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A positron is odd under charge conjugation and parity reversal but nevertheless even with respect to time reversal. Is a theoretical positron which would be odd under all three symmetries (C, P, T) physical?

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What do you mean by a "positron is -c -p"? I think you should read up a little bit on CPT and other symmetries in a QFT book such Peskin and Schroeder to clarify the notion of discrete symmetries and their action on elementary particles. – user346 Jan 16 '11 at 0:53
@space_cadet: Go easy on the guy. He's asking a genuine question, even if the notation isn't conventional. :) – Noldorin Jan 16 '11 at 1:07
I edited your question to clarify it for other posters. Hopefully I understood your meaning correctly. If not, you can certainly edit it yourself. – David Z Jan 16 '11 at 1:56
I'm sorry if I sounded harsh. That was not my intention. – user346 Jan 16 '11 at 6:47
  1. first of all, "even" and "odd" are different words than "positive" and "negative" - although both pairs of adjectives form a $Z_2$ group. So a positron is not "odd under charge conjugation". Instead, the charge conjugation of a positron is an electron. Because an electron is not "minus positron" (in the sense of vectors in the Hilbert space - they are two totally different vectors), it is not true that the positron is odd under C. A combination of the electron state and a positron state could be odd under C. However, if I remember well, the operator $C^2=-1$ for Dirac fermions, so the eigenvalues of $C$ are equal to $+i$ and $-i$. None of them is $-1$ so no fermionic state can really be "odd" under C.

  2. Second, C is the only one among the three transformations C,P,T that changes the "character" of the particle: it turns electrons into positrons and vice versa. On the other hand, the transformations P,T, whenever they exist in the first place, are meant to to change the positions of particles in the space and time only. P is the parity, moving objects from $(x,y,z,t)$ to $(-x,-y,-z,t)$, while T is the time conjugation, changing the sign of $t$.

  3. The action of $T$ on a specific state of an instant of time is simple; effectively, one only changes the sign of all the velocities. The result is that with these velocities, if T is a symmetry, the particles will be moving in the opposite direction than previously. So the evolution into the future will be equivalent to the previous evolution in the past. Such a transformation T doesn't "really" commute with the Hamiltonian: it turns it into minus itself because the Hamiltonian is really the generator of translations in time - and time changes the sign under T. So T is not really a symmetry of objects in space: even when it is unbroken, it is really a symmetry of histories in the whole spacetime. You shouldn't think that T acts on objects as a symmetry: it acts on their whole histories in spacetime if you want to see that it is a symmetry.

  4. T is not a linear operator but an antilinear operator which is related; $T\lambda|\psi\rangle$ is equal to $\lambda^* T|\psi\rangle$, including the complex conjugation of the $\lambda$.

  5. Contrary to some claims in this thread, CPT - which is also an antilinear operator because it includes T - is not the identity. It is a nontrivial operation and a nontrivial operation is the same thing as "not identity". CPT is a symmetry but a symmetry is something completely different than "identity". A symmetry is an element of a symmetry group; the identity is just an element of a symmetry group that doesn't do anything and that differs from all the genuine symmetries. It's kind of interesting why so many people still can't get this simple point.

So if one uses the word "odd" in the proper physics sense, there doesn't exist any particle that is odd under P, and probably C, because these two operators square to minus one. Even if they or some of their products squared to plus one, the eigenstates of the operation would be a strange combination of the positron state and the electron state. We don't usually like to combine states with different electric charges because they belong to different superselection sectors: they can't possibly evolve from the other one so without a loss of generality, one may assume that the initial state is an element of one superselection sector, and it is guaranteed that no mixture of another superselection sector will ever be added by the evolution.

What your question may have actually wanted to ask was whether the actual states obtained by C,P,T,CP, and CPT transformations exist. For electrons and positrons, they always do. However, for neutrinos, the C-transformed and P-transformed states don't exist at all. That's because neutrinos are left-handed, antineutrinos are right-handed, but there are no right-handed neutrinos and left-handed antineutrinos which is what you would get from the first two by C (or P). The transformation of a state by T always exists - because the transformation by CP exists as well and CPT is a symmetry (whose transforms therefore exist as well). T only changes the velocities of particles - not the positions or charges (or other internal characteristics) of the particles.

