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When reading about Scattering Amplitudes Notes, the text said the following:

Under little group scaling of each particle $i = 1, 2, . . . , n$ , the on-shell amplitude transforms homogeneously with weight $−2h_i$, where $h_i$ is the helicity of particle $i$:

$$A_n( \{ |1\rangle |1], h_1\}, . . . , \{t_i |i\rangle, t^{−1}_i|i], h_i\}, ...) = t^{−2h_i}_i A_n(... \{|i\rangle, |i], h_i\} . . .)\tag{2.93}$$

An on-shell 3-points amplitude $A$ (...) Let us suppose that it depends on angle brackets only.

We can then write a general Ansatz:

$$A_3(1^{h_1}2^{h_2}3^{h_3})= c \langle 12 \rangle ^{x_{12}} \langle 13 \rangle ^{x_{13}} \langle 23 \rangle ^{x_{23}}\tag{2.94} $$

The little group scaling (first equation above) fixes:

$$-2h_1 =x_{12}+x_{13},$$ $$-2h_2 =x_{12}+x_{23},$$ $$-2h_3 =x_{13}+x_{23}. \tag{2.95}$$

I don't understand how the first equation was used to obtain these last 3 equalities. How can this be used to calculate the MHV of particles?

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1 Answer 1

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In eq. (2.10) of Elvang and Huang it is explained that square/angle spinors has helicity $\pm 1/2$, respectively. (If the particle 3-momentum is along the $z$-axis, then the generator of the helicity/little group-scaling is given by $\sigma_z$.) This leads to the scaling $t^{\mp 1}$ for the square/angle spinors, respectively, cf. eq. (2.92).

Using eq. (2.92) on the RHS and eq. (2.93) on the LHS of the Ansatz (2.94) yields the sought-for eq. (2.95).

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  • $\begingroup$ I understand that those equations are all used, but I don't see how (eq(2.95)) are obtained. (2.94) is of the same form as (2.93). And I believe the rhs of (2.93) is obtained from (2.92). But I still don't get how we can go from these to (2.95) $\endgroup$ Mar 20, 2020 at 21:46
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Mar 22, 2020 at 13:58

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