Tag Info

Hot answers tagged

14

Spin is not about stuff spinning. (Confusing, I know, but physicists have never been great at naming things. Exhibit A: Quarks.) Spin is a purely quantum mechanical phenomenon, it cannot be understood with classical physics alone, and every analogy will break down. It has also, intrinsically, nothing to do with any kind of internal structure. ...


7

The magnetic moment of the electron is a magnetic moment, so the right magnetic field around it is $$ \mathbf{B}({\mathbf{r}})=\nabla\times{\mathbf{A}}=\frac{\mu_{0}}{4\pi}\left(\frac{3\mathbf{r}(\mathbf{\mu}\cdot\mathbf{r})}{r^{5}}-\frac{{\mathbf{\mu}}}{r^{3}}\right). $$ The world is quantum mechanical – and so is any viable description of the spin – so we ...


6

"The electron has no known internal structure", but since it does have a spin, does that mean that we know the electron has an internal structure but we just don't know what it is? An electron has no known internal structure simply means that nobody knows if the electron has an internal structure. So far they know none and therefore they suppose it ...


6

Elementary particles have quantum mechanical spin. This induces a spin magnetic moment, independent of the presence (or, indeed, absence) of a (net) electric charge. This is how the neutron attains its magnetic moment (as you already mentioned). The case of the neutrino magnetic moment is slightly confusing, as they are not completely understood yet. ...


5

You already received several answers. However the fundamental physical reason is elementary: In classical, quantum and relativistic physics the physical laws describing an isolated physical system in an inertial reference frame are the same (are invariant) if you rotate (with an element of $SO(3)$) the system (there are many other symmetries depending on the ...


2

Just like $j=1/2$ (spin-half) particles have two-complex-component wave functions, spinors, particles with spin $j$ have $(2j+1)$-dimensional wave functions describing the spin degrees of freedom. The dimension is what it is because $j_z$ always goes from $-j$ to $+j$ with the spacing equal to one. All the transformation rules under rotations may be ...


2

As per Rob's suggestion, I decided to make this an answer. The Answer The "vanilla" Schrodinger's equation (from non-relativistic QM) does not describe a spin-1/2 particle. The plain, old Schrodinger's equation describes a non-relativistic spin-0 field. Case Studies If we pretend the wave function is a classical field (which happens all the time during ...


2

One can make sense of the introduction of $\mathrm{SO}(3)$ into quantum mechanics as follows: Consider a physical system in three spatial dimensions which we'll think of as residing in a box sitting on a table, like a table-top experiment. Suppose that we prepare the system in a particular way, so that the system is measured to be in a (pure) state ...


2

Spin is a wave property. It exists in classical relativistic wave theories as well. A circularly polarized wave carries an angular momentum that's related to the spin of the field. A gravitational wave (spin-2) can carry twice the angular momentum of a classical electromagnetic wave (spin-1). Being "pointlike" is a particle property. You can think of the ...


1

The best way to understand spin is actually to consider the Dirac Equation $$ i\hbar \frac{\partial }{\partial t}\Psi=\left[c\sum_i{\alpha_i p_i}+mc^2\beta\right]\Psi $$ or more compactly: $$(i\gamma^\mu\partial_{\mu}-m)\psi=0$$ The solutions to the Dirac equation are collections of complex valued fields called spinors. The spinor solution actually ...


1

First of all, you need to understand from which root cause the spin appears in general. This root cause is a symmetry of the physical space-time. Particles with different spins (I mean, spin-0 particles, spin-½ particles, spin-1 particles, and so on) use different representations of a symmetry group to map geometry of the space-time to their quantum spin ...


1

Although I see already two answers, I’ll add mine. The situation, as original poster described, is such for a massive spin-½ particle and is not symmetric in this sense for a massless spin-½ particle (some people say this case is properly named “helicity”, but it is a question of terminology). Why does it matter? Because a massive particle has its ...


1

The question is "Why are the possible outcomes the same for all directions?" It happens also for observables of classical physics! QM does not matter here, the truly relevant idea is the fact that in a inertial system physics appears to be isotropic. In practice, it is not possible to physically distinguish different directions with physical experiments. ...


1

The Lie algebras and $\mathfrak{so}(3)$ $\mathrm{su}(2)$ are isomorphic, but the Lie groups $\mathrm{SO}(3)$ and $\mathrm{SU}(2)$ are not. In fact $\mathrm{SU}(2)$ is the double cover of $\mathrm{SO}(3)$; there is a 2-1 homomorphism from the former to the latter. How is this possible when every so(3) irrep can get raised to one of SO(3) as described ...


1

Spin in (non-relativistic) QM is fairly ad hoc, it has no deep reason. The underlying reason is that, in a relativistic setting for QM/QFT our states/fields must transform in some representation of the Lorentz group $\mathrm{SO}(1,3)$ if we want the amplitudes/Lagrangians to be Lorentz invariant (since something cannot be invariant if there's no rule how its ...



Only top voted, non community-wiki answers of a minimum length are eligible