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The 4D SUSY algebra can be written as

$$\{ Q_{\alpha}^{A} , Q_{\beta}^{B \dagger} \} = 2 m \delta^{AB} \delta_{\alpha \beta} + 2 i Z^{AB} \Gamma^0_{\alpha \beta}, \tag{B.2.37} $$

in a particular reference frame. One can find this formula in the Appendix B, page 448 of Polchinski's String Theory vol.II.

I am confused with the $'i'$ before the central charge. If we do a Hermitian conjugate on both side:

$$\{ Q_{\alpha}^{A \dagger} , Q_{\beta}^{B} \} = 2 m \delta^{AB} \delta_{\alpha \beta} - 2 i Z^{AB} (\Gamma^0_{\alpha \beta})^* $$

and then exchange $(A,\alpha)$ with $(B,\beta)$, the LHS is invariant. But the RHS is

$$2 m \delta^{AB} \delta_{\alpha \beta} - 2 i Z^{BA} (\Gamma^0_{ \beta \alpha})^* = 2 m \delta^{AB} \delta_{\alpha \beta} - 2 i Z^{BA} (\Gamma^0)^{\dagger}_{ \alpha \beta}.$$

Since $Z_{AB}$ is anti-symmetric and $(\Gamma^0)^{\dagger} = -\Gamma^0$, It seems that we have the wrong sign before the central charge term:

$$2 m \delta^{AB} \delta_{\alpha \beta} - 2 i Z^{AB} (\Gamma^0)_{ \alpha \beta}.$$

I think I made a mistake but I can not figure out where is it.

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    $\begingroup$ Is $Z_{AB}^\dagger=+Z_{AB}$ or $Z_{AB}^\dagger=-Z_{AB}$? $\endgroup$ – AccidentalFourierTransform Apr 10 '18 at 14:17
  • $\begingroup$ $Z_{AB}$ is real and anti-symmetric, therefore $Z_{AB}^{\dagger} = - Z_{AB}$. But I don't think the dagger on the both side will involve the indices $A$ and $\alpha$. $\endgroup$ – JQ Skywalker Apr 10 '18 at 14:42
  • $\begingroup$ I believe that in this convention Z is imaginary and antisymmetric and therefore hermitian. In conventional field theoretic settings where the $Γ^0$s are hermitian, Z is real antisymmetric and the i is missing. But you choose the opposite, so Z must be imaginarry and hermitian, somewhat unconventionally. It is iZ which is antihermitian! The first and second term on the rhs have the same hermiticity properties. $\endgroup$ – Cosmas Zachos Jan 10 '19 at 17:53
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It is important to remember that operator order gets reversed under Hermitian conjugation: $$(ST)^{\dagger}~=~T^{\dagger}S^{\dagger}.$$ Therefore a Hermitian conjugation on the LHS of eq. (B.2.37) effectively exchanges indices $(A,\alpha)\leftrightarrow (B,\beta)$. The same should happen on the RHS. This is implemented by choosing the central charges $Z_{AB}$ to be anti-Hermitian and the gamma matrix $\Gamma^0_{\alpha\beta}$ to be Hermitian.

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  • $\begingroup$ I know that $Z_{AB}$ is anti-hermitian but I think the $\Gamma^0$ is also chosen to be anti-hermitian throughout his book, you can check that in (B.1.7a) where $\Gamma^0 = \left[ \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right] \otimes \left[ \begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array} \right] $ in 4D. $\endgroup$ – JQ Skywalker Apr 11 '18 at 1:41

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