# What is actually spinning in quantum spin?

• B
Summary:
If elementary particles themselves that are not literally spinning, what is? Is it their wave function that is spinning?
Some texts say quantum spin is analogous to the spin of a planet in that it gives a particle angular momentum and a magnetic moment. However, as subatomic particles are tiny, the surfaces of charged particles would have to be moving faster than the speed of light in order to produce the measured magnetic moments!
If the particle is not physically spinning, is anything spinning?
Is the particle's wave function spinning?

Nugatory
Mentor
Nothing is spinning.
Saying that it’s “analogous” means that it’s just an analogy - and although this is a matter of personal taste I think it’s a pretty crummy one.

Hamiltonian299792458 and Stevexyz
vanhees71
Gold Member
I think a better heuristics for spin is to think in terms of the magnetization of the particle associated with its spin, i.e., $$\hat{\vec{\mu}}=g q/(2m) \hat{\vec{s}}$$, i.e., it's like a dipole moment of a permanent magnet. In fact a permanent magnet (ferro magnet) is a material, where a macroscopic number of particles have their magnetic moments aligned.

Stevexyz
Dale
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2020 Award
it gives a particle angular momentum
That is it. Spin is intrinsic angular momentum. It has no need to be dressed up in classical terms or analogies or descriptions. It is just angular momentum.

hutchphd, Stevexyz, vanhees71 and 1 other person
atyy
Some texts say quantum spin is analogous to the spin of a planet in that it gives a particle angular momentum and a magnetic moment. However, as subatomic particles are tiny, the surfaces of charged particles would have to be moving faster than the speed of light in order to produce the measured magnetic moments!
If the particle is not physically spinning, is anything spinning?
Is the particle's wave function spinning?

It is analogous, but not in the sense of a rotation in space. The analogy is by the form of the algebraic relations.
http://farside.ph.utexas.edu/teaching/qmech/Quantum/node86.html
http://farside.ph.utexas.edu/teaching/qmech/Quantum/node87.html

vanhees71
In string theory how a string vibrates determines its spin.
It is possible that we are sending strings through Stern-Gerlach type experiments that we, for lack of a better description at this point, are labeling 'electrons' or 'photons' with intrinsic spins.

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That is it. Spin is intrinsic angular momentum. It has no need to be dressed up in classical terms or analogies or descriptions. It is just angular momentum.
Wasn't it originally thought to be a result of actual, mechanical spin? I believe that to be true for electrons at least. Science is riddled with misnomers due to misconceptions at the time of discovery.

Summary:: If elementary particles themselves that are not literally spinning, what is? Is it their wave function that is spinning?

Some texts say quantum spin is analogous to the spin of a planet in that it gives a particle angular momentum and a magnetic moment. However, as subatomic particles are tiny, the surfaces of charged particles would have to be moving faster than the speed of light in order to produce the measured magnetic moments!
If the particle is not physically spinning, is anything spinning?
Is the particle's wave function spinning?
Doesn’t Spin have enormous consequences, in addition to telling us about a particle’s intrinsic angular momentum? Particles with integral Spin (bosons) have symmetric wavefunctions whereas particles with Spin ½ (fermions) have symmetric wavefunctions. As consequence spin ½ wave functions lie on top of each other and cancel out giving zero possibility of this situation being permitted (Pauli Exclusion Principle). Is this correct?

PeroK
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2020 Award
Doesn’t Spin have enormous consequences, in addition to telling us about a particle’s intrinsic angular momentum? Particles with integral Spin (bosons) have symmetric wavefunctions whereas particles with Spin ½ (fermions) have symmetric wavefunctions. As consequence spin ½ wave functions lie on top of each other and cancel out giving zero possibility of this situation being permitted (Pauli Exclusion Principle). Is this correct?
A particle's intrinsic angular momentum (spin) determines its statistical properties. Spin is scientifically equivalent to intrinsic angular momentum. There is, by definition, no difference.

