In What Space Does Quantum Spin Take Place

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Quantum spin is orientable so it takes place in a space with an even number of dimensions. What is that space?
Quantum spin is orientable so it takes place in a space with an even number of dimensions. What is that space?

If the space had an odd number of dimensions, then spin in that space wouldn't be orientable. But quantum spin is orientable.

We could say that it is Minkowski space, but that space is hyperbolic so a cyclic rotation like spin is not possible.
 

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  • #3
Spin is also called intrinsic angular momentum as it is generally the fixed total spin of a specific particle or system of particles. Spin is quantized and can be described by a half-integer, namely: ##0, \frac1 2, 1, \frac 3 2 \dots##.

A measurement of spin can be made about any spatial axis, although spin about the coordinate axes, for example, are incompatible observables. A measurement of spin about any axis always returns a value of ##\pm m##, where ##m## is a half integer value less than or equal to ##s##

The spin state itself can be described by a vector in a complex ##2s+1## dimensional Hilbert space. This state determines the probability of getting a measurement of ##\pm m## along any chosen axis.
 
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  • #4
it takes place
What do you mean by "takes place"? Quantum spin does not mean an actual little ball is spinning about an axis.
 
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  • #5
What do you mean by "takes place"? Quantum spin does not mean an actual little ball is spinning about an axis.
Yes I know that. Am I correct in thinking that quantum "spin", whatever it may be, takes place in a space of some nature that is unknown other than that it has to be even dimensional?
 
  • #6
Yes I know that. Am I correct in thinking that quantum "spin", whatever it may be, takes place in a space of some nature that is unknown other than that it has to be even dimensional?
What's your reference for that claim?
 
  • #7
Yes I know that. Am I correct in thinking that quantum "spin", whatever it may be, takes place in a space of some nature that is unknown other than that it has to be even dimensional?
Neither correct nor incorrect, until you explain what you mean by “takes place”.

We wouldn’t say that the electric charge of a particle “takes place” in some space, we just say the particle has an electric charge. Why should we treat spin differently?
 
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  • #8
So two complex dimensions plus time. Are the complex numbers that represent the spin perchance constrained to be of modulus one?
 
  • #9
So two complex dimensions plus time. Are the complex numbers that represent the spin perchance constrained to be of modulus one?
for spin 1/2 particles you use the 2x2 Pauli matrices as generators
 
  • #10
Yes I know that. Am I correct in thinking that quantum "spin", whatever it may be, takes place in a space of some nature that is unknown other than that it has to be even dimensional?
In quantum mechanics the observables are described as self-adjoint operators in a Hilbert space, which obey a certain (Lie) algebra. This Lie algebra is determined by some symmetry. From classical physics we take over the description of space and time.

For simplicity let's restrict ourself to non-relativistic quantum mechanics based on the Galilei-Newtonian space-time model. The detailed analysis is more complicated than most textbooks suggest, but the simple heuristic version goes as follows:

In classical mechanics we have Noether's theorem. In Hamiltonian formulation it says that for each one-parameter symmetry there is a generator (describing an infinitesimal symmetry transformation) that is a conserved quantity and vice versa.

Applying this to the symmetry of Galilei-Newton spacetime, you get the usual observables that are conserved for a closed system: energy due to time-translation invariance, 3 momentum components due to translation invariance in space, 3 angular-momentum components due to rotational invariance, and center-of-mass velocity due to invariance under Galilei boosts.

In quantum theory you must realize the corresponding Lie algebra in terms of a representation on Hilbert space with self-adjoint operators. For momentum this is, e.g., realized by the operator ##\hat{p}_j=-\mathrm{i} \hbar \partial_j## acting on square-integrable wave functions ##\psi(\vec{x})##.

If you now investigate the representations of the rotation group using its Lie algebra, this implies that the generators of rotations are the angular-momentum operators obeying the commutation relations
$$[\hat{J}_j,\hat{J}_k]=\mathrm{i} \hbar \epsilon_{jkl} \hat{J}_l.$$
From this algebra alone and the self-adjointness of the operators you can derive that any irreducible representation of the rotation algebra is characterized by the eigenvalue of ##\hat{\vec{J}}^2## which can take values ##J(J+1)\hbar^2##, where ##J \in \{0,1/2,1,\ldots \}## and ##\hat{J}_z## which can take eigenvalues ##m \hbar## with ##m \in \{-J,-J+1,\ldots,J-1,J \}##.

One way to realize the algebra is, of course, to simply take the analogon from classical mechanics, orbital angular momentum,
$$\hat{\vec{L}}=\hat{\vec{x}} \times \hat{\vec{p}}.$$
It turns out that because of this special relation here the angular-momentum quantum numbers can only be integers, i.e., ##\hat{\vec{L}}^2## has eigenvalues ##\hbar^2 \ell (\ell+1)## with ##\ell \in \{0,1,2,\ldots \}## and correspondingly also the "magnetic quantum number" ##m## can only be integers, ##0,\pm 1,\pm 2,\ldots##.

Now it would be very suprising if nature wouldn't also make use of the half-integer realizations, which are however not one-to-one representations of the rotation group but of its socalled "covering group" SU(2). Nevertheless, indeed you can realize also these half-integer realizations of the angular-momentum algebra (and in fact also each integer realization too) in terms of a finite-dimensional Hilbert space. For each ##J## you have an irreducible representation on a ##(2J+1)##-dimensional Hilbert space, spanned by the eigenvectors of ##\hat{J}_3##. These realizations we call "spin", because they have nothing to do with orbital angular momentum and instead of ##J## we call the quantum number ##s## and instead of ##m## we call the quantum number ##\sigma##. For each ##s \in \{0,1/2,1,3/2,\ldots \}## again ##\sigma \in \{-s,-s+1,\ldots,s-1,s \}##.

A particle with spin ##s## is then described by a socalled spinor field, i.e., a function with ##(2s+1)## complex components ##\psi_{\sigma}(\vec{x})##. The total angular momentum then is ##\hat{\vec{J}}=\hat{\vec{L}}+\hat{\vec{s}}##, where ##\hat{\vec{s}}## are ##(2s+1) \times (2s+1)## self-adjoint matrices acting on the spinor components.

Physically the spin manifests itself usually in terms of a magnetic moment of the particle. It's operator is ##\hat{\vec{\mu}}=g_s q \hat{\vec{s}}/(2m)##, where ##g_s## is the gyromagnetic factor related to spin (also known as Lande factor). For elementary spin-1/2 particles like the electron ##g_{s} \simeq 2##.

In non-relativistic physics spin is pretty simple, because the spin operators commute with ##\hat{\vec{x}}## and ##\hat{\vec{p}}## and thus also with ##\hat{\vec{L}}##, i.e., spin and orbital angular momentum can take simultaneously determined values (i.e., ##\hat{\vec{L}}^2##, ##\hat{\vec{s}}^2## and ##\hat{L}_z## and ##\hat{s}_z##).

In relativistic physics spin is a bit more complicated. Particularly there's no way to define a unique physical split of total angular momentum in terms of orbital and spin angular momentum, and thus only total angular momentum is a proper observable, but that's another story.
 
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  • #11
Well dang, that's about as succinct a summary of QM I have ever to see.
 

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