# Quantum spin vs spin

quantum spin vs ....spin

I keep hearing people say, "comparing particle spin to a spinning top will lead you in the wrong direction." I'm curious.. what exactly is the difference between the two?

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Nugatory
Mentor
I keep hearing people say, "comparing particle spin to a spinning top will lead you in the wrong direction." I'm curious.. what exactly is the difference between the two?
That's the wrong question. The right question is "what exactly is the similarity between the two?", and the answer is "not much, except that by historical accident they both have the same name".

well what is spin? haha

HallsofIvy
Homework Helper
A spinning top can be spinning "clockwise" or counter-clock wise". Spin of an electron or other fundamental particle has two eigenvectors so is given the same name. Similarly, there is a property of fundamental particles that has three eigenvectors, just as our eyes can see best in the red, green, and blue wavelengths so that we can think of all colors as made of those three, so we call that property "color". That certainly doesn't mean we are actually talking about the color of the fundamental particle! There are also, then, similar properties called "iso-spin" and "hyper-color".

Thanks! That makes sense. I'm trying to visualize electrons and it's pretty difficult haha.

That's the wrong question. The right question is "what exactly is the similarity between the two?", and the answer is "not much, except that by historical accident they both have the same name".
I would argue. Quantum spin and rotational motion actually have much in common. They obey similar maths, they share the same unit. Spin of compound particles is the sum of spin of its components PLUS their orbital angular momentum. All current theories do not exclude the possibility that spin and rotational motion may be interchanged. I.e. a particle with spin could decay into a system of particles without spin but with angular momentum. In Einstein-Cartan theory, spin is equivalent to rotational motion in the limit of zero rotation radius.

There are of course dissimilarities, but I would not ignore the similarities.

• I keep hearing people say, "comparing particle spin to a spinning top will lead you in the wrong direction." I'm curious.. what exactly is the difference between the two?

To add to your confusion - quantum spin can be separated from fundamental particles(somehow quantum spin can be reduced to quasiparticles).

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vanhees71
Gold Member
2019 Award
I disagree with the statement quantum spin would have nothing to do with classical spin. Of course, as usual, one must not use classical analogies to quantum theory in a careless way. The intuition on the behavior of quanta gained by learning modern quantum theory (the only consistent and very successful theory we have about the behavior of matter today) is necessarily not in a direct sense from extrapolating our everyday experience with macroscopic matter that behaves according to the laws of classical physics.

So to understand "spin" it's indeed pretty misleading to think about, say, an electron as a little classical ball that is rotating. Rather an electron is described by a quantum field.

To get an idea, how "spin" is described in quantum theory we must get an idea from classical physics that is adequate to be generalized to the quantum realm. There is no strictly logical path to this, because quantum theory is more comprehensive than classical mechanics, which is an approximate description of quantum mechanics for macroscopic quantities that can be understood as coarse grained averages over a lot of microscopic degrees of freedom. So there is always some heuristics involved when one likes to "quantize" a classical model.

It has turned out that the key tool to understand this "quantization procedure" are symmetry principles. In physics by a symmetry we mean some transformation of a set of observables that keep the dynamical equations for this observables unchanged. It's like symmetry in geometry: You can rotate a sphere around its center around any axis by any angle (that's the operation) without changing it in any way (that's making the operation to a symmetry).

Very important symmetries are those of the space-time description. Here we stick to the non-relativistic (Galilei-Newton) space-time description. It describes space as a three-dimensional Euclidean manifold, known from school geometry, and time as a directed line. Space is symmetric under translations, i.e., the geometry is everywhere in space the same (no distinguished point) and under rotations around any point (no direction is distinguished in any way). Further, the fundamental physical laws do not change with time, i.e., physics is also invariant under time translations. Finally, we can not determine any dependence of the physical laws on the absolute velocity against any fixed reference frame, i.e., there exists the class of inertial frames, where Newton's 2nd Law holds (a body moving freely will move in a straight line with constant velocity or stay at rest), and one cannot distinguish the motion with constant speed of one such inertial frame against another one. The whole set of symmetry transformations build a mathematical structure called a group, i.e., with two transformations also their composition is again a symmetry transformation and for each symmetry transformation we can find another symmetry transformation bringing us back to the original situation.

