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.

Many thanks in advance


  • $\begingroup$ @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. $\endgroup$ – Diracology Jul 3 '16 at 14:53

When dealing with theories containing both commuting and anti-commuting (even and odd) variables, physicists often use the square bracket notation to denote both commutators and anti-commutators according to the following rule:

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: $$[c, d] = c^a t_a d^b t_b + d^b t_b c^a t_a = c^a d^b (t_a t_b - t_b t_a ) = f_{ab}^c c^a d^b t_c$$

  • $\begingroup$ David - many thanks for your answer. Just to confirm I understand.... $\endgroup$ – Alan Hardgrave Jul 4 '16 at 15:46
  • $\begingroup$ Fermion ghost fields are part of the Lie algebra for the gauge field so can be expanded in terms of the lie algebra basis {t}. What makes them fermions rather than bosons is that the coefficients are grassman numbers rather than normal numbers - which effectively turns commutation relations into anti-commutation ones. The convention re brackets that you describe above, then means that most formula involving square brackets are then the same irrespective of whether they involve fermions or bosons. Is that right? $\endgroup$ – Alan Hardgrave Jul 4 '16 at 16:15
  • $\begingroup$ @Alan Hardgrave you are correct. We can use the same notation of square brackets and all we need to remember is the degree of the fields inside the bracket. $\endgroup$ – David Bar Moshe Jul 5 '16 at 10:28

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