# Tag Info

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By no means not a complete answer, more a criticism of @luksen’s one. It is posted here because the text is too long to fit in the comment field. First of all, the spin is not a well-defined concept for composite particles. More precisely, whether the spin of a particle is defined depends on how the “particle” is defined. Look at an atom: it has the nucleus ...

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Sorry for reiteration of some points already made by Luboš, but in short, the spin operator of a massless particle isn’t a vector and hence, one can’t define its spatial projection. To understand the difference it is important to point out why said operator for a massive particle is a vector. It is so for the reason explained by Luboš: the center-of-mass ...

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For massive particles the intuition of thinking of spin as a rotation is correct. In the rest frame, a massive ($M^2>0$) particle has momentum $$p_0^\mu=(M,0,0,0).$$ Remember that the quantity $p^2=p_\mu p^\mu$, for arbitrary $p_\mu$ is invariant under Lorentz transformations. In the case above the subgroup of the Lorentz group leaving $p_0^\mu$ ...

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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 ...

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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. ...

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For what it's worth, I've always had the same feeling that the spin should have some sort of reason behind it. It seems so unsatisfying to be told more or less that "it just came that way." Is there any deeper sort of explanation at all? I recall a paper in AJP from years ago called "What is spin?" by Ohanian, but I didn't put in the effort to follow it. I ...

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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 ...

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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. ...

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"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 ...

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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. ...

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There are two possible answers, one for formula-writers and another for conceptually-thinking physicists. First: the “spin $x$” is just $\sigma_x$ and its exact configurations are merely eigenvectors, like the “spin $z$” is $\sigma_z$ and its exact configurations are simply the standard basis vectors. Don’t ask what does it mean because it doesn’t mean ...

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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 ...

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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 ...

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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 ...

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Schrödinger simply does not account for spin. For spin 1/2, you need the Pauli equation or Dirac. A spinless particle, meaning spin zero such as the Higgs boson or pions (ignoring their internal quark/gluon structure) is described by the Klein-Gordon wave equation. I've read a claim that, to the extent that it does describe a particle with spin, ...

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It is possibly a reference to the Pauli equation or Schrödinger–Pauli equation: the formulation of the Schrödinger equation for spin-½ particles, which takes into account the interaction of the particle's spin with an external electromagnetic field. It is the non-relativistic limit of the Dirac equation and can be used where particles are moving at speeds ...

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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 ...

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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 ...

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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 ...

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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 ...

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The very important group in physics is the Lorentz group $SO(3,1)$. It is built from the $SO(3)$ group and Lorentz boosts. The algebra of the Lorentz group generators (boosts $L_{i}$ and 3-rotations $R_{i}$) doesn't have the separation on $SO(3)$ and boosts' parts. But by introducing the "new" generators $J^{i}_{\pm} = \frac{1}{2}(R_{i} \pm iL_{i})$ we may ...

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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 ...

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