Questions (very) closely related to this one have been posted dozens of times, but the joint information (including even lecture notes and books) is incredibly contradictory, so I state my specific doubt again.

Take for instance the line of thought that ends with the conclusion that the decay $\pi^0 \to e^+ e^-$ is chirality suppressed:

Taking the rest frame of the pion, conservation of momentum and angular momentum tell us that the helicity of both end-state particles must be the same, so consider, for example, they to be left-handed

$$ \text{Hel}(e^+)=\text{Hel}(e^-)=L $$

Here, since $M_\pi \gg m_e$, the electrons are ultra-relativistic, so that helicity and chirality become interchangable.

Now, I have read/listened many times the statement that "Left-chiral anti-particles are Right-handed and vice-versa" -- I was never able to understand what's the exact math behind the semantics here although I'm aware of where it comes from. In any case, if this is to be true, then we have that the chiralities in our example must be

$$ \text{Chir}(e^+)=R, \;\;\;\; \text{Chir}(e^-)=L. $$

But we expect the possible main contributions to the process to be vector-mediated. We also know, however, that a vector current is of the form

$$ \bar{e}_L\gamma^\mu e_L, $$ (where $\bar{e}_L\equiv (P_L e)^\dagger \gamma^0 $) from which we should be able to see that the process is chirality-suppressed.

My question: does the symbol $\bar{e}_L$ (as defined above) represent a left or right-chiral anti-particle? If the reasoning above is correct, then it should represent a left-chiral one, for the process to be suppressed. Another possibility is that it represents a right-chiral anti-particle, and what is wrong is my operational interpretation of the statement quoted 2 paragraphs above.

The problem is a very simple one but which has troubled me for quite some time.

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    $\begingroup$ Have you compared this electromagnetic decay with the simpler suppression of $\pi^+\to e^+\nu_e$, where the weak current means that only one chirality is involved? $\endgroup$
    – rob
    Oct 22, 2021 at 0:22

1 Answer 1


Rant and introduction

There are a lot of wrong statements in lecture notes about chirality and helicity, and plenty of wrong answers on this site as well. I've personally written too many answers on this subject (1, 2, 3, 4), but here's a fifth that lays out the ideas again, in case people want it for future reference.

  • Quantum field theory deals with two basic things: particles and fields. Particles are physical objects with momentum, spin, helicity, and so on. They can be created or annihilated using operator-valued fields. Describing dynamics with field operators is nice because, e.g. you can make locality manifest using a local Lagrangian.
  • Suppose your theory contains a massless right-helicity particle of electric charge $1$. If it is a relativistic theory, then the CPT theorem implies that your theory must also contain a left-helicity particle with the opposite internal quantum numbers, i.e. it has electric charge $-1$). This other particle is typically called an antiparticle.
  • This set of two degrees of freedom can be described with a left-chiral massless Weyl spinor field $\psi_L$ of charge $1$. In general, a left-chiral Weyl spinor field annihilates a left-helicity particle with the same internal quantum numbers, and creates a right-helicity antiparticle with the opposite internal quantum numbers.
  • The complex conjugate of this field is a right-chiral Weyl spinor field. It has the roles of the particles reversed: it annihilates the right-helicity antiparticle, and creates the left-helicity particle.

Therefore, "the chirality of a particle" is meaningless. If you have a left-helicity massless particle, for example, it could be annihilated by a left-chiral Weyl spinor field or created by a right-chiral Weyl spinor field. The particle doesn't care which kinds of fields you use to describe it. Any Lagrangian containing Weyl spinors, for instance, can be trivially rewritten in terms of only left-chiral or right-chiral Weyl spinors (or any mix of the two) by sprinkling in complex conjugates.

The statement "chirality is the same thing as helicity for massless particles" is also meaningless. When people say this, they're trying to convey that a massless left/right-chiral field annihilates a left/right-helicity particle. But it also creates a particle of the opposite helicity. That's the origin of your second confusing statement, "left-chiral anti-particles are right-handed", which contains an honestly impressive number of mistakes in just seven words. The closest correct analogue of this statement is "a right-chiral field creates left-helicity particles with the opposite internal quantum numbers".

Why do people seem to get this wrong, time and time again? One reason is that these arguments are not "load bearing". Many people know how to calculate cross sections, and you don't really need to understand the difference between chirality and helicity to turn the crank correctly. At the end you get the expected result, so there's no need to think harder. Another reason is that often, people who understand the difference perfectly well will elide the distinction between particles and fields to make the subject easier for new students. (I don't think it helps though, because it tends to get them very confused later, losing time on net.)

Answering the question

Anyway, with that rant out of the way, let's focus on your example. For simplicity, let's suppose the electron is massless. Let's focus on the part of the current you pointed out, $$\bar{e}_L \gamma^\mu e_L.$$ In this notation, $e$ is a four-component Dirac spinor field, $e_L$ is the left-chiral part, and $\bar{e}_L$ is its conjugate. Therefore, using the rules laid out above,

  • $e_L$ annihilates a left-helicity electron and creates a right-helicity positron
  • $\bar{e}_L$ creates a left-helicity electron and annihilates a right-helicity positron

We want to use this vertex to create two particles, so it must create a right-helicity positron and a left-helicity electron. But this violates angular momentum conservation. You can also run the same argument with the other part of the current $\bar{e}_R \gamma^\mu e_R$, in which case the helicities flip and the same conclusion applies. Therefore, the process cannot occur for a massless electron, so it is helicity suppressed. (In reality, the decay $\pi^0 \to e^+ e^-$ occurs through a loop diagram with two copies of the electromagnetic vertex, but the same logic applies.)

  • 1
    $\begingroup$ Despite admittedly not nearly fully grasping the difference between field (the mathematically well-defined entity) and particle (the well-localized physical excitation), the picture regarding the specific operational confusion is what I had in mind, but I needed to read out loud the '$\bar{e}_L$ anihilates a right-HELICITY positron' to be confident after absorbing so many non-coherent expositions. Thank you! $\endgroup$
    – GaloisFan
    Oct 22, 2021 at 3:03

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