I'm reading wikipedia's page on the 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 $m$?


1 Answer 1


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 $z_0$.


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