I found the following explanation for chirality for spin 1/2 particles here

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What happens when you rotate a left- vs right-chiral fermion 360 degree about its direction of motion. Both particles pick up a -1, but the left-chiral fermion goes one way around the complex plane, while the right-chiral fermion goes the other way. The circle on the right represents the complex phase of the particle's quantum state; as we rotate a particle, the value of the phase moves along the circle. Rotating the particle 360 degrees only brings you halfway around the circle in a direction that depends on the chirality of the fermion.

My question is: Isn't the way the phase of the wavefunction changes related to the direction of rotation of the particle, which is an external parameter depending on how you rotate it? Is there a better explanation for chirality?

  • $\begingroup$ Oddly, in Ryder, QFT, page 46,formula $2.110$, the rotation transformation is the same for left-handed component and right-handed component (the difference is in a boost transformation). So, I begin to have a doubt about the article. $\endgroup$
    – Trimok
    Nov 15, 2013 at 18:46
  • $\begingroup$ Hmm, yea, that doesn't seem to make sense to me. If different chiral states were affected differently during rotations then if you had a state that's a superposition of equal parts left- and right-chirality, then by simply rotating by $90^o$ you would get rid of the state (the left-chiral state would pick up a $-i$ while the right-chiral state would pick up $i$, cancelling each other out). That makes no sense to me. $\endgroup$ Nov 19, 2013 at 16:13

1 Answer 1


Tarek (OP) e-mailed me to contribute to this thread. Here's the response that I gave him (slightly edited for clarity).

I see why this was confusing, my apologies! I was perhaps too glib in the post. Iwas implicitly talking about a chiral rotation but wanted to present it somewhat more intuitively.

Let me try to spell it out more carefully, and hopefully I can also explain why this is the same as what is being explained in QFT textbooks.

The point is the difference between helicity and chirality. Helicity has to do with angular momentum, i.e. spin up or spin down relative to the momentum of the particle. In the massless limit, helicity and chirality are identical. For our purposes, let's treat "chirality" as a label: I have a left-chiral fermion and a right-chiral fermion, just treat these as two different fermions. (This is probably not a great use of the term 'chirality', but hopefully it elucidates the point.)

A left-chiral fermion comes pre-packaged with a left-helicity and a right-helicity state. The anti-particle of a left-helicity, left-chiral fermion is a right-helicity, left-chiral fermion. (Again, you could argue what we mean by chirality here---but for our purposes, it's just a label for this fermion field and is distinct from helicity.)

Suppose I have a left-chiral fermion and a right-chiral fermion. These are two completely different types of fields. With some foresight, let's call the left-chiral fermion an electron and the right-chiral fermion a positron. We will clarify later the connection to the physical electron and positron (its antiparticle), but for now these are just names.

We have a total of four kinds of particles:

  1. left-chiral (electron) left-helicity
  2. left-chiral (electron) right-helicity
  3. right-chiral (positron) left-helicity
  4. right-chiral (positron) right-helicity

To emphasize once again, "left/right-chiral" is just part of the fermion's name. We could now ask how do #1 and #3 differ, since "chirality" is just some name. Note that this is the relevant comparison, if we want to compare apples-to-apples, then we should compare objects of the same helicity. A left-helicity state is obviously different from a right-helicity state, but how do the two left-helicity states above differ?

In the Ryder formula referenced in the comments to your question, you're comparing the left-helicity and right-helicity transformations. Indeed, both left-helicity and right-helicity transform with the same phase under a z-axis rotation. (Where z is assumed to be collinear with the momentum.)

But as I said, that's not the apples-to-apples comparison we're interested in. If we want to compare left-helicity states, then we should compare the left-chiral left-helicity state (left-helicity electron) with the right-chiral left-helicity state (left-helicity positron). But the "right-chiral left-helicity" state (the left-helicity positron) is the antiparticle of the "right-chiral right-helicity" (right-handed positron) state, so this thing picks up a complex conjugation.

