# Tag Info

## New answers tagged spin

4

"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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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 ...

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 ...

0

What comes into my mind is the three-magnon process, in which an optical magnon with $k=0$ and $\omega=\omega_0$ splits into two acoustic magnons, one $k=k_1, \omega=\frac{\omega_0}{2}$, the other $k=-k_1, \omega=\frac{\omega_0}{2}$. Note that in this process energy and momentum are conserved. This 3-magnon process can lead to a finite life time. Not sure.

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 ...

1

The W is massive so can be in the spin 0 state (or $s=-1, 0, 1$ in general). The photon is massless so does not have this "longitudinal" polarization. For the massive vector boson, the relevant symmetry group is the little group $SO(3)$, and for the photon it is $SO(2,1)$.

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1.) The easiest way to count the energy is as that of domain walls. Both of the configurations you have drawn have two domain walls. Each domain wall costs energy $E\Delta/2$, so both configurations have energy $E\Delta$ above the ground state. I don't understand either how one would count the spin of a domain as 1. Perhaps that is the spin/site? 2.) To see ...

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 ...

4

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 $\vec{S}\cdot\hat{n}$. $$\vec{S}\cdot\hat{n}=\frac{\hbar}{2}\begin{pmatrix} \cos{\phi} & ... 1 Spin eigenvectors are the same for the electron and the positron. The transition amplitude between the singlet state s, and for instance, a state up for electron and down for positron may be written (up to a complex unit phase) : A = ((up_1)^\dagger \otimes (down_2)^\dagger) s \\=((\chi_+(\theta_1, \phi_1))^\dagger \otimes \chi_-(\theta_2, ... 0 I think the best way to understand spin is to look at it from a complete abstract point of view: do not try to find classical analogs to it. What you find when you perform experiments with electrons is that they interact with a magnetic fields as if they had an intrinsic magnetic moment with two possible values: a positive one and a negative one (loosely ... 1 It is true, you have to "rotate twice" (or by 720^\circ) to recover the original state. You can prove this in the following way. Let$$|a\rangle=|+\rangle\langle+|a\rangle+|-\rangle\langle -|a\rangle$$be a general ket. Consider now a rotation by a finite angle \theta around the z axis. I remind here that if a ket of a spin 1/2 system is ... 0 Splitting happens when the two states have different couplings to the interaction hamiltonian -- such as the classical case where an external magnetic field will differentially couple to different values of the orbital and spin angular momentum of electrons. If the ineteraction Hamiltonian treats both degenerate states identically (say, if you added a ... 2 @PPG: Well (−1,−1) is not a solution of your equation, but (1,−1) is... – Adam What he said was: Finding the eigenvectors You did right... but:$$(\hbar/2)\alpha+(\hbar/2)\beta=0\implies\alpha+\beta=0\implies\beta=-\alpha  Then, the corresponding eigenvector is: $c\left[\begin{array}{c} 1 \\ -1 \end{array}\right]$.

0

If you want to do ordinary non-relativistic quantum mechanics with electrons without having to put spin in "by hand", you would start with the Dirac equation $(i\gamma^\mu \partial_\mu - m) \psi = 0$ and use this to derive the Schrödinger equation as the $|\mathbf{p}|^2 \ll m^2$ limit. The field $\psi$ is a 4-component spinor, although the number 4 is ...

1

It is not true that "we" consider spin only in spatial directions, except "we" means something like "undergraduate students", maybe. Instead relativistic physics is all controled by spacetime spinors, namely by representatins of the double cover of the Poincaré Lie group of spacetime translations, rotations and boosts. Maybe the best way to get an ...

0

The thing with spin is that it does not quite have a physical interpretation, it is rather a mathematical formulation for the intrinsic angular momentum of a particle. Spin corresponds to an orthogonal translation of a Lorentz Group (a lie group), which one can usually represent in terms of spinors. Spinors have very interesting, and complex, properties, ...

2

It depends on the non-trivial topological solution of the equations of motion under consideration. The most famous example is the 't Hooft–Polyakov monopole (see http://en.wikipedia.org/wiki/%27t_Hooft%E2%80%93Polyakov_monopole) which has the following solution: $$\phi = h \frac{x^a} {r} t^a$$ A_0=0 ...

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