# Is angular momentum always quantized?

pellman
For a bound particle we have that quantized energy levels while a free particle has a continuous spectrum of energy. Similarly for momentum.

Thus electrons in an atom have definite energy levels. They also have orbits of discrete angular momentum. But what about free particles?

One might think angular momentum for free particles would also be continuous. But any wavefunction can be expressed as a superposition of spherical harmonics with integer (m,l). Measuring the angular momentum amounts to reduction to a single spherical harmonic and yielding $$m\hbar$$ with m of the associated spherical harmonic.

Do I have this right? And is this because the range of angles if bounded whether a particle is bounded or not, while momentum and energy for a free particle are unbounded?

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Even free particles have discrete values of angular momenta with respect to a perticular (arbitrary choosen) point in space. That is what I've been thaught.

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Thus electrons in an atom have definite energy levels. They also have orbits of discrete angular momentum. But what about free particles?

A free particle basically moves on a straight line (besides from its wave-function
spreading out) A particle on a straight line does have an angular moment with
respect to any given point but it is not quantized. This is not possible mathe-
matically in the first place since it has a different angular momentum with respect
to each point in the universe and they can't be all quantized.

A practical requirement for quantization to occur would be that the particle's wave
function is non-zero over an entire closed trajectory. An electron circulating
the Earth or an accelerator ring doesn't generally fulfill this requirement. It is,
in general, localized.

Regards, Hans

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But you are now saying that a free particle moves on a straight line plus its wave function is spread out? Classically, a free particle moves in a straight line, but QM, we must use the plane wave - wave function for a free particle. We can not say "here is the path of the particle", then we wouldn't use the concept of wave functions, which is the probablity to find a particle at a certain point at a certain time.

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But you are now saying that a free particle moves on a straight line plus its wave function is spread out? Classically, a free particle moves in a straight line, but QM, we must use the plane wave - wave function for a free particle. We can not say "here is the path of the particle", then we wouldn't use the concept of wave functions, which is the probablity to find a particle at a certain point at a certain time.

But infinite plane waves do not have an angular momentum regardless of the question
if they really represent a free physical particle or not.

In an angular momentum atomic orbit the value L = p x r is constant, where p is the
gradient of the phase of the amplitude. In a plane wave the value p x r is different
everywhere and effectively undefined when integrated over the entire wave function.
That is, the integral depends on which point you associate with the center of the wave
function.

Regards, Hans

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maybe one should specify what is meant by angular momentum in QM ?

lbrits
But infinite plane waves do not have an angular momentum regardless of the question
if they really represent a free physical particle or not.

I'm going to disagree on this one. A plane wave admits the following partial wave expansion in terms of discrete angular momentum eigenstates.

$$\psi_\vec{k}(\vec{r}) = \frac{1}{(2\pi)^{\frac{3}{2}}} e^{i \vec{k}\cdot \vec{r}} \to \frac{1}{(2\pi)^{\frac{3}{2}}} \sum_\ell {i}^\ell\, (2\ell+1)\, \frac{i}{2k} \left[ \frac{1}{r} e^{-i \left( k r-\frac{\ell\pi}{2} \right) } -\frac{1}{r}e^{i \left( k r-\frac{\ell\pi}{2} \right) } \right]\, P_\ell(\cos\theta)\,$$

In any event, the angular momentum always depends on which point you take as the origin. For example, if you do perturbation theory on the Helium atom, then you have to write eigenstates of electrons around one center in terms of those around the other center. A messy calculation with lots of 3j symbols, but it can be done (in principle, and in practice).

So while a plane wave isn't in an eigenstate of L, L is most certainly quantized. You just end up with a superposition of eigenstates (interestingly, a discrete sum).

pellman
In a plane wave the value p x r is different
everywhere and effectively undefined when integrated over the entire wave function.
That is, the integral depends on which point you associate with the center of the wave
function.

