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It seems that some people liked this question so I shall post my thoughts so far. I don't have a definitive answer, but I did get some interesting results. Let $\rho_m(\boldsymbol r)$ and $\rho_e(\boldsymbol r)$ be the mass and charge densities of the electron. The $g$ factor is given by $$g=\frac{m}{e}\frac{\int\mathrm d\boldsymbol r\ r^2\sin\theta\ ... 0 Yes you can do that; the space and spin parts just have to have opposite symmetry characteristics so that the total wavefunction is antisymmetric. 0 After realizing that the electron energy does not depend on the projection of the spin it seems obvious that the energy of the electron doesn't change when it passes a magnetic field (or SG). But then something very peculiar immerges: 1.the state of something changes without any energy exchange 2.the magnet induces a change in the electron but it doesn't ... -1 www.physicspages.com/2013/04/11/magnetic-dipole-moment-of-spinning-spherical-shell/ My search gives \mu = \frac{e\omega R^2}{3} This gives g = 5/3 = 1.667 Did not you provided link given below? https://en.wikipedia.org/wiki/Electron_magnetic_moment#The_classical_theory_of_the_g-factor Which explains that non-uniform charge distribution can explain ... -1 I wonder why we so seldom mention, when discussing these things, that you cannot answer this question without adopting a human convention with respect to the combination of spin states. If we take the +/- z direction to be the north and south poles, then any state with equal amplitude-squared in both components will correspond to a spinor pointing somewhere ... 0 The multiplicity 2S+1 actually tells you how many degenerate spin states there are, each labelled with the total spin projection quantum number M_S (this is from the total spin projection operator \hat{S_z}(conventionally taken to be in the z-direction) whose eigenvalues are \hbar M_S). The possible values of M_s are -S\le M_S\le S in integer ... 1 First, let's clarify the expression of |\phi\rangle. The kets |+\rangle_z and |-\rangle_z are eigenvectors of \hat{S}_z such that$$ \hat{S}_z |+\rangle_z = +\frac{\hbar}{2}|+\rangle_z \\ \hat{S}_z |-\rangle_z = -\frac{\hbar}{2}|-\rangle_z $$This means that in the \{|+\rangle_z = ... 2 Your feeling looks very misguided. Whatever you do, stay away from SU(3) for rotations. The rotation group and its Lie algebra are always linked to SO(3) ~ SU(2), to avoid formal forays into double covers and half angles. Read up on the spin matrices for any representation of the very same group (any spin). There are, in fact, simple systematic ... 1 The system can be separated, but not necessarily in nice form. For instance, the time derivative of the first eq. reads$$ i\hbar {\ddot c}_1 = - B {\dot c}_1 - {\dot V}c_2 - V {\dot c}_2 $$Now remove c_2 using again the first eq.,$$ c_2 = -\frac{i\hbar}{V} {\dot c}_1 - \frac{B}{V} c_1 $$and {\dot c}_2 using the second eq., {\dot c_2} = ... 0 First, observe that the Slater determinant you have written down is the linear combination of the singlet state and the z-spin-0 state of the triplet. Vice versa, you can produce the singlet and triplet states as linear combinations of Slater determinants. Whether you prefer the Slater determinant or the singlet/triplet formalism for writing down your ... 1 Peter is right. The tensor nature requires it to be a spin 2 field, and the graviton is its presumed quanta. But there have been and are theories of gravity that include a spin 0 field. Brans-Dicke theory was one (I think mostly or fully disporved), and some theories for dark energy are spin 0 - quintessence is one, it assumes the cosmological constant is ... 2 I assume you are familiar with Wigner's classification in d=4, as you might be implying. The mâ†’0 limit is best appreciated on the PoincarÃ© sphere, but let us skip that here to count particle states. So, reviewing Wigner, for a massive state, we can Lorentz-transform the momentum to the rest frame, (m,0,0,0) so the little group is SO(3) and its vector rep ... 2 Within the context of first quantization, spin itself is not characterized by the wave function: it exists as a separate (Hilbert) space. The total state of the particle is then a composite of its wave function (often projected onto the configuration or momentum basis) and spin state. -1 Anna V gave the correct explanation. Only when weak bosons are created on mass shell, e.g. at collisions @ Ecm = M, can you apply total angular momentum conservation at a single vertex (production and decay). On the other hand, even for off mass shell bosons chirality (read helicity for ultra relativistic particles) imposses costraints at each vertex. 