Srednicki QFT chapter 67, LSZ formula

[physicus]

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
$\langle f|i \rangle = i \int d^{4}x_1e^{ik_1x_1}(\partial^{2}+m^{2})\ldots \langle 0|T\phi(x_1)\ldots|0\rangle$
This is supposed to be equivalent to:
$\langle f|i \rangle = \lim_{k_i\to m^2} (-k_1^{2}+m^{2})\ldots \langle 0|T\widetilde{\phi}(k_1)\ldots|0\rangle$
where $\widetilde{\phi}(k) = i \int d^4x e^{ikx} \phi(x)$ an $k^2=m^2$ is not fixed.

Homework Equations

None

The Attempt at a Solution

Especially, I don't understand where the limes comes from. Here my attempt:
$\langle f|i \rangle = i \int d^{4}x_1e^{ik_1x_1}(\partial^{2}+m^{2})\ldots \langle 0|T\phi(x_1)\ldots|0\rangle$
$=\int d^{4}x_1e^{ik_1x_1}(\partial_1^{2}+m^{2})\ldots \langle 0|T \int\frac{d^4q_1}{(2\pi)^4}e^{-iq_1x_1}\widetilde{\phi}(q_1)\ldots|0\rangle$
$=\int d^{4}x_1\int\frac{d^4q_1}{(2\pi)^4}e^{i(k_1-q_1)x_1}(-q_1^{2}+m^{2})\ldots \langle 0|T\widetilde{\phi}(q_1)\ldots|0\rangle$
$=\int{d^4q_1}\delta^4(k_1-q_1)(-q_1^{2}+m^{2})\ldots \langle 0|T\widetilde{\phi}(q_1)\ldots|0\rangle$
$=(-k_1^{2}+m^{2})\ldots \langle 0|T\widetilde{\phi}(k_1)\ldots|0\rangle$
$=\ldots$

So, I am missing the limes in the last expression. Can it simply be introduced in the end since the on shell condition fixed $k_1^{2}=m^{2}$ before ?
Why isn't $-k_1^{2}+m^{2}=0$ true here? Is it because we are considering an interacting theory?

Very best, physicus

 

[king vitamin]

I'm not 100% certain, but here's my guess.

If you take $$k_i^2 = -m^2$$ through all of the steps you've shown, you get a problem here:

$$\langle f|i \rangle =\int{d^4q_1}\delta^4(k_1-q_1)(-q_1^{2}+m^{2})\ldots \langle 0|T\widetilde{\phi}(q_1)\ldots|0\rangle$$

where as you integrate through all values of q_1, you get zero from the delta function unless q_1 = k_1, but when q_1 = k_1, then the term

$$-q_1^{2}+m^{2}$$

still makes the integrand go to zero. In fact, the product of the delta and the above term at q_1 = k_1 is sort of like an indefinite form infinity*0 inside the integrand, so you sort of need to choose a prescription for dealing with it.. In your steps, you let the delta function take precedence, but then your last line must be identically zero for on-shell momenta. Srednicki works through this by letting the momenta be a little off-shell.

If you follow through the steps after 67.4, you'll see that Srednicki is also using this sort of limit thing to show how he's considering these effectively "singular terms" in the correlation function to be the only contributors to the scattering amplitude, so you let the momenta go on-shell eventually and take residues.