- #1
physicus
- 55
- 3
Homework Statement
I would like to know how to get from eq. (67.3) to (67.4) in Srednicki's book on QFT. The problem is the following:
Given the LSZ formula for scalar fields
[itex]\langle f|i \rangle = i \int d^{4}x_1e^{ik_1x_1}(\partial^{2}+m^{2})\ldots \langle 0|T\phi(x_1)\ldots|0\rangle[/itex]
This is supposed to be equivalent to:
[itex]\langle f|i \rangle = \lim_{k_i\to m^2} (-k_1^{2}+m^{2})\ldots \langle 0|T\widetilde{\phi}(k_1)\ldots|0\rangle[/itex]
where [itex]\widetilde{\phi}(k) = i \int d^4x e^{ikx} \phi(x)[/itex] an [itex]k^2=m^2[/itex] is not fixed.
Homework Equations
None
The Attempt at a Solution
Especially, I don't understand where the limes comes from. Here my attempt:
[itex]\langle f|i \rangle = i \int d^{4}x_1e^{ik_1x_1}(\partial^{2}+m^{2})\ldots \langle 0|T\phi(x_1)\ldots|0\rangle[/itex]
[itex]=\int d^{4}x_1e^{ik_1x_1}(\partial_1^{2}+m^{2})\ldots \langle 0|T \int\frac{d^4q_1}{(2\pi)^4}e^{-iq_1x_1}\widetilde{\phi}(q_1)\ldots|0\rangle[/itex]
[itex]=\int d^{4}x_1\int\frac{d^4q_1}{(2\pi)^4}e^{i(k_1-q_1)x_1}(-q_1^{2}+m^{2})\ldots \langle 0|T\widetilde{\phi}(q_1)\ldots|0\rangle[/itex]
[itex]=\int{d^4q_1}\delta^4(k_1-q_1)(-q_1^{2}+m^{2})\ldots \langle 0|T\widetilde{\phi}(q_1)\ldots|0\rangle[/itex]
[itex]=(-k_1^{2}+m^{2})\ldots \langle 0|T\widetilde{\phi}(k_1)\ldots|0\rangle[/itex]
[itex]=\ldots[/itex]
So, I am missing the limes in the last expression. Can it simply be introduced in the end since the on shell condition fixed [itex]k_1^{2}=m^{2}[/itex] before ?
Why isn't [itex]-k_1^{2}+m^{2}=0[/itex] true here? Is it because we are considering an interacting theory?
Very best, physicus