I have this problem with the sign of the spinor for the antiparticle.

In the chiral basis, a spinor is represented by $\psi =(\psi_{L},\psi_{R}$). Now, we consider a particle with mass = 0, so Dirac's equation amounts to $i\gamma^{\mu} \partial_{\mu}\psi = 0$.

The equation for the right-handend component is $(i \partial_{t} + i \vec{\sigma} \cdot \nabla) \psi_{R} = 0 $ and the solutions are something like $(E - p \sigma^{3}) u_{R} = 0$. If we consider a plane wave moving in the z direction, the solution with positive energy is $\psi_r = \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} e^{-iEt + i \vec{p}\cdot \vec{x}}.$

This spinor is moving in the +z direction with spin + $\frac{1}{2}$. The helicity is $\frac{1}{2}$.

Now, the problem is that the corresponding spinor $v$ that describes the antiparticle should have negative helicity. But, for the antiparticle, $E = -p$, so this particle is moving in the -z direction, and the 2-component spinor that describes the antiparticle is now $v^{T} = (0,1)$, the spin is $- \frac{1}{2}$, and so the helicity is $\frac{1}{2}$.

Where did I lose the sign?

  • $\begingroup$ Near duplicate. $\endgroup$ Dec 14, 2021 at 22:55
  • 1
    $\begingroup$ for readers in the future : I have founded the solution of the problem: in the Feynmann-stueckelberg interpretation of negative energy, the spin operator for the antiparticle takes a minus sign! This is necessary to the theory to obtain the correct sign of the energy the antiparticle case. I suggest, the Thomson book where the problem is well explained. $\endgroup$ Mar 12, 2022 at 19:21


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