# Lie Algebra for fermion fields

A key identity (e.g. when deriving BRST symmetry for gauge fields) is that:

$$[c,d]_a =f_{abc}c_b d_c$$

where $c$ and $d$ are both Fermion Fields.

How do I derive this from the lie algebra expansion $[t_a,t_b] = f_{abc}t_c$ ?

It seems obvious for Boson fields...

i.e. $[X,Y] = [X_a t_a, Y_b t_b] = X_a t_a Y_b t_b -Y_b t_b X_a t_a = X_a Y_b [t_a, t_b] = X_a Y_b f_{abc} t_c$

as $X_a$ and $Y_b$ commute with each other. But for fermion fields I thought that $c_a$ and $d_b$ anticommute so surely the equivalent calculation would lead to $[c,d]=c_a d_b\{t_a, t_b\}$ which does not equal $f_{abc}c_b d_c t_a$.

I'd be really grateful for an answer - I'm a retired person trying to teach myself Quantum Field Theory just from books and the internet - and this is really confusing me.

Alan

• @ACuriousMind The Fadeev-Popov ghosts do anti-commute since they are Grassmann valued fields. They do not obey spin-statistics theorem because they are anti-commuting spin $0$ fields. – Diracology Jul 3 '16 at 14:53

The brackets are commutators unless both variables are odd, in this case they are anti-commutators, please see footnote no. 3 in Mañes Stora and Zumino: Algebraic study of chiral anomalies . Thus in your example: $$[c, d] = c^a t_a d^b t_b + d^b t_b c^a t_a = c^a d^b (t_a t_b - t_b t_a ) = f_{ab}^c c^a d^b t_c$$