# Does Supersymmetry really assures that the Higgs self-energy contributions cancel out?

The well-known "Supersymmetry Primer" from Stephen Martin explains in his introduction that for the cancellation of the Higgs self-energy diagrams (Fig. 1.1 a + b)

generated by fermions and their supersymmetric scalar counterparts the equality

$$\lambda_S = |\lambda_f|^2$$

is necessary where $$\lambda_f$$ is the Yukawa-coupling of the Higgs to a (specific) fermion $$f$$ according the coupling term: $$-\lambda_f H\bar{f}f$$ and $$\lambda_S$$ the 4-coupling of the Higgs to the corresponding sfermion $$-\lambda_S |H|^2 |S|^2$$.

I was searching in the "Primer" for a hint how supersymmetry indeed can provide such a relation and I haven't found much about it. There is something about it in section 3.2 on "Interactions of chiral multiplets", in particular p.23, but I find it unsatisfactory. Let me show it: In the Lagrangian (3.2.19) and the superpotential (3.2.18) appear two terms:

$$-\frac{1}{2}y^{ijk} \phi_i \psi_j \psi_k \quad \text{and}\quad \frac{1}{4}y^{ijn}y^{\ast}_{kln}\phi_i\phi_j \phi^{k\ast}\phi^{l\ast}$$

which seem to be the right candidates. Let's take the case of the contribution of the left top chiral multiplet to self energy of the Higgs. I will denote members (in particular indices running over them) of the left top chiral multiplet $$t$$ and the Higgs (and indices running over it) $$H$$, moreover assume that $$\psi$$ describes fermions and $$\phi$$ scalars (Higgs-scalar and stops typically). Specialized on this particular case the tricoupling term looks like (corresponding to fig. 1.1a):

$$-\frac{1}{2}y^{Ht\bar{t}} \phi_H \psi_t \psi_{\bar{t}}$$

whereas the quadruple-coupling terms looks like corresponding to Fig. 1.1b:

$$\frac{1}{4}y^{tHn}y^{\ast}_{tHn}\phi_t\phi_H \phi^{t\ast}\phi^{H\ast}$$

Actually, it seems indeed possible that we get $$\lambda_S = |\lambda_f|^2$$ if there were not this unspecified index $$n$$. The most obvious would be now to choose $$n=\bar{t}$$, and the desired equality could be realised. But apparently implicit summation over double indices (i.e. Einstein's summation convention) is applied, so $$n$$ could run over all particles of the MSSM which would lead to a long series of coupling constants and instead one would get something like

$$\sum_n |\lambda^f_n|^2$$

so I would have no idea how the initial required equality of the couplings of stops and tops could be realised. It is curious the "Primer" does not provide an explicit proof (at least I have not found it) that with the introduction of Supersymmetry the original problem is indeed solved. It would be interesting to know if my approach is about correct or if i missed something significantly.

• The $y$ in your 4-coupling are still the Yukawa coupling between Higgs, top and "some other" particle. For what particle except the $\bar{t}$ do you think this term is non-zero in the MSSM, i.e. why does the sum over $n$ not simply collapse to the $\bar{t}$ term you expect? – ACuriousMind Apr 18 at 13:01
• @A CuriousMind: That's a good point. Thank you! If you like the question give it a vote. – Frederic Thomas Apr 18 at 13:14