\left(e^{-V_{U}}\right)\bar{E}\vert_{\theta\theta\bar{\theta}\bar{\theta}}$$

Adding them we get a SUSY Lagrangian without the Higgs. But I don't know how to add the Higgs (Yukawa terms).

• Where's the superpotential? Jun 9 '17 at 12:22
• If I'm not wrong, what you call superpotential is $V_{W},V_{V},V_{U}$ .
– 7919
Jun 9 '17 at 12:27
• $W_{\alpha}=-\frac{1}{4}\bar{D}\bar{D}\left(e^{-V_{W}}(D_{\alpha}e^{V_{W}})\right)$. Analogous relation you have for $V_{\alpha} ,U_{\alpha}$
– 7919
Jun 9 '17 at 12:32
• Hmm. Not sure. But Yukawas will be in superpotential, for sure. Jun 9 '17 at 12:56

$V_W$, $V_V$, $V_U$ are vector superfields, not superpotentials. Superpotential is of the form $$W=a_i\Phi_i+m_{ij}\Phi_i\Phi_j+y_{ijk}\Phi_i\Phi_j\Phi_k$$ where $a_i$, $m_{ij}$, $y_{ijk}$ are coupling constants ($m$ is mass, $y$ is Yukawa). $\Phi_i$ are chiral superfields. To add these terms to the Lagrangian, $$\int d^2\theta\;W + h.c.\;\subset\mathcal{L}$$