# Tag Info

## New answers tagged quantum-spin

2

Things have intrinsic spin. There's no "class of objects" that has spin and another class of objects that doesn't. Everything is a quantum object with a quantum state, and spin is a number that tells you how the state of the object transforms under rotations. It is different from "classical" angular momentum in that spin is not the operator associated to ...

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What does and does not have intrinsic spin? An electron's got intrinsic spin, and so has a proton. And a neutron, which will decay into an electron and a proton and an antineutrino. So anything made of matter has got it. Matter as we know it Jim. Wikipedia Spin (Physics) https://en.wikipedia.org/wiki/Spin_(physics) says: "In quantum mechanics and ...

4

The fundamental difference between an electron's spin and that of a baseball is that the electron is (as far as we know) a point particle. It therefore cannot rotate in the usual sense, where individual parts move relative to the center of mass; we say that its angular momentum is intrinsic. The magnitude $\lvert\vec{S}\rvert^2$ of a particle's intrinsic ...

1

Entanglement is a quantum mechanical phenomenon. It is a shorthand to saying " aspects of the wavefunction for these particles are completely known", i.e. the particles are entangled by the wavefunction describing their probabilistic behavior. A laser beam emerges from zillions of coherent photons, i.e. their phases with respect to each other and the beam ...

0

I'm not sure your question is as well posed as you think it is. In order to be 100% sure to have an entangled spin state, one would have to measure it, but can entangled states be eigenvectors of Hermitian operators (= results of measurements) other than the trivial one? If you know something is in one of various orthogonal states then in principle the ...

0

Just to close shop here. $H$ and $\sigma_i p_i$ do not commute, because $H$ does not necessarily include $\sigma_i p_i$, because $\alpha$ is free. Under the constraint that $\alpha$ is purely imaginary, the term does turn into $\sigma_i p_i$ and the commutation is valid. So, both of you are right. $H$ and $\sigma \cdot p$ should commute, which they only do ...

0

Maybe you should take into account that $(\sigma_i p_i)^2=p_i p_i$.

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The three generators of right-handed spinor rotations are given by $\left\{- i\sigma_x,-i\sigma_y,-i\sigma_z\right\}$, see for instance Peskin & Schroeder page 44, and the rotation matrix for a spinor rotation over an angle $\phi$ around a unit vector $\hat{s}$ is given by: $R~=~ \exp\left(-i\frac{\phi}{2}~\hat{s}\cdot\vec{\sigma}\right) ~=~ ... 4 When labeling states of the hydrogen atom, one doesn't refer to the z component of the angular momentum, but rather to the total angular momentum. The total angular momentum is positive, but, as you've stated, there are two states for$J=\frac{1}{2}$with$L=0$, and those are$J_z=\pm\frac{1}{2}$(Or some linear combination of them) As to why this is, ... 0 In your first paragraph you describe a Stern-Gerlach device as one with a magnetic field in the$\hat z$direction. And then later you talk about having a large homogeneous component of the magnetic field. I'm not sure you have an accurate physical model of a Stern-Gerlach device. The Hamiltonian for a Stern-Gerlach has magnetic fields components combined ... 3 You are looking for the Wigner d-functions. They relate angular momentum eigenstates through rotation. As you can see in the link the definition is $$d^{(j)}_{m,m'}(\theta) = \langle jm|e^{-i\theta J_y}|j m' \rangle$$ where$e^{-i\theta J_y}$is a unitary rotation operator. We'll have two sets of states:$|jm;0\rangle$for the original basis and ... 0 You do this mainly because we want to calculate the spin-orbit interaction precisely for the electron. And we do this because we observe that movement of the electrons (within the validity of this view) is decoupled from the movement of the nucleon. This makes sense because the nucleus mass is about N*2000 times larger than that of the electron, where N is ... 2 Think about the kinetic energy observable$T$for a spin-1/2 particle in 3D space. The particle's Hilbert space is technically${\mathfrak H} = L^2({\mathbb R}^3) \otimes {\mathbb C}^2$, yet the kinetic energy operator$T$is first defined on$L^2({\mathbb R}^3)$, where its eigenfunctions$\Psi_{\bf p}({\bf x})$are easily found. The extension of$T$to ... 1 If the spin is an actual magnetic moment, then its behavior under time reversal is simply similar to that of classical magnetization, which changes sign. Think of magnetic fields and dipoles as generated by electric currents. Under time reversal the currents reverse direction and so do the corresponding magnetic fields or dipoles. At quantum level, spin ... 0 If you define spin angular momentum as$\Sigma^{i}= 1/2 e^{ijk} \sigma_{jk}$and$\sigma_{jk}= e^{ajk} \sigma_{a} $then there is some inconsistency about when you place upper versus lower indices but you get: $$\Sigma^{i}=\frac{ 1}{2 }e^{ijk} \sigma_{jk}= \frac{ 1}{2 }e^{ijk} e^{ajk} \sigma_{a} =\frac{ 2\delta^i_a}{2}\sigma_{a} =\sigma_i$$ It really ... 1 The hyperphysics site you mention states spin rate of some$10^{32}$radian/s would be required to match the observed angular momentum. Classical angular momentum is calculated as$I\omega$, where$I = \frac{2}{5}mr^2$for a sphere. The mass of an electron is$9.11\times10^{-31}$kg and the site mentions an upper limit of$10^{-3}$fermis or$10^{-18}\$ ...

1

The Stern-Gerlach experiment uses an inhomogeneous magnetic field on a particle with a spin that is proportional to the particle's magnetic moment. An inhomogeneous magnetic field does increase the speed of the particle, not merely bend it. The link you cited does not claim otherwise.

1

Spin-resolved current in the context of scanning probe methods means that due to a finite magnetisation of probe and sample the current consists of electrons of one spin in a larger quantity than of the other. Spin current usually refers to current that consists exclusively of electrons of one spin direction.

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