In classical mechanics the most adequate description of all these symmetry ideas is to work in phase space, i.e., for a set of $N$ point particles you use their positions and momenta to describe their behavior, which makes together $3N$ position coordinates and $3N$ momentum coordinates. So phase space is a $6N$ dimensional space. The equations of motion can be derived from Hamilton's principle of least action, leading to the socalled Poisson brackets. The change of a quantity in phase space with time is governed by equations like
$$\mathrm{d}{\mathrm{d} t}F(\vec{q},\vec{p})=\{F(\vec{q},\vec{p}),H(\vec{q},\vec{p}) \}=\frac{\partial F}{\partial \vec{q}} \frac{\partial H}{\partial \vec{p}}-\frac{\partial F}{\partial \vec{p}} \frac{\partial H}{\partial \vec{q}},$$
where $H$ is the Hamilton function, given the total energy of the system as function of the phase-space coordinates.

Then a symmetry can be described as an action on the phase-space variables, which leaves the fundamental Poisson brackets
$$\{q_j,q_k \}=\{p_j,p_k \}=0, \quad \{q_j,p_k \}=\delta_{jk}$$
invariant, the socalled canonical transformations.

Applied to the symmetry transformations of spacetime this leads to Noether's theorem: Any continuous symmetry is generated by some phase-space quantity, and that quantity is conserved. In this way one is lead to the definition of the conserved quantities by their role as generators of the corresponding symmetry transformations. For Galilei-Newton spacetime that's

temporal translations -> energy
spatial translations -> momentum
spatial rotations -> angular momentum
Galilei boosts -> center-of mass position

The Galilei boosts are a bit special, but we don't need to discuss them further, although that's a pretty interesting subject in itself.

Here we concentrate on angular momentum and rotations. In quantum theory we have a pretty similar structure although that's not so obvious on first glance. First of all the (pure) states are represented by rays in a Hilbert space, and the observables by self-adjoint operators on Hilbert space. One way to define the observables is to look for realizations of the symmetry group of space-time in terms of such self-adjoint operators as generators for the corresponding unitary transformations that represent the symmetry transformations, building a socalled unitary representation of this group. Now there's a entire mathematical theory behind such group representations by unitary operators on Hilbert space.

Now you can start simply with "quantization" of a classical particle moving freely by introducing operators for position and momentum and demand that instead of Poisson brackets the corresponding relations hold for the commutators (up to a factor $\mathrm{i}$):
$$[\hat{q}_j,\hat{q}_k]=[\hat{p}_j,\hat{p}_k]=0, \quad [\hat{q}_j,\hat{p_k}]=\mathrm{i} \hbar \delta_{jk}.$$
As it turns out this is consistent with the idea that the momentum operators are the infinitesimal generators for translations in space, and one can build the entire quantum theory for a particle on this basis. The angular-momentum operators, defined as generating rotations around the corresponding axes, are simply given as in classical physics by $\vec{L}=\vec{x} \times \vec{p}$ (here as an operator relation).

Now this is, however, not the only possibility! Quantum theory admits somewhat more sophisticated realizations of the Galilei symmetry, in that in addition to the above given orbital angular momentum $\vec{L}$ there can also be an additional spin-angular momentum. To understand its meaning one must refer to the fact that unitary operators of course act on the Hilbert-space vectors and not only as unitary transformations on the operators describing observables. Now consider a particle at rest. The corresponding vector in Hilbert space would be a momentum eigenstate with eigenvalues 0 for all three momentum components. Such a thing doesn't really exist but only as a socalled "generalized eigenstate", which is not normalizable to 1 but only to a Dirac $\delta$ distribution, but this we neglect here just as physicists use to do it.