And, yes, the exclusion principle applies to all fermions.

Stevexyz
vanhees71
Gold Member
I think one should not bother too much to find a classical picture for spin since it's a specific quantum-theoretical notion. As I said above the most fitting classical analogue is to rather think about the magnetic moment of the particle which is proportional to its spin, i.a., an axial vector associated with the particle at rest. Of course the spin itself is an angular momentum which is present also for a particle at rest (i.e., in the momentum eigenstate for ##\vec{p}=0##). For sure, it cannot be understood as some orbital angular momentum ##\vec{L}=\vec{x} \times \vec{p}##.

Dale
AndreasC
Gold Member
Wasn't it originally thought to be a result of actual, mechanical spin? I believe that to be true for electrons at least. Science is riddled with misnomers due to misconceptions at the time of discovery.
Yeah, kind of. When Pauli came up with it, he realized it wasn't describable classically. Ralph Kronig was the one who thought it must be because of the electron spinning around an axis, but then Pauli explained to him that if that were the case, the electron would have to spin faster than light to produce the observed magnetic moment, as long as a whole host of other issues with that picture so he didn't publish. Then two other physicists independently came up with the same idea and published it. It was clear from early on it has significant issues, nevertheless, people still use it as a heuristic picture.

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Yeah, kind of. When Pauli came up with it, he realized it wasn't describable classically. Ralph Kronig was the one who thought it must be because of the electron spinning around an axis, but then Pauli explained to him that if that were the case, the electron would have to spin faster than light to produce the observed magnetic moment, as long as a whole host of other issues with that picture so he didn't publish. Then two other physicists independently came up with the same idea and published it. It was clear from early on it has significant issues, nevertheless, people still use it as a heuristic picture.
It's not really a wonder that people do poorly quantum if they are tought wrong things from the beginning. This is more in regards to chemistry, though. I'm sure physics students do well in their quantum classes.

AndreasC
Gold Member
It's not really a wonder that people do poorly quantum if they are tought wrong things from the beginning. This is more in regards to chemistry, though. I'm sure physics students do well in their quantum classes.
I don't think that model hurts people very much, it's kinda like the orbital model, people are taught it to have an intuitive crutch, but they are also taught that it isn't really true.

PhDeezNutz
Summary:: If elementary particles themselves that are not literally spinning, what is? Is it their wave function that is spinning?

Some texts say quantum spin is analogous to the spin of a planet in that it gives a particle angular momentum and a magnetic moment. However, as subatomic particles are tiny, the surfaces of charged particles would have to be moving faster than the speed of light in order to produce the measured magnetic moments!
If the particle is not physically spinning, is anything spinning?
Is the particle's wave function spinning?

As mentioned by several people before, the classical analogy of a "spinning" planet is misleading, as with point particles there is no "surface" with "elements" spinning around a common centre.

In the end, point particles at rest are determined by exactly two spacetime quantities: mass and spin. It is sometimes (and in this thread too) mentioned that spin is essentially a quantum property, but I would in general argue against it. Yes: spin-1/2 is quantum, but otherwise spin is as much a quantum property as mass: either they both are or neither, but both contribute to macrosocopic angular momentum and mass. Moreover, it is not more problematic to accept a pointlike particle with spin than to accept a pointlike particle with mass only. If the latter would be visualized classically, you end up with a singular density of mass. Therefore pictures like this just don't work.

Interestingly enough, both mass and spin not only follow as the two fundamental spacetime properties of point particles in both nonrelativistic and relativistic physics (see the representation theory of Galilei/Bargmann and Poincare groups). They also are the two fundamental spacetime properties of black holes as well.

Therefore I would argue to treat them both the same way.

vanhees71 and Stevexyz
vanhees71
Gold Member
Ok, in this sense you can say that spin is a field rather than a particle property. Of course relativistic physics, no matter whether classical or quantum, is most naturally described as a (quantum) field theory to begin with.

