In equation 16.47 in Peskin & Schroeder, it is claimed that $$ -\frac{1}{2}g^2f^{abc}f^{cde}\left(A_{\mu}\,^{b}c^{d}c^{e}+A_{\mu}\,^{d}c^{e}c^{b}+A_{\mu}\,^{e}c^{b}c^{d}\right) ~=~ 0 \tag{16.47}$$

using the Jacobi identity $$ f^{ade}f^{bcd}+f^{bde}f^{cad}+f^{cde}f^{abd}~=~0, \tag{15.70}$$

where $A$ is the gauge field and $c$ is the ghost Grassmann field.

I was trying to prove this assertion and failed.

This is what I've tried:

1) Write $ f^{abc}f^{cde}=-f^{dbc}f^{cea}-f^{ebc}f^{cad} $ using the given Jacobi identity.

2) Relabeling some indices and using the anti-commutation of the ghost fields I am able to rewrite the expression as $$ +g^2f^{dbc}f^{cea}\left(A_{\mu}\,^{b}c^{d}c^{e}+A_{\mu}\,^{d}c^{e}c^{b}+A_{\mu}\,^{e}c^{b}c^{d}\right) $$

Now I am stuck.



  1. The left-hand side $$-\frac{1}{2}g^2f^{abc}f^{cde}\left(A_{\mu}^{b}c^{d}c^{e}+{\rm cycl}(b,d,e)\right)$$ of eq. (16.47) can be relabelled as $$-\frac{1}{2}g^2\left( f^{abc}f^{cde}+{\rm cycl}(b,d,e)\right)A_{\mu}^{b}c^{d}c^{e}.$$

  2. P&S assume that the structure constants $f^{abc}$ are totally antisymmetric, cf. text below eq. (15.79).


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