ismaili
Jul28-08, 06:36 AM
I would like to derive the explicit formula of the BRST charge.
http://en.wikipedia.org/wiki/BRST_formalism
(Bottom of Wiki link, I copy the formula here)
Q = c^i\left(L_i - \frac{1}{2}f_{ij}{}^kb_jc_k\right)
where c is the ghost field, and b is the antighost field, L_i is the gauge group generator.
Actually, in wiki's article, right above the formula of BRST charge, there is a Lagrangian. I tried to use Noether theorem to calculate the charge, but in vain. J^\mu_a\sim \frac{\partial\mathcal{L}}{\partial\psi_{,\mu}}\delta\psi_a, replacing the \psi with the ghost c, I ends up with
J^0 = \dot{b}\,\delta c = \dot{b}\left(-\frac{1}{2}f_{ij}{}^kc^ic^j\right) whose volume integral looks different as the correct answer.
In there anybody who can help me or give me some hints?
Many thanks!
http://en.wikipedia.org/wiki/BRST_formalism
(Bottom of Wiki link, I copy the formula here)
Q = c^i\left(L_i - \frac{1}{2}f_{ij}{}^kb_jc_k\right)
where c is the ghost field, and b is the antighost field, L_i is the gauge group generator.
Actually, in wiki's article, right above the formula of BRST charge, there is a Lagrangian. I tried to use Noether theorem to calculate the charge, but in vain. J^\mu_a\sim \frac{\partial\mathcal{L}}{\partial\psi_{,\mu}}\delta\psi_a, replacing the \psi with the ghost c, I ends up with
J^0 = \dot{b}\,\delta c = \dot{b}\left(-\frac{1}{2}f_{ij}{}^kc^ic^j\right) whose volume integral looks different as the correct answer.
In there anybody who can help me or give me some hints?
Many thanks!