Canonical definition of the integrand in planar $\mathcal{N}=4 \ \mathrm{SYM}$ theory - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-07-21T17:55:12Z https://physics.stackexchange.com/feeds/question/222392 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/222392 3 Canonical definition of the integrand in planar $\mathcal{N}=4 \ \mathrm{SYM}$ theory giulio bullsaver https://physics.stackexchange.com/users/41757 2015-12-05T16:30:47Z 2016-11-20T00:10:56Z <p>According to page 101 of <a href="http://arxiv.org/abs/1308.1697" rel="nofollow"><em>Scattering Amplitudes</em></a> (Elvang, Huang), one can use the zone variables $y$ to define a unique integrand, in the planar case.</p> <p>This is done by saying that the momenta associated to an internal line is $y_a - y_b$ where $y_a$ and $y_b$ are the zone variable associated to the two zones adjacent to the particular line. </p> <p>In this scheme, is it necessary to also devise a scheme to canonically label the interior zone variables, in order to obtain a well-defined integrand function? </p> <p>To clarify, if one considers a double box integral, and swaps the two zone variables associated to the internal faces, one obtains two different contributions to the integrand. In fact, there will be two internal lines whose propagators are $(y_4 - y_a)^{-2} (y_2 - y_b)^{-2}$ which is not the same function if one swaps $a$ with $b$.</p> https://physics.stackexchange.com/questions/222392/canonical-definition-of-the-integrand-in-planar-mathcaln-4-mathrmsym-t/293667#293667 1 Answer by Physicsworks for Canonical definition of the integrand in planar $\mathcal{N}=4 \ \mathrm{SYM}$ theory Physicsworks https://physics.stackexchange.com/users/4237 2016-11-19T23:48:43Z 2016-11-19T23:48:43Z <p>Good question. The answer is that one can always (completely) symmetrize the integrand with respect to $L!$ loop variables. For instance, the two-loop </p> <pre><code> 1 2 \ x1 / \_________/ | | | x4 | y1 | y2 | x2 |____|____| / \ / x3 \ 4 3 </code></pre> <p>integral in dual coordinates is (the numerator is chosen so that the integral is dual-conformally invariant): $$\int \frac{d^4y_1 d^4 y_2}{2!} \frac{(x_1-x_3)^4(x_2-x_4)^2}{(y_1-x_3)^2(y_1-x_4)^2(y_1-x_1)^2(y_2-x_1)^2(y_2-x_2)^2(y_2-x_3)^2(y_1-y_2)^2} + (y_1 \leftrightarrow y_2)$$</p> <p>People usually write these integrals in momentum-twistor variables. As you remember, to each of the loop variables $y_i$ there is an associated pair of twistors $(Z_{A_i},Z_{B_i})$. The above integral turns out to be $$\int_{(A_1 B_1, A_2 B_2)} \frac{\langle 1234 \rangle^2 \langle 2341 \rangle}{\langle A_1 B_1 41 \rangle \langle A_1 B_1 12 \rangle \langle A_1 B_1 23 \rangle \langle A_2 B_2 23 \rangle \langle A_2 B_2 34 \rangle \langle A_2 B_2 41 \rangle \langle A_1 B_1 A_2 B_2 \rangle},$$ where $(A_1 B_1, A_2 B_2)$ means that the integration measure has a factor of $\frac{1}{2!}$ in it from the symmetrization $(A_1 B_1 \leftrightarrow A_2 B_2)$.</p>