# Mass in the LSZ reduction formula

I'm reading wikipedia's page on the LSZ reduction formula

https://en.wikipedia.org/wiki/LSZ_reduction_formula

For the scalar LSZ reduction formula, after performaing a Fourier transform on the $$n$$-point correlation function, we get

$$\langle p_1,...,p_n, \text{out}|q_1,...,q_n,\text{in}\rangle=\prod_{i=1}^n\bigg\{-\frac{i(p_i^2-m^2)}{(2\pi)^{3/2}Z^{1/2}}\bigg\}\prod_{j=1}^n\bigg\{-\frac{i(q_j^2-m^2)}{(2\pi)^{3/2}Z^{1/2}}\bigg\}\Gamma(p_1,...,p_n,-q_1,...,-q_m).$$

My question is: why isn't this zero? Since $$p_i^2-m^2=0$$ and $$q_j^2-m^2=0$$; since $$m$$ is the mass and $$p_i,q_j$$ are the in, out four momentum vectors. Is this because we are supposed to use a different mass $$m$$?

This is because the Fourier transform of the time-ordered correlation function, which you denote as $$\Gamma(p_1,\dots, p_n,-q_1,\dots, -q_m)$$, has poles when the momenta go on-shell, i.e., $$p_i^2\to m^2$$ and $$q_j^2\to m^2$$. In that case multiplying by $$p_i^2-m^2$$ and $$q_j^2-m^2$$ cancels these poles and yields a finite result when the limit is taken. In fact this is exactly what the LSZ reduction formula is telling you: time-ordered correlation functions have poles corresponding to the mass-shell and the residues at those poles are the scattering amplitudes.
To give you a very basic example of this kind of thing, notice that you cannot say that $$\lim_{z\to 0}zf(z)$$ is zero. If, for example, $$f(z)=\frac{\alpha}{z}$$ then $$\lim_{z\to 0}zf(z)=\alpha$$ is finite. This happened because $$f(z)$$ had a simple pole at $$z=0$$ with residue $$\alpha$$. In fact, more generally, $$\lim_{z\to z_0}(z-z_0)f(z)$$ is the definition of the residue of $$f(z)$$ at a simple pole $$z_0$$.