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The magnetic moment of the electron is a magnetic moment, so the right magnetic field around it is $$\mathbf{B}({\mathbf{r}})=\nabla\times{\mathbf{A}}=\frac{\mu_{0}}{4\pi}\left(\frac{3\mathbf{r}(\mathbf{\mu}\cdot\mathbf{r})}{r^{5}}-\frac{{\mathbf{\mu}}}{r^{3}}\right).$$ The world is quantum mechanical – and so is any viable description of the spin – so we ...

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Spin-0 can be either massive or massless. Examples of known massive spin-0 particles are the pion $\pi^+$, kaon $K^+$, and also the recently discovered Higgs boson $H$. No known spin-0 particles are exactly massless, but the Goldstone boson arising from the spontaneous breakdown of a continuous internal symmetry is a good theoretical example. Spin-2 can ...

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In addition to Qmechanic's interesting answer, you might want to explicitly see that $\mathrm{SU(2)}$ only has three independent variables. Let us start by writing the $U \in \mathrm{SU(2)}$ as: $$U=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$ with $a,b,c,d \in \mathbb{C}$. Since we are looking at the special unitary ...

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I) Quantum mechanically, the Lie group associated with rotational symmetry is $$G~:=~Spin(3)~\cong~ SU(2),$$ which is a double cover of $SO(3)$, and has a 3-dimensional real Lie algebra $$L~:=~so(3)~\cong~ su(2)$$ with generators $J_i$ satisfying $$[J_i,J_j]~=~i\hbar\sum_{k=1}^3\epsilon_{ijk} J_k.$$ II) In quantum mechanics, we are interested in Lie ...

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It is not surprising that you found one of the Pauli matrices as the generator of your rotation. Let us see how it can be seen algebraically that it is to be expected: Observe that 2D rotations embed naturally into 2D unitary matrices, as $\mathrm{SO}(2) \subset \mathrm{SU}(2)$, corresponding to the subgroup of real matrices. Now, as we know, the ...

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In the special case where $\ell^2$ or $s^2$ has eigenvalue zero, then $j^2$ is fixed. Otherwise you must know the projections $m_\ell,m_s$ to find $j$.

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There seems to be no way to proceed unless more information is given. In fact, from what you have above (correcting the typo pointed out by Bernhard and ticster): $$\vec{j}=\vec{l}+\vec{s} \quad \Rightarrow \quad \vec{j}^2=\big(\vec{l}+\vec{s}\big)^2 = \vec{l}^2+\vec{s}^2 + 2\, \vec{l}\cdot\vec{s}\,,$$ meaning that knowledge of the eigenvalues of ...

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As the previous post mentioned, forget about the concept that the electron is actually spinning. Spin, like rest mass and electric charge, is an intrinsic property of subatomic particles. Yes, it's angular momentum. No, nothing is spinning. Although many physicists today do not like this explanation, special relativity introduces a useful analogy with mass. ...

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Note that $J_{xy}$ and $J_z$ have the dimensions of energy while $J$ and $K$ have dimension of energy/(area x time). In the process of making the continuum Hamiltonian into a discrete one, you will need to choose short distance scales for both space as well as (Euclidean) time. You can choose a square lattice of length say, $a$ for space and split the time ...

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First, the electron isn't actually spinning. Physical objects made up of collections of electrons and protons (and neutrons) can have angular momentum because they rotate; the electron does not get its angular momentum for the same reason. Second, the magnetic moment of an object with angular momentum L is proportional to $$\mu \propto \frac{qL}{M}$$ ...

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First, note that we are quite sure what the overall nuclear spin is; we are not sure how to obtain it mathematically from available models. Due to the phenomenon of color confinement, there are no gluons at low energies in QCD (the theory underlying nuclear physics). Importantly, you can't say there are this or that many gluons in any proton or neutron. ...

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