Mandl & Shaw QFT: Commutation Rules on Page 263

[Vic Sandler]

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]

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).

 

[Vic Sandler]

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.

 

[Vic Sandler]

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:

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]

This is badly worded by M&S; it is not correct as written. I believe that all they are trying to say that if you move a Grassmann derivative through an even number of Grassmann variables, then the overall sign is +.

 

[andrien]

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?

yes,you are misinterpreting the words.He is saying that two terms like [θ(x₁)k(x₁,y₁)θ^-(y₁)] and [θ(x₂)k(x₂,y₂)θ^-(y₂)] are bilinear(and so on).So they commute with each other i.e. you can write them in reverse order because every θ will pass through even number of θ's so there will not be any change of sign.