2 Let's say we could perform the experiment with W^\pm bosons. These particles are similar to electrons, but the possible spin states are -1,0,+1, that is, three different possibilities. The magnetic moment of these bosons is, therefore,$$ \mu_z=\begin{cases} -\mu_W\\\phantom{+}0\\+\mu_W\end{cases} where \mu_W=6\ 10^{-6}\ \mu_B is the W magneton. ... 2 For force carriers the interacting field theory determines the spin. A scalar field yields spin 0; the Higgs is the only example; a vector field yields spin 1, the photon, W, and Z are examples; a tensor field yields spin 2. Since gravitational field theory requires a tensor field for General Relativity, quantized gravity, in the weak-field, linearized ... 2 Courtesy of its spin the electron has a magnetic dipole moment. That means if we place it in a magnetic field the two states aligned with and against the magnetic field have different energies. The magnitude of the energy difference depends on the strength of the field and the size of the magnetic dipole moment, which in turn depends on the spin. So by ... 2 This is the prototypical example of a superselection rule. The operator U(2\pi) commutes with all observables (because it represents a full rotation, and is hence physically a "do nothing" operator), and yet is not a multiple of the identity (because it is -1 on the fermionic and 1 on the bosonic parts of the Hilbert space). Therefore, the representation ... 2 The spin s of a particle characterizes how the rotation generators act on it. In D dimensions, you represent the little group SO(D-1) for massive particles and SO(D-2) for massless ones. In fact, you really need to consider its universal cover \textrm{Spin}(n) which happen to be just its double cover. Now, you can define the spin to be the largest ... 1 The company explanation is wrong, really except for the first sentence. The correct hand waving explanation would be much longer. Materials are composed of atoms which contain electrons, and electrons have intrinsic magnetic moments and specific orbitals around atoms that are not affected by temperature. The molecular bonds are affected at high ... 1 As Praan confirmed, you are fine. Your result is the n=4 case of the general expression for composing spin 1/2 doublets, (so here denoted by their dimensionality, 2), {\mathbf 2}^{\otimes n} = \bigoplus_{k=0}^{\lfloor n/2 \rfloor}~ \Bigl( {n+1-2k \over n+1} {n+1 \choose k}\Bigr)~~({\mathbf n}+{\mathbf 1}-{\mathbf 2}{\mathbf k})~,  where $\lfloor n/2 ... 0 There is an interesting detail about the intensity distribution behind edges from photons and electrons. The shadow from photons is smaller the "geometrical" shadow. The shadow behind an edge, built from electrons is always wider than geometrical shadow. This could be explained by the interaction between the electric field of the sharp edge (the surface ... 4 The double-slit experiment is a one-body experiment, meaning that one is only looking at interferences of one particle with itself. Thus the Bose or Fermi statistics does not play a role in that case. What the OP has in mind in the Hong-Ou-Mandel effect, which for bosons implies that there is an increased probability that two identical bosons will be ... 0 When adding angular momentum$S$and$L$to get$J$,$J$can range from absolute value of$|L-S|$to$L+S$in integer steps. Then once you have$J$you can figure the allowed values of$m_j$just like you do for$S$and$m_s\$

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Yes, the state is a soverposition, with amplitudes x and y still to define, of the eigenstates up and down. You would just impose normalization for the state |r>, by performing a scalar product of |r> with itself. Also, you want the two probabilities to be the same ,i.e. the two projections of the state |r> over the 2 basis vectors up and down are equal.

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In the case that photons are coherent (having the same wavelength and oscillating in phase), they still separable into two states. This has to do with the two field components of the photon, the electric dipole moment and the magnetic dipole moment. Dipole moment means that at some time there is an electric field with a plus-minus direction and there is at ...

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The average separation of the electrons (integrated over time) in parahydrogen state is more than the average separation of the electrons in orthohydrogen state. And whenever electrons are away from each other, by nature, they would feel less Coulomb force and hence a lower energy state. This handwaving explanation can be justified in quantum mechanics more ...

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The spin quantization was first detected by the Stern-Gerlach experiment. This one ingenious experiment proved that the spin is an observable. Refer to the experiment. I prefer the book: Modern Quantum Mechanics by J.J.Sakurai. chapter 1

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