The naive statement for the "CPT theorem" one usually finds in the literature is "relativistic theories should be CPT invariant". It is clear that this statement is not true as written, e.g. topological theories are typically not invariant under CPT. A much more precise statement of CPT is found e.g. in Freed's "Five lectures on SUSY", namely (paraphrased)

In a local QFT the CPT theorem states that representations of the connected component of the Poincaré group can be lifted to representations of the whole group (i.e., containing reflections and time-inversions).

This is much better, because it explicitly excludes topological theories (inasmuch as these have no propagating degrees of freedom, i.e., the Hilbert space does not contain irreps of Poincaré). It also deals with the Hilbert space directly, and so it applies to e.g. non-lagrangian theories.

That being said, I am still unsure what the "theorem" is really doing for us. Is it really a theorem, or rather an axiom? Are we to impose it when constructing theories, or should it follow automatically?

The main reason I am confused can be illustrated by considering the standard construction of supermultiplets. For example, if we take a massless multiplet whose highest weight has helicity 0, and act on the latter with the SUSY generators, we also find states of helicity 1/2 and 1. At this point, every book says that, by CPT, the correct multiplet must contain the CPT conjugate, i.e., states of helicity -1/2 and -1. One thus obtains the standard vector multiplet. This application of CPT exactly follows Freed's statement: the first half 0,1/2,1 is a good irrep of the connected component of (super)Poincaré, but does not lift by itself; we are to enlarge it by its conjugate so that the result does lift.

It seems that here we are imposing CPT invariance, rather than observing that it holds. In other words, what if I refused to include the CPT conjugate in the multiplet? Then CPT would be violated, and so the theorem is not really a theorem, for I can construct theories where it does not hold. Instead, it seems that, in constructing theories, I should impose CPT, i.e., it is an axiom. Is this understanding correct? Or perhaps it turns out that if I tried to construct a theory with the half multiplet only, i.e., helicity 0,1/2,1 (and no conjugate), the result ends up being pathological for some reason?

A similar situation is found when constructing non-supersymmetric states. Here a state of helicity +1 is typically packaged together with its CPT conjugate -1, but this is done for phenomenological reasons: as Weinberg explains (page 73), electromagnetic phenomena is observed to be invariant under parity, and so the existence of a state of helicity +1 requires the existence of one with helicity -1. But if we are interested in QFT for purely theoretical reasons, then it is perfectly sensible to try and construct theories of particles of helicity +1 that violate parity symmetry -- this is specially so for SUSY, where no phenomenological data exists!

  • $\begingroup$ Interesting and well formulated question! Sadly I don't have anything to contribute, other than the humorous note that "if we are interested in QFT for purely theoretical reasons"(Weinberg) then I would append, "then we are mathematicians, not physicists!" :P Just a joke of course, I come from theory myself. $\endgroup$ May 5, 2020 at 14:16
  • $\begingroup$ You may cook up notional theories without Lorentz invariance or CPT, to hone your skills, but reality banishes such from the scene. There is a minuscule cottage industry seeking Lorentz and CPT violations in nature, but they are not a discourageable lot. $\endgroup$ May 5, 2020 at 15:10
  • $\begingroup$ @CosmasZachos What if I want to violate CPT, but not Lorentz? is that possible? Do you happen to know any reasonable reference? $\endgroup$ May 5, 2020 at 15:59
  • $\begingroup$ Sorry I know of no references (that you wouldn't be aware of). I always understood the theorem as excluding that. Possibly related. $\endgroup$ May 5, 2020 at 16:00
  • $\begingroup$ Well, this one might be relevant. $\endgroup$ May 5, 2020 at 16:09

4 Answers 4


As you have mentioned, when we Wick rotate to Euclidean signature, the four-component Lorentz group $O(d,1)$ becomes the two-component $O(d+1)$. Let's suppose we have infinitesimal Lorentz symmetry. Then our Euclidean signature correlation functions enjoy the full symmetry of $SO(d+1)$, which is the connected component of the identity. Not all of these symmetries will descend to operators on the Hilbert space. However, transformations that fix a spatial slice will define such operators for us. An example of such an operator is a $\pi$ rotation in a plane containing one direction of space and one direction of time. This will give us our CRT symmetry (involving just a single reflection of space---we can get CPT for odd $d$ by combining with some space rotations).

You can decide for yourself whether or not to regard this as a proof. However, I do not know any counterexamples. You mention that some TQFTs don't have CPT symmetry. I suppose you are talking about chiral theories, but note that while CPT (or CRT) is anti-unitary, it also reverses the orientation of space, so something like a Chern-Simons term is actually invariant. Maybe you mean something else though?

By the way C, R, and T are all meaningless on their own (without extra assumptions on their existence). The way I like to think about the theorem is that it says: reflect your system somehow. Now reverse the direction of time. The theorem says there is guaranteed to be some internal transformation "C" which if we now apply C we have performed a symmetry. For free complex fermions for instance C happens to be charge conjugation if R and T are the usual ones. For real free fermions C is the identity.

