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Solution for Dirac equation with zero mass.

  1. Feb 2, 2010 #1
    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?
    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.
     
  2. jcsd
  3. Feb 4, 2010 #2
    This paragraph from the Chirality article at Wikipedia seems to hold the answer:
     
  4. Feb 5, 2010 #3
    Thanks for the reply.

    I agree with those quoted statements from wikipedia.

    For example if you boost [tex] \left(\begin{array} (E+p_z \\ 0 \end{array}\right) e^{i(-Et+p_z z)} [/tex] we get [tex] \left(\begin{array} (E^'+{p^'}_z \\ 0 \end{array}\right) e^{i(-E^'t+{p^'}_z z)} [/tex] where [tex]E^' [/tex]and [tex] {p^'}_z [/tex] are the new energy and momentum in the boosted frame. Now the helicity operator which is [tex] \left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right) [/tex] gives a helicity of +1 for the above solution before and after the boost. Everything is fine. Now in the same way the second solution [tex] \left(\begin{array} (0 \\ -E+p_z \end{array}\right) e^{i(-Et+p_z z)}[/tex] has helicity -1 which is unchanged by boost. But both these are the solutions of the equation [tex] i \tilde{\sigma}^\mu\partial_\mu \psi_R= 0 [/tex]. So both are right chiral.

    Now wikipedia says
    "For massless particles — such as the photon, the gluon, and the (hypothetical) graviton — chirality is the same as helicity; a given massless particle appears to spin in the same direction along its axis of motion regardless of point of view of the observer."

    I agree with the second part " a given massless particle appears to spin in the same direction along its axis of motion regardless of point of view of the observer." , but direction of spin has nothing to do with chirality. Chirality is defined based on the lorentz transformation property of the solution.

    Am I wrong or is the wikipedia wrong?
     
    Last edited: Feb 5, 2010
  5. Feb 6, 2010 #4
  6. Feb 21, 2010 #5
    Thanks for the link. It doesn't contradict what I thought. I think the wikipedia article wrongly attribute chirality to direction of spin.
     
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