Is Lorentz symmetry broken if SUSY is broken?
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I have seemingly convinced myself that the entire Poincare group is spontaneously broken if one of the supersymmetric charges is spontaneously broken. We know that if one of the supersymmetric charges is spontaneously broken, then a vacuum with zero three-momentum MUST have a nonzero energy. There is no way to re-scale the Hamiltonian since the supersymmetry algebra provides an absolute scale. Let's suppose the vacuum is an eigenstate of $P^{mu}$ , then we have $$P^{mu}|Omegarangle=p^{0}delta^{mu}_{0}|Omegarangle$$ If we lorentz transform this equation with the unitary operator $U(Lambda)$ , we find that a new state $U(Lambda)|Omegarangle$ solves the eigenvalue equation: $$P^{mu}U(Lambda)|Omegarangle=(Lambda^{-1})^{mu}_0p^0U(Lambda)|Omegarangle$$ Since $U(Lambda)P^{mu}U^{-1}(Lambda)=...
