I do not understand this bra-ket notation equality for BCFW recursion

Question

Background:

When learning about BCFW recursion I am shown the deformation equation:

$$\hat{p_1}=p_1 -zq \hspace{5mm}; \hspace{5mm} \hat{p_n} = p_n +zq$$

This deformation represents a $\langle1n]$ shift and $q$ is defined as $q=\lambda^\dot{\alpha}_1 \lambda^\alpha _n $

To check under which circumstances $\lim _{z \to \infty} \hat{A_n}(z)=0$, we consider deforming the 4-point MHV amplitude $A_4(1^- 2^- 3^+ 4^+)$.

Supposing we do a $\langle 1^- 2^-] $ shift then:

$$\hat{A_4}^{--}= \frac{\langle \hat{1}\hat{2}^3 \rangle}{\langle \hat{2}3 \rangle \langle 34\rangle \langle 4\hat{1} \rangle}$$

Problem:

In the notes, I am told to note that:

$$\langle \hat{1} \hat{2}\rangle= \left( \langle1|-z\langle 2|\right)|2\rangle =\langle 12\rangle $$

where $\langle 22\rangle=0 $ has been used.

How is this last equation true? I don't understand the middle term, how it was worked out or how it makes $\langle \hat{1} \hat{2} \rangle = \langle 12 \rangle$ I am pretty sure the first two equations are used in it but I don't see how.

Answer 1

When doing a complex shift in order to enact the BCFW transformation, we take, as you pointed out,

$$ \hat{p}_1 = p_1 - zq~,~~~~~ \hat{p}_2 = p_2 + zq$$

Choosing to do a $\langle 12]$ shift means we make a clever choice for $q$ and work with spinors, which gives $$\hat{p}_1 = |1\rangle|1] - z|2\rangle|1] = (|1\rangle - z|2\rangle)|1] = |\hat{1}\rangle|1].$$ The choice of $q$ is determined from $\hat{p}_1^2 = 0 = \hat{p}_2^2$.

This can be seen by considering $\hat{p}_i^2 \propto 2p_i\cdot q = \langle iq\rangle[iq]$: we have to choose either $q \propto |1\rangle[2|$ or $q \propto |2\rangle[1|$.

From the above, we have then that $|\hat{1}\rangle = |1\rangle - z|2\rangle$, which is the origin of the middle line. We should also note that $|\hat{2}\rangle = |2\rangle$, which you can check by shifting $p_2$ by the same $q$ as above.