First of all let me note that you are talking about spontaneous SUSY breaking which is usually disscussed as a consequence of our dissatisfaction with various aspects of soft SUSY breaking (usually associated with the MSSM which has around 100 unwanted parameters and other pathologies like non-conservation of flavor symmetry). Therefore let us consider how we can spontaneously break SUSY in a "cousin" way of the way the electro-weak symmetry is broken in the SM. That is, we want to find some supersymmetric Lagrangian whose ground state breaks SUSY. Now, there are various conditions that must be respected in order our theory to be consistent and one of these is that the Hamiltonian (the Legendre transform of the SUSY respecting Lagrangian), which can be written as a functional of the supercharges, is positive definite (bigger or equal to zero). From this we can extract the conclusion, and this is what you must keep in mind, that a state will be SUSY invariant, i.e., $Q_{\alpha} | \text{state} \rangle = 0$, iff its energy is zero, i.e., $H | \text{state} \rangle =0$. Here $Q_{\alpha}, H$ is one of the supercharges and the Hamiltonian respectively. We conclude that the vacuum energy is an order parameter for SUSY breaking and that a vacuum expectation value for a given superfield can be an order parameter for SUSY breaking as well. To understand better, what this actually means is that we want $Q_{\alpha}\langle 0 | S | 0 \rangle = Q_{\alpha}\langle S \rangle = 0$ (and complex conjugate). How can this happen? As probably know $S$ can be written in an expansion of the Grassman variables $\theta, \bar{\theta}$. The previous requirement takes place when the higher components of the superfield's VEV are non-vanishing. Now we can manage SUSY breaking in two ways. In a SUSY gauge theory the scalar potential can be written as $$V= F^{i*}F_i + \frac{1}{2}D^aD^a,$$ and if we can find models such that $Fi=0$ and $D^a=0$ are not simultaneously solved then we will have a spontaneously broken SUSY! By the way, this is where F-term and D-term SUSY breaking comes from. I am gonna say a few words on F-term and D-term SUSY breaking (taken from various notes and books).
F-term. These models of SUSY breaking are not very hard to understand. Consider the Lagrangian $$\mathcal{L}_F = \int d^4 \theta Q^{\dagger}Q + \int d^2 \theta W(Q) + \text{h.c.},$$ and note that in order to get a vacuum where SUSY is unbroken we require $\langle F_i^{\dagger} \rangle= \Big\langle \frac{\partial W}{\partial Q^{i}} \Big\rangle=0$. Now, everything depends on the form of the superpotential $W$. Maybe it is such that there are no solutions as we asked before and SUSY is indeed spontaneously broken. The order parameter is the F component of a chiral superfield (which if you remember we add in the Lagrangian in order to make SUSY manifest and well defined off-shell). Let us look at a specific superpotential. Take for example $$W_F = -k^2\Phi_1 + m\Phi_2 \Phi_3 + \frac{y}{2}\Phi_1\Phi_3^2$$. It is not hard to find that the scalar potential (which is what we are so interested about) can be written as $$V = |F_1|^2 + |F_2|^2+|F_3|^2 = |k^2 - \frac{y}{2}\phi_3^{*2}|^2 + |m\phi_3^*|^2 +|m\phi_2^*+y\phi_1^* \phi_3^*|^2$$. Whatever you might try to this eqaution you will not find a solution wjere both $F_1=F_2=0$. Now, if you remember something from the stuff on moduli space (even for $\mathcal{N}=1$ SUSY) you can see that by allowing the parameter $m$ to be large enough we can get the minimum of the potential at $\phi_2=\phi_3=0$. At this minimum you can also show that the scalar potential takes the value $k^4$. Furthermore, let us take the undetermined $\phi_1$ to be zero. Then we find that mass spectrum of the scalars which is $(0,0,m^2, m^2, m^2-yk^2, m^2+yk)$ while the mass spectrum of the fermions is $(0,m,m)$. Do an excerise now, let $k$ be zero. You see that SUSY is not broken if this is the case? On the other hand, for broken SUSY we have to take into account mass corrections of the complex scalar $\phi_1$ which we took to be massless (but only at the classical moduli space point 0). If you try to compute the Feynman diagrams which correct the propagator (to leading order in $k^2$) SUSY breaking comes as mass insertions in the $\phi_3$ propagator and only in that one! Other contributions do not relate to the breaking of the supersymmetry. Finally note that SUSY breaking does not affect the massless fermion $\psi_1$ which is the Goldstone boson of our spontaneously broken theory, the goldstino. This model is also called O'Raifeartlaigh model.
D-term. Now, let us focus on the gauge-theory related D-term (which is introduced in the theory for reasons similar to the ones we introduce the F-term). This method was pioneered from Fayet and Jean Iliopoulos (whom I had the chance and honor to meet last month and is a really amusing and interesting personality). They proposed to add a linear term in the Lagrangian of the form $\mathcal{L}_{FI} = u^2D$ where $[u]=$mass and is just a constant parameter. Then we find that the scalar potential is given by $$V = \frac{1}{2}D^2 - u^2D + gD\sum_{j} q_j \phi^{j*}\phi_j$$ and by integrating $D$ out we find that $$D = u^2 - g\sum_{j} q_j \phi^{j*}\phi_j$$. So just like before, we first find out how we can write the scalar potential and now we have an expression for$D$ in terms of the scalar components of the superfield. Now, if the scalar components $\phi_i$ have large positive mass squared terms then their VEV is zero $D=u^2$ from the previous equation. SUSY is broken since, at least if we consider in the MSSM, such a mechanism would directly give mass to the squarks and sleptons because these operators cannot have superpotential mass terms. Again, this breaks SUSY.
This type of SUSY breaking can be gauge or gravity mediated. Now, since I am no expert at all and I have written these down to remind them to myself as well let me mention that we do not like these models as much. You see, these models say completely nothing for the SUSY breaking scale. SUSY breaking is expected to occurs dynamically in the so-called hidden sector and we need to go beyond to non-renormalizable interactions to understand how it is being communicated. Can someone else to carry on with dynamical SUSY breaking? To end with, let me mention where I have learned a few details from and where I run back and rely on
- A video on SUSY breaking from Seiberg.
- Berolini's webpage
- From the books of Wess Bagger, Terning, Van Proyen and Freedman
- Further TASI videos
- Google
Well, I hope this helps a little bit. It surely heped me remember some of the basic of SUSY breaking )
Q&A (4871)
