Equation 5.45 in Elvang and Huang scattering amplitudes

[nrqed]

If anyone is familiar with the calculation of scattering amplitudes using momentum twistors. I am working through the book "Scattering Amplitudes in Gauge Theory and Gravity" by Elvang and Huang.

I am completely stumped by one step that should be simple. My question is about Eq. (5.45). My question is on the last step, which is

$$\biggl( |i \rangle^{\dot{b}} ~\langle i-1|_{\dot{a}} ~ - ~ |i-1 \rangle^{\dot{b}} \, \langle i |_{\dot{a}} \biggr) y_i^{\dot{a} a} = \langle i-1, i \rangle \, y_i^{\dot{b}a}$$

I am stumped, conservation of momentum or the Schouten identity does not help here.

I can provide more details with the various quantities here, but probably someone already quite familiar with the notation will be able to help.

 

[MathematicalPhysicist]

Do the authors of the book know how to infer this identity?

 

[nrqed]

Do the authors of the book know how to infer this identity?

I assume they do :-) . But even if they don't, it should be provable. The problem is that I don't even see how it is true, let alone how to prove it. If we just pick some values for the indices $a, \dot{b}$, I don't see how the two sides are equal. But they use it in the following to obtain key equations, so I am confused.

 

[nrqed]

If anyone is familiar with the calculation of scattering amplitudes using momentum twistors. I am working through the book "Scattering Amplitudes in Gauge Theory and Gravity" by Elvang and Huang.

I am completely stumped by one step that should be simple. My question is about Eq. (5.45). My question is on the last step, which is

$$\biggl( |i \rangle^{\dot{b}} ~\langle i-1|_{\dot{a}} ~ - ~ |i-1 \rangle^{\dot{b}} \, \langle i |_{\dot{a}} \biggr) y_i^{\dot{a} a} = \langle i-1, i \rangle \, y_i^{\dot{b}a}$$

I am stumped, conservation of momentum or the Schouten identity does not help here.

I can provide more details with the various quantities here, but probably someone already quite familiar with the notation will be able to help.

I think I have figured it out. Plugging in specific values for the indices actually gives that the two sides are equal, I was making mistakes with the sign conventions of the epsilon tensor.And to prove this, one can do the following: Let's consider the expression inside the parenthesis. First, I contract with an arbitrary bra $\langle r |_{\dot{b}}$ on the lhs to get

$$\langle r ,i \rangle ~\langle i-1|_{\dot{a}} ~ - ~ \langle r,i-1 \rangle\, \langle i |_{\dot{a}}$$

Now , after using $\langle r, i \rangle = - \langle i,r \rangle$ and using the Schouten identity, this is

$$\langle i-1,i \rangle~ \langle r |_{\dot{a}}$$
Now I write this as

$$\langle i-1,i \rangle~\delta_{\dot{a}}^{ \dot{b}} \, \langle r |_{\dot{b}}$$
So finally,

$$\biggl( |i \rangle^{\dot{b}} ~\langle i-1|_{\dot{a}} ~ - ~ |i-1 \rangle^{\dot{b}} \, \langle i |_{\dot{a}} \biggr) <br /> = \langle i-1,i \rangle~\delta_{\dot{a}}^{ \dot{b}}$$

This completes the proof.

Cheers

 

[MathematicalPhysicist]

I assume they do :-) . But even if they don't, it should be provable. The problem is that I don't even see how it is true, let alone how to prove it. If we just pick some values for the indices $a, \dot{b}$, I don't see how the two sides are equal. But they use it in the following to obtain key equations, so I am confused.

Well mistakes can and will happen...
But happy for you that you found a derivation.

 

[nrqed]

Well mistakes can and will happen...
But happy for you that you found a derivation.

I see your point. This is why I had checked that they had used exactly that expression in their following steps to obtain other results.

Cheers!