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Weak interactions with $W$ and $Z$ gauge bosons violate parity simply because the righ-handed and the left-handed fermions coupled differently to $W$ and $Z$. For example, the $W$'s couple only to the left-handed fields. A parity inviariant dynamics would require that both left- and right- fields couple in the same way to the gauge vector since they get ...

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I guess it is because you first of all change sign of $\vec x$ to $- \vec x$ in physical space(this is parity transformation in a nutshell). All this peculiar algebra concerning left and right chirality fields comes from $J = 1/2$ Lorenz group representation, so transformation rules are defined as representatives of parity transformation of physical space.

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Okay, I think I have an idea why the terminology is used, but I think this argument makes little sense: The Lagrangian term describing weak interactions is of the form $$\bar \Psi \gamma_\mu P_L \Psi W^\mu$$ Under parity transformations $\Psi \rightarrow \gamma_0 \Psi$ and $\bar \Psi \rightarrow \bar \Psi \gamma_0$, therefore  \bar \Psi ...

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For anyone with similar problems: The following observation has helped me immensly: We have in fact four particles directly related to an electron: A left-chiral electron $\chi_L$, with isospin $-\frac{1}{2}$ and electric charge $-e$, A right-chiral anti-left-chiral-electron $(\chi_L)^c=\chi_R$ with isospin $\frac{1}{2}$, electric charge $+e$ A ...

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Roughly sketched, for the quantized Dirac field one has: $$\hat\psi(x)\sim \int d\mathbf{p}\, \sum_r \bigg[ u^{r}(p)\, \hat a^r_\mathbf{p}\,e^{-ipx}+v^{r}(p)\, {\hat b^r_\mathbf{p}}^\dagger e^{ipx}\bigg],$$ where $r=\pm1$ denotes helicity. The ${\hat a^r_\mathbf{p}}^\dagger$ operator creates a helicity-$r$ particle state when it ...

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Consider a one-particle state of a relativistic quantum field theory, and let this state be an eigenstate of the 4-momentum operator, $\hat P^\mu |p^\mu\rangle = p^\mu|p^\mu\rangle$. Other than the 4-momentum, what other quantum numbers can the state have, and how should they transform? That is, if there are any quantum numbers collectively labeled by $s$, ...

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Indeed, the photon doesn't have spin, what it has is POLARIZATION. For a photon emitted from an atom the polarization is CIRCULAR. That means, if we look in a fixed plane perpendicular on its motion, we see that the electric vector does not have a fixed position, it rotates. If so, if the electric vector rotates counter-clockwise the circular polarization ...

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We then talk about a left-chiral electron we do it in an informal way, you are correct that a massive particle cannot be inherently chiral. To see this, let us remember that he handedness of an elementary particle depends on the correlation between its spin and its momentum (helicity). If the spin and momentum are parallel, the particle can be said to be ...

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