There are of course many similarities concerning spin in non-relativistic and relativistic physics but also some subtle differences: While in non-relativistic QT spin is just an independent additional fundamental observable besides momentum and position and also commuting with both the momentum as well as the position operators. This is no longer true for (massive) relativistic particles, where usually only the total angular momentum is an observable, while the split into orbital and spin angular momentum parts is not unique. For massless particles you have helicity rather than spin anyway (and the necessity to have a gauge theory for particles with spin ##\geq 1##).

Concerning mass the non-relativistic case is a bit special. While it's just a Casimir operator of the proper orthochronous Poincare group, it's a non-trivial central charge of the Galilei group, such that the only successful non-relativistic QTs are based on a central extension of the covering group of the Galilei group, while the Poincare group has no non-trivial central extensions.

otennert
Ok, in this sense you can say that spin is a field rather than a particle property. Of course relativistic physics, no matter whether classical or quantum, is most naturally described as a (quantum) field theory to begin with.

There are of course many similarities concerning spin in non-relativistic and relativistic physics but also some subtle differences: While in non-relativistic QT spin is just an independent additional fundamental observable besides momentum and position and also commuting with both the momentum as well as the position operators. This is no longer true for (massive) relativistic particles, where usually only the total angular momentum is an observable, while the split into orbital and spin angular momentum parts is not unique. For massless particles you have helicity rather than spin anyway (and the necessity to have a gauge theory for particles with spin ##\geq 1##).

Concerning mass the non-relativistic case is a bit special. While it's just a Casimir operator of the proper orthochronous Poincare group, it's a non-trivial central charge of the Galilei group, such that the only successful non-relativistic QTs are based on a central extension of the covering group of the Galilei group, while the Poincare group has no non-trivial central extensions.

Correct. And the centrally extended Galilei algebra even has a third Casimir operator, constituting the "internal / rest energy" of a system proportional to E-p^2/(2m), which however can be "gauged" away so to speak by redefining the absolute zero level of the energy E. This is a consequence of the fact that while in relativistic phyics (rest) energy and mass are the same and define an absolute zero level, in nonrelativistic physics they are two separate concepts.

vanhees71
DaveC426913
Gold Member
A more recent thread has been redirected here, so I'll ask my questions here:

Does this intrinsic-angular-momentum-that-is-not-mechanical-spin give rise to secondary properties, such as poles and gyroscopic precession?

Dale
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2020 Award
Absolutely, yes. That is the whole basis of MRI

DaveC426913
Meir Achuz
Homework Helper
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Classical descriptions of quantum properties lead to confusion.

vanhees71
DaveC426913
Gold Member
Classical descriptions of quantum properties lead to confusion.
Sure but I think the problem is that the term 'angular momentum' is used in both of them.
The question then being: what's the difference between the two flavours of AM?

Nugatory
Mentor
The question then being: what's the difference between the two flavours of AM?
None, if we define angular momentum carefully: it is the conserved quantity associated with rotational symmetry.

jtbell
Mentor
A more recent thread has been redirected here, so I'll ask my questions here:

Does this intrinsic-angular-momentum-that-is-not-mechanical-spin give rise to secondary properties, such as poles and gyroscopic precession?
This may or may not be along the lines of what you're looking for, but...

If you align the spins of electrons in a magnetizable material, then "flip" them, the material acquires a macroscopic angular momentum in order to conserve total angular momentum.

See the Feynman Lectures:

http://www.feynmanlectures.caltech.edu/II_37.html#Ch37-F3

in particular Fig. 37-3 and the paragraph preceding it (you'll probably have to scroll the page up a bit).

This is known as the Einstein-de Haas effect.

vanhees71
Gold Member
None, if we define angular momentum carefully: it is the conserved quantity associated with rotational symmetry.
Yes! That's the key, and this is the total angular momentum, including both orbital angular momentum and spin. As relativistic QFT teaches us, there's no unique split into orbital and spin angular momentum, and what's observable is total angular momentum.