Sure. A plane wave is not an eigenstate of angular momentum. But it is a superposition of eigenstates of angular momentum. The question amounts to, is it always a sum over a countable number angular momentum eigenstates

$$\Sigma \alpha_{ml}f_{ml}(r)Y^{m}_l(\theta,\phi)$$

or do we ever have to integrate over continuous variables corresponding to m and l?

maybe one should specify what is meant by angular momentum in QM ?

Pick an axis, define $$\phi$$ in a right-handed manner. The component of angular momentum in that direction is $$-i\hbar\frac{\partial}{\partial\phi}$$. That's what I am talking about. Can we ever have an eigenfunction of this operator whose eigenvalue is not an integer times $$\hbar$$?

lbrits
Sure. A plane wave is not an eigenstate of angular momentum. But it is a superposition of eigenstates of angular momentum. The question amounts to, is it always a sum over a countable number angular momentum eigenstates or do we ever have to integrate over continuous variables corresponding to m and l?

Pick an axis, define $$\phi$$ in a right-handed manner. The component of angular momentum in that direction is $$-i\hbar\frac{\partial}{\partial\phi}$$. That's what I am talking about. Can we ever have an eigenfunction of this operator whose eigenvalue is not an integer times $$\hbar$$?

To the first question, if you examine my post, is that the plane wave is a sum over discrete eigenvalues. There's no integral there. This is because the angular momentum operators only have discrete eigenvalues (there's no continuum to sum over, even if you wanted to).

The discreteness has to do with the compactness of the group SO(3), and nothing to do with what particles are doing. This is one of the reasons that special relativity sort of forces us to do quantum field theory, because the relevant group there is SO(1,3), which is non-compact. It has a continuum of eigenvalues corresponding to boosts.

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A plane wave admits the following partial wave expansion in terms of discrete angular momentum eigenstates...So while a plane wave isn't in an eigenstate of L, L is most certainly quantized. You just end up with a superposition of eigenstates (interestingly, a discrete sum).

One can decompose a plane wave always into an infinite series of discrete angular
momentum eigenstates because the latter form mathematically a complete set of
orthogonal eigenstates. The decomposition is different for each and every point
one takes as the center of rotation. But what physical meaning is left with such
an arbitrariness?

Regards, Hans

pellman
The decomposition is different for each and every point
one takes as the center of rotation. But what physical meaning is left with such
an arbitrariness?

That's true if you're talking about the angular momentum vector. But for just one component all you need is an axis.

I would think it would show up in scattering experiments if the target is very massive. Plane wave in, spherical waves out. The direction of the plane wave defines your z-axis about which there is a symmetry with respect to $$\phi$$.

peter0302
Angular momentum (spin) is always quantized. A free particle is represented only by its wave function, and you cannot know precise values for its spin in x,y,z. But once you isolate it by, say, sending it through a Stern-Gerlach magnet, you can know its spin value for particular vector which the magnet is aligned to, and that value is always quantized. If the electron emerges from the S-G magnet, you know its spin value for that direction is +1/2 hbar, and its spin value for the orthogonal vector is -1/2 hbar (you still can't know the third axis). So for the free particle spin is merely uncertain, but it's still quantized w/r/t any axis you choose to measure it by.

lbrits
Angular momentum (spin) is always quantized. A free particle is represented only by its wave function, and you cannot know precise values for its spin in x,y,z...

Well, both orbital ($$L$$) and spin ($$S$$) angular momentum are quantized. I think the OP was concerned with orbital, just for clarification.

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That's true if you're talking about the angular momentum vector. But for just one component all you need is an axis.

For me the way a system is quantized always depends on the boundary conditions.

There are infinitely many sets of eigenfunctions in which you can decompose
any arbitrary function, both in classical physics as well as in QM, but it's the
boundary conditions which determine which set of eigenfunctions has a physical
meaning.

Spherical harmonics in case of a central potential, Gaussian Hermite functions
in case of non relativistic linear harmonic oscillator, confluent hypergeometric
functions in case of a non-relativistic spherical harmonic oscillator, imaginary
exponentials in case of a square box, and so on.

