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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

$$P^{\mu}|\Omega\rangle=p^{0}\delta^{\mu}_{0}|\Omega\rangle$$

If we lorentz transform this equation with the unitary operator $U(\Lambda)$, we find that a new state $U(\Lambda)|\Omega\rangle$ solves the eigenvalue equation:

$$P^{\mu}U(\Lambda)|\Omega\rangle=(\Lambda^{-1})^{\mu}_0p^0U(\Lambda)|\Omega\rangle$$

Since $U(\Lambda)P^{\mu}U^{-1}(\Lambda)=\Lambda^{\mu}_{\nu}P^{\nu}$.

Therefore we have a whole family of vacua which are orthogonal and related by a lorentz transformation.

Is there something I am missing here? Is this even a bad thing?

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No, Lorentz symmetry is not broken if SUSY is broken. All you have to do is add a constant to the energy; then the four-momentum of the vacuum is zero, as it must be. This is a completely standard thing to do. For instance, it's how we subtract out the divergent vacuum energy contribution around the second week of a first quantum field theory course.

I can hear you complaining that this messes up the SUSY algebra since $\{Q, Q\} \sim H$, but who cares? The fact that SUSY is broken means there does not exist a set of operators satisfying the SUSY algebra and annihilating the vacuum. Now forget about SUSY; does there exist a set of operators satisfying the Poincare algebra and annihilating the vacuum? Yes, by adding a constant to $H$. So Lorentz symmetry is not broken here.

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  • $\begingroup$ That makes so much sense! $\endgroup$
    – fewfew4
    Apr 4, 2019 at 23:37

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