# Is all angular momentum quantized?

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?

• 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 Mar 26, 2012 at 1:10
• 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) Mar 26, 2012 at 1:47
• Also, in 2D there is no need for quantised angular momentum. Mar 26, 2012 at 10:54

## 3 Answers

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).

• 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.” Mar 26, 2012 at 15:48
• 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. Mar 27, 2012 at 8:54

For those interested in an algebraic approach to this problem, it is possible to prove that angular momentum is quantized using only the commutation relations $$\mathbf{J}\times\mathbf{J}=i\hbar\mathbf{J}.$$ From these, one can build the operator $$\mathbf{J}^2$$, called the total angular momentum operator. One can also build the raising and lowering operators $$J_{\pm} = J_x \pm iJ_y.$$ These operators commute with $$\mathbf{J}^2$$ (and therefore leave its eigenvalue unchanged), and raise/lower the eigenvalue of $$J_z$$ by one unit of $$\hbar$$ (all this can be proved from the commutation relations). It is conventional to define the eigenvalue $$j$$ using $$\hbar^2 j(j+1)=a$$, where $$a$$ is the actual eigenvalue $$\mathbf{J}^2$$. It is also conventional to use $$m=b/\hbar$$, where $$b$$ is the eigenvalue of $$J_z$$.

With these definitions, and the commutation relations, we can prove the inequality $$j(j+1)\geq m^2$$ must hold for any simultaneous eigenstate of $$\mathbf{J}^2$$ and $$J_z$$. Intuitively, this just the quantum mechanics version of the fact that the $$z$$-component of angular momentum cannot be bigger than the total angular momentum.

Suppose $$|\psi\rangle$$ is an eigenstate of $$\mathbf{J}^2$$ and $$J_z$$. Then $$(J_+)^n|\psi\rangle$$ is also an eigenstate of these operators with the same value of $$j$$, but with $$m$$ increased by $$n$$. Therefore, for the inequality to always hold, we must eventually reach a termination point where $$(J_+)^n|\psi\rangle=0$$. Let $$|\psi_\text{max}\rangle$$ be the state such that $$J_+|\psi_\text{max}\rangle=0$$. We can determine the eigenvalues of this state using the fact that $$J_-J_+|\psi_\text{max}\rangle=J_-0=0.$$ You can show that $$J_-J_+=\mathbf{J}^2-J_z^2-\hbar J_z$$. Therefore, we have $$(\mathbf{J}^2-J_z^2-\hbar J_z)|\psi_\text{max}\rangle=(\hbar^2 j(j+1)-\hbar^2 m_\text{max}^2-\hbar^2 m_\text{max})|\psi_\text{max}\rangle=0$$ $$\implies j(j+1)=m_\text{max}(m_\text{max}+1)\implies j=m_\text{max}.$$ Now we just need to use the fact that $$m$$ cannot be lowered indefinitely without violating the inequality. This implies there is $$|\psi_\text{min}\rangle$$ with $$0=J_+J_-|\psi_\text{min}\rangle=(\mathbf{J}^2-J_z^2+\hbar J_z)|\psi_\text{min}\rangle=(\hbar^2j(j+1)-\hbar^2m^2_\text{min}+\hbar^2m_\text{min})|\psi_\text{min}\rangle$$ $$\implies j(j+1)=m_\text{min}(m_\text{min}-1).$$ Combining this with the relation we got above, we get $$m_\text{min}^2-m_\text{min}=m_\text{max}^2+m_\text{max}.$$ From this, we infer $$m_\text{max}=-m_\text{min}$$. Now if we start raising the minimum state, we must eventually get to the maximum state or we will be able to raise $$m$$ indefinately, violating the inequality. Therefore, $$m_\text{max}=m_\text{min}+n$$ for some integer $$n$$. Combining this with the previous equation gives $$j=m_\text{max}=\frac{n}{2}.$$ Therefore, $$j$$ can only be an integer or a half-integer.

I'm going to disagree with the other answers: I think that the angular momentum of macroscopic "classical" objects is not quantized.

