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Why is a Fayet-Iliopoulos term $-kD$ irrelevant or subdominant in the in the MSSM (Minimal Susy Standard Model)?

According to Martin (A Supersymmetry Primer, p.70) it's because squarks and sleptons don't have a mass term in the superpotential. Considering that we want a positive value for the scalar potential (in order to have susy breaking):

$$V=\sum_i |m_i|^2 |\phi_i|^2 +1/2 (k-g\sum_i q_i |\phi_i|^2)^2$$

It seems to me that we can achieve, with every possible VEV for the scalar fields, a positive value for the scalar potential, even without the mass term. What I'm missing?

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Not only must supersymmetry be broken, it must be broken in a way that doesn't lead to phenomenological disasters. In the MSSM, $D$-term breaking with a Fayet-Iliopoulos term achieves neither: it doesn't break supersymmetry but it does lead to phenomenological disasters!

Consider again your potential $$ \begin{align} V&=\sum_i |m_i|^2 |\phi_i|^2 +1/2 (k-g\sum_i q_i |\phi_i|^2)^2,\\ &=\sum_{i=u,d} |\mu|^2 |H_i|^2 + 1/2 (k-g\sum_i q_i |\phi_i|^2)^2, \end{align} $$ with a sum running on the Higgs, squarks and sleptons. In the second line, I've made it explicit that in the MSSM, only the Higgs doublets are permitted a mass term form the superpotential. A minima at $V=0$ can only be achieved with $H_i=0$ and at least one $\phi\neq0$ cancelling the $\kappa$ in the second term, such that each term is independently zero.

This is bad news: if $V=0$ we haven't broken supersymmetry, but if $\langle \phi \rangle \neq 0$, we have broken colour or electromagnetism. There is a slight loophole, here, though, because it could be a sneutrino that obtains the vev, and breaks only $SU(2)\times U(1)_Y$ but preserves EM.

I haven't thought about this possibility before - it would break $R$-parity and lepton number, and lead to neutrino-bino/wino mixing. I think, then, that it's ruled out by limits on gaugino and neutrino masses (and in any case, it doesn't break supersymmetry).

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Maybe I've understood the problem. In the minimum we have (only one scalar field for simplicity):

$$\frac{dV}{d\phi}=0=\phi [ m^2 -kgq+g^2q^2 \phi^2]$$

If $m=0$ we are we are forced to choose a mexican hat potential with one maximum in $\phi=0$ and two degenerate minima. So we are forced to have a non zero vev for the scalar fields. If these scalar fields are sleptons and squarks this implies that color and EM must be broken (and we don't want this)

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