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For massless spinors case we can decompose momentum into Weyl sub-parts as

$$p = \lambda_{a}\tilde \lambda_{\dot a}.$$

But for the case of massive fermions can I do something like this? Decompose them into Weyl subparts with some additional terms? If so, how?

Why do I need it? I am performing a twistor transform for the equation of the process $q\bar q \to gg$ so I have to write the amplitude and the 4d delta function of the momentum and then Fourier transform the $\lambda$ and $\tilde \lambda$ separately.

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Ref - Formulae 2-1 and 2-19

Suppose a given light-like momentum $q$.

Then, for each momentum $k$, such as $k.q \neq 0$, there exist a light-like momentum $k^l$, such as :

$$k = k^l + \frac{k^2}{2 k.q}q$$

So you can write :

$$k_{\alpha \dot\alpha } = k^l_\alpha ~k^l_{\dot\alpha} + \frac{k^2}{2 k.q} q_\alpha ~q_{\dot\alpha}$$

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  • $\begingroup$ shouldnt there be $2 k^l .q$ in the denominator $\endgroup$ – sol0invictus Jun 29 '13 at 11:31
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    $\begingroup$ $k.q = k^l.q$, because $q^2=0$ $\endgroup$ – Trimok Jun 29 '13 at 11:35
  • $\begingroup$ yeah!! sorry my bad :) $\endgroup$ – sol0invictus Jun 29 '13 at 12:27
  • $\begingroup$ okay one more quick question how do i simplify $<pq>$ where p is for a massive fermion and q is for massless gluon. In weyl representation $<pq> = p_a q^a $ in weyl representation. The notation you gave me in this for massive particles $k_{a\dot a}$ cannot be decomposed into $k_a$ and $k_{\dot a}$ so how do i express $<pq>$ $\endgroup$ – sol0invictus Jun 29 '13 at 14:21
  • $\begingroup$ Frankly, the whole formalism seems not so easy, and I am not a specialist of it.I can only give you the following reference - Chapter 2.2 (2.2 Spinors and polarisation vectors) - page 4 . But there is some work to do to completely understand the formalism. $\endgroup$ – Trimok Jun 29 '13 at 18:09

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