This complex conjugation is precisely what flips the phase.

Thus: the left-helicity positron transforms with an opposite phase as the left-helicity electron.

If you want, this really all boils down to the following (which may be more confusing):

You have electrons (left-chiral) and positrons (right-chiral). Under a rotation about the z-axis, the left-helicity electrons pick up a phase that is opposite to the phase the left-helicity positrons pick up.

This may be confusing because it makes you pause and say "wait, i thought electrons and positrons were antiparticles of one another?" Here we need to clarify the words that we used. The `left-chiral' fermion (electron) and the right-chiral fermion (positron) mix when you introduce mass term. Just look at which two-component spinors are connected when you write out a Dirac mass. Indeed, you can think of a mass term as a 2-point vertex where a left-helicity electron goes in and a left-helicity "anti-positron" goes out. In the Dirac picture there's a total of four different states--just look at the four components of a Dirac spinor compared to the two components of a Weyl spinor. The four components of a Dirac spinor are the marriage of two Weyl spinors. Each Weyl spinor has a left- and right-helicity piece. Alternately, this is why you have plane wave spinors called u, ubar, v, and vbar.

Does this clarify things? (probably not, I know this is a bit of a thorny issue.)

It might help if you looked at the chiral particle content of the Standard Model. (Or, e.g. the particle content of the MSSM, which is most naturally written with respect to `chiral superfields' which contain Weyl spinors). Usually people write the particles as: Q, uR*, dR*, L, eR*, where the star indicates a conjugate so that all five fields are written as left-chiral fields. Note that the doublet L contains both a left-chiral electron (eL) and a left-chiral neutrino. The eR* has the quantum numbers of a positron. The fields eL and eR each contain left- and right-helicities. But the left-helicity component of eR picks up an opposite phase under rotations as the left-helicity component of eL.

You can also see the two-component bible for a more thorough discussion.

  • $\begingroup$ Thanks Flip for the answer. It seems from the first part of your answer that you consider chirality just as another quantum number that has different values for particles and anti-particles. Is that true? Otherwise, is it possible to explain chirality without referring to anti-particles at all? $\endgroup$
    – Tarek
    Nov 21, 2013 at 14:44
  • 1
    $\begingroup$ Calling the two states of chirality left and right gives the impression that it has an intuitive meaning similar to helicity, and this intuitive meaning is what I couldn't grasp till now. In nonrelativistic quantum mechanics, the phase associated with rotating around the z-axis depends only on the angle of rotation (the sense of rotation and hence the sign of the phase is encoded in the angle). In relativistic QM, I don't know how this works since I don't know how a relativistic spin operator is defined. $\endgroup$
    – Tarek
    Nov 21, 2013 at 14:46
  • $\begingroup$ As the Wikipedia article on chirality explains: helicity is a label for angular momentum state, while chirality is a label for representation under the Poincare group. $\endgroup$
    – Flip
    Nov 21, 2013 at 18:05
  • $\begingroup$ The discussion based on anti-particles makes use of more familiar phrases ("antiparticle") rather than referring to the conjugate spin-1/2 representation of SO(3,1). However, it is useful to see how the representation theory of the Poincare group gives you this structure--i.e. why there are (1/2,0) and (0,1/2) fermions. I believe that the Ryder book that @Trimok mentioned above discusses this a bit, as does the first few chapters of Weinberg's QFT volume I. $\endgroup$
    – Flip
    Nov 21, 2013 at 18:08
  • $\begingroup$ If you're confused about QM vs QFT, then I think the relevant reference is the beginning of Weinberg's book where he really goes through the construction of representations of the Poincare group. It can be a little difficult to slog through at first---feel free to use other references, but make sure you're looking into the method of induced representations (buzzwords are "Wigner decomposition", "little group," etc.). $\endgroup$
    – Flip
    Nov 21, 2013 at 18:14

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