# Question on derivation of Ward identity

I'm currently reading these notes about the Ward identity (pages 259 - 261). I will repeat some of the steps to make the question self-contained.

Let us consider a local transformation on the field $\phi$: $$\phi(x) \rightarrow \phi'(x) = \phi(x) + \delta \phi(x) \tag{1}$$ where: $$\delta \phi(x) = i \epsilon^a(x) t^a \phi(x)$$ where $t^a$ are the generators of the transformation and $\epsilon$ is a the space-time dependent parameter characterizing the field transformation. Then, the Noether current is given by: $$j_\mu^a = i \frac{\partial \mathcal{L}}{\partial \left(\partial_\mu \phi \right)} t^a \phi$$ and the variation of the action is: $$\delta S = \int \mathrm{d}^4 x \; j_\mu^a \partial^\mu \epsilon^a$$ provided $\delta S = 0$ for global transformations.

Let us now consider the usual generating functional:

$$Z[J] = \int \mathcal{D}\phi \; \exp\left(i S \left[\phi, \partial_\mu \phi \right] + i \int\limits \mathrm{d}^4 x \; J \phi \right)$$

If we subsequently perform the (local) change of variables (see equation $(1)$), and assume that the integration measure is invariant, then we get:

$$Z[J] = \int \mathcal{D}\phi \; \exp\left(i S \left[\phi, \partial_\mu \phi \right] + i \delta S \left[\phi, \partial_\mu \phi \right] + i \int\limits \mathrm{d}^4 x \; J \phi - \int\limits \mathrm{d}^4 x \; J \epsilon^a t^a \phi \right)$$ $$= \int \mathcal{D}\phi \; \exp\left(i S \left[\phi, \partial_\mu \phi \right] + i \int \mathrm{d}^4 x \; j_\mu^a \partial^\mu \epsilon^a + i \int\limits \mathrm{d}^4 x \; J \phi - \int\limits \mathrm{d}^4 x \; J \epsilon^a t^a \phi \right) \tag{2}$$

This is equation (10.170) in the aforementioned notes (up to a minus sign). Now, according to the same notes, we can expand the above equation at the first order in $\epsilon^a$ as follows: $$Z[J] = \int \mathcal{D}\phi \; \exp\left(i S \left[\phi, \partial_\mu \phi \right] + i \int\limits \mathrm{d}^4 x \; J \phi \right)\left(1 + i \int \mathrm{d}^4 x \; j_\mu^a \partial^\mu \epsilon^a - \int\limits \mathrm{d}^4 x \; J \epsilon^a t^a \phi \right) \tag{3}$$ Now, my question is:

To get from equation $(2)$ to $(3)$: why can we just use normal Taylor series to get expand the exponential? Shouldn't we be using the Baker–Campbell–Hausdorff formula?

Of course, if we are only considering Abelian gauge transformations, then we can just use the normal Taylor series. However, they are not mentioning this and now I am confused.

• It seems your formulas are not entirely consistent with those in the link you gave, everything on the exponential is scalar, not matrix, look carefully at equation 10.170, all indices, including the matrix entry index of $t$, are contracted. – Jia Yiyang Mar 19 '14 at 1:52
• @JiaYiyang hmm.. I'm not sure what you mean. In the formula I wrote, every thing is contracted as well as far as I'm aware. Which term according to you is not contracted? Note the index $a$ is a `color' index, so it does not matter whether I write $j^a \epsilon^a$ or $j_a \epsilon^a$. – Hunter Mar 19 '14 at 1:56
• I'm saying for example you last term should be $J_i\epsilon^a t^a_{ij}\phi_j$, which is just a scalar, since everything is scalar on exponential, surely it will be a simple Taylor expansion. – Jia Yiyang Mar 19 '14 at 2:02
• @JiaYiyang aha, that makes sense, thanks! Now I feel silly :). If you want to, you can post it as an answer, and I will accept. – Hunter Mar 19 '14 at 2:05
• Minor comment to the post (v4): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. – Qmechanic Oct 10 '14 at 21:55

The confusion possibly comes from the casual notation, for example the last term in equation (3) in its full form ought to be $J_i\epsilon^a t^a_{ij}\phi_j$, which is just a number; while in the original notation $J \epsilon^a t^a \phi$ it might lead you to think it is a matrix because of the presence of $t^a$. One quick way to check the mistake is to notice that this term is added to the action $S$ in equation (2), and action must be scalar, hence the term also has to be scalar.