Whenever I learn about anything involving fermions and the path integral, I get confused about Grassmann numbers. I'm currently following Weigand's notes, specifically the section on BRST symmetry. The BRST symmetry transformation is defined by $$\delta_\epsilon \Phi = \epsilon\, S \Phi$$ where $\epsilon$ is a Grassmann parameter, $\Phi$ is a field, and $S\Phi$ obeys $$S A_\mu = - D_\mu c = - (\partial_\mu c + i g [A_\mu, c]), \quad Sc = \frac{i}{2} g f^{abc} c^b c^c t^a, \quad S \bar{c} = - B, \quad SB = 0.$$ First off, what is the Grassmann parameter $\epsilon$ here? I get the feeling that for the manipulations to go through, it must be independent of the Grassmann numbers $c(x)$ and $\bar{c}(x)$ for any $x$. But that means that to define the BRST transformation at all, we must enlarge the space of Grassmann numbers in our theory, which doesn't make sense to me.

So far we've been working semiclassically. Next, Weigand quantizes the theory, giving a BRST charge operator $\hat{Q}$ satisfying $$[\epsilon \hat{Q}, \hat{X}] = i \delta_\epsilon \hat{X}.$$ Now I don't understand what $\epsilon$ is in this context. If it's still Grassmann, then we now have a Hilbert space where the scalars include Grassmann numbers, which doesn't make any mathematical sense to me. Alternatively, is $\epsilon$ supposed to be an operator that anticommutes with all the other operators? In that case, why doesn't it have a hat over it?

I'd really appreciate clarification on what $\epsilon$ means in both these contexts!


1 Answer 1


That's a very good question. At the pragmatic level:

  • The field of complex numbers $\mathbb{C}$ has been replaced by the set of supernumbers.

  • Hilbert spaces and operators are not just (conjugate) linear wrt. $\mathbb{C}$ but are $\mathbb{Z}_2$-graded (conjugate) linear wrt. supernumbers.

  • The BRST parameter $\epsilon$ is an independent infinitesimal Grassmann-odd supernumber (aka. a Grassmann number).

  • It should be stressed that (soul-valued) supernumbers belong to an intermediate mathematical construct, and that they cannot be part of/have been integrated out/have been differentiated out/ in physically measurable quantities at the end of the calculation. This fact should be fairly obvious in the path integral formalism. In the operator formalism, if a creation operator creates a Grassmann-odd variable, then the corresponding annihilation operator behaves like a Grassmann-odd differentiation operator (wrt. that variable), or vice-versa.

For more details, see also e.g. this & this Phys.SE posts; this & this Math.SE posts, and links therein.

  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Mar 24, 2018 at 10:26
  • $\begingroup$ Do you have a reference that explains more about your final point, "In the operator formalism..."? $\endgroup$
    – knzhou
    Mar 24, 2018 at 11:42

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