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The separation of the different spin states in the Stern-Gerlach experiment depends on the strength of the magnetic dipole moment. For an electron, this is approximately twice the Bohr magneton: $$ \mu_e\approx 2\mu_B=2\left(\frac{e\hbar}{2m_e}\right) $$ The nuclear magnetic moment depends on the exact nucleus in question, but it tends to be on the order of ...


3

$S_x$, $S_y$, and $S_z$ are components of a vector operator $\mathbf{S}$. It's refered to as a vector operator because, when you do a rotation, the operator components of $\mathbf{S}$ transform just like the components of a normal vector.


3

The last step is incorrect. The sequence is Atoms are split into $|\uparrow \rangle$ and $|\downarrow \rangle$, but only the $|\uparrow \rangle$-atoms are kept (=atoms possessing a spin which points in the $|\uparrow \rangle$ direction). Atoms are split into $|\leftarrow \rangle$ and $|\rightarrow \rangle$, but only the $|\leftarrow \rangle$ atoms are kept....


3

I think the physical context of your question is spin angular momentum of elementary particles (for example: an electron) and rotations of this particle in space. Let's begin with spin angular momentum $\bf{S}$ (a vector with 3 components $S_1$, $S_2$, $S_3$). Physical experiments showed that the measured spin component $S_j$ of the particle is either $+\...


3

$\hbar=\frac{h}{2\pi}$ is proportional to the Planck constant $h$ and has the dimension of angular momentum, so the observables consist of Pauli matrices multiplied by a $\hbar$ correspond to the angular momentum, which is not just real numbers, but have direct physical meaning. Sometimes physicists would ignore the dimension of an obervable and only ...


3

"I tried to compute it explicitly by using the Pauli matrices, but was unable to derive the scalar product of the two direction vectors.". This is a possibility: Let measurement system A defined by matrix $$ \sigma\mathbf{a}=\begin{pmatrix}a_z & a_x-ia_y \\ a_x+ia_y & -a_z\end{pmatrix}$$ where $|\mathbf{a}|=1$. This matrix has normalized ...


2

In metrology, spin-coherent states are classical in the sense of not reaching the Heisenberg limit, while spin-squeezed states reach. In terms of the $SU(2)$ Wigner function both sets are "classical", since their Wigner functions are positive, as it happens in the harmonic oscillator case with standard coherent states and squeezed states. Squeezed ...


2

They do, for a short time. However, a spin which is precessing has higher magnetic potential energy than a spin which is completely aligned with the field, so the precessing state is unstable. The mechanism by which spins "decay" from a precessing state to a state in which they are aligned with the external field is provided by the spin-phonon ...


2

Your question is written as if Mach's principle were true, but in fact it's false, in the sense that general relativity doesn't obey Mach's principle, and GR is verified by experiment, while more machian theories like Brans-Dicke gravity are falsified by experiment. So if we were to answer your question, we would have to (1) come up with some new theory of ...


2

Briefly, because the photon is a spin 1 boson and the graviton is spin 2 boson, so they can not be the same particle. If the photon was the particle responsible for gravitation, then the classical limit of the theory would not be general relativity or anything like it. And yet we have a great deal of evidence that GR works in practice, extremely well, as ...


1

The tensor part of the nuclear force contains a term $-(\vec{s_1}\cdot\vec{s_2})$. This means that the spin triplet state is at a lower energy than the spin singlet state; in fact, due to the deuteron's miniscule binding energy, the spin singlet state is unbound, making the spin triplet the only bound state.


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Classical material: a material that can be described both classically and quantum mechanically. Quantum material: a material that can only be understood quantum mechanically. Case in point: magnetism.


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In the systems that I know about that have the Spin orbit coupling term that you have (the Rashba effect in solids) it is of the form ${\bf E}\cdot ({\boldsymbol \sigma}\times {\bf p} )$ where ${\bf E}$ is an electric field. In that situation the field is usually constant. I conjecture, however, that the Rashba term is hermitian provided that $\nabla \...


1

One big difference between QM and classical physics is the concept of a system’s state. In classical physics a state is a set of values or attributes (position, energy, momentum etc.) which we can, in principle, measure to any degree of accuracy that we choose. In QM a state is a set of probabilities (or, to be more precise, a complex-valued function that ...


1

Let's break down the sentence "Invariance with respect to rotation of the electron’s spin, or SU(2) symmetry, leads to conservation of spin polarization". The invariance means that the Hamiltonian commutes with total spin, $[H,\vec{S}] =0$. (You can show this also by requiring that $H$ is invariant under arbitrary spin rotation $U= e^{i \vec{s}\...


1

Formally, we consider that a quantum mechanical system is symmetric with respect to a continuous transformation if its Hamiltonian commutes with associated symmetry generator. In this case, the generator of the spin rotation transformation is simply the spin operator. If the Hamiltonian commutes with the spin operator $$ [H,\vec{S}] = 0 $$ and if $\left | \...


1

Now my question is, since we cannot measure spin in all three dimensions simultaneously, then how can we define the dot product $S\cdot \hat{n}$ (which clearly shows the sum of spins in all three directions for a particular $\hat{n}$)? The definition $S\cdot \hat{n}$ does not allow you to simultaneously measure the spin along different axes. You can check, ...


1

Tuning $S$ is equivalent to tuning the representation of SU(2), i.e. one considers spin-1/2 [spin operators represented by $2 \times 2$ matrices], spin-1 [spin operators represented by $3 \times 3$ matrices], spin-3/2 [spin operators represented by $4 \times 4$ matrices] , etc. Tuning $N$ takes us away from SU(2) to SU(N) spins. Mathematically, a key ...


1

There is a simple answer to your question. First, in $\mathrm{Cl3}$ (geometric algebra of 3D Euclidean vector space), the elements of the even part of the algebra are just quaternions. Second, the simplest representation of orthonormal vectors in $\mathrm{Cl3}$ are the Pauli matrices. In addition, we have a geometric meaning of both the Pauli matrices and ...


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