Ward Identity in Srednicki QFT book

1. Jul 10, 2014

wphysics

Hello guys, I am working on Ch22 "Continuous symmetries and conserved currents" of Srednicki QFT book.

I am trying to understand how to prove the Ward-Takahashi identity using path integral method, done in page 136 of Srednicki.

I understood everything up to Equation 22.22, which is
$$0 = \int \mathcal{D}\phi e^{iS} \int d^4 x \bigg[i \frac{\delta S}{\delta \phi_a (x) }\phi_{a_1}(x_1) \cdots \phi_{a_n}(x_n) + \sum_{j=1}^n \phi_{a_1}(x_1)\cdots \delta_{aa_j} \delta^4 (x-x_j) \cdots \phi_{a_n}(x_n) \bigg]\delta\phi_a (x)$$

Here, $\delta\phi_a(x)$ is arbitrary, so we can drop this and integration over x. So, what we get is
$$0 = \int \mathcal{D}\phi e^{iS}\bigg[i \frac{\delta S}{\delta \phi_a (x) }\phi_{a_1}(x_1) \cdots \phi_{a_n}(x_n) + \sum_{j=1}^n \phi_{a_1}(x_1)\cdots \delta_{aa_j} \delta^4 (x-x_j) \cdots \phi_{a_n}(x_n) \bigg]$$

Let's consider the free real scalar field and n=1. Then, $\frac{\delta S}{\delta \phi_a (x) }$ is just $(\partial_x^2-m^2)\phi(x)$. So, the above equation becomes
$$0 = \int \mathcal{D}\phi e^{iS}\bigg[i(\partial_x^2-m^2)\phi(x) \phi(x_1) + \delta^4(x-x_1) \bigg]$$
So, this is equivalent to
$$(-\partial_x^2 + m^2) i <0|T\phi(x)\phi(x_1) |0> = \delta^4(x-x_1)$$
Here comes my question.
The author argued that the Klein-Gordon wave operator should sit outside the time-ordered product is clear from the path integral form of Eq(22.22).
I guess that this is due to that we can pull the Klein-Gordon wave operator out of path integral. Is this guess right? Even though this is right, I would like to have clearer explanation for this.

My second question is what kinds of path integral I need to calculate
$$i <0|T(\partial_x^2 - m^2) \phi(x)\phi(x_1) |0>$$
where the Klein-Gordon wave operator is inside of time ordered operator.
Because if my guess is right, the Klein-Gordon wave operator can always be pull out of path integral, so it seems that there is no way to calculate this from the path integral method.
Here, I assume that $\phi(x)$ is not a solution of classical equation, so the Klein-Gordon wave operator acting on this field can be non zero.