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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 ...


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 ...


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 ...


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 ...


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 = ...


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 ...


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 ...


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 ...


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 ...


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 ...


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 ...


2

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. ...


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 ...


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 ...


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 ...


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

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 ...


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 $\vec S_1 \cdot \vec S_2$. ...



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