Consider an automobile tire spinning in a wheel well. On the tire is a device that triggers whenever a certain point on the wheel crosses a certain point on the well, adding one to an internal counter if it passes it clockwise and subtracting one if it crosses it counterclockwise. (Alternately, you could dispense with the wheel well and say that the device tracks its own position by inertial navigation.) The state of this system can be described by a value $$θ\in\mathbb R$$, where the current value of the counter is $$\lfloor θ/2π\rfloor$$ and the angle of the wheel is $$θ\text{ (mod }2π\text{)}$$. In the absence of external forces, the Hamiltonian of the system is essentially that of a free particle in $$\mathbb R$$, and the spectrum of angular momenta is continuous just like the free particle's momentum spectrum.

That's a 2+1 dimensional system. In 3+1 dimensions, there's the Dirac belt trick to worry about. Does it matter? I don't think so. There's no reason to limit the device to holding a single integer, or to being reversible. It could simply store the entire history of its orientation readings internally, or broadcast them by radio, indelibly recording them in the universal wave function. That's a very noncompact state space, and it's an accurate enough model of bodies like the earth.

The angular momentum operator on this monstrosity obviously violates the assumptions of any proof of the quantization of angular momentum, but that's no reason not to call it angular momentum. We do call it angular momentum, and it's what the question was about.

In response to comments I'll try to clarify my answer.

These are quantum systems, but the earth system is "classical" in the sense of being a quantum system with emergent classical behavior.

The reason that high temperature systems behave classically is that they constantly leak which-path information into the environment. If you do a double-slit experiment with the earth, it will emit different patterns of light going through one slit than through the other. You can literally see which slit it goes through, but even if you don't look, the which-path information is there in the patterns of light, or in patterns of heat if the light is absorbed by the walls of the lab, and that's all that's necessary to make the final states orthogonal and destroy the interference pattern.

It's sometimes said that you can't see an interference pattern in the earth double-slit experiment simply because its de Broglie wavelength is so small. That would be correct for a supermassive stable particle that doesn't radiate, but it's wrong for the earth. For the earth there's no interference pattern at all, for the same reason there's no interference pattern when there's a detector at one of the slits. Earth's thermal radiation is the "detector".

The case of rotation is similar. Different rotations are different paths through the state space (it's the state space, not physical 3D space, that the wave function is defined on and which matters here). If you consider two different paths ending in the same physical orientation (analogous to the same position on the screen in the double-slit case), these paths will interfere if no information about which path was taken is recorded anywhere. In the case of the earth, this means they'll interfere if there's no way for anyone to tell whether the earth rotated around its axis or not. If there's any record of it – if any animals remember the day-night cycle, or don't but could in principle, or if aliens see it rotate through a telescope, or don't but could in principle – then there's no interference.

The proof that angular momentum is quantized depends on the compactness of the space of orientations. This is fine if the space of orientations is the phase space, i.e., if the system is memoryless. If it has a memory, rotating the system through $$2π$$ or $$4π$$ doesn't leave it in the same state as not rotating it.

The tire example in the second paragraph may have been a mistake since it seems to have only caused confusion. But it's a perfectly good quantum system in the abstract, and its state space is $$\mathbb R$$, not $$S^1$$.

• Quantization is a quantum effect, so you are completely correct that it doesn't happen if we take the classical limit. However, every object should be treated, technically speaking, with the laws of quantum mechanics, so all angular momentum is quantized, even if those effects are not noticeable. Aug 13, 2020 at 23:58
• The system you've described is not essentially like a free particle in $\mathbb R$, it's essentially like a free particle in a loop (i.e. in $S^1$ rather than $\mathbb R$). The momentum spectrum of a free particle in a loop is not continuous, and neither is the angular momentum spectrum here.
– Chris
Aug 14, 2020 at 0:04
• @JoshuaTS I am treating these objects as quantum mechanical. Aug 14, 2020 at 1:05
• @Chris Every value of $θ\in\mathbb R$ maps to a distinct state of the system. The system includes the counter and its stored count. Aug 14, 2020 at 1:06
• @benrg The stored count clearly has nothing to do with the physical motion of the wheel. Keeping a count of that would require an extra term in the Hamiltonian anyway- if it's truly a free particle there's no way to update the count.
– Chris
Aug 14, 2020 at 1:10