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2

Let us make clear that the problem If proton spin emergence from quarks and gluons is mysterious, why is silver atom spin not? is a modelling problem. The spin both of the proton and the silver atom is measured and known to identify them. John's answer covers it, the energy carried by the virtual quarks and gluons within the proton are much larger ...


3

I'm not altogether sure what you are asking, but I suspect the following may help. To represent rotations, spins and vectors in $SU(2)$ we work as follows. Rotations live in $SU(2)$. Vectors (in the physicist's sense) live in the algebra $\mathfrak{su}(2)$. The position vector $(x,\,y,z)$ is: $$X =x\,\hat{s}_x+y\,\hat{s}_y+z\,\hat{s}_z = ...


1

I'm not so sure, if this is really, what you're looking for, but you can of course solve this easy problem analytically. To do this, it is clever to first analyze the easier Hamiltonian $H_0 = 2g (\vec L \cdot \vec S)$, where the $L_i$ and $S_j$ fulfill independent $SU(2)$-algebrae $$ [L_i, L_j] = i \epsilon_{ijk} L_k\\ [S_i, S_j] = i \epsilon_{ijk} S_k. ...


0

The entanglement does not depend on the velocity vectors of the particles. The SG apparatus will determine the spin state of the measured particle, not necessarily the whole system. By performing a measurement on one particle in a known entangled state you would know what the state of the other particle is as well. You do not need to measure it.


2

Everything depends on how your fields (vectors and spinors are fields in the classical theory, and when you quantize in QFT, they become operator-valued fields) transform when you make a Lorentz transform: An scalar is a field that doesn't change at all: $\phi'(x') = \phi(x)$. Examples are the Higgs and pions. A vector field is a field that transform like ...


1

Magnetic moment, in classical physics, is related to current in a loop, which in turn can be connected to angular momentum of a charged particle. Thus, in classical physics, magnetic moment and angular momentum are connected. In fact, they are proportional with the constant of proportionality being the gyromagnetic ratio. Moving to quantum mechanics, some ...


6

That spin follows the angular momentum algebra is no accident - like angular momentum, it is part of the conserved quantity - the Noether charge - associated to rotations. The reason why the $\mathfrak{so}(3)$ transformations of spin should be indeed those associated to the $\mathfrak{so}(3)$ of spatial rotations is not answerable in QM alone - you have to ...


6

The binding energy of the electrons in a silver atom is far less than the rest energy of an electron, so there is no ambiguity about the number of electrons in a silver atom. That makes adding up the spins a straightforward business. By contrast, the combined mass of the two up and one down quarks in a proton is about 10MeV (it isn't precisely known) but ...


2

$\newcommand{\ket}[1]{\left| #1 \right>}$ $\newcommand{\o}{\mathbf 1}$ Physicists are lazy people we all are! When you see something like $S_{1z}+S_{2z}$ you should really think of the following: $$S_{1z}+S_{2z} \equiv S_{1z} \otimes \mathbf 1+ \mathbf 1 \otimes S_{2z}$$ Since you get tired of writing it over and over you just shorten it by an addition ...


4

You appear confused by how spin is introduced in ordinary QM. It is rather ad hoc: Given a Hilbert space without spin degrees of freedom of a particle $\mathcal{H}_0$, and the spin $s$ of the particle, we take the total space of states of the particle to be $\mathcal{H}_0\otimes \mathcal{S}_s$, where $\mathcal{S}_s$ is a $2s+1$-dimensional complex Hilbert ...


0

Be careful: The spin constribution to the number current is proprotional to $\nabla \times S$ where $S$ is the spin density. The spin contribution to the momentum density is $(\nabla\times S)/2$ because the $g=2$ gyromagnetic ratio makes spin twice as effective at contributing to the number (electric) current as to the momentum density. See my contribution ...


0

I would say that the question is still not very well defined, as it is important if in the high-temperature (classical) limit spin is conserved or not. If spin is conserved (think strong magnetic fields $B\gg T$), then in the classical limit spinful quantum gases become a mixture of classical gases, as coherent superpositions between spin states are not ...


2

Your final result looks right to me. Everything should be half-integers. A basic rule of combining two quantized angular momenta is that the quantum number of the resultant can be anywhere between the sum of the original quantum numbers and the absolute value of the difference of them, in integer steps. Consider $\ell_1$ = 1 combining with $\ell_2$=3. The ...


