# Mandl & Shaw page 263

In the second edition of Mandl & Shaw QFT, on page 263, below eqn (12.75) he says (I am freely paraphrasing)

$\frac{\delta}{\delta\theta(z)}$ commutes with $\theta(x_1)K(x_1,y_1)\tilde{\theta}(y_1)$ because $\theta(x_1)K(x_1,y_1)\tilde{\theta}(y_1)$ is bilinear in the Grassmann fields $\theta$ and $\tilde{\theta}$

By eqn (12.72a), $\frac{\delta}{\delta\theta(z)}$ anti-commutes with $\tilde{\theta}(y_1)$, so it would be sufficient to show that it also anti-commutes with $\theta(x_1)$. However, by eqn (12.70) they do not anti-commute. Instead we have $[\frac{\delta}{\delta\theta(z)}, \theta(x_1)]_+ = \delta^{(4)}(z - x_1)$.

What am I missing?

dextercioby
Homework Helper
They commute as Grassmann variables, i.e. the diff. operator has parity 1 and the product has parity 0, thus it doesn't matter on which of the products of 3 in the infinite product you decide to act with the derivative first. You can start from the leftmost term, or with the rightmost one, or you can squeeze it in the middle (jokingly if you can find the middle term of an infinite product :D).

it doesn't matter on which of the products of 3 in the infinite product you decide to act with the derivative first. You can start from the leftmost term, or with the rightmost one, or you can squeeze it in the middle.
I can do that by noting that the products of 3 commute. All I have to do is commute them one at a time to the left and then act on the leftmost one with the derivative. That's how I was able to derive eqn (12.76). However, that's different from saying that the derivative commutes with the product of 3. In other words, I can get

$$\frac{\delta}{\delta\theta(z)}\theta(x_1)K(x_1,y_1)\tilde{\theta}(y_1) \theta(x_2)K(x_2,y_2) \tilde{\theta}(y_2) = \frac{\delta}{\delta\theta(z)}\theta(x_2)K(x_2,y_2)\tilde{\theta}(y_2) \theta(x_1)K(x_1,y_1)\tilde{\theta}(y_1)$$

but how do I get

$$\frac{\delta}{\delta\theta(z)}\theta(x_1)K(x_1,y_1)\tilde{\theta}(y_1) \theta(x_2)K(x_2,y_2)\tilde{\theta}(y_2) = \theta(x_1)K(x_1,y_1)\tilde{\theta}(y_1)\frac{\delta}{\delta\theta(z)} \theta(x_2)K(x_2,y_2)\tilde{\theta}(y_2)$$

when $\frac{\delta}{\delta\theta(z)}$ anticommutes with $\tilde{\theta}(y_1)$ but not with $\theta(x_1)$?

Avodyne
Your 2nd equation is wrong. Is this eq in M&S? I suspect you are over-interpreting their words (I don't have the book to check), and that all they are doing is keeping track of relative minus signs.

Your 2nd equation is wrong. Is this eq in M&S? I suspect you are over-interpreting their words (I don't have the book to check), and that all they are doing is keeping track of relative minus signs.
The exact wording is:

Mandl & Shaw said:
Note that the order of the factors $\theta(x_i)K(x_i,y_i)\tilde{\theta}(y_i)$ on the right-hand side of this equation is unimportant since, being bilinear in the Grassmann fields, they commute with each other. For the same reason, they also commute with the functional differential operators $\frac{\delta}{\delta\theta(z)}$ and $\frac{\delta}{\delta\tilde{\theta}(z)}$.
I agree that my second eqn is wrong, and I also suspect that I am mis-interpreting their words. But what is the correct interpretation?

Avodyne