5
$\begingroup$

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.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

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 :

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

Improve this answer
$\endgroup$
5
  • $\begingroup$ shouldnt there be $2 k^l .q$ in the denominator $\endgroup$
    – sol0invictus
    Commented Jun 29, 2013 at 11:31
  • 1
    $\begingroup$ $k.q = k^l.q$, because $q^2=0$ $\endgroup$
    – Trimok
    Commented Jun 29, 2013 at 11:35
  • $\begingroup$ yeah!! sorry my bad :) $\endgroup$
    – sol0invictus
    Commented Jun 29, 2013 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
    Commented Jun 29, 2013 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
    Commented Jun 29, 2013 at 18:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.