Why does the scalar field still have a VEV in a vacuum state of a SUSY-theory?

In a supersymmetric theory the vacuum state is Lorentz invariant, so all the vacuum expectation values (VEVs) of fields that transform non-trivially under Lorentz transformations should be zero. Thus, only the scalar field can have a non-zero VEV. But the vacuum is not only Lorentz symmetric, but also supersymmetric and the scalar field does not transform trivially under supertransformations. It gets mapped to a fermionic field. So why shouldn't the VEV. of the scalar field be zero as well?

• Have you tried explicitly calculating the variation of the fermion field? – knzhou Jun 26 '18 at 10:40
• Do you mean the variation of the fermionic field under Lorentz transformations? Because that is zero for sure, by definition. Or do you mean the transformation of the scalar field under susy that becomes a fermionic field? That depends a bit on dimension and what kind of multiplet you're in but for example in 4d N=1 in a chiral multiplet the transformation is $\delta \sigma \propto \epsilon^{\alpha} \psi_{\alpha}$ where $\sigma$ is the scalar field, $\psi$ the fermionic field and $\epsilon$ the spinorial infinitesimal parameter. (See lectures by Bertolini on SUSY, 4.59) – Graphite Jun 26 '18 at 11:22

Consider a single chiral multiplet with scalar and spinor components $\phi$ and $\xi$. Schematically, the SUSY transformations are $$\begin{split}\delta \phi &\sim \bar \xi\epsilon\\\delta \xi&\sim \bar \epsilon\gamma^\mu\partial_\mu \phi \,.\end{split}$$ (Note that I have neglected auxiliary fields and Hermitean cojugates.)
The key point is that $\delta\text{(scalar)}\sim\text{spinor}$ and $\delta\text{(spinor)}\sim\text{derivative of scalar}$. Hence, a vacuum configuration $\left(\phi,\xi\right)=\left(\phi_0,0\right)$ with constant $\phi_0$ is invariant under SUSY transformations.