0
$\begingroup$

Reading about Dirac's equation in higher dimensional space-times I have read that the gamma matrices are $2^{[D/2]}\times{}2^{[D/2]}$.

So, if we have $D=11$, for example, how is this formula supposed to be understood?

$\endgroup$
2
$\begingroup$

You should understand it as rounded down. For example, in $D = 3$, you should have $2^1\times 2^1$ matrices. Indeed in $D = 3$ the Dirac matrices are the familiar Pauli matrices. $$ \sigma_1 = \frac{1}{2}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \quad \sigma_2 = \frac{1}{2}\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \quad \sigma_3 = \frac{1}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$$ (you can verify that they satisfy the anticommutation relation required with respect to the Euclidean metric).

$\endgroup$
  • $\begingroup$ How is 2+3/2= 3.5 rounded down 1? $\endgroup$ – Yossarian Apr 28 '14 at 17:52
  • $\begingroup$ The $+2$ shouldn't be there. Surely in $3+1$ dimensions we do not have $16\times 16$ Dirac matrices. $\endgroup$ – Robin Ekman Apr 28 '14 at 17:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.