And if this is what you asked: The CPT transformation of a positron at $(x,y,z)$ moving in the direction $(v_x,v_y,v_z)$ is an electron at $(-x,-y,-z)$ moving in the direction $(v_x,v_y,v_z)$. Note that the signs of the velocities didn't change. Under CP, they would change the sign, but T changed the sign again. So if the positron is moving towards the origin, the CPT-transformed particle, the electron, will be moving away from the origin.

Best wishes Lubos

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Good answer largely. One point: I think you misunderstand the meaning of "identity" in this case. I simply use it in an every-day sense, not a mathematical one. (I'm not a big one for mathematical jargon. ;) – Noldorin Jan 16 '11 at 22:39

T reversal simply reverses momenta. So as long as you can fire antiparticles in any direction you want, applying a T operation is trivial.

Applying CP turns your left-handed particle into a right-handed antiparticle. Applying T to it then flips its momentum.

What the CPT theorem tells you then is that the physics of any process involving particles and antiparticles of particular handedness is identical to the physics of the process in reverse with particles swapped for their antiparticles and all handedness reversed.

For example, the probability that an electron neutrino will oscillate into a tau neutrino after a certain distance (See Neutrino Oscillations) is equal to the probability that a tau antineutrino will oscillate into an electron antineutrino within the same distance. (Where the handedness of each has been reversed)

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This is a needlessly complicated answer, in my view. Perhaps you could extract your main point? – Noldorin Jan 16 '11 at 1:04
I left the neutrino oscillation example in because I don't think T-reversal symmetry (or lack of it) can be fully appreciated by just considering what happens to a single particle. The power of the CPT symmetry is in predicting consequences for experimentally observable situations - as in the example of probabilities for oscillation. – dbrane Jan 16 '11 at 1:16
Fair enough. I do still think there's a really simple answer to this question, which I've presented directly in my answer. :) – Noldorin Jan 16 '11 at 1:18

Thank you all for your answers. I apologize if the question was a bit unclear. "No matter, never mind" if you misunderstood my original question which was “Real” antimatter i.e. -c -p -t , is it unphysical? A positron is -c -p but nevertheless +t. Is a theoretical -c -p -t positron physical?

Dear Lubos Motl,

I was visualizing the end result geometrically (without mathematics) in 3D like you simply and clearly and with enough details show using mathematics which I do not however master: An electron moving forwards in time in the form of a rightwards expanding logarithmic spiral and a leftwards expanding logarithmic positron coming from the exactly opposite position and moving backwards in time. From one observer's point of view (at x,y,z) both particles expand/rotate to the right from the observer's at (−x,−y,−z)  point of view they both rotate to the left. (I would like to add that if this visualization is correct than I have myself understood it!). Is it correct that a particle moving on a Moebius band surface undergoes these three transformations when it has moved one time around and restoring to the same particle when moving two times around as one sometimes can see in popular notes? My original question was colored by the known fact that we do not detect the same amount of antimatter as matter in nature. It has to do with an asymmetry of "histories in the whole spacetime"?

Many thanks for taking the time to answer!

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Welcome to physics.SE. Please don't ask questions as an answer, the idea of this site is "one question and its answer(s)". You can either edit your question (if the meaning is clarified but not changed) or ask a new question. Also you can leave comments on answers to your own questions without the 50 rep requirement so that persons gets notified – Tobias Kienzler Jan 17 '11 at 15:52
Yes, I see. I intended to comment below the answer there was not any comment's form ("add comment") visible. The only one active just now is this comment form. – curious Jan 17 '11 at 22:13

Actually, what you're describing - the transformation $(-c, -p, -t)$ - gives you exactly the same particle! (Not the antiparticle, not a "real" antiparticle - whatever that is...). In other words, a "time-reversed" positron is precisely an electron (and vice-versa).

CPT symmetry by definition implies that the transformation $(-c, -p, -t)$ is the identity transformation.

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Has anybody seen a time-reversed positron? – curious Jan 16 '11 at 0:47
Yes, it's called an electron. :) – Noldorin Jan 16 '11 at 0:50
By CPT definition yes. It comes to mind that Feynman spoke about time-reversed electrons as identical to positrons. Am I missing something? – curious Jan 16 '11 at 1:01
And the positrons we detect are going forwards in time. – curious Jan 16 '11 at 1:05
One comment on your last sentence: the statement of symmetry is that two different physical states, related by the symmetry, have equal probability amplitude. This is different from the statement that the two physical situations related by the symmetry are identical. – user566 Jan 16 '11 at 7:31

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