Let's have chirality projection operator $$ \hat {C}_{\pm} = \frac{1 \pm \gamma^{5}}{2}. $$ We introduce it and called it chirality, because $$ \hat {C}_{+}\psi = \begin{pmatrix} \psi_{\alpha} \\ 0 \end{pmatrix}, \quad \hat {C}_{-}\psi = \begin{pmatrix} 0 \\ \eta^{\alpha} \end{pmatrix}, $$ which accords to the definition of left- and right-handed spinor (by the classification of irreducible representations of the Lorentz group).

Then let's have helicity projection operator: $$ \hat {h} = \frac{ 1 \pm \left(\hat {\mathbf S} \cdot \hat {\mathbf p}_{\mathbf 1}\right)}{2}, \quad \hat {\mathbf p}_{1} = \frac{\hat {\mathbf p}}{|\mathbf p|}. $$ It can be showed, that for the case $v = c$, where $v$ is the speed of the particle, these operators acts on bispinor the same way. So for bispinor formalism in a massless case helicity and spirality are the same.

But I don't know how to show analogical thing for the particles of different spines (and, according to this, different irreducible Lorentz reps). Maybe, the answer is in the words below?

We can construct different reps of Lorentz group by using spinor formalism. We can construct rep $\left(\frac{1}{2}, \frac{1}{2}\right)$ by $\left(\frac{1}{2}, 0\right)\times \left(0, \frac{1}{2}\right)$, rep $\left(0, 0\right) +\left(1, 0\right)$ by $\left(\frac{1}{2}, 0\right) \times \left(\frac{1}{2}, 0\right)$. So, maybe, we can generalize the calculations with chirality and helicity for the bispinor case on arbitrary cases? If yes, how exactly we can do it?

Related questions: Chirality, helicity and their relationship for the massless case , Why helicity is proportional to the spin of particle and has two values? .

  • $\begingroup$ For a massless particle sign of helicity is independent of Lorentz frame. This can be understood by the usual argument that you can't overtake a particle moving at the speed of light and so if projection of its spin in one frame is found to have positive (respectively negative) value, then in any other frame too it will be found to be positive (negative). $\endgroup$ – user10001 Aug 15 '13 at 14:57
  • $\begingroup$ So now if we consider space of eigenstates of helicity operator for a massless particle then since Lorentz transformations do not change the sign of helicity so the eigenstates with positive and negative eigenvalues will separately form a representation of Lorentz group. You may call them two chiral representations if you want. $\endgroup$ – user10001 Aug 15 '13 at 14:59
  • $\begingroup$ @user10001 . These arguments don't forbid helicity values $s - 1, ..., -(s - 1)$. So at least in the massless case helicity can't be identified with chirality. $\endgroup$ – user8817 Aug 15 '13 at 18:02
  • $\begingroup$ sorry I didn't understand. Is there anything wrong with above argument? $\endgroup$ – user10001 Aug 15 '13 at 18:29
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    $\begingroup$ sorry I still don't understand how absense or presence of helicity values s-1,..., -(s-1) is relevant to the question of relation between chirality and helicity. $\endgroup$ – user10001 Aug 15 '13 at 19:45

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