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The decay of a neutral $\pi^0$ to three photons would indeed violate charge conjugation. The charge conjugation argument goes as follows: The reaction $$\pi^0 \to 3 \gamma$$ is mediated by electromagnetism. QED has a charge conjugation symmetry, so you should be able to apply a charge conjugation to both sides of the equation. Under charge conjugation, ...

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I don't know if I'm right and it's not directly my topic, but naively guessing I think that sea quarks emerge from fluctuations in the vacuum, for example as a pair of a quark and its antiquark. And as such, because of conservation of angular momentum, each of these pairs then should have zero total angular momentum.

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I want to refer you to Weinberg QFT1, if you have not read it yet. Below is my attempt to answer your question in the formalism addressed there. You are classifying particles by representations of the Poincaré group. A one particle state has to be transformed under an element of this group. Part of this transformation just changes the momentum, the other ...

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@1. As for the $m \rightarrow 0$ limit. I don't think that will give you a massless particle. For any positive $m \gt 0$, however small, it is fundamentally different from a massless particle -- since you can boost into its rest frame, and the number of states will be the same as for any massive particle. In the calculus notion of a limit, there is no ...

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For two electrons your are right! The absolute must is for the total state to be anti-symmetric. Technically, this means that the total wave-function must belong to a one-dimensional representation of the permutation group such that each permutation is represented by either +1 or -1 depending on it's parity. You can ask about permutational symmetry ...

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There's no related exception for tachyons. Tachyons' statistics must be determined a priori. Most typically, tachyons have to be bosons – and under certain additional assumptions, they have to be scalar (spin-zero) bosons. They differ from massive bosons just by the fact that the mass term $m^2\phi^2/2$ has the opposite sign – opposite sign of their $m^2$. ...

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Halzen, Martin: Quarks and Leptons has a good introduction into that matter in Chapter 2. $|uuu\rangle$ is actually an abbreviation of the idea that we keep an eye on three quarks in particular order, call them "first", "second" and "third"; and then first quark is in $u$ flavour state, of $SU(2)$ or $SU(3)$ flavour group, that is, ...

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First of all, I want to comment that, at least in 3D case, the statement that we impose on some momentum-like commuting operators $S_i$ the relation $S_x^2+S_y^2+S_z^2=S(S+1)I$ is more or less tautologic, since in general it follows from the commutation relations that the RHS can be cast (by a change of basis) to block-diagonal form, where each block has ...

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I believe you can, if you try to follow the path of finding representations of the $SO(n)$ group over a given Hilbert space. I really haven't done the calculation, but if it is the same, you would have something like this: $H=L_2(\mathbb R^n,\mathbb C)$ would be the Hilbert Space that would correspond to spin 0 particles, and the representation of the ...

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You wrote: Basis an intuition around, all matter/space/time is expanding outward in a similar fashion from the start point (not a fixed point in space I realize) of the universe. And then: So does that mean that all matter inherently spins around a central axis from which the universe expanded from? But these are contradictory statements. ...

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First of all, states of the spinless oscillator are $|n\rangle$ such that: $$H_0|n\rangle=E_n|n\rangle=\hbar\omega(n+1/2)|n\rangle$$ Then you introduce the spin $S_z$, so you now have a different space of states -- you have more degrees of freedom, not only should you specify the energy, but also the spin. You can choose a basis in this space ...

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Here is the precise treatment for determining the eigenvectors of the full Hamiltonian. You will probably find the two physics.SE posts I link to at the end useful for understanding this stuff (which basically boils down to understanding tensor products): Let $\mathcal H_0$ denote the harmonic oscillator Hilbert space and $\mathcal H_{1/2}$ denote the spin ...

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No, elements of $Spin(n)$ don't obey the Clifford algebra. Instead, it's the gamma matrices that obey it. And no, the commutator of the $Spin(n)$ Lie algebra isn't the commutators of the elements of the group but elements of the Lie algebra. Now positively. The spinor representation is the representation on which the generators $J_{ij}$ (the basis of the ...

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