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The Dirac equation for a particle with charge $e$ is $$\left[\gamma^\mu (i\partial_\mu - e A_\mu) - m \right] \psi = 0$$ We want to know if we can construct a spinor $\psi^c$ with the opposite charge from $\psi$. This would obey the equation $$\left[\gamma^\mu (i\partial_\mu + e A_\mu) - m \right] \psi^c = 0$$ If you know about gauge transformations $$\... 17 With one minor qualification, the answer is yes: the CPT theorem holds for all spacetime dimensions. The qualification is that the P in CPT should be interpreted as a reflection of an odd number of spatial dimensions. (The simplest choice is to reflect just one spatial dimension.) If the total number of spatial dimensions is odd, then this is the same as ... 12 Conservation of energy follows from invariance under translation in time, not inversion. This symmetry states that no matter when you do your experiment, it will give the same results. All isolated systems obey this symmetry (and therefore conserve energy) and no violation of it has ever been detected. (Needless to say, it would be a huge event if it were.) ... 12 As you've already noted, we cannot detect cosmic antimatter from its spectrum. As the answers at How would we tell antimatter galaxies apart? indicate, there are two ways we could detect cosmic antimatter. Firstly, we would see the tell-tale 511 keV gamma ray signature of electron + positron annihilation reactions coming from the border of the antimatter ... 10 How about just testing the two different cases? I.e. if \mu\not=0 then the LHS becomes \begin{equation} (\gamma^\mu)^\dagger= (\gamma^i)^\dagger= -\gamma^i \tag{see below} \end{equation} while the RHS becomes \begin{equation} (\gamma^\mu)^\dagger=\gamma^0\gamma^i\gamma^0 = -\gamma^0\gamma^0\gamma^i=-\gamma^i~~~~~~~~ (\text{OK}). \end{equation} For \... 10 Mathematical physicists will tell you the question you're asking has no answer: only CPT as a whole has a rigorous definition. That means that practicing physicists, who consider concrete problems, are free to define it however they want! So while I don't know the mathematical niceties, let me lay out what I think particle physicists usually mean when they ... 10 It is time reversible, you just have to remember to time reverse the field too. Absorption is the time reverse of emission. In the fully time reversed situation, the charge absorbs coincidentally perfectly arranged incoming EM radiation, which causes it to spiral outward. In the standard derivation of EM radiation, you find that to satisfy the equations, ... 9 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 ... 8 Antihydrogen atoms have been created in the lab and their basic spectral characteristics confirmed as identical to hydrogen. So we cannot tell by directly observing an object. But we infer it from the fact that no gross interactions between matter and antimatter have been observed. A full answer is given at How would we tell antimatter galaxies apart?, but ... 6 Surely someone has mulled over why the universe might exhibit such a non-intuitive and thus interesting asymmetry? As always, when it comes to valid and important physical theories, the reason why the Universe has non-intuitive features is simply that the intuition is wrong. Arguments based on wrong intuition are irrational and unscientific. Rationally ... 6 Two basic misunderstandings here: 1) under time reversal the velocity changes sign but not the acceleration or the force, which remains attractive, and 2) symmetry of the laws does not imply that a particular situation is symmetrical, only that the transformed situation (which may be different) is also a valid solution of the laws. Which it is - there is no ... 6 There is not, because the combined transformation CPT is a symmetry of all Lorentz-invariant systems. The P-violating decay distribution observed by Wu et al. is also a C-violating distribution, because polarized anti-cobalt would have had the opposite sign of asymmetry. (However no one has ever made, or probably will ever make, polarized anti-cobalt,... 6 So this is really the twist in the construction of QFT that is still sometimes misunderstood. Historically, people like Dirac noticed that solutions to relativistic field equations such as the spinor equation for \psi(x) admitted apparent negative mass-energy densities. However, once you go into QFT, either from a "second quantization" or a Fockian bottom-... 5 In classical mechanics, it is often possible and convenient to describe a system with an object called a Lagrangian (in that it governs a system's behaviour, the Lagrangian is similar to a Hamiltonian). Like the Hamiltonian, the Lagrangian ought to be real - and any terms inside the Lagrangian ought to be Hermitian. In quantum field theory (QFT), the ... 5 Let us here just consider the classical case (\hbar=0). I) Noether's theorem does not work for discrete symmetries like time reversal symmetry,$$\tag{1} T: t ~\longrightarrow~ -t, $$cf. e.g. this Phys.SE post. II) Instead, energy conservation follows from time translation symmetry$$\tag{2} t ~\longrightarrow~ t + a, \qquad a~\in~ \mathbb{R}, $$... 5 The spin-statistics thing isn't a problem, it is a theorem (a demonstrably valid proposition), and it shouldn't be addressed, it should be understood and celebrated. The Higgs field gives us interactions between chiral fermions and the Higgs, yh\cdot \chi_\alpha\eta^\alpha which produces mass terms m \chi_\alpha\eta^\alpha if the Higgs field has a ... 5 I am not very familiar with details of the proof of the CPT theorem, but could it be that T is anti-unitary? For example consider a bosonic QFT with a Klein-Gordon field \phi and a vector field A^\mu, and take the interaction Lagrangian$$\mathcal L_\text{int} = \frac{1}{M^2} \epsilon^{\mu\nu\sigma\rho} (\partial_\nu A_\mu) (\partial_\rho \phi) (\...

