Bogomol'nyi–Prasad–Sommerfield state

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

${\displaystyle {\begin{aligned}\{Q_{\alpha }^{A},{\bar {Q}}_{{\dot {\beta }}B}\}&=2\sigma _{\alpha {\dot {\beta }}}^{m}P_{m}\delta _{B}^{A}\\\{Q_{\alpha }^{A},Q_{\beta }^{B}\}&=2\epsilon _{\alpha \beta }\epsilon ^{AB}{\bar {Z}}\\\{{\bar {Q}}_{{\dot {\alpha }}A},{\bar {Q}}_{{\dot {\beta }}B}\}&=-2\epsilon _{{\dot {\alpha }}{\dot {\beta }}}\epsilon _{AB}Z\\\end{aligned}}}$

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

Take linear combinations of the above generators as follows:

${\displaystyle {\begin{aligned}R_{\alpha }^{A}&=\xi ^{-1}Q_{\alpha }^{A}+\xi \sigma _{\alpha {\dot {\beta }}}^{0}{\bar {Q}}^{{\dot {\beta }}B}\\T_{\alpha }^{A}&=\xi ^{-1}Q_{\alpha }^{A}-\xi \sigma _{\alpha {\dot {\beta }}}^{0}{\bar {Q}}^{{\dot {\beta }}B}\\\end{aligned}}}$

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

${\displaystyle {\begin{aligned}(R_{1}^{1}+(R_{1}^{1})^{\dagger })^{2}\psi &=4(M+Re(Z\xi ^{2}))\psi \\\end{aligned}}}$

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

In particular the strongest result from this is

${\displaystyle {\begin{aligned}M\geq |Z|\\\end{aligned}}}$

• Supersymmetric black hole entropies[2]

• Bogomol'nyi–Prasad–Sommerfield bound
• Short supermultiplet
• Wall-crossing