The point is that there exist unitary representations of the Galilei group where the generalized momentum eigenstates for vanishing momentum change under rotations. The generators for these rotations introduce new observables, unknown for classical point particles, that behave like a contribution to total angular momentum, which now is given as
$$\vec{J}=\vec{L}+\vec{S}.$$
In nonrelativistic quantum theory the spin operator commutes with momentum, and thus one can diagonalize also $\vec{S}^2[itex] and [itex]S_z$ together with momentum. An "elementary" particle is described by definition by an irreducible representation of the Galilei group, and thus one must have a fixed eigenvalue ofr $\vec{S}^2$, which must take the value $s(s+1) \hbar^2$ with $s \in \{0,1/2,1,3/2,\cdots \}$. For $s=0$ the spin components are also always 0, and we are back to a particle described in the naive way by quantization the classical particle. For $s \neq 0$ a particle at rest can have also a determined $s_z$ spin component with $s_z=\sigma_z \hbar$ with $\sigma_z \in \{-s,-s+1,\ldots,s-1,s \}$.

Usually a particle with spin carries a magnetic moment that is described by the operator
$$\vec{\mu}=\frac{q}{2m} g \vec{S}.$$
Here $q$ is the charge of the particle, $m$ its mass, and $g$ the socalled gyromagnetic factor (which is close to 2 for an electron but can take different values for composite particles like the proton).

In this way spin was observed for the first time in the famous Stern-Gerlach experiment, although at this time the above described modern notion of spin was not known, and it has been misinterpreted using old quantum theory (the Bohr-Sommerfeld model for atoms). Funnily enough there are two mistakes in this outdated model compensating each other such that it could be interpreted as proving the "quantization of the direction" predicted by the Bohr-Sommerfeld model of the magnetization of the electron.

• 1 person
Great post, but can you elaborate on this?

To understand its meaning one must refer to the fact that unitary operators of course act on the Hilbert-space vectors and not only as unitary transformations on the operators describing observables. Now consider a particle at rest. The corresponding vector in Hilbert space would be a momentum eigenstate with eigenvalues 0 for all three momentum components. Such a thing doesn't really exist but only as a socalled "generalized eigenstate", which is not normalizable to 1 but only to a Dirac δ distribution, but this we neglect here just as physicists use to do it.

The point is that there exist unitary representations of the Galilei group where the generalized momentum eigenstates for vanishing momentum change under rotations. The generators for these rotations introduce new observables, unknown for classical point particles, that behave like a contribution to total angular momentum

jtbell
Mentor
There is experimental evidence that the intrinsic angular momentum ("spin") of electrons contributes to the total macroscopic angular momentum of a magnetized object. I first read about this in the Feynman Lectures on Physics many years ago:

http://www.feynmanlectures.info/docroot/II_37.html#Ch37-F3

See Fig. 37-3 and the paragraph following it, on the page above. This is known as the Einstein-de Haas effect. When Einstein and de Haas studied this in the mid 1910s, electron spin was unknown (along with the rest of quantum mechanics as we know it!), so they interpreted this phenomenon in terms of the "bound currents" associated with magnetization (polarization of magnetic dipoles) in classical electrodynamics.

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vanhees71
Gold Member
2019 Award
Another funny historical side remark is that de Haas and Einstein got the wrong gyro factor of 1 in their experiment first, because since at the time (around 1915) the idea of spin wasn't known and consequently also the Lande g factor was not considered. This means that they had the old Ampere idea of molecular currents for the explanation of magnetic moments in mind. There the gyro factor is always 1, and due to this theoretical prejudice they also measured this gyro factor. The true value is 2 (which follows from the minimal coupling of the electromagnetic field to the correct current given by the Dirac field of the electron) modulo QED radiative corrections, leading to an "anomalous magnetic moment" of the electron, which is one of the best agreements between theory and experiment ever.

The conclusion from the story of the wrong g-value in the Einstein-de Haas measurement is that one must be very careful with experimental results obtained by theorists or under sufficiently strong influence of theorists ;-)).

vanhees71
Gold Member
2019 Award
Great post, but can you elaborate on this?
Sure, but what do you want to know?

