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:

\begin{align} u^{\uparrow}(p) &= \sqrt{2E} \begin{pmatrix} 0\\0\\1\\0 \end{pmatrix}, &h&=1 &&\Rightarrow \text{Right-handed}, \\ u^{\downarrow}(p) &= \sqrt{2E} \begin{pmatrix} 0\\1\\0\\0 \end{pmatrix}, &h&=-1&&\Rightarrow \text{Left-handed}, \end{align} 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:

\begin{align} v^{\uparrow}(p) &= \sqrt{2E} \begin{pmatrix} 0\\1\\0\\0 \end{pmatrix}, &h&=-1&&\Rightarrow \text{?-handed}, \\ v^{\downarrow}(p) &= \sqrt{2E} \begin{pmatrix} 0\\0\\-1\\0 \end{pmatrix}, &h&=1 &&\Rightarrow \text{?-handed} \end{align}

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?


4 Answers 4


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.

  • $\begingroup$ So, does this mean that both neutrino and antineutrino are left helicity($h=-1$)? $\endgroup$ Commented Mar 4, 2017 at 9:16
  • 1
    $\begingroup$ 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
    Commented Mar 4, 2017 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
    Commented Mar 5, 2017 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
    Commented Mar 5, 2017 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
    Commented Mar 5, 2017 at 8:22

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.


For clarity one might avoid using the words "right" and "left" altogether, and talk about positive and negative chirality; positive and negative helicity.

Helicity is +ve if $\sigma \cdot p > 0$ and negative if $\sigma \cdot p < 0$.

For a Weyl spinor the chirality has one sign if the spinor matrix elements $\langle s| \sigma^a | s \rangle = U^a$ give the contravariant (as opposed to covariant) components of a 4-vector, and the chirality of $|s\rangle$ has the other sign if $\langle s| \sigma^a | s \rangle = U_a$, i.e. covariant components. But it is a matter of convention which case is said to be of positive or negative chirality, and it is a matter of convention whether positive chirality is said to be "right" or "left". I think the convention is that "right" is associated with "positive" which is in turn associated with "contravariant".

In this convention, a positive chirality Weyl spinor, when written as a two-component complex vector, transforms as $$ s \rightarrow \Lambda s $$ under a change of reference frame, where the Lorentz transformation $$ \Lambda = \exp(i \sigma \cdot \theta/2 - \sigma \cdot \rho/2) $$ in which $\sigma$ is the three-vector of Pauli matrices, $\theta$ is a three-vector defining an axis and amount of rotation, and $\rho$ is the rapidity three-vector.

(I included this information about $\Lambda$ in order to be precise about the terminology in this answer.)

Coming now to Dirac spinors, I think a common but not universal convention in the chiral representation is to put the negative chirality Weyl spinor at the top, and the positive chirality Weyl spinor at the bottom. Thus one sees things such as: $$ \left( \begin{array}{c} \phi_L \\ \chi_R \end{array} \right). $$ This is the convention implied by the answer here provided by octonion. However I think it may be useful to note that some literature may adopt the other possible convention, and put $\chi_R$ at the top.

Under the above convention the Dirac matrices read $$ \gamma^0 = \left( \begin{array}{cc} 0 & I \\ I & 0\end{array} \right), \;\; \gamma^i = \left( \begin{array}{cc} 0 & \sigma^i \\ -\sigma^i & 0\end{array} \right) $$ but I have also seen them written in the version transposed from this, which implies the Dirac spinor is then being written the other way up.

I think the only way to get all these signs right in ones own work is to realise that it is not easy, and then painstakingly track your own expressions through to the point where you can do a Lorentz transformation and see what you think happened to the momentum of your particle.


For a particle, the helicity and chirality are identical while for an antiparticle, the helicity and chirality are opposite in the massless case. So, for massless particles has left-handed chirality will also have left-handed helicity. For massless antiparticle has left-handed chirality will have right-handed helicity. This fact is based on using the Dirac equation for massless particles/antiparticles.


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