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I'm really confused by the helicity and handeness of antiparticles.

Consider the particle case, the plane wave solution is $\psi(x) = u(p)e^{-ip\cdot x}$, where $u^s(p) = \begin{pmatrix} \sqrt{p\cdot \sigma}\xi^s\\ \sqrt{p\cdot \bar{\sigma}}\xi^s\end{pmatrix} $. Assuming the particle is ultra-relativistic and moving along the $+\hat{z} $ direction, if the particle spins up, then: $$u^{\uparrow}(p) = \sqrt{2E} \begin{pmatrix} 0\\0\\1\\0 \end{pmatrix},h=1\Rightarrow Right-handed$$ , and $$u^{\downarrow}(p) = \sqrt{2E} \begin{pmatrix} 0\\1\\0\\0 \end{pmatrix},h=-1\Rightarrow Left-handed$$ , everything is quite simple.

The antiparticle case, $\psi(x) = v(p)e^{ip\cdot x}$, where $v^s(p) =\begin{pmatrix} \sqrt{p\cdot \sigma}\eta^s\\ -\sqrt{p\cdot \bar{\sigma}}\eta^s\end{pmatrix} $ with $\eta^{\uparrow} = \binom{0}{1}$ and $\eta^{\downarrow} = \binom{1}{0}$. Again with the assumptions of the particle is ultra-relativistic and moving along the $+\hat{z} $ direction: $$v^{\uparrow}(p) = \sqrt{2E} \begin{pmatrix} 0\\1\\0\\0 \end{pmatrix},h=-1\Rightarrow ?handed$$ , and: $$v^{\downarrow}(p) = \sqrt{2E} \begin{pmatrix} 0\\0\\-1\\0 \end{pmatrix},h=1\Rightarrow ?handed$$ I think that the spin up state should be left-handed and the spin down state should be right handed, but the spin of the spin state seems to be parallel with the momentum, i.e. right handed. Which is correct?

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If the spinor is only non-zero in the upper two components it has left chirality. If the spin is opposite to its momentum it has left helicity.

As an example only neutrino fields with left chirality participate in the weak interaction. But as you just saw left chirality means right helicity for antiparticles. That is why you'll sometimes hear people say neutrinos are left handed and antineutrinos right handed. They are talking about helicity in that case.

So to be clear, it's ambiguous what to put in place of the question marks in your question. If we are talking about chirality, $v^\uparrow$ is left and $v^\downarrow$ is right. If we are talking about helicity, $v^\uparrow$ is right and $v^\downarrow$ is left.

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  • $\begingroup$ So, does this mean that both neutrino and antineutrino are left helicity($h=-1$)? $\endgroup$ – LY3000 Mar 4 '17 at 9:16
  • $\begingroup$ No they both have left chirality. Neutrinos have left helicity, and anti-neutrinos right helicity. I agree the terminology is confusing when people say right and left and don't specify which concept. $\endgroup$ – octonion Mar 4 '17 at 19:30
  • $\begingroup$ One thing to note is that if we quantize the left-chiral Weyl field, it becomes a linear combination of a term that annihilates a left-handed particle and another term that creates a right-handed antiparticle. So saying that both the particle and the antiparticle are left-chiral doesn't sound right to me. $\endgroup$ – higgsss Mar 5 '17 at 7:31
  • $\begingroup$ In fact, nothing prevents us from constructing a right-chiral quantum Weyl field out of antiparticle annihilation operators and particle creation operators, although it will be quite awkward to represent the weak interaction using this object. $\endgroup$ – higgsss Mar 5 '17 at 7:37
  • $\begingroup$ @higgsss, In my answer I said that it is only the left chiral field that participates in the weak interaction. Which is true. Yes that involves a convention on which is the particle and antiparticle. But sometimes when answering a question I think it helps to keep the answer simple, although I understand if you don't like applying the word chiral to states rather than fields or spinors. $\endgroup$ – octonion Mar 5 '17 at 8:22
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Dirac's hole theory interpretation is useful here. The solution $v^{\uparrow}(p) e^{ipx} = v^{\uparrow}(p) e^{-i kz}e^{i\omega t}$ with $\omega = \sqrt{m^2 + k^2}$ is a negative-energy particle state of momentum $-k\hat{z}$, energy $-\omega$, and $S_{z} = -1/2$. One should annihilate this state from the vacuum to get the corresponding positive-energy antiparticle state.

The result of annihilating momentum $-k\hat{z}$, energy $-\omega$, and $S_z = -1/2$ from the vacuum is a state of momentum $k\hat{z}$, energy $\omega$, and $S_z = 1/2$. So the antiparticle state in question is right-handed.

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