# Why does unbroken supersymmetry imply the absence of tachyons?

Just a quick question, same as in the title. I'm trying to understand stable D-branes.

-

## 1 Answer

In theories with unbroken supersymmetry, the energy can be written as $$E=\sum_i c_i Q_i Q_i$$ where $Q_i$ are some Hermitian supercharges and the coefficients are positive. This is a sum of squares of Hermitian operators which is why it's positively semidefinite. It can't be negative.

Tachyons obey $E^2-p^2=m^2$ for a negative $m^2$ so the 4-momentum (or $d$-momentum) is a spacelike vector, assuming it is real. For spacelike vectors, one may always change $E$ to a negative number by a Lorentz transformation, but no such states with $E\lt 0$ may exist due to the positivity above, which means tachyons can't exist at all.

Spacetime supersymmetry was of course the feature that eliminated the spacetime tachyons of bosonic string theory when it was superseded by superstring theory in the early 1970s.

-
thanks for your clear answer! –  Ryan Thorngren Jan 22 '13 at 6:05