# Explicit supersymmetry breaking fermion mass terms

I hope you can clear up my following confusions.

In Girardello's and Grisaru's paper (Nuclear Physics B, 194, 65 (1982)) where they analysed the most general soft explicit supersymmetry breaking terms, they explicitly mentioned that $\mu \psi \psi$ is not soft, where $\psi$ is the fermionic component of a scalar superfield. This means that such a term will give rise to quadratic divergences.

However, in Stephen Martin's SUSY primer, on page 49, he explictly says the following:

One might wonder why we have not included possible soft mass terms for the chiral supermultiplet fermions, like $L = −\frac{1}{2} m_{ij}\psi^i\psi^j + c.c.$. Including such terms would be redundant; they can always be absorbed into a redefinition of the superpotential and the terms ...

This would seem to suggest that adding explicit chiral fermion mass terms is not problematic, since it is equivalent to a redefinition of the superpotential and the other soft breaking terms (e.g. scalar masses), both of which don't give rise to quadratic divergences.

I seem to see a contradiction here, but I am sure it is due to some subtlety I am too blind to notice.

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+1 Great question! – JeffDror Jun 19 '14 at 15:37

Martin is right. It is very simple to see it: just consider for a moment to add this extra superpotential term $$\delta W(\Phi_i)=\frac{1}{2}m_{i j}\Phi_i \Phi_j$$ where $\Phi_i$ are chiral superfields. This extra $W$ corresponds to an extra lagrangian term $$\delta\mathcal{L}=-\frac{1}{2}m_{i j}(\psi_i\psi_j+\text{h.c.})- \bar{\phi}_iM^2_{ij}\phi_j\qquad M^2_{ij}=\bar{m}_{ik}m_{jk}\,.$$ Therefore, starting with $W$ and adding a soft mass for the fermions $\delta\mathcal{L}_{\mathrm{soft}-\psi}=-\frac{1}{2}m_{i j}(\psi_i\psi_j+\text{h.c.})$ $$W(\Phi)\rightarrow \mathcal{L}=\mathcal{L}_{SUSY}+\delta\mathcal{L}_{\mathrm{soft}-\psi}$$ is just the same as doing starting with $W+\delta W$ (with $\delta W$ in the first equation above) and add a soft mass terms for the scalars, $\delta\mathcal{L}_{\mathrm{soft}-\phi}=+\bar{\phi}_iM^2_{ij}\phi_j$ $$W(\Phi)+\delta W(\Phi)\rightarrow \mathcal{L}=\mathcal{L}^\prime_{SUSY}+\delta\mathcal{L}_{\mathrm{soft}-\phi}=\mathcal{L}_{SUSY}+\delta\mathcal{L}_{\mathrm{soft}-\psi}$$

In other words, it is intuitive that only the mass splitting between fermions and boson that matters and should be considered. Adding soft masses that respect susy, that is that do not split fermions and bosons, it is just like adding a mass term to $W$. Therefore, the mass term for the fermion is soft, in the sense that does not give rise to quadratic divergences, since it is equivalent to a new superpotential with soft scalar masses that we know do not generate quadratic divergences. Perhaps, the only possible subtle point left that I see would be about the vacuum stability, since the soft masses for the scalars come flip in sign. But in any case, this is not about their being soft.

Extra Edit about possible exceptions: prompt by a comment below ,I should point out that this reasoning works only when the trilinear non-holomorphic couplings induced by the mass term, schematically of the form $$\delta L_{soft-\psi} \propto m^*_{ij} y_{ikl} \phi^*_j \phi_l \phi_k+h.c.,$$ involves no gauge singlets. I think that Girardello and Grisaru (as well as most of textbooks on susy) assume this is the case , so that such a term is just another soft term which doesn't worsen the UV behavior. When instead gauge singlets are around, such a trilinear may generate quadratic divergences (I think from generating tadpoles).

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This isn't entirely correct - the SUSY breaking fermion masses can only be reabsorbed by shifting, amongst other things, non-holomorphic SUSY-breaking trilinears. Those terms are ONLY soft if the model contains no gauge singlets. Girardello assumes that there are gauge singlets around. – innisfree May 2 '15 at 14:23
In other words, there's no contradiction between Martin and Girardello – innisfree May 2 '15 at 14:24
@innisfree I have shown in my answer that adding only fermion masses is equivalent to a new superpotential and soft scalar masses. If you disagree with that, can you please point to a specific flaw, if any, in the argument? Note that I am not reabsorbing away the fermion masses, I am just claiming the equivalence with another theory that is supersymmetric but deformed by soft scalar masses. – TwoBs May 3 '15 at 9:26
The specific flaw is that you've missed out $m y^* \phi \phi \phi^*$ operators generated by a mass-term in the superpotential. To compensate for them, you need non-holomorphic breaking operators $\phi^*\phi\phi$, which are NOT in general soft (only soft if no gauge singlets) – innisfree May 3 '15 at 10:37
@innisfree ah, ok, now I understand your criticism. You are absolutely right: this argument works iff the non-holomorphic trilinear term is soft as it happens when no gauge singlets are involved. It is somehow rooted in my mind that this is always the case and that the trilinear is indeed soft. Thank you for pointing this point out, I will add an extra remark into my answer. – TwoBs May 3 '15 at 19:19