2012-06-07 03:21:26 8 Comments

Why is it that unlike other quantum properties such as momentum and velocity, which usually are given through (probabilistic) *continuous* values, spin has a (probabilistic) *discrete* spectrum?

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## 3 comments

## @anna v 2012-06-07 06:29:08

I will take a hand waving guess at this.

Nature is quantum mechanical , i.e. it is ruled by quantum mechanical equations that define motion etc. The classical Lagrangians are a limiting case mostly for large dimensions.

Quantization appears when the variables are constrained , for example in the confines of a potential well. One finds that only quantized values are allowed, so in a confining potential also momentum will be quantized as long as there are discrete energy levels.

So the question can only be correct if one considers the unconfined particles and becomes:

My intuitive answer is: probably because spin is a rotation and rotations are limited by the $0$ to $2\pi$ confinement of the values of phi, a finite constraint, in contrast to momentum which can go from zero to infinity. Constraints are conditions for quantization.

As a help in intuition look at section 14 of Schiff's quantum mechanics, separation in spherical coordinates of the schroedinger equation for spherically symmetric potentials. . The angular equations have no dependence on the potential and their solutions

arequantized.## @anna v 2012-06-07 08:39:20

see also physics.stackexchange.com/q/1

## @Arnold Neumaier 2012-06-08 15:01:11

The deeper reason is that the components of the spin (angular momentum) vector generate the group of rotations. This group is compact, which means that a rotation perpendicular to an arbitrary direction necessarily closes. This implies for mathematical reasons (valid for every compact Lie group) that its representations as operators in a Hilbert space come in discrete batches only, and the eigenvalues of any component, in general functions of the label of the representation, must in the compact case be discrete.

In contrast, position and momentum generate the noncompact Weyl group (a central extension of the phase space translations), and a translation laong an arbitrary phase space direction never closes. This implies that the eigenvalues vary continuously.

## @Steve Byrnes 2012-06-07 12:57:25

I recently was writing about this on wikipedia. The most intuitive way to see why an operator like $S_z$ has discrete values is based on its relation to rotation operators:

$R_{internal}(\hat{z},\phi) = \exp(-i\phi S_z / \hbar)$

where the left side means rotation of angle $\phi$ about the $z$-axis, but only rotating the "internal state" of particles not its spatial position (see wikipedia article for details). Since a rotation of $\phi=720^\circ$ [see below] is the same as no rotation at all (i.e. the identity operator), you conclude that the eigenvalues of $S_z$ can only be integers or half-integers.

...much like how a standing wave on a circular string has to have an integer number of wavelengths.

--

Wait, why did I say $720^\circ$ not $360^\circ$?? Well, there are two mathematical groups that could plausibly correspond to rotation in the real world: $SO(3)$ and $SU(2)$. In $SO(3)$ but not $SU(2)$, rotating $360^\circ$ is the same as not rotating at all. In BOTH of them, rotating $720^\circ$ is the same as not rotating at all. So we can be totally sure that the $720^\circ$ rotation operator is the identity operator, whereas for $360^\circ$ it would just be a guess based on extrapolating from classical physics intuition. As long as there are fermions present, the guess is wrong! Rotating a fermion by $360^\circ$ corresponds to flipping the sign of its wavefunction.

## @anna v 2012-06-08 11:50:36

Nice answer, but maybe you should add a paragraph for why the momentum of a free particle is not quantized, for completeness in the answer.