BPS state and annihilation of SUSY charges As we all know that anticommutator of one set of supercharges in massive extended supersymmetry is something like $$\{b_\alpha, b_\beta^\dagger \} = \delta_{\alpha \beta} (M-\sqrt{2} Z).$$
My question is that everyone says that it is obvious that BPS states annihilate half of the supersymmetric charges. Why is this so? It may be trivial but I don't know how. May be it is just because the annihilation operator acts on the lowest energy state and annihilate BPS States? But why only half? Can anyone clear that up.
 A: Take for example $\mathcal{N}=2$ supersymmetry. The algebra, including a central charge $Z$, is given by
$$\{Q_\alpha^a,Q^\dagger_{\dot{\alpha}b}\}=2\sigma^\mu_{\alpha\dot{\alpha}}P_\mu \delta^a_b$$
$$\{Q_\alpha^a,Q_\beta^b\}=2^{3/2}\epsilon_{\alpha\beta}\epsilon^{ab}Z$$
$$\{Q^\dagger_{\dot{\alpha} a},Q^\dagger_{\dot{\beta} b}\}=2^{3/2}\epsilon_{\dot{\alpha}\dot{\beta}}\epsilon_{ab}Z.$$
We can now define
$$a_\alpha=\frac12\left(Q_\alpha^1+\epsilon_{\alpha\beta}\left(Q_\beta^2\right)^\dagger\right)$$
and
$$b_\alpha=\frac12\left(Q_\alpha^1-\epsilon_{\alpha\beta}\left(Q_\beta^2\right)^\dagger\right),$$
which reduces the algebra to 
$$\{a_\alpha,a_\beta^\dagger\}=\delta_{\alpha\beta}(M+\sqrt{2}Z)$$
and
$$\{b_\alpha,b_\beta^\dagger\}=\delta_{\alpha\beta}(M-\sqrt{2}Z).$$
A BPS-state satisfies $M=\sqrt{2}Z$, hence the second part of the algebra reduces to 
$$\{b_\alpha,b_\beta^\dagger\}=0.$$
This tells us that the operators in the second half of the algebra, which now vanishes, only generate states of zero norm. This is how you should understand the statement you were asking about. 
