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
[tex]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) [/tex]
Here, [itex] \delta\phi_a(x) [/itex] is arbitrary, so we can drop this and integration over x. So, what we get is
[tex]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][/tex]
Let's consider the free real scalar field and n=1. Then, [itex]\frac{\delta S}{\delta \phi_a (x) } [/itex] is just [itex](\partial_x^2-m^2)\phi(x) [/itex]. So, the above equation becomes
[tex]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][/tex]
So, this is equivalent to
[tex](-\partial_x^2 + m^2) i <0|T\phi(x)\phi(x_1) |0> = \delta^4(x-x_1) [/tex]
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
[tex]i <0|T(\partial_x^2 - m^2) \phi(x)\phi(x_1) |0> [/tex]
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 [itex]\phi(x) [/itex] 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.
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
[tex]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) [/tex]
Here, [itex] \delta\phi_a(x) [/itex] is arbitrary, so we can drop this and integration over x. So, what we get is
[tex]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][/tex]
Let's consider the free real scalar field and n=1. Then, [itex]\frac{\delta S}{\delta \phi_a (x) } [/itex] is just [itex](\partial_x^2-m^2)\phi(x) [/itex]. So, the above equation becomes
[tex]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][/tex]
So, this is equivalent to
[tex](-\partial_x^2 + m^2) i <0|T\phi(x)\phi(x_1) |0> = \delta^4(x-x_1) [/tex]
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
[tex]i <0|T(\partial_x^2 - m^2) \phi(x)\phi(x_1) |0> [/tex]
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 [itex]\phi(x) [/itex] 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.
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