Is the Majorana condition $$ \psi = \psi^c = C \overline{\psi}^T, $$ general?

The point is often made that Majorana particles should be defined by CPT symmetry and not C as generally theories do not have C symmetry. Does this mean that the Majorana condition is wrong and we ought to use $$ \psi = \psi^{CPT}. $$ In which case, why does so much literature, for instance neutrino theory papers use the former definition and not the latter?

  • $\begingroup$ "The point is often made that Majorana particles should be defined by CPT symmetry and not C as generally theories do not have C symmetry." Can you give an example of such a claim, with more detail? $\endgroup$
    – knzhou
    Jan 23, 2019 at 19:27
  • 1
    $\begingroup$ The Majorana condition only proposes that particles are their own antiparticles, hence the definition. However, since QFT theories are defined by CPT symmetry then QFT for Majorana particles still preserve that symmetry, with the additional constraint specified in your first equation. $\endgroup$
    – Charlie
    Jan 23, 2019 at 19:48
  • $\begingroup$ @knzhou The issue is brought up in journals.aps.org/prd/abstract/10.1103/PhysRevD.26.3152 and various papers since. $\endgroup$
    – Kris
    Jan 23, 2019 at 20:34

1 Answer 1


I think this question mixes up notions involving particles and fields in quantum field theory.

Recall that at the classical level, charge conjugation may be defined for fields simply by complex conjugation, up to a possible additional linear transformation. If a theory has such a symmetry, it is usually straightforward to produce a quantum operator $\hat{C}$ that acts on particle states, and we define the antiparticle of a one-particle state to be its image under $\hat{C}$.

If the Lagrangian does not have the classical charge conjugation symmetry for the fields, it may be difficult or even impossible to define an appropriate $\hat{C}$ operator at the quantum level. (Sometimes one can define a $\hat{C}$ that isn't conserved, $[\hat{C}, \hat{H}] \neq 0$, but in many cases, such as in the Standard Model, there's just no way to define a $\hat{C}$-like operator.)

The paper you linked points out that this is problematic if $\hat{C}$ is used to define antiparticles. However, we always have the operator $\widehat{CPT}$, so we may use it to define the word "antiparticle" at the quantum level. This is a completely separate issue from the definition of a Majorana fermion at the level of classical fields. Unlike $\hat{C}$, complex conjugation is always defined, so there's no problem.

For much more about discrete symmetries, see here. For specifics on charge conjugation, see here.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.