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## Hot answers tagged spin

5

The term canonical gives it away. The canonical ensemble density matrix $\rho$ is defined as follows in terms of the Hamiltonian $H$ and inverse temperature $\beta = 1/kT$: \begin{align} \rho(\beta) = \frac{1}{Z(\beta)}e^{-\beta H}, \qquad Z(\beta) = \mathrm{tr}(e^{-\beta H}) \end{align} Then the canonical ensemble average of any observable $O$ is given ...

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"Total spin conservation" means global $SU(2)$ spin-rotation symmetry (a continuous symmetry) of the Heisenberg model, and "spin wave" indicates an ordered ground state that spontaneously breaks the spin-rotation symmetry. Thus, according to Goldstone theorem, there must be a gapless mode for spin wave.

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Take a spin $1/2$ particle with its spin pointing along $\hat{n}$ defined by $$\hat{n}=(\sin{\phi}\cos{\theta},\sin{\phi}\cos{\theta},\cos{\phi})$$ We are measuring the spin along $\hat{n}$ and the operator corresponding to this observable is $\vec{S}\cdot\hat{n}$. $$\vec{S}\cdot\hat{n}=\frac{\hbar}{2}\begin{pmatrix} \cos{\phi} & ... 2 Here is the diagram you are discussing: It seems you are worried by the angular momentum carried by the W+. The W+ is a virtual particle in this reaction. In virtual paths the particle is off mass shell and its mass is unphysical, and angular momentum as a part of a four vector will be a complicated function also having unphysical measure, so ... 2 The number of neutrons is even, so it indeed means that they contribute spin zero and positive parity. The spin and parity comes from the "last proton" because the number of protons is odd. The dependence of the energy on the angular momentum is such that the pairs at a high value of J are preferred (lower in energy) due to the special, spin-dependent ... 2 This link seems to be along the required path of thought. Please note the "axiomatic" facts: experimental inputs in value of S^2, raising and lowering operators, desirability of hermitian operators... that go inside the derivation. Also, once having chosen them, note that the 3 Pauli matrices along with the 2d identity matrix can be used as a basis to ... 1 Suppose that your decay does not violate parity, and let us take the example of strong interaction process involving mesons. Looking at this wiki paragraph and array, you will understand that each meson has a total angular moment, and a parity, noted J^P, and there are several possibilities for S and L, for a given J^P (J^P is a characteristic ... 1 In the optics regime, every time a wave impinges on a surface it is modifying the angular momentum of an electron. Since the electron PE is usually comparable to the visible regime. As for measurements with coherent waves. I don't think this is an easy task, though I think it is possible. I mention waves, because depending on the energy level of your ... 1 Spin eigenvectors are the same for the electron and the positron. The transition amplitude between the singlet state s, and for instance, a state up for electron and down for positron may be written (up to a complex unit phase) : A = ((up_1)^\dagger \otimes (down_2)^\dagger) s \\=((\chi_+(\theta_1, \phi_1))^\dagger \otimes \chi_-(\theta_2, ... 1 It is true, you have to "rotate twice" (or by 720^\circ) to recover the original state. You can prove this in the following way. Let$$|a\rangle=|+\rangle\langle+|a\rangle+|-\rangle\langle -|a\rangle be a general ket. Consider now a rotation by a finite angle $\theta$ around the $z$ axis. I remind here that if a ket of a spin $1/2$ system is ...

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