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?


2 Answers 2


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).


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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