A relation in "Scattering Amplitudes in Gauge Theory...", Elvang et al

  • A
  • Thread starter nrqed
  • Start date
  • #1
nrqed
Science Advisor
Homework Helper
Gold Member
3,572
192

Main Question or Discussion Point

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

[tex] \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} [/tex]

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.
 

Answers and Replies

  • #2
MathematicalPhysicist
Gold Member
4,189
168
Do the authors of the book know how to infer this identity?
 
  • #3
nrqed
Science Advisor
Homework Helper
Gold Member
3,572
192
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 [itex] a, \dot{b} [/itex], 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.
 
  • #4
nrqed
Science Advisor
Homework Helper
Gold Member
3,572
192
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

[tex] \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} [/tex]

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 [itex] \langle r |_{\dot{b}} [/itex] on the lhs to get

[tex]
\langle r ,i \rangle ~\langle i-1|_{\dot{a}} ~ - ~ \langle r,i-1 \rangle\, \langle i |_{\dot{a}} [/tex]

Now , after using [itex] \langle r, i \rangle = - \langle i,r \rangle [/itex] and using the Schouten identity, this is

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

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

[tex]
\biggl( |i \rangle^{\dot{b}} ~\langle i-1|_{\dot{a}} ~ - ~ |i-1 \rangle^{\dot{b}} \, \langle i |_{\dot{a}} \biggr)
= \langle i-1,i \rangle~\delta_{\dot{a}}^{ \dot{b}}
[/tex]

This completes the proof.

Cheers
 
  • #5
MathematicalPhysicist
Gold Member
4,189
168
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 [itex] a, \dot{b} [/itex], 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.
 
  • #6
nrqed
Science Advisor
Homework Helper
Gold Member
3,572
192
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!
 

Related Threads for: A relation in "Scattering Amplitudes in Gauge Theory...", Elvang et al

Replies
2
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
3
Views
3K
Replies
2
Views
1K
  • Last Post
Replies
2
Views
5K
  • Last Post
Replies
9
Views
3K
Top