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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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$\newcommand{\spinup}{\vert{\uparrow\rangle}} \newcommand{\spindown}{\vert{\downarrow\rangle}}$ The "spin property" you mention is false. Since spin up and spin down by definition have different eigenvalues of the spin operator, they must be orthogonal states. Therefore for a general state, there is no relation between $\psi(r,s)$ and $\psi(r,-s)$ other ...


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Ok--I think I understand where my confusion was coming from. The change in basis is simply convenient when attacking problems from a variational point of view. The factors of $\cos \varphi$ and $\sin \varphi$ are simply there to guarantee that the basis is orthonormalized. With Dicke states as eigenstates of the total spin squared ...


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Gravitational waves are transverse but their possible polarizations are described not by a transverse vector but by a transverse tensor. Electromagnetic waves moving in the $z$ direction may have two possible polarization vectors $x$ or $y$, or their (complex) linear combinations – vectors perpendicular to the $z$ axis. Gravitational waves in 3+1 ...


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As you correctly notice, this does not follow from strictly physical considerations, and it is mostly for convenience that we do this. Essentially, this simplifies quite a bit the calculations of the coefficients of your state in a given basis. Suppose, for example, that you have a basis $\{\chi_+,\chi_-\}$ which is orthogonal but not necessarily ...



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