- #1
ismaili
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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)
[tex]Q = c^i\left(L_i - \frac{1}{2}f_{ij}{}^kb_jc_k\right)[/tex]
where [tex]c[/tex] is the ghost field, and [tex]b[/tex] is the antighost field, [tex]L_i[/tex] 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. [tex]J^\mu_a\sim \frac{\partial\mathcal{L}}{\partial\psi_{,\mu}}\delta\psi_a[/tex], replacing the [tex]\psi[/tex] with the ghost [tex]c[/tex], I ends up with
[tex]J^0 = \dot{b}\,\delta c = \dot{b}\left(-\frac{1}{2}f_{ij}{}^kc^ic^j\right)[/tex] 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)
[tex]Q = c^i\left(L_i - \frac{1}{2}f_{ij}{}^kb_jc_k\right)[/tex]
where [tex]c[/tex] is the ghost field, and [tex]b[/tex] is the antighost field, [tex]L_i[/tex] 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. [tex]J^\mu_a\sim \frac{\partial\mathcal{L}}{\partial\psi_{,\mu}}\delta\psi_a[/tex], replacing the [tex]\psi[/tex] with the ghost [tex]c[/tex], I ends up with
[tex]J^0 = \dot{b}\,\delta c = \dot{b}\left(-\frac{1}{2}f_{ij}{}^kc^ic^j\right)[/tex] whose volume integral looks different as the correct answer.
In there anybody who can help me or give me some hints?
Many thanks!
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