32

The point is that whether we call it 'positive' charge or 'negative' charge makes no difference, as long as we are consistent. If we decided to label the charge of a proton as 'negative' then, to be consistent, we must also relabel the charges of the quarks (i.e. d would become +1/3e, and u would become -2/3e). In which case your question is void.


11

As for your question on how to get negative charges in the nucleus, this is easy. Just use anti-protons (and anti-neutrons). The formers quark content is simply $\bar{u}\bar{u}\bar{d}$, so it consists of two anti-up and one anti-down quark. The charge of an anti-up quark is the opposite of that of an up-quark, so -2/3 and, analogously, +1/3 is the charge of ...


9

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 ...


6

A Dirac spinor can be written as the sum of two Weyl (chiral) spinors, eigenstates of $\gamma_5$, $$ \psi= \psi_L + \psi_R,\tag{1} $$ but also, alternatively, as the sum of two Majorana (self-conjugate) spinors, eigenstates of C (and $i\gamma_2$) $$ \psi= \chi + i\omega= \chi^c + i\omega^c,\tag{2} $$ where each Majorana component is real in the Majorana ...


6

The definition of a real representation is not merely that it is isomorphic to its conjugate. Both real and pseudoreal representations are isomorphic to their conjugates. This isomorphy forces the existence of an equivariant anti-linear map $$ J : V \to V, $$ where $V$ is the representation space, which simply is the isomorphism $\phi : V\to V^\ast$ ...


6

Answer to part 1 There is no natural pairing between particles and anti-particles at the level of individual states (any such pairing is purely a matter of convention), but there is a natural pairing between particle species — that is, between species and their anti-species. That's because Poincaré symmetries don't permute species, even though they ...


5

Representation of $SU(2)$ is pseudo-real. Which means, if $\mathbf{[2]}$ and $\mathbf{[\bar{2}]}$ are the fundamental and anti-fundamental representation of $SU(2)$, then there exists an anti-symmetric matrix $\cal{C}$, which connect both of them, as $\cal{C}\mathbf{[2]}\cal{C}^{-1}=\mathbf{[\bar{2}]}$. Another way of saying this, both $\mathbf{[2]}$ and $\...


5

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,...


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) (\...


4

Your equations (1) (2), saying $\delta \psi =-\frac{1}{4}\lambda ^{\mu \nu}\gamma _{\mu \nu}\psi$ with or without $^c$, just says that both $\psi$ and $\psi^c$ are in the same representation, namely $(1/2,0) + (0,1/2)$. The third equation (3), saying $(P_L\psi)^c=P_R \psi^c$, just says that the charge conjugation swaps the two irreducible components of ...


4

When investigating spinorial representations of the Lorentz group, one finds that if $\Psi$ is a left-handed Dirac spinor, then $\Psi^c = -i\gamma^2\Psi^*$ is a right-handed Dirac spinor. At that moment, however, the physical meaning of the operation is latent. Having quantized the Dirac spinor, $$ \Psi(x) = \int \frac{d^3p}{(2\pi)^3\sqrt{2E_p}} \sum_{s=1,2}...


4

The short answer to the question, "What happens to the Lagrangian of the Dirac theory under charge conjugation?" is, "Nothing." It is invariant with respect to charge conjugation. Before getting to the longer exposition, I'd like to point out a potential misunderstanding about the nature of invariance of the equations of motion under symmetry ...


4

Here is a nice late-undergrad or early-grad-school lab report on the determination of spin states in nuclear decays. The references to that paper (from 1940 and 1950) are relatively accessible, too. As your pullquote says, you get $J^{PC}$ from measuring angular correlations. If a particle at rest decays into two daughters, the angular correlation is ...


4

The positive and negative charges assigned to protons and electrons respectively are by convention. There is no specific reason for making the electron negative. Just like gluons have colour charge, similarly in order to show that there are particles similar to the electron which rebel each other, they've been marked as negative charge. They could have been ...


4

The operation that maps particles to antiparticles is just $C$. (This is somewhat of a simplification. A better thing to say is that in theories with $C$ symmetry, you can pair particle states with the same spacetime quantum numbers but the opposite internal quantum numbers. When $C$ is violated, there may exist no pairing that gets the quantum numbers right....