Also on that note what we call CRT is possibly ambiguous up to internal symmetries (as well as up to rotations in space). In a recent paper of ours, we needed to actually pin down a "canonical" CRT, which is basically the one which comes "directly" from the argument I made above, for some anomaly matching reasons. We were able to do this by breaking all internal symmetries by hand, but you can also think about the analytic continuation more carefully to write an expression for the CRT transformation in terms of the boost matrix $M$ in the $x^0, x^1$ plane where we do our rotating. It is

$$O(x^0,x^1,...) \mapsto (i^F e^{i \pi M}O(-x^0,-x^1,...)e^{-i\pi M})^\dagger,$$

where $F$ is the fermion parity of $O$.

  • 1
    $\begingroup$ Indeed, your paper is one of the things that inspired this question; it's been on my mind ever since it came out. Can I ask about your remakr about TQFTs? I'm not sure what you mean by these always having CRT as a symmetry. For example, the semion-fermion $U(1)_2\times\{1,f\}$ is known to be time-reversal invariant, but the semion only $U(1)_2$ is not. Would you say $U(1)_2$ is invariant under CPT? Most people would say that such system does not have any anti-unitary symmetries at all, so it sounds weird to me to say that it does. (1/2) $\endgroup$ May 6, 2020 at 11:32
  • $\begingroup$ (2/2) It seems to me that one can definean anti-unitary operation, but such operation is not a symmetry -- it does not leave the observables invariant. If it is really a symmetry, then what does one gain when including the fermion $\{1,f\}$? If it was time-reversal invariant before, what's the point of the fermion? I thought that only with the latter does the system become time-reversal invariant? $\endgroup$ May 6, 2020 at 11:32
  • $\begingroup$ @AccidentalFourierTransform Thanks for your interest! I agree with you about time reversal being a special symmetry only certain Chern-Simons theories (and other theories) enjoy. For the Chern-Simons functional, the problem is that it depends on the orientation of spacetime, so either a time-reversal or a reflection will negate it. If we do both, it's invariant, in accordance with CRT. $\endgroup$ May 6, 2020 at 14:04
  • $\begingroup$ By the way I think the point of the fermion is so that some anyons can be tensored with it when we time reverse them. I learned this from Delmastro and Gomis' nice paper arxiv.org/abs/1904.12884 . You can consider theories like $U(1)_2$ as examples where R, and T are not symmetries, but RT is. $\endgroup$ May 6, 2020 at 14:06
  • $\begingroup$ Hmm I'm confused: if you can undo R using T, then RT is "trivial", but this is still an anti-unitary operator, right? so you are saying that in TQFTs the anti-unitary "do nothing" operation is a symmetry? i.e., RT is an anti-unitary symmetry which essentially fixes everything. Wouldn't such a symmetry, for example, require the path-integral to be real? Surely this is not always true, e.g. in the semion theory on the three-sphere $Z=i$, I believe. What am I missing? $\endgroup$ May 6, 2020 at 14:50

What makes CPT special

As an analogy, consider Noether's theorem. The justification for calling a conserved quantity "energy" doesn't come from considering any single theory by itself. It comes from the idea that time-translation symmetry combined with the action principle always gives a conserved quantity, along with a recipe for constructing the conserved quantity in terms of those general ingredients. In other words, the justification for calling it "energy" comes from looking at a whole family of theories.

Similarly, the justification for calling a symmetry CPT (instead of just PT-like) comes from the idea that a certain list of conditions (Lorentz symmetry, microcausality, ...) always imply such a symmetry. In other words, it comes from considering a whole family of theories, some of which might have more than one PT-like symmetry (modulo the full Poincaré group). The theorem is what picks out one of those PT-like symmetries as special, and that's the one we call CPT.

...but not as special as sometimes advertized

What is the most general formulation of the CPT theorem? I don't think that dust has settled yet,$^{[1]}$ but one condition that seems to be essential is Lorentz symmetry. CPT symmetry is not required for a QFT to be consistent, just like Lorentz symmetry is not required for consistency.$^{[2]}$ CPT symmetry, like Lorentz symmetry, is presumably something that we should only expect to hold as an approximation in a sufficiently small region of spacetime, in a QFT with a Lorentzian background metric.

${[1]}$ I can't rule out the possibility that somebody will discover some natural generalization of the CPT theorem that doesn't rely on any concept of spacetime symmetry but that implies the usual CPT theorem in the case of Lorentz-symmetric theories.

${[2]}$ We can always start with a lattice QFT, and the difficult question of the existence of a nontrivial continuum limit is beside the point here.

Theorem or axiom?