The representation theory of the Galilei group was worked out by Inönü and Wigner. I've once taught this in a QM 2 lecture, but the manuscript is in German. A good treatment can be found in

L. Ballentin, Quantum Mechanics - A modern development.

The same analysis for the Poincare group can be found more often, also going back to Wigner. It's very well described for the most general case of arbitrary spin in

S. Weinberg, The Quantum Theory of Fields, Vol. I.

My own attempt to explain it can be found in my QFT manuscript (Appendix B),

http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

ChrisVer
Gold Member
They have some things in common, but they are not the same at all.
They are not the same, because they have different definitions. For example the angular momentum (that would correspond to a classically spinning body) is the cross product of the distance with the momentum:
$L= r \times p$
This thing does not work for the quantum spin for many reasons- you can't find such a thing.
Also their eigenvalues are different- the angular momenta can take integer eigenvalues (when you treat quantum problems with the classically defined angular momenta that's what you get) while the spin eigenvalues can also be half integers.
Their similarities come from the fact that their groups' algebras are isomorphic. The spin group is SU(2) and the angular momentum is SO(3). That brings the similarities but also keeps the differences (the well known/popular difference between them, is that for spin 1/2 particles you need 4π rotations to coincide to your initial state while for angular momenta you need 2π).

DrChinese
Gold Member
Great post, but can you elaborate on this?
It was a great post, I've seen papers shorter than the post.

So I laughed when you asked him to elaborate... ChrisVer
Gold Member
I do have one question though on this topic, so I guess I should post it here -no need to open a new thread-, concerning the 4 and 2 π rotations.
In the phase space (p-x axis) the angular momentum L is a normal vector that coincides with itself under 2π rotations, so I get a feeling that the phase space is like a sphere.
Now for the spin if I could somehow "define" a new type of momentum and position, let's call that P-X, and I wanted the spin to be given by $S= X \times P$ I would have to get the same "normal vector" under 4π rotations. That would mean that the phase space of those P-X is like a moebius manifold.
Wouldn't that somehow allows us to generalize a momentum and position vectors that would also give the spin in the same way our already known ones give us the angular momentum , just by changing the topology of the phase space? and how could we do that?

WannabeNewton
Great post, but can you elaborate on this?
It's explained in section 7.2 of Ballentine.

WannabeNewton
$L= r \times p$
Classically a planar rigid body has a total angular momentum that can be decomposed into the orbital angular momentum about the origin of a coordinate system, which is ##\vec{L} = \vec{r}\times \vec{p}##, and the spin angular momentum about the center of mass, which is ##\vec{S} = I_{CM} \vec{\omega}## where ##I_{CM}## is the moment of inertia about the center of mass. What you wrote down is the orbital angular momentum which can easily be quantized as ##\hat{L} = \hat{Q}\times \hat{P}## which in the coordinate representation is just ##-i\hbar (r \times \nabla)## (see reference above). It's the spin angular momentum which can't straightforwardly be quantized in such a trivial manner but, as vanhees noted, one can still decompose the angular momentum operator as ##\hat{J} = \hat{L} + \hat{S}## analogous to the classical decomposition of the angular momentum of a planar rigid body and in the same light the spin operator ##\hat{S}## is an intrinsic contribution to the angular momentum (as opposed to ##\hat{L}## which is a contribution to angular momentum about the origin of some coordinate system as ##-i\hbar (r \times \nabla)## makes clear).

dextercioby
Homework Helper
[...] The true value is 2 (which follows from the minimal coupling of the electromagnetic field to the correct current given by the Dirac field of the electron) [...]
Small addendum: you don't need to use Dirac's equation to show that g_el = 2. There's a common misconception that special relativity is needed to show that g_el = 2.

ChrisVer
Gold Member
Small addendum: you don't need to use Dirac's equation to show that g_el = 2. There's a common misconception that special relativity is needed to show that g_el = 2.
how?
I mean Dirac's Equation is relativistic itself XD
but how does special relativity do that itself?

dextercioby
Homework Helper
how?
I mean Dirac's Equation is relativistic itself XD
but how does special relativity do that itself?
You didn't understand my words. Reread them. And find the connection between the 1st sentence and the 2nd.

vanhees71
Gold Member
2019 Award
The gyro factor 2 can be derived also from "gauging" the Schrödinger equation for a spin-1/2 particle, but it's an ad-hoc procedure. You just must do the "gauging" in the right way to get this result. Gauging the free Dirac equation, however, gives uniquely g=2.