3

You missed a rather important conservation law: $$m_nc^2 < m_pc^2 + m_\pi c^2$$ But in general, yes, the only way to really confirm that a reaction is allowed is to check all the conservation laws. This is why we have tables of decay modes. Other people have checked the conservation laws (and done experiments to back it up) so you don't have to. I mean, ...


3

Complex conjugation has nothing to do with charge conjugation. Charge conjugation flips quantum numbers, which don't appear at all in the standard Schrodinger equation. The actual symmetry related to complex conjugation is time reversal. However, to actually perform time reversal, you must also replace $i$ with $-i$, so the time reversed Schrodinger ...


3

In the standard model, there is no elementary spin 0 boson being electrically charged (but there are many charged spin 0 composite particles). However, in many extensions such as supersymmetry, there are such particles: the scalar partner of the electron, the selectron carries the same charge as the electron. The anti-selectron is the spin 0 partner of the ...


3

Explicit expressions for the Euclidian signature are given in the following Hitoshi Murayama lecture notes (Section 1.3). The expressions are given in the Pauli matrix tensor product basis.


3

As to your question, yes, the QED Lagrangian is indeed invariant under charge conjugation. You may have found differently because your transformations under charge conjugation are faulty. The prefactors are correct, however, under charge conjugation $\psi$ goes to $\bar{\psi}$ and vice versa, i.e. $$ \hat{C} \, \psi \, \hat{C} = -i(\bar\psi \gamma^0 \gamma^...


3

Yes, you can show this using only the fact that the Clifford Algebra has a unique representation up to similarity transformation in any dimension. This is shown in the first few pages of http://arxiv.org/pdf/hep-th/9811101.pdf Then you observe that if $\gamma^\mu$ obeys the clifford algebra, then so does $-(\gamma^\mu)^T$. $\mathcal{C}$ is then defined as ...


3

When you ask "how can we determine...", that sounds like an experimental question. But when you ask "don't we expect infinitely many soft photons?", that sounds like a theoretical question. I'll focus on the the experimental question. The facetious answer to how you determine the number of photons in a decay is that you measure them. That's obviously not ...


3

Majorana field can not be charged under the U(1) group. However, it can be charged under the $\mathbb{Z}_2$ subgroup of U(1). Under the $\mathbb{Z}_2$ transformation, $\psi\to\psi'=-\psi$, any Majorana fermion Hamiltonian (such as the Majorana mass $m\psi\psi$) should be invariant under this $\mathbb{Z}_2$ symmetry, which is also called the fermion parity ...


3

Recall that in an elementary Linear Algebra course, you proved that for two matrices $A,B$, $(AB)^T=B^T A^T$. Do this in component notation. For this proof you will commute the component of $A$ past the component of $B$. In the case where the components of $A,B$ are anti-commuting (Grassmann) numbers, the same step will give you an extra minus sign. Hence, ...


3

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^...


3

In QED, for example, charge conjugation commutes with the Lorentz group. It's an "internal" symmetry, not part of Lorentz (or Poincaré) symmetry. However, a different kind of connection exists between charge conjugation and the Lorentz group, via the CPT theorem. The CPT theorem says that every relativistic QFT (satisfying certain axioms) has a symmetry ...


3

Weinberg vol 1 Eqs (2.6.7) to (2.6.12) for $P$ and $T$.


2

I think it's a matter of choice. If you look through several books you'll see all the possible combination $C\Psi(x)C$, $C\Psi(x)C^{-1}$, $C\Psi(x)C^{\dagger}$ (and the same for $P$ and $T$). I think it all comes down to the representation you are using. Like it is said in the book of Sterman (page 524) :"The precise nature of $T$ depends on the ...


2

Yes, when we want to obtain the equation of motion using Euler-Lagrange equation, we should treat $\psi$ and $\psi^c$ independent, but $\overline{\psi}$ and $\psi^c$ dependent. The reason for this is that we can simply expressed $\psi^c$ in terms of $\overline{\psi}$ by $$ \psi^{c}=C\overline{\psi}^{T}, $$ where $C=-i\gamma^{2}\gamma^{0}$ is the charge ...


2

I) The Dirac spinor $\psi$ and its complex conjugate $\psi^*$ are not independent variables, but in some calculations one can treat them as such. For the similar question about a complex scalar field $\phi$ and its complex conjugate $\phi^*$, see e.g. this Phys.SE post. II) The charge conjugated field $\psi^{c}=-i\gamma^{2}\psi^{*}$ is tied to the complex ...


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