# BPS state

In theoretical physics, BPS states are massive representations of an extended supersymmetry algebra with mass equal to the supersymmetry central charge Z. Quantum mechanically, if the supersymmetry is not broken, the mass is exactly equal to the modulus of Z. Their importance arises as the multiplets are shorter than for generic massive representations, the states are stable and the mass formula is exact.

## d=4 N=2

The generators for the odd part of the superalgebra have relations:[1]

\begin{align} \{Q_\alpha^A , \bar{Q}_{\dot{\beta} B} \} & = 2 \sigma_{\alpha \dot{\beta}}^m P_m \delta^A_B\\ \{Q_\alpha^A , Q_\beta^B \} & = 2 \epsilon_{\alpha \beta} \epsilon^{A B} \bar{Z}\\ \{ \bar{Q}_{\dot{\alpha} A} , \bar{Q}_{\dot{\beta} B} \} & = -2 \epsilon_{\dot{\alpha} \dot{\beta}} \epsilon_{AB} Z\\ \end{align}

where: $\alpha \dot{\beta}$ are the Lorentz group indices, A and B are R symmetry indices.

Take linear combinations of the above generators as follows:

\begin{align} R_\alpha^A & = \xi^{-1} Q_\alpha^A + \xi \sigma_{\alpha \dot{\beta}}^0 \bar{Q}^{\dot{\beta} B}\\ T_\alpha^A & = \xi^{-1} Q_\alpha^A - \xi \sigma_{\alpha \dot{\beta}}^0 \bar{Q}^{\dot{\beta} B}\\ \end{align}

Consider a state ψ which has 4 momentum $(M,0,0,0)$. Applying the following operator to this state gives:

\begin{align} (R_1^1 + (R_1^1)^\dagger )^2 \psi & = 4 ( M + Re(Z\xi^{2}) ) \psi\\ \end{align}

But because this is the square of a Hermitian operator, the right hand side coefficient must be positive for all $\xi$.

In particular the strongest result from this is

\begin{align} M \geq |Z|\\ \end{align}

## Example applications

• Supersymmetric black hole entropies[2]