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  Fermion propagator is not a Grassmann-odd object?

+ 0 like - 0 dislike

Is the following differentiation correct: $$ \frac{\delta}{\delta\eta\left(z\right)}\int d^{4}yS_{F}\left(z-y\right)\eta\left(y\right) = S_F\left(z-z\right)$$

where $\eta$ is a Grassmann-valued field and $S_F$ is the Fermion propagator, or is the result actually with a minus sign?

This post imported from StackExchange Physics at 2014-04-13 14:30 (UCT), posted by SE-user Psycho_pr
asked Apr 11, 2014 in Theoretical Physics by PPR (135 points) [ no revision ]

1 Answer

+ 2 like - 0 dislike

The bounds of the integral have no dependence on any of the variables, and hence we may move the differential operator into the integrand,

$$\frac{\delta}{\delta \eta (z)} \int \mathrm{d}^4 y \, S_F (z-y) \eta(y) = \int \mathrm{d}^4 y \, S_F (z-y) \delta^{(4)}(z-y)$$

Evaluating the integral using the standard delta distribution identity, we obtain your result, namely $S_F(z-z)$. In this case, the final answer does not pick a minus sign, even though $\eta$ is Grassmann-valued. See Peskin and Schroeder's text on QFT for a summary of Berezin/Grassmann integration.

This post imported from StackExchange Physics at 2014-04-13 14:30 (UCT), posted by SE-user JamalS
answered Apr 11, 2014 by JamalS (895 points) [ no revision ]

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