Ward Identity in Srednicki QFT book

[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.

Thank you for you guys answer in advance.

 

[eljose79]

Hello! Thank you for sharing your progress and questions on your current work. I am also familiar with the Ward-Takahashi identity and its proof using the path integral method, and I would be happy to provide some clarification and additional information on your questions.

Firstly, your understanding of the first part of the proof is correct. The reason why we can drop the term \delta\phi_a(x) and integrate over x is because it is arbitrary and does not affect the integral. This is a common technique in path integral calculations, where we can manipulate the integrand as long as we do not change the boundary conditions or the overall value of the integral.

To answer your first question, yes, you are correct in your guess that the reason why the Klein-Gordon wave operator can be pulled out of the path integral is because of its linearity. This property allows us to manipulate it and pull it out of the integral without changing the overall value. This is also why we can calculate the path integral of the Klein-Gordon operator inside the time-ordered product, as it can be pulled out and treated as a constant factor.

For your second question, you are correct in assuming that there is no way to calculate the path integral of the Klein-Gordon operator inside the time-ordered product. This is because, as you mentioned, the Klein-Gordon operator acting on a non-classical field can result in a non-zero value. Therefore, we cannot simply pull it out of the path integral and treat it as a constant factor. In these cases, we need to use other methods, such as perturbation theory, to calculate the path integral.

I hope this answers your questions and helps clarify your understanding of the Ward-Takahashi identity proof using the path integral method. If you have any further questions or need any additional clarification, please don't hesitate to ask. Good luck with your work!