I am learning the Faddeev–Popov path integral formlism with Schwartz's QFT textbook. In the section 25.4.2 "BRST invariance", I came across the Lagrangian as: $$\mathcal{L}=-\frac{1}{4} F_{\mu \nu}^{2}+\left(D_{\mu} \phi_{i}^{\star}\right)\left(D_{\mu} \phi_{i}\right)-m^{2} \phi_{i}^{\star} \phi_{i}-\frac{1}{2 \xi}\left(\partial_{\mu} A_{\mu}\right)^{2}-\bar{c} \square c$$ And it says the equation of motion of the fermionic fields satisify: $$\square c=\square \bar{c}=0$$ I am curious and surprised with the derivative of the fermionc field(which certainly is a grassmann-valued function with respect to space-time coordinates.) But the whole Schwartz's textbook doesn't mention how to define the derivative of the grassmann-valued function with respect to certain real-valued variables(There is indeed a section about Grassmann Algebra, but it doesn't introduce how to define this!).

So my question is it correct to define the derivative as similar to ordinary calculus: $$c'(x)=\lim_{\epsilon \to 0}\frac{c(x+\epsilon)-c(x)}{\epsilon}$$ I suspect that the above "definition" is not well-defined. Could anyone give a physical-intution explanation to the derivative, or any reference to the strange derivative?


1 Answer 1


OP is asking a good question. However, the mathematical foundation of supernumber-valued fields is quite intricate, cf. e.g. this Phys.SE post. The first issue is that a Grassmann-number is an indeterminate that does not actually take a value! For a full rigorous explanation, consider to consult a textbook in supermathematics or to ask on Math.SE/MO.SE instead. The upshot is that differentiation of supernumber-valued fields do make sense when cast in the appropriate mathematical setting.

As always when considering a path integral (whose rigorous mathematical definition is an open problem in general) physicists often envision spacetime as a continuum limit of a discrete spacetime lattice. Spacetime derivatives of supernumber-valued fields then get replaced by their corresponding finite differences. Whether the continuum limit makes sense is an open problem not just for supernumber-valued field theories.

  • $\begingroup$ Thanks, do you mean for convenience we might treat the contiunous limits as the lattice finte-difference expression, such as a discrete matrix formalism. Maybe I shouldn't dig into much the details of the mathematics. $\endgroup$
    – Hawk Kou
    May 26, 2020 at 12:45
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    May 26, 2020 at 13:00

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