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The Higgs is part of a complex scalar doublet in the standard model. It. carries both hypercharge and weak charge. So we have discovered charged scalars. Now perhaps you are only interested in ELECTRIC charge. So does the Higgs doublet carry this? Well once the Higgs picks up a vev, then some parts do and some parts don't. The parts that do carry electric ...

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Yes, spin direction is also reversed. An "up" electron spin state simply becomes a "down" spin state. Electron spin remains fundamentally a type of angular momentum, even though the rules get a bit odder at the quantum scale. Incidentally, when time is reversed all particles also turn into their antimatter equivalents. So a bit of safety advise: Never, ever ...

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The Lagrangian $$\mathcal L = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + \mathcal L_\textrm{free} + eA_\mu J^\mu \tag{1}$$ where $A_\mu$ is the 4-potential, $F_{\mu\nu} = \partial_{[\nu}A_{\mu]}$ is the field tensor, $\mathcal L_\textrm{free}$ describes fields other than $A_\mu$, and $J^\mu$ is the 4-current density expressed in these other fields, describes a ...

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Spin is an intrinsic property of quantum objects that, unlike a particle's orbital momentum, does not depend on the frame of reference you are considering. Another intrinsic quantity of that kind would be charge, which is also just a fundamental number you assign to a particle, no matter its state of motion. One possible source of confusion when talking ...

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The standard model is very successful in its group structure in ordering all observed particles. To introduce a particle with charge and zero spin, you will need a different model that would also accommodate the symmetries observed experimentally and fitted by the standard model. So the answer to "why" is "because" we have not seen any and can model well ...

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If $S_\text{tot}^2$ is a symmetry i.e., commutes with the Hamiltonian, then one can choose a basis in where it is diagonal. This corresponds to going to the coupled basis, the triplet (total spin=1) and singlet (total spin=0) combinations of the two spins. Those are the four choices that you have listed for the `spin functions' -- the first three are the ...

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This is because there are just two possible values to the spin in any direction, $-\frac{\hbar}{2}$ and $\frac{\hbar}{2}$, the just differ in a sign, so when you square it you get a single value $\frac{\hbar^2}{4}$. Think about this, the only possible value when you measure the square of $S_z$ is $\frac{\hbar^2}{4}$ for any state, so  ...

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OP asks: Is there any physical meaning to this? Yes, the Pauli matrix $\sigma_j$ represents (up to a proportionality factor) the spin in the $j$th direction of a spin $\frac{1}{2}$ system. Such system has only two spin states: $\uparrow$ and $\downarrow$, with opposite eigenvalues. The square $\sigma_j^2$ can no longer see the sign, so it only has one ...

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It's important that the spin need not be up or down, but may be in some superposition of the two, $|\psi\rangle = a |\uparrow\rangle+b|\downarrow\rangle$. If you choose to measure it in the $z$-direction, the squared magnitudes of $a$ and $b$ will give relative probabilities of getting each answer. But given any direction, you can always rewrite the state ...

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Up or down, obviously. However, the representation of the x-up state or the x-down state using z-up and z-down states depends on how you represent the spin operator. For the most favoured choice $S_x=\frac{1}{2}\left( \begin{array}{cc} 0 & 1\\ 1 & 0 \end{array} \right)$ gives up/down state as (not normalised) \$\left|x_{up}\right>=\left( ...

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By no means not a complete answer, more a criticism of @luksen’s one. It is posted here because the text is too long to fit in the comment field. First of all, the spin is not a well-defined concept for composite particles. More precisely, whether the spin of a particle is defined depends on how the “particle” is defined. Look at an atom: it has the nucleus ...

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