- #1
Vic Sandler
- 4
- 3
In the second edition of Mandl & Shaw QFT, on page 263, below eqn (12.75) he says (I am freely paraphrasing)
[itex]\frac{\delta}{\delta\theta(z)}[/itex] commutes with [itex]\theta(x_1)K(x_1,y_1)\tilde{\theta}(y_1)[/itex] because [itex]\theta(x_1)K(x_1,y_1)\tilde{\theta}(y_1)[/itex] is bilinear in the Grassmann fields [itex]\theta[/itex] and [itex]\tilde{\theta}[/itex]
By eqn (12.72a), [itex]\frac{\delta}{\delta\theta(z)}[/itex] anti-commutes with [itex]\tilde{\theta}(y_1)[/itex], so it would be sufficient to show that it also anti-commutes with [itex]\theta(x_1)[/itex]. However, by eqn (12.70) they do not anti-commute. Instead we have [itex][\frac{\delta}{\delta\theta(z)}, \theta(x_1)]_+ = \delta^{(4)}(z - x_1)[/itex].
What am I missing?
[itex]\frac{\delta}{\delta\theta(z)}[/itex] commutes with [itex]\theta(x_1)K(x_1,y_1)\tilde{\theta}(y_1)[/itex] because [itex]\theta(x_1)K(x_1,y_1)\tilde{\theta}(y_1)[/itex] is bilinear in the Grassmann fields [itex]\theta[/itex] and [itex]\tilde{\theta}[/itex]
By eqn (12.72a), [itex]\frac{\delta}{\delta\theta(z)}[/itex] anti-commutes with [itex]\tilde{\theta}(y_1)[/itex], so it would be sufficient to show that it also anti-commutes with [itex]\theta(x_1)[/itex]. However, by eqn (12.70) they do not anti-commute. Instead we have [itex][\frac{\delta}{\delta\theta(z)}, \theta(x_1)]_+ = \delta^{(4)}(z - x_1)[/itex].
What am I missing?