# SUSY vacuum has 0 energy?

This is related to Modern Supersymmetry: Dynamics and Duality by Terning.

Consider $$N=1$$ SUSY. $$\{Q_a,\bar{Q}_{\dot{a}}\}=2\sigma_{a\dot{a}}^\mu P_\mu$$. Sum over $$a=\dot{a}=1,2$$ and this yields $$4P^0=\sum_{a=\dot{a}=1,2}\{Q_a,\bar{Q}_{\dot{a}}\}$$.

Suppose vacuum $$|\Omega>$$ is supersymmetric.(According to the book, $$Q_a|\Omega>=0$$ pg 6 equation (1.24). The book did not say $$\bar{Q}_{\dot{a}}|\Omega>=0$$.) Here I treated $$\bar{Q}_{\dot{a}}$$ as creation operator by boosting to the rest frame and I am assuming massive states.

I compute $$<\Omega|4P^0|\Omega>=\sum_{a=\dot{a}=1,2}<\Omega|Q_a\bar{Q}_{\dot{a}}|\Omega>$$ where the other part drops out by $$Q_a|\Omega>=0$$. The computation can be further simplified. One can check $$\{Q_2,\bar{Q}_2\}=0$$ by rest frame and compute this anti-commutator. Hence $$<\Omega|4P^0|\Omega>=<\Omega|Q_1\bar{Q}_{\dot{1}}|\Omega>$$. And I still do not see this vanishes.

$$\textbf{Q:}$$ I do not have SUSY vacuum energy $$0$$ here. If I have $$\bar{Q}_{\dot{a}}|\Omega>=0$$, it goes through but $$\bar{Q}_{\dot{a}}$$ is creation operator for massive fermionic state. Where are mistakes with thought process?

• The vacuum is invariant under all supersymmetries involved, and so it is, in fact, annihilated by all four Q s. Jan 2 '19 at 1:40

## 1 Answer

The definition of the vacuum in $$N=2$$ SUSY is from

$$| \Omega \rangle = Q_1 Q_2|m, j\rangle.$$

where $$m$$ is the mass eigenvalue of the casimir $$P^2$$ and the $$j$$s are the eigenvalues of the casimir $$J^2$$ where

$$J_i = S_i - \frac{1}{4m} \bar{Q}\bar{\sigma}_iQ.$$

From the SUSY algebra one can immediately see that

$$Q_i \Omega \rangle = 0 \qquad i=1,2$$

and that if we act with the $$\bar{Q}$$s on the clifford vacuum that

\begin{align*} \{|\Omega \rangle, \bar{Q}_1 |\Omega\rangle, \bar{Q}_2 |\Omega\rangle, \bar{Q}_2\bar{Q}_1 |\Omega \rangle\} \end{align*}

are the only irreducible representations (for the massive case).