A CPT theorem singles out one PT-like symmetry (modulo Poincaré transformations) as being special, even in theories that have more than one. As an axiom, we might as well just call it a PT axiom, because the axiom doesn't care if a theory overachieves by having more than one PT-like symmetry, as long as it has at least one.

Consider Freed's description of the CPT theorem as stating that representations of the connected component of the Poincaré group can be lifted to representations of the whole group (at least the subgroup generated by an even number of reflections). Is the lift always unique? If not, then we have the same issue as before: we might as well just call it a PT-axiom, because the axiom doesn't care if a theory overachieves by admitting more than one such lift.


If membership in a (super)multiplet is supposed to be governed by spacetime symmetries, then we need to decide — as a matter of convention — whether or not that government should include PT-like symmetries. That's only an option if the theory actually has any PT-like symmetries, which we could enforce either by directly imposing the existence of such a symmetry as an axiom, or by imposing the conditions of a CPT theorem as axioms.

what if I refused to include the CPT conjugate in the multiplet? Then CPT would be violated...

If a theory doesn't have any states that otherwise could be used to "complete" the multiplet (C)PT-style, then it must violate one of the conditions of a (C)PT theorem — but not Lorentz symmetry, because that's implied by SUSY. So it must violate some other condition, like microcausality, and that would normally be called "pathological." I haven't tried to contrive any examples of this.

  • 2
    $\begingroup$ Thanks, I think your last paragraph hits the nail on the head. I still think the standard proof of the CPT theorem is slightly circular, because it already assumes that helicity +1 states are always accompanied by their CPT conjugates, states with helicity -1. (This is enforced by using vector fields, which always create both types of particles). But perhaps I am looking at simplified proofs, and I need to check a more rigorous proof? Anyway, I still have a lot of thinking to do. $\endgroup$ May 6, 2020 at 11:36

I'd say it's a theorem, or at least a physics theorem. The standard proofs (e.g., Weinberg Vol I) always have loopholes in them, but they cover the typical cases pretty well. Generally if you have a relativistic QFT with local degrees of freedom that violates CPT, you expect it to have trouble with causality. Maybe one also wants perturbativity hypotheses, so that one can safely reduce to the free fields.

EDIT: Misunderstood what OP meant by multiplets. When people are writing down examples, as in your SUSY example, they insert CPT conjugates to avoid these problems. I think if you apply the argument from Weinbergt to the free version of the SUSY model without CPT conjugates, you'll see that the space-like commutators don't vanish.

  • 1
    $\begingroup$ The issue with Weinberg's proof is that, to some extent, he is already assuming a big piece of what's he's trying to prove. He says that helicity +1 is always accompanied by helicity -1 for phenomenological reasons. This allows him to write the interpolating field for these particles as a vector field. Once you assume that, the rest of his argument follows. But I see no reason to impose parity invariance eg in SUSY. Thus, I don't know how to continue his argument. Of course, it is hard to imagine how to construct an interpolating field for helicity +1 only; a vector field will not do the job. $\endgroup$ May 5, 2020 at 18:57
  • $\begingroup$ Nonetheless, I think you can take the fields you're looking at in your example and compute their commutators. It's just an operator calculation. $\endgroup$
    – user1504
    May 5, 2020 at 18:59
  • $\begingroup$ That's the issue: which fields? If you impose parity invariance, then you need CPT conjugates, and the field is a standard vector field, from which you can indeed compute commutators. But if you do not impose parity invariance, then you can consider a particle of helicity +1 with no conjugate, and so you cannot write down a vector field for it. (A vector field automatically creates both +1 and -1; thus, any argument that requires vector fields already assumes the conclusion). In order to run your argument, one needs a field that creates only +1 states. Is there any field that does that? $\endgroup$ May 5, 2020 at 19:07
  • $\begingroup$ If there aren't any fields, is it a field theory? Something has to encode local causality. Which reminds me of something: The CPT theorem is supposed to hold in the context of S-matrix theory, where local causality is expressed via analyticity instead of fields. If I've traced the literature correctly, the argument comes from HP Stapp, Derivation of the CPT Theorem and the Connection Between Spin & Statistics From the Postulates of the S-matrix Theory, available at escholarship.org/uc/item/1nb798td $\endgroup$
    – user1504
    May 6, 2020 at 1:32
  • $\begingroup$ I don't have anything against fields; indeed, they better be there! The problem is that I don't know which field should create states of helicity +1 only. A vector field automatically creates both +1 and -1. So we need a different type of field -- perhaps something like a constrained vector field? But anyway, the reference looks nice, I'll give it a look, thanks! $\endgroup$ May 6, 2020 at 11:25

CPT invariance is specifically a theorem in quantum field theory (your second statement is of course much better). Basically it says that antiparticles (the T part in the Feynman-Stuckelberg interpretation) have opposite charge and parity to the corresponding particles. There is a better statement, an outline argument, and numerous references at this wikipedia page


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