3

$\newcommand{\ket}[1]{\left| #1 \right>}$ Note that $ l=0$ has only one state $m=0$. Therefore the tensor product of $l=1$ and $l=0$ can be written as: $$ (l=1)\otimes (l=0) = \left\{ \begin{array} &\ket{l= 1,m=1} \otimes \ket{l=0,m=0} \\ \ket{l=1,m=1} \otimes \ket{l=0,m=0} \\ \ket{l=1,m=1} \otimes \ket{l=0,m=0} \\ \end{array} \right\}=(l=1) $$ As ...


0

Clebsch-Gordan coefficients let you treat n spins (or generaly - any n particles with arbitrary angular momentum) as a single composite system. The coefficients are simply the matrix element of basis transformation from seperated to composite system. For 2 particles with total angular momentum eigenvalues $ l_1,l_2 $, such that for example $ ...


1

A spin-spin interaction is really a magnetic moment - magnetic moment interaction, where the magnetic moment of each particle is proportional to spin. [Of course, it might be a chromomagnetic moment - chromomagnetic moment interaction if two quarks are interacting, as they are here.] In any case, the interaction term goes like $\vec S_1 \cdot \vec S_2$. ...


0

For solutions of the massive Dirac equation the standard Gordon decomposition is $$ \bar\psi \gamma^\mu\psi =\frac{i}{2m} (\bar \psi \partial^\mu \psi -(\partial^\mu \bar \psi) \psi)+\frac{1}{2m} \partial_\nu(\bar\psi \Sigma^{\mu\nu}\psi). $$ What happens when $m=0$? For the both the massive and massless case assume that $\psi({\bf r},t)=\psi({\bf ...


1

It's not uncommon that popular science articles will refer to paradoxes in physics. However it is extremely important to understand that there are no paradoxes in physics. Our current theories of physics are self consistent and do not contain paradoxes (though there are some conditions not covered by any of our existing theories). Non-physicists tend to use ...


2

You're probably used to the convention where a hat is used to denote that something is an operator. But that convention is not universal. In many cases, when it's clear from the context whether something is an operator or not, we just write it without a hat either way. For this case in particular, $\boldsymbol{J}$ is defined to be an operator. The fact that ...


1

For a famous example of a nucleus with internal orbital angular momentum, consider the deuteron. Considerations of exchange symmetry, spin, and isospin demand that the deuteron have unit spin, rather than zero spin. However the pion-nucleon interaction, gleaned from neutron-proton scattering and deuteron formation, suggests that about 4% of the deuteron ...


1

When someone says that spin measured about different axis can't both be known, they mean that whatever state you pick will have variability in at least one of the possible spin measurements you can do. So that is what you will get when measure the spin, you will get variable results. This happens even with entanglement with even just one particle. With ...


0

We know that the "sourceful" fields like gravity or electric static field due to charge are all curl=0. But both E/M fields have curl and they mutually "curling" into each other, making them kind of "sourceful" in a very different way. This gets tricky, but IMHO there is a way to understand it via a water analogy. See how AV23 said the most fundamental ...


2

The reason is not really spin - we get circulating magnetic fields due to currents in straight wires, where spin is effectively absent. Many would argue that the most fundamental quantity in electromagnetism is not the fields themselves, but what is called the electromagnetic vector potential, $\mathbf{A}$. This potential is directly 'sourced' from the ...


0

And if I have to guess, I would say it is spin. You guess right. Electric field comes from electric charges from protons and electrons. Magnetic field comes from magnetic dipole moment from protons, electrons and neutrons too. And the magnetic dipole moment is connected 1 to 1 to the intrinsic spin.


0

The notion that there is a curl involved is vector calculus voodoo. You can write Maxwell's equations in special relativity, and there is no curl involved. But either way, there is only the EM field and its source, which is current. Spin has nothing to do with it.


0

In a Hydrogen atom, the "orbital angular momentum of electron" is in fact the relative orbital angular momentum. The nucleus (proton) turns around the common atomic center too, but in a smaller orbit. The atomic electron and the nucleus do not have certain individual orbital angular momenta. They are in mixed states. See my explanations here and here.


8

When I first started to study quantum mechanics, my physics text book told that particles have spin of either 1/2 or -1/2. That's wrong. Particles can have any integer or half-integer spin. (There are some deeply technical reasons that fundamental particles are expected to have spin ranging from -2 to 2, but if you include composite particles, any ...


0

Susskind says (in his Stanford lectures on string theory) that subatomic particles are not point particles, they have a spatial extent. This does not depend upon his contention that they are really extended string objects. It's just a consequence of the fact that even the electron is "fuzzy" due to being embedded in a little cloud of virtual photons and ...



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