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Angular momentum is definitely quantized in elementary particles and electrons in atoms. Molecules also have characteristic rotation spectra.

  1. Is it true that all angular momentum is quantized, including big things like automobile tires, flywheels and planets?

  2. If so what is the largest object for which this quantized rotation has been verified/observed/measured?

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    $\begingroup$ I'm quite sure it's true even for macroscopic objects, albeit obviously not measurable since $\hbar\ll L$. But what's the biggest objects for which it is measurable would be interesting to know $\endgroup$ – leftaroundabout Mar 26 '12 at 1:10
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    $\begingroup$ I would submit the angular momentum of neutron stars as the answer, but someone who really knows (I worry it might be apocryphal) should write it up. Most of the neutron star should be some form of superfluid, in which angular momentum must be contained in vortices (similar to en.wikipedia.org/wiki/Abrikosov_vortex). As the rotation slows due to radiation/energy loss the vortices leave one by one from the core, and it is possible to observe the spikes in the rotation rate: en.wikipedia.org/wiki/Glitch_(astronomy) $\endgroup$ – genneth Mar 26 '12 at 1:47
  • $\begingroup$ Also, in 2D there is no need for quantised angular momentum. $\endgroup$ – genneth Mar 26 '12 at 10:54
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The angular momentum only has quantized eigenvalues; this statement is valid quite generally for all bodies. For example, $J_z$ has to be a multiple of $\hbar/2$ because $$ U = \exp(4\pi i J_z) $$ is the rotation by $4\pi$ and such a rotation brings every state to itself and has to be identity. (For a $2\pi$ rotation, the state changes the sign if it contains an odd number of fermions.) Therefore, we have $$\exp(4\pi i j_z) = 1\quad \Rightarrow\quad j_z\in\{0,\frac 12, 1, \frac 32, \dots\}$$ Can the quantization of $j_z$ be actually measured? Well, one may only measure a sharp value of $j_z$ if the object is an eigenstate. Eigenstates of $j_z$ are rotationally symmetric with respect to rotations around the $z$-axis, up to an overall phase. So if we have a non-axially-symmetric object, its sharp $j_z$ eigenvalue obviously can't be observed because it's a linear superposition of many states with different $j_z$ eigenvalues.

For atoms, the angular momentum may be observed; these are the usual quantum numbers associated with the electrons. In the same way, the total angular momentum may obviously be measured and shown to be quantized for nuclei.

Larger systems are molecules. For some molecules, the quantized nature of the angular momentum may be measured. To add some terminology, we measure the rotational quantum numbers of these molecules by observing transitions in the rotational spectrum and the method is the rotational spectroscopy:

http://en.wikipedia.org/wiki/Rotational_spectroscopy

It only applies to molecules in gases because in solids and liquids, collisions constantly distort the angular momentum. Also, one can't have a well-defined quantized $j_z$ for "true solids" i.e. crystals because crystals aren't symmetric under continuous rotations; they're only kept invariant by the discrete crystalline subgroup of the rotational group.

So the maximum size for which the quantization may be verified are "rather large" molecules of gases and the maximum size is getting larger as the progress goes on (and as people are able to reduce the temperature and improve the accuracy).

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  • $\begingroup$ I did find halexandria.org/dward156.htm about superconductors: "Although quantum mechanical behavior is usually thought of as being restricted to the microscopic scale of an atom or molecule, superconductivity operates at a macroscopic quantum level; pairs condense into a single large-scale quantum state, which has long-range order and can be described as if it was a giant molecule with a single wavefunction.” $\endgroup$ – anna v Mar 26 '12 at 15:48
  • $\begingroup$ That's surely right and excitations in superconductors may have quantized spin. Still, it is hard to find that the background superconducting "medium" would be an angular momentum eigenstate. This won't really happen easily. $\endgroup$ – Luboš Motl Mar 27 '12 at 8:54

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