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$PT-, T-, P-$ transformations refer to subgroup of discrete transformations of the Lorentz group. They transform connected components of the Lorentz group between each other ($PT$ transformation transforms $L^{\uparrow}_{+}$ representation to $L^{\downarrow}_{+}$). In general, they can't be represented as the special case of rotation, which refer to subgroup ...

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A partial answer, is that supposing the gamma matrices, block-diagonal , as $\begin{pmatrix}A&\\&\epsilon A\end{pmatrix}, \begin{pmatrix}&A\\\epsilon A&\end{pmatrix}$, where $A$ is hermitian or anti-hermitian, and $\epsilon =\pm1$, give constraints on $A$ and $\epsilon$ due to $(\gamma^0)^2= \mathbb Id_4, (\gamma^i)^2= - \mathbb Id_4$. For ...

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I slightly deviate from your notation and use $\phi$ to denote the scalar field as its more standard. Also I should point out that quantum fields are operators and thus under a transformation they get acted on from both the left and the right. The complex scalar field is given by, \begin{equation} \phi (x) = \int \frac{ \,d^3p }{ (2\pi)^3 } \frac{1}{ \...

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Electric charge is not special. Every charge is replaced by its opposite under the C conjugation. For instance, electric charge goes from positive to negative and vice versa. Color charge goes from blue to antiblue and vice versa. Mutatis mutandis with green/red color charge. Every charge is sent to its anti-charge.

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It looks like Terning and Weinberg are talking about answering slightly different question. Terning is first answering the question, is there an N=2 hypermultiplet that is CPT self conjugate? The answer is yes, if the fermions transform in a real representation of the gauge group. If the gauge group representation is complex we have to add the CPT conjugate ...

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From your link: Dirac came out of his depression when he received a phone call from his friend Abdus Salam, saying: "Relax Paul, my friend Nino Zichichi has discovered the antideuteron". This is the opposite of the summary in your question (v1) which suggests that Dirac believed for some reason that the antideuteron shouldn't occur and was upset when it ...

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Space-time symmetries are symmetries with respect to coordinate transformations. In non-relativistic physics, space and time is assumed to be Galilean invariant. Galileo symmetries form a ten parameters continuous group. These parameters label three space translations, time translation, three rotations and three Galileo's boosts. It also includes discrete ...

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C, P, and T need not all exist in a quantum field theory, and they may not even be unique. Only CPT is guaranteed in a general unitary QFT. In the standard model for instance, $CP$ and $T$ are not symmetries but their composition is. A simple example, consider a 2-component real fermion $\psi$ in 1+1D. The massless free Lagrangian for this field is i \psi^...

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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 ...

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Let me try to clear things out. You are mixing the symmetries of spacetime with the symmetries of the interactions of the Standard Model. The first, are symmetries of spacetime to which according to special relativity (in relation to the Standard Model) is the Poincaré group of symmetries. So spacetime itself is translationally invariant, isotropic, parity ...

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The sentence in Peskin's and Schroeder's book that "the weak interactions preserve CP and T" is a bit misleading but there is a sense in which it is right. Experimentally, CP and T is known to be violated and CPT is always a symmetry. Theoretically, CPT is always a symmetry, too – it's proven by the CPT theorem. The CPT transformation is effectively a ...

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surely someone has mulled over why the universe might exhibit such a non-intuitive and thus interesting asymmetry? Oh yes, definitely. I have for one (though I haven't made a significant contribution to the question)! :) There are a number of "left-right symmetric" models out there which usually involve a group like $SU(2)_L \times SU(2)_R$ where the $SU(2)... 3 If you define$C:q\rightarrow -q$,$P:(x,y,z)\rightarrow (-x,-y,-z)$and$T:t\rightarrow -t$, then all the Maxwell eaquations are invariant under$C$,$P$,$T\$ or any combination of them. To see this you just have to notice how the transformations act on sources and coordinates. The charge conjugation acts non trivially as \begin{align} C\rho&=-\rho,\\ ...

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