So, somebody who says: "Orbital angular momentum is quantized" should really
say: "Orbital angular momentum is quantized in case of a central potential
boundary condition"

Regards, Hans

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If I interpret lbrits' statement about compactness correctly, what is being said is that the discreteness comes from the fact that angular momentum is related to what happens when you rotate a system, and the rotation symmetry has a periodicity (over $$2 \pi$$ in the usual coordinatization). That periodicity implies discreteness in the eigenstates associated with rotational symmetry, in a manner similar to how Bohr originally conceptualized the quantization of angular momentum in an atom. Translational symmetry, on the other hand, in an unbounded universe is not periodic, and gives rise to continuous momentum eigenvalues.

In other words, there is a continuously infinite class of functions that map into themselves under translations in infinite space, but there is only a countably infinite class of functions that can map into themselves under rotations, i.e., that conform to the symmetry. So this implies that we need a 'less dense' set of solutions that can fully represent the rotational degrees of freedom satisfying that symmetry. I think this all says that if you did not quantize angular momentum, you would be including solutions that do not conform to the basic rotational symmetry that says some spin-related integer number of $$2 \pi$$ rotations must bring you back to where you started. In short, rotation has a fundamentally different topology than translation of a plane wave.

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In the case of a non-central potential -- spin orbit interactions are an example; or a multipole potential, scattering from two central potentials - angular momentum is indeed quantized. That's because of the imposition of the notion that physics from 0-2pi is the same as physics from 2npi to 2(n+1)pi, which of course is the basic reason why we say total angular momentum is conserved.

However, a particle in a non-central potential will generally not be in an eigenstate of angular momentum.This becomes very evident in partial wave expansions, particularly for nuclear reactions and relativistic scattering problems. Generally such problems require considerable use of angular momentum theoretic tools; multipole expansions for, say, solutions of the Dirac Equation, various addition theorems, the Wigner-Eckart Thrm. The Jacob and Wick approach is particularly powerful in such problems -- allows one to figure out, in finite time, the invariant amplitudes for, say elastic scattering of a spin 0 particle and a spin1 vector boson mediated by the exchanges of spin one and two particles, which then allows the partial wave amplitudes to be determined(See Atkinson, Phys. Rev. Vol 142,#4,1966, and Atkinson and Everett, Phys. Rev. Vol 154,#5, 1967. Probably, Phys Revs that old have been put in the basement, or a round file.)

Basically, partial wave amplitudes are the equivalent of a multipole expansion of an arbitrary charge distribution's potential. Most often, such an expansion has a natural center point or origin -- the center of charge of a charge distribution, the position of the lab-frame target in a scattering situation. And, various addition theorems allow the use of differing origins. All inertial observers will agree that a particular event has occurred, but will "see things" with differing coordinate systems; this includes observers connected by rotations and linear displacements, according to the Poincare Group -- Weinberg, in his QFT Vol 1. book discusses the Poincare Algebra, and also partial wave amplitudes. Particularly for angular momentum, some coordinate systems are more equal than others in their descriptive simplicity. Nonetheless, every observer will undoubtedly suppose that his or her angular momentum is legitimate, but might be excessively complex, which then could lead to the use of a unitary transformation to make life somewhat more simple.

Sommerfeld in his PhD thesis, used the notion of complex angular momentum in order to enhance the convergence of Legendre expansions -- he dealt with radiation from an antenna on a conducting spherical earth, a rather nasty problem. The idea came from work of Watson(1918), of Bessel Function fame, on replacing a discrete sum by a contour integral, which enhances convergence. This was rather obscure in the physics community until Tulio Regge used this approach to invent "Regge Poles ' in particle physics in the 1960s, the heroic days of S-Matrix Theory.

I have not thought this through entirely, but I wonder if General Relativity allows the lifting of the usual periodicity condition -- that an angular displacement of 2pi is not necessarily an identity transformation for a spin zero particle. Hmm.

(I've said much the same things as have others; my apologies for omitting acknowledgments .I've tried to tie various comments together. )
Regards,
Reilly

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