(This question comes from exercise 3.5 of Elvang's and Huang's "Scattering amplitudes in Gauge Theory and Gravity" book. This is not for a class, this is to learn a new technique; albeit I am already close to the full solution.)
Consider the following Lagrangian for scalar QED (SQED) \begin{equation} \mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} - (\partial\phi)^2 + ieA_\mu\left((\partial_\mu\phi^*) - \phi^*\partial_\mu\phi\right) - e^2A_\mu A_\mu \phi^*\phi - \frac{1}{4}\lambda|\phi|^4 \end{equation} We have 2 4-point vertices namely $-i\lambda$ for 4 scalars and the scalar-scalar-photon vertex $-2ie^2\eta_{\mu\nu}$. The main goal of exercise 3.5 is to compute $A_4(\phi\phi^*\gamma^+\gamma^-)$ (where I will be using the notation $A_4(1^02^03^+4^-)$ instead). In order to compute this amplitude, we will use the BCFW recursion relation, but in doing so we will need the 3-point amplitudes. These are completely determined by units, locality, and little group scaling namely (equation (2.99) from the text) \begin{equation} A_3(1^{h_1}2^{h_2}3^{h_3}) = c\langle 12\rangle^{h_3-h_1-h_2}\langle 13\rangle^{h_2-h_1-h_3}\langle 23\rangle^{h_1-h_2-h_3} \end{equation} This equation is valid when $\sum_i^3 h_i<0$. The equation above becomes all square brackets when $\sum_i^3h_i>0$ (otherwise if you only use angle brackets then your answer must have arisen from a non-local term at the level of the Lagrangian). For SQED we have two 3-point amplitudes we care about \begin{equation} A_3(\phi\phi^*\gamma^-) = \tilde{e}\frac{\langle 13\rangle\langle 23\rangle}{\langle 12\rangle}\:\:\text{and}\:\: A_3(\phi\phi^*\gamma^+) = \tilde{e}\frac{[13][23]}{[12]} \end{equation} where $\tilde{e}\equiv \sqrt{2}e$. Note: these are not colored ordered diagrams... Armed with these, the BCFW recursion relation is given by \begin{equation} A_n = \sum_{h_I}\sum_{\text{diagrams}\:I}\hat{A}_L(z_I)\frac{1}{P_I^2}\hat{A}_R(z_I) \end{equation} Since we are working with SQED, then intermidate particle for every possible channel in the set $I$ has helicity 0 (scalar) and thus the helicity sum drops out (for our case). The number of possible diagrams are 2 with $I = \{13,23\}$ (or equivalents using conservation of momentum $\sum_i^4p_i = 0$). We are also consider a $[4,3\rangle$ shift implying that we are shifting the vectors (or lines as used in the text) with the following relations \begin{equation} |\hat{4}] = |4] + z|3],\:|\hat{3}] = |3],\:|\hat{4}\rangle = |4\rangle,\: |\hat{3}\rangle = |3\rangle - z|4\rangle \end{equation}
The sum of the two diagrams then are the following \begin{equation} A_4(1^02^03^+4^-) = \tilde{e}^2\frac{[13][\hat{P}_{13}3]}{[1\hat{P}_{13}]}\frac{1}{\langle 13\rangle[13]} \frac{\langle\hat{P}_{13}\hat{4}\rangle\langle 2\hat{4}\rangle}{\langle\hat{P}_{13}2\rangle} + \tilde{e}^2\frac{[\hat{P}_{23}\hat{3}][2\hat{3}]}{[\hat{P}_{23}2]}\frac{1}{\langle23\rangle[23]}\frac{\langle 1\hat{4}\rangle\langle\hat{P}_{23}\hat{4}\rangle}{\langle1\hat{P}_{23}\rangle} \end{equation} where I have already removed the hats off of some of the momentum vectors since they don't change under the $[4,3\rangle$ shift. Lets only focus on the second term (since my question is on the first). Notice that we can pair the brackets containing $\hat{P}_{23}$ in the numerator and denominator and using the massless Weyl equation to reduce the term in the numerator to (modulo a minus sign since I am being sloppy with them) \begin{equation} \langle 4\hat{P}_{23}\rangle [\hat{P}_{23}3] =\langle 4|p_2+\hat{p}_3|3] = \langle 4|2|3] = \langle 24\rangle[23] \end{equation} and in a similar manner for the denominator \begin{equation} \langle 1\hat{P}_{23}\rangle[\hat{P}_{23}2] = \langle 1\hat{3}\rangle[23] \end{equation} Then the second term becomes \begin{equation} A_4(1^02^03^+4^-)\supset \tilde{e}^2\frac{\langle 14\rangle\langle 24\rangle}{\langle 1\hat{3}\rangle\langle 23\rangle} \end{equation} Now, this is where I am a bit fuzzy on the exact details, but the answer given in the book is $A_4 = g^2\langle 14\rangle\langle 24\rangle/\langle 13\rangle\langle 23\rangle$ which we can see is very close to my answer, but without the hat. But, I forgot to evaulate the complex pole. So, recall that \begin{equation} 0 = \hat{P}_{23}^2 = \langle 2\hat{3}\rangle[23] = [23]\langle 2|(|3\rangle -z_{23}|4\rangle)\implies z_{23} = \langle 23\rangle/\langle 24\rangle \end{equation} is where the pole lies. But, this is where I get stuck, namely where do I go from here in evaluating the poles?
The first part of the 4-point amplitude comes to be \begin{equation} A_4(1^02^03^+4^-)\supset \tilde{e}^2\frac{\langle 14\rangle\langle 24\rangle[13]}{\langle 13\rangle[12]\langle 2\hat{3}\rangle} \end{equation} which I do not know if I need it, or it should have vanished in the first place.
Any help or suggestions or comments are great (or better yet a final answer!). (I will be posting many more questions like this in the future... like one with gravitons very shortly...)
