The dirac equation for massless particles can be decoupled into separate equations for left and right handed parts. [tex]i \tilde{\sigma}^\mu\partial_\mu \psi_R= 0[/tex] and [tex]i \sigma^\mu\partial_\mu \psi_L= 0[/tex]. Now we can have four solutions for each of the above equations. For the equation [tex]i \tilde{\sigma}^\mu\partial_\mu \psi_R= 0[/tex], the solutions are [tex]\left(\begin{array} (E+p_z \\ 0 \end{array}\right) e^{i(-Et+p_z z)}, \left(\begin{array} (0 \\ -E+p_z \end{array}\right) e^{i(-Et+p_z z)}, \left(\begin{array} ( -E+p_z \\ 0 \end{array}\right) e^{i(Et+p_z z)} and \left(\begin{array} ( 0 \\ E+p_z \end{array}\right) e^{i(Et+p_z z)}[/tex] with [tex]E^2-p_z^2 = 0[/tex].From my understanding the above four solutions represent particle with positive helicity, particle with negative helicity, antiparticle with positive helicity and antiparticle with negative helicity. But all the four are right chiral. But people say for massless particles helicity and chirality are the same. How is this possible? Am I making some mistake somewhere?(adsbygoogle = window.adsbygoogle || []).push({});

You may say that massless fermions doesn't exist. But even in standard model at very high energies above the symmetry breaking, all fermions are massless.

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# Solution for Dirac equation with zero mass.

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