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In the second example, only the 3d shell is not completely filled. Using the aufbau principle to fill this shell with the 7 valence electrons leaves three unpaired electrons resulting in a total electron spin of $S=3/2$, so you arrive at a quarted state. The electron orbital angular momenta of the five possible states in the d orbital are -2,-1,0,1, and 2. ...

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I dont think we should consider the energy of the electrons. For both para- and ortho- hydrogens, the two electrons are sitting in the same ground state (but with opposite spins) of the hydrogen molecule, so I dont think there is a difference. The energy of a hydrogen molecule arises from its translational, rotational and vibration kinetic energy. At room ...

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No, gravitational waves are not emitted isotropically. In the weak-field limit (i.e. far from the sources), the radiation emitted by a gravitational system is determined by the third time derivative of its quadrupole moment, which, being a tensor, needs to be projected along the line of sight to yield a (scalar) energy flux. This projection is what gives the ...

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Let´s go: [1] From the DIS (Deep Inelastic Scattering) of eletron-proton, we can imagine that the photon exchanged in te process "sees" a parton (possible constituent of the proton) distribuition. We can imagine a cross-section of photons and that constituents of the proton. And we can analyze two situations: From a cross-section of longitudinal (scalar) ...

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The Spin of a particle is better understood from a group-theoretic point of view. It is just telling you how a particle, i.e. an asymptotically free state of your theory, transforms under the symmetry of your theory, Lorentz symmetry. Well, actually under its double cover as Weinberg explains in his first book, that is why we are allowed to have spinors. ...

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Your group theory text probably betrayed you if it did not spend much time contrasting the two cases. A possibly related question is 254461. People use loose language and symbols that aggravate the confusion. Talking abstractly without explicit hands-on formulas clinches it (the confusion)! Let me stick to your 4-dimensional matrices and vectors, all tensor ...

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Spin is connected to the intrinsic magnetic dipole moment of the particle, this is what makes the particle capable of interacting with an external magnetic field. Namely, the intrinsic dipole magnetic moment $\vec\mu$ of a particle with spin $\vec{S}$ can be found through: $$\vec\mu = g\left(\frac{q}{2m}\right)\vec{S}$$ where $g$ is the g-factor, and $q$ ...

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If we model a particle as a uniform ball spinning around its own centre (which it probably isn't) then: $$I = \frac{2mr^2}{5}$$ $$L = I\omega$$ The speed of a point at the edge of the ball is: $$v = r\omega$$ We know the spin of fundamental particles so: $$rL = \frac{2mr^2v}{5}$$ $$rv = \frac{5L}{2m}$$ We know that for the electron the spin angular ...

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The surrounding system is in a very large very mixed state, and that washes out the distinction in angular momentum. Trivial Example Consider the trace distances between the following nearly-maximally-mixed density matrices: \begin{align} D \left( \frac{1}{2} \begin{bmatrix} 1& & \\ &1& \\ & &0 \end{bmatrix}, \frac{1}{2} ... 2 You are right that the magnetic field, or rather the apparatus used to generate it, absorbs some angular momentum from the quantum spin. This apparatus includes, for example, some coils of wire containing more than 10^{23} electrons, as well as many other macroscopic components. The upshot is that increasing the angular momentum of this apparatus by an ... 1 Neutral charged objects are made of charges. Although net charge is zero, imperfections in spatial arrangement of charges, causes them to interact with electromagnetic fields. Edit: 1) Neutrino's are formed during nuclear reactions. Once created, they sustain their direction of spin which does not change during course of time. 2) Photons don't change their ... 0 Instead of counting all possible permutations of the spin configuration you could try this: 1.) what is the maximal energy that the system can have? 2.) what is the minimal energy? 3.) what is the "energy step", \Delta E, that can be realised in this situation? The highest/lowest energy states are the ones were all spins are aligned. If you flip one ... 2 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\ ...

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Yes you can do that; the space and spin parts just have to have opposite symmetry characteristics so that the total wavefunction is antisymmetric.

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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 ...

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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 ...

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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 ...

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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 ...

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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} = ...

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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 ...

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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 ...

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I assume you are familiar with Wigner's classification in d=4, as you might be implying. The m→0 limit is best appreciated on the Poincaré 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 ...

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