3
$\begingroup$

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.

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ Is $Z_{AB}^\dagger=+Z_{AB}$ or $Z_{AB}^\dagger=-Z_{AB}$? $\endgroup$
    – AccidentalFourierTransform
    Commented Apr 10, 2018 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
    Commented Apr 10, 2018 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
    Commented Jan 10, 2019 at 17:53

1 Answer 1

Reset to default
3
$\begingroup$

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.

Improve this answer
$\endgroup$
1
  • $\begingroup$ 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
    Commented Apr 11, 2018 at 1:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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