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Unitarity and amplitudes

Physics Asked by Nikita on December 15, 2020

In Bootstrap and Amplitudes: A Hike in the Landscape of Quantum Field Theory there are few statements about analytical structure of amplitudes.

I want to understand statement:

In a local theory of massless scalars, they can have simple poles and the residue of such a pole is, by
unitarity, a product of lower-point amplitudes.

I understand this property from Feynman rules for tree level amplitude, but I didn’t get role of unitarity.

I wanna understand, how unitarity related to such property? How residues will change, if we lose unitarity?

One Answer

Unitarity ($SS^dagger = 1$) dictates that begin{equation}label{key} T-T^dagger = iTT^dagger end{equation} For tree-level $2rightarrow 2$ scattering, we have then that $$ langle p_1,p_2|T|p_3,p_4rangle - langle p_1,p_2|T|p_3,p_4rangle^* = ilangle p_1,p_2|TT^dagger|p_3,p_4rangle. $$ Using $langle p_1,p_2|T|p_3,p_4rangle = (2pi)^4delta^{(4)}(p_1 + p_2 - p_3 - p_4)mathcal{A}[12rightarrow 34]$, and inserting a complete set of one particle states (to stay at tree-level), we can write this as begin{align} 2text{Im}(mathcal{A}[12rightarrow 34]) &= 2pisum_kint frac{d^3k}{2E_k}delta^{(4)}(p_1+p_2 - k)mathcal{A}[12rightarrow k]mathcal{A}^*[krightarrow 34]\ &= 2pisum_kint d^4kdelta(k^2)delta^{(4)}(p_1+p_2 - k)mathcal{A}[12rightarrow k]mathcal{A}^*[krightarrow 34] end{align} Now, the left hand side is the imaginary part of the 4pt amplitude, which we will take to have numerators $n_i$ and propagators $p_i^2 + iepsilon$, where $i$ labels the ways we can arrange a particle exchange (the $s,t,u$ channels). Thus, we have begin{equation} 2text{Im}left(sum_kfrac{n_k}{k^2+iepsilon} + text{contact}right) = 2pisum_kint d^4kdelta(k^2)delta^{(4)}(p_1+p_2 - k)mathcal{A}[12rightarrow k]mathcal{A}^*[krightarrow 34], end{equation} In a local theory of massless scalars, we can write the imaginary part of the propagator as $$ text{Im}left(frac{1}{p^2 + iepsilon}right) = frac{1}{2i}left(frac{1}{p^2 + iepsilon} - frac{1}{p^2 - iepsilon}right) = frac{-epsilon}{p^4 + epsilon^2}. $$ This seems like it vanishes for $epsilon rightarrow 0$, which, by the optical theorem, means that your amplitude must be zero for real external momenta. However, this is misleading and only true when the propagator is off-shell. Recognising the fact that the last term above is the nascent dirac delta function, we learn that $$ lim_{epsilonrightarrow 0}frac{-epsilon}{p^4 + epsilon^2} = pidelta(p^2). $$

Plugging this in, we find that, as the propagator goes on shell, we have begin{equation} 2pisum_kn_kdelta(k^2) = 2pisum_kint d^4kdelta(k^2)delta^{(4)}(p_1+p_2 - k)mathcal{A}[12rightarrow k]mathcal{A}^*[krightarrow 34]. end{equation} Or, in other words, the numerators of the tree-level amplitudes factorize into two lower-point amplitudes (the residues) as the propagator goes on-shell begin{equation} n_k = int d^4kdelta^{(4)}(p_1+p_2 - k)mathcal{A}[12rightarrow k]mathcal{A}^*[krightarrow 34] end{equation} We note now a major problem: the right hand side is actually zero due to momentum conservation, and this is probably the reason most books don't discuss the optical theorem at tree-level. This is due to the fact that Lorentz-invariant three-particle amplitudes vanish on-shell by virtue of the fact that $p_icdot p_j = 0$ for all $i,j$ due to momentum conservation. However, this is not true if we use spinor helicity variables and assume complex momentum, which is exactly what the bootstrapping amplitudes program does.

Correct answer by Akoben on December 15, 2020

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