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Spin is determined from the representation of the Lorentz group the quantum field transforms in. The projective finite-dimensional representations of the Lorentz group are labeled by two half-integers $(s_1,s_2)$. The spin of a field is the sum $s = s_1+s_2$. For example, a scalar transforms in $(0,0)$, a vector field in $(\frac{1}{2},\frac{1}{2})$, a Dirac ...


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Why, yes, they do! Probably the most decisive way to show this involves "handedness" of neutrinos. I won't get too much into the details here, but neutrinos are nearly massless and travel at nearly the speed of light; and when something travels at the speed of light you can describe its spin as either "right-handed" (momentum and spin are in-line) or ...


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Since you know about $SU(2)$ characters, this is a doable exercise. Let me remind you that the character of spin-$j$ is $$\chi_j(t) = \sum_{m=-j}^j e^{2imt} = \frac{\sin (2j+1)t}{\sin t}.$$ The crucial property is that the character of a tensor product is the product of characters, i.e. $\chi_{j \otimes j'}(t) = \chi_j(t) \chi_{j'}(t).$ In your case, ...


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The important thing here is that a magnetic dipole, like a permanent magnetic or induced magnetism in ferrous material, produces a nonuniform field. The potential energy of a magnetic dipole $\vec\mu$ in a magnetic field $\vec B$ is $$ U = -\nabla( \vec \mu \cdot \vec B).$$ Most frequently (as in anna v's answer) this is used to explain the torque which ...


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I fear your question is not very precise. Unfortunately my reputation is not high enough to allow me to comment instead of answer. What exactly do you mean by the term "exciton"? The way it is normaly understood, is a pair of one electron and one hole. So there are no multiple electrons whose spins could line up parallel oder antiparallel. In a biexciton ...


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It is true that two electrons can't have identical quantum numbers, but spin itself is a quantum number. That is, the state with quantum numbers 111 can hold two electrons: one of spin up and one of spin down. So when you are finding the ground state, for example, find the four lowest energy eigenstates (ignoring spin), and the ground state of eight ...


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This is not my answer, it's one of the answers you can find here Is there a reason why the spin of particles is integer or half integer instead of even and odd? I just wrote here (and re-posted) the work of @Siva which I found a very good answer. However, follow the link to read more interesting useful answers The "spin" tells us how the wavefunction ...



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