Posted:

**Jun 10, 2012 12:27 pm**Post 13 MHV Amplitudes

Spinor Helicity Formalism

A quick recap - I didn't have LaTeX last time and the notation was horrendous! We're using this stuff prinicipally for modelling the scattering of gluons off each other. The energies involved are so large that we can ignore the gluon rest masses and treat them as zero-rest-mass particles.

The conventional way to characterize the motion of such a particle is to give:

1) Its momentum. This is a null vector [math]

2) Its angular momentum [math] (antisymmetric in [math]) with respect to some arbitrary origin in spacetime.

If we form the spin-vector [math], then positive helicity gluons satisfy [math] and negative helicity gluons [math]. The positive/negative character determines the handedness of their spin relative to their direction of travel.

So that (momentum, helicity) is all we need to label a particle heading towards the collision or away from it. Well, actually, no, gluons have colour as well, but that won't make a significant difference to our deliberations so we'll ignore it. (This means that we're considering "color-ordered" scattering amplitudes).

An amplitude for scattering of n gluons, labelled 1..n, where two of them, gluons j and k, have helicity opposite to the rest is denoted by [math]. To compute such a scattering amplitude, you follow the rules - add up all the contributions from the relevant Feynman diagrams. The ROFL moment comes when you realise how many diagrams you need to add up:

4 Gluons - needs 4 diagrams

5 Gluons - needs 25 diagrams

6 Gluons - needs 220 diagrams

7 Gluons - needs 2485 diagrams

8 Gluons - needs 34300 diagrams

9 Gluons - needs 559405 diagrams

AND, the contribution from even a single diagram is complex to evaluate.

Amazingly some people have actually done this (at least up to 6 gluon level - they may have gone further - I'm not sure). Thank goodness for symbolic mathematics packages. Anyway just like when you've done a really complex calculation in an exam and got a nice simple answer, you know you haven't made a mistake, they got a nice simple answer. Namely, they demonstrated that, at least as far as n=6, the amplitude is given by

[math]

where [math]

The delta function just has the sum of the four-momenta of all the particles in it, so it forces momentum to be conserved. A null four momentum can be written as a product of a primed and an unprimed spinor, hence the lambdas inside the delta function.

Even before it was computed for the low n cases, this formula was guessed by Parke and Taylor. Nima Arkani-Hamed has repeatedly voiced the opinion that the fact that its such a pain in the ass to compute using Feynman diagrams points to the fact that the route that such diagrams take, namely via space time isn't the right one. Space-time isn't the right concept to describe this process. The Feynman diagram approach enforces manifest locality and Lorentz invariance at every step, and for the first time in our lives, we're beginning to doubt that this is the right thing to do. To put it in Nima's words "spacetime is doomed".

Conformal Invariance of Scattering Amplitudes

The spacetime approach to scattering amplitude calculation is a pain in the ass - is there another way? To motivate the one [math] approach to this piece of physics, consider the action of the conformal group on amplitudes. We know the algebra of the conformal group (last post), we just need a representation of the various conformal generators acting on the scattering amplitudes. Here they are:

Lorentz Boosts:

[math]

[math]

If you're asking yourself where the antisymmetric pair of vector indices in the Lorentz generator, remember I can write a vector as a primed/unprimed spinor index pair and decompose an antisymmetric vector as

[math]

Translations:

This is easy, it's just the multiplicative factor

[math]

Special Conformal Transformations

These have scaling dimension -1, meaning [math]. The only possibility to represent it is

[math]

Dilatations:

Spinors have scaling dimension +1/2, meaning [math]

This is achieved if we represent the dilatation operator by

[math]

At first, the "2" in the formula could be an arbitrary constant, but it gets fixed to the value 2 by imposing the commutation reltion between the special conformal generator and the translation generator.

So the conformal group has a defined action on the scattering amplitudes. It's not a nice action though - it's a right mixed bag: a multiplicative operator, a first order differential operator, a second order differential operator....

"Twistor Quantization" and a Nicer Conformal Group Action

In the "classical" picture, post 3 , a twistor [math] describes the worldline of a zero rest mass particle with momentum [math] and trajectory [math] such that [math]

The dual twistor is [math]

So [math]

Hence, if we take the exterior derivative

[math]

The RHS is just the symplectic form (conventionally written [math]), i.e. the differential form preserved by Hamiltonian time evolution. (First) quantization normally proceeds by making the momentum and position operators satisfy the commutation relation

[math]

where in the position representation

[math]

(we set Planck's constant to unity). So in the twistor picture, the natural thing to do seems to be to set

[math]

[math]

and we have the twistor relation

[math]

This is analogous to the position/momentum commutator

[math]

Having done this, we can define the generators of the conformal group on twistor space:

[math]

[math]

[math]

[math]

[math]

(note these conventions are the ones chosen by Witten and would correspond to a twistor incidence relation

[math]

compared to my previous convention

[math] ).

These are much nicer. They're all first order differential operators.

Scattering Amplitudes in Twistor Space

Our expression (1) for the gluon scattering amplitude is a function of [math], and [math] is the gluon (null) momentum, so the scattering amplitude is a function on momentum space. We can now see what happens if we transform it to twistor space.

Twistor space is the space of unprimed, primed spinor pairs [math] (note the literature seems to favour [math] and [math], rather than [math] and [math], so I'll swap to that convention). To get there, we make a Fourier transform in the [math] variable:

Writing (1) in the form

[math]

we first of all note that, actually, for the MHV case, [math] is only a function of [math], there are no [math] terms in it. I'm leaving the spinor indices off the spinors - the indices i.j,k etc are the gluon labels.

The Fourier transform is

[math]

Now, I can do the usual trick of replacing the delta function by its own little Fourier integral

[math]

inserting this, we end up with

[math]

Now this is interesting, because [math] forces the twistors to lie on a (complex projective)line in twistor space. Recall from post 3 that a line in twistor space is just the Riemann sphere's worth of light rays through a point in spacetime. The point in question is the spacetime point where all the gluons must meet. In Feynman diagram terms, we are talking about an interaction vertex.

A line is a genus zero, degree 1 curve in twistor space. The moduli space of such curves is Minkowski space. ("Moduli space" just means "space of parameters").