Ward Takahashi identities from Z invariance

I'm trying to get Ward-Takahashi identities using the approach used in Ryder's book (pages 263-266). I like that he starts from demanding gauge invariance of Z in a explicit way and them explores the consequences of that to functional generators of vertex functions. But the actual calculation is bugging me out.

The author seems oblivious to the fact that the fermionic fields and sources ($\psi$ and $\eta$) are Grassmann variables and keeps commuting them out with no regard for my sanity. For instance, equation (7.102) has a term:

$$i e (\bar{\eta}\psi-\bar{\psi}\eta)$$

that promply becomes (exchanging the fields by ${1\over i}$ times the derivatives on the sources, acting on Z to the right):

$$e (\bar{\eta}{\delta \over \delta \bar{\eta} }-\eta{\delta \over \delta \eta })$$

In my opinion that should be:

$$e (\bar{\eta}{\delta \over \delta \bar{\eta} }+\eta{\delta \over \delta \eta })$$

This one has no consequences because he commutes them again right after. But when I try to calculate it being careful with the Grassmann variables, I can never get the right signs in (7.111). I'm specially troubled by the derivatives below (this is what I'm getting, but one of them should have a different sign in order to get the right WT identities):

$${\delta \over \delta \bar{\psi}(x_1) } {\delta \over \delta \psi(y_1) } \left[{\delta \Gamma \over \delta \psi(x) }\psi(x)\right]_{\psi=\bar{\psi}=0}=-\delta^4(x-y_1){\delta^2 \Gamma \over \delta \bar{\psi}(x_1) \delta\psi(x) }$$

$${\delta \over \delta \bar{\psi}(x_1) } {\delta \over \delta \psi(y_1) } \left[\bar{\psi}(x) {\delta \Gamma \over \delta \bar{\psi}(x) } \right]_{\psi=\bar{\psi}=0} =\\=-\delta^4(x-x_1){\delta^2 \Gamma \over \delta \psi(y_1) \delta\bar{\psi}(x) } = \delta^4(x-x_1){\delta^2 \Gamma \over \delta \bar{\psi}(x) \delta\psi(y_1) }$$

Does anybody ever did this calculation in detail and has some pointers? Are there any other references that follow this same approach?

EDIT: Just a shameless bump: I still looking for some light on this. Any reference on where this is done in detail would help.

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Ryder knows the answer and is making mistakes in intermediate steps. It is hard to answer without looking at the book. –  Ron Maimon Sep 2 '12 at 22:02
Just be sure: so I'm right about this commutation business being all wrong? Also: is the link I posted to the book itself (on Google books) working? –  Forever_a_Newcomer Sep 3 '12 at 0:40
Yes, the commutations are wrong. The link works. I just didn't rederive the WT identity. –  Ron Maimon Sep 3 '12 at 2:24

Disclaimer: I don't have a copy of Ryder's book and don't know what conventions it uses.

But: it's not true that products of fermions always anticommute. For instance, suppose we are studying two-component spinors, $\psi_\alpha$, where $\alpha$ takes values 1 or 2. Now, it's true that $\psi_1$ is a Grassmann variable. But when taking the product of spinors, we usually mean something like $\psi \chi = \psi^\alpha \chi_\alpha$, where raising an index is defined as $\psi^\alpha = \epsilon^{\alpha \beta} \psi_\beta$. Here $\epsilon$ is the antisymmetric symbol, $\epsilon^{12} = -\epsilon^{21} = 1$. In other words:

$$\psi \chi = \psi_1 \chi_2 - \psi_2 \chi_1 = -\chi_2 \psi_1 + \chi_1 \psi_2 = \chi \psi$$

In the middle step, we used the Grassmann nature of the variables, and the other steps are just the definition of the spinor product. So, two-component spinors commute (when multiplication is the product defined using the $\epsilon$ symbol) even though their components are anticommuting variables.

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I understand what you are saying, that's what you do when writing supersymmetric models in terms of 2-spinors. But these are usual 4-component Dirac spinors and their sources. I never seen their scalar product defined in any other way than just the sum of the product of their components. In fact this seems to indicate just that. –  Forever_a_Newcomer Aug 30 '12 at 13:26

I remember going through this same section and having concerns about the signs. I think Ryder is probably wrong here.

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