# Spin sign for antiparticle

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?

• Near duplicate. Dec 14, 2021 at 22:55
• 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. Mar 12, 2022 at 19:21