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We know that in 4D a Dirac spinor has 4 complex components or 8 real components meaning 8 real off shell degrees of freedom (please correct me if I say something wrong here). When we go on-shell i.e impose Dirac equation on the spinor, half of them are killed thereby leaving only 4 real on-shell degrees of freedom. I want to see explicitly why is the Dirac equation killing half of the off-shell degrees of freedom.

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  • $\begingroup$ Have you tried solving the Dirac equation? Maybe you can express some components in terms of others? $\endgroup$ – innisfree May 11 '15 at 19:11
  • $\begingroup$ aahh of course!! you are right! Thanks!! So by the same argument this should hold true in any number of dimensions right? My goal is to show is in any dimension $\endgroup$ – Orbifold May 11 '15 at 19:19
  • $\begingroup$ the reduction happens because the Dirac equation is first-order; $\pi_\psi= dL/d\psi = i\psi^\dagger$, for example (it won't happen for a scalar) I think in any number of dimensions the dof will be halved, but idk for sure $\endgroup$ – innisfree May 11 '15 at 19:23
  • $\begingroup$ @innisfree I can see how it explicitly works out in the answer posted below but I can't seem to understand your argument exactly. Can you elaborate a bit as to why the equation of motion being of the first order implies halving of the number of degrees of freedom? If we take a free theory of (massive) complex vector bosons then also the same thing would happen? I seem to think so because $\pi_\psi\propto \psi^\dagger$ would even hold for free (massive) complex vector bosons, right? Thank you! :-) $\endgroup$ – Dvij Mankad May 2 '19 at 14:24
  • $\begingroup$ I suppose some gauge invariance would need to be sacrificed for having a massive complex vector boson theory but I think that is irrelevant here, right? $\endgroup$ – Dvij Mankad May 2 '19 at 14:26
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Dirac equation $\left( i \gamma^\mu \partial_\mu - m\right) \psi=0$ gives in momentum space $$ \left(\gamma^\mu p_\mu - m\right)\psi(p)=0 $$ which becomes in the rest frame $p^\mu= (m,\vec{0})$ $$ \left(\gamma^0 - 1\right)\psi(p)=0 $$ Now you can show that $\gamma^0 - 1$ is a projection operator of rank $D/2$, or more explicitly work in the Dirac representation where $$\gamma^0 = \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)\Rightarrow \gamma^0 - 1\propto \left( \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array}\right)$$ Then you can see how in the Dirac representation and the $CoM$ frame, the dirac equation just sets the lower half of the spinor's components to zero. Then since the counting of $DoF$ is independent of both the Lorentz frame and the Clifford algebra representation you get the desired result.

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  • $\begingroup$ True! Also by the same argument one can show that the same result holds for m=0. $\endgroup$ – Orbifold May 14 '15 at 7:07

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