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I was reading the book Supersymmetry, Theory, Experiment and Cosmology by P. Binétruy, and on page 25 the author goes from

$$ 1)[Q_r,Q_t]_+ \gamma^{0}_{ts}=2\gamma^{\mu}_{rs}P_{\mu} $$ $$ 2)\mbox{Contracting with $ \gamma^{0}_{sr}$, } \sum_{r,t}[Q_r,Q_r]_{+}(\gamma^{02})_{tr} = 2Tr(\gamma^{0}\gamma^{\mu})P_{\mu} $$ $$ 3) \sum_{r}Q^{2}_{r}=4P^{0}$$ $$ 4) H=\frac{1}{4}\sum_{r}Q^{2}_{r} $$

I don't really understand how to go from stage 1 to 2 or 3 to 4, can anyone help?

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The step from 1 to 2 is: multiply by $\gamma^0$ and trace over the result.

2->3: $Tr((\gamma^0)^2)=Tr(1)=1$ on the left hand side and $2Tr(\gamma^0\gamma^\mu)P_\mu=8\eta^{0\mu}P_\mu=8P_0=8P^0$ on the right hand side. Moreover the anticommutator gives $2Q_r^2$ so you can just strip off a factor of 2 on both sides.

3->4: $P^0$ is the total energy by definition, as the right hand side is in fact mulitplied by an unit operator (as $Q$ are operators) one has $P^0 1 = H$ giving the final result.

For some magic involving the gamma matrices see the wikipedia article http://en.wikipedia.org/wiki/Gamma_matrices

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  • $\begingroup$ Am i correct in thinking that $[Q_r,Q_t]_+ = 0$ if r doesn't equal t? $\endgroup$ – user21119 Mar 1 '13 at 21:30

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