The "gauging" for the Schrödinger equation works as follows. You start from the Hamiltonian for a free particle,
$$\hat{H}=\frac{\hat{\vec{p}}^2}{2m}.$$
The trick now is to rewrite this as
$$\hat{H}=\frac{[(\vec{\sigma} \cdot \hat{\vec{p}}]^2}{2m},$$
where $\vec{\sigma}$ are the three Pauli matrices.

Now from classical mechanics you know that at presence of a magnetic field the canonical momenta are given by $\vec{p}=\vec{p}_{\text{mech}}+q \vec{A}$. In QT the momenta are the canonical momenta. So we have to write
$$\hat{H}_{\text{em,mag}}=\frac{(\vec{\sigma} \cdot (\hat{\vec{p}}-q \hat{\vec{A}})^2}{2m}.$$
For the electric field you add the corresponding potential,
$$\hat{H}_{\text{em}}=\hat{H}_{\text{em,mag}} + q \hat{\Phi}.$$
Multiplying out the square in $\hat{H}_{\text{em,mag}}$ indeed gives the correct gyro factor.

Written in position representation you get the Pauli equation:
$$\hat{H}_{\text{em}} \psi(t,\vec{x})=\left [-\frac{\hbar^2 (\vec{\nabla}-q \vec{A})^2}{2m} -\frac{q}{2m} g_S \frac{\hat{\vec{s}} \cdot \vec{B}}{2m}+q \Phi \right ]\psi(t,\vec{x}) \quad \text{with} \quad g_S=2.$$
Of course, had we not used the trick to multiply $\hat{\vec{p}}$ first with the Pauli matrices and then made the minimal substitution for the vector potential, we'd had missed the $\vec{s} \cdot \vec{B}$ term. So in some way this procedure is somewhat ad hoc.

For the Dirac equation, starting from the covariant free-particle Lagrangian and gauging it, leads uniquely to the correct gyro factor, and you can derive the Pauli equation as the non-relativistic approximation of the Dirac equation with the same result given above. So the relativistic way to derive the magnetic moment for a non-relativistic spin-1/2 particle is somewhat more convincing than the non-relativistic one given above.

dextercioby
Homework Helper
Hendrik, I didn't mean that 'derivation', but the true derivation of Levy-Leblond described neatly in the 13th chapter of Greiner's 'Introduction to QM' from his theoretical physics series.

@vanshees71
Sure, but what do you want to know?
You said:
The point is that there exist unitary representations of the Galilei group where the generalized momentum eigenstates for vanishing momentum change under rotations.

Could you please provide an explicit construction of a field with zero momentum eigenstates, yet changing under rotations? Or provide some other explanation. Or at least point me to the place where I can find it.

vanhees71
Take a non-relativistic spin-1/2 particle. Then you have the common generalized eigenstates $|\vec{p},\sigma_z \rangle$ (since in non-relativistic physics the spin commutes with the momentum operator, you can diagonalize spin and momentum simulaneously). Choosing the basis properly, under rotations you have
$$\hat{U}(\vec{\varphi}) |\vec{p},\sigma_z \rangle=\sum_{\sigma_z' \in \{\pm 1/2 \}} |\hat{R}(\vec{\varphi}),\sigma_z' \rangle \exp(-\mathrm{i} \hat{\vec{\sigma}} \cdot \vec{\varphi}/2)_{\sigma_z',\sigma_z}.$$
Here, $\hat{R}(\vec{\varphi})$ is the usual SO(3) rotation matrix and $\hat{\vec{\sigma}}$ the three Pauli matrices. As you see, also for $\vec{p}=0$ the basis vectors transform in a non-trivial way under rotations.