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Ok, this is my reasoning. I am probably making some wrong assumptions here, pls tell me where I am going wrong.

Spin as a quantum phenomenon: Quantum phenomena disappear as the Planck constant goes to zero and since the intrinsic spins are proportional to the Planck constant, it is a quantum phenomenon.

Spin as a relativistic phenomenon: On other the hand, we know that the elementary particles are the irreducible representations(irreps) of the Poincaré group which is a purely relativistic symmetry group of the spacetime and since the spins show up in these irreducible representations. Therefore, the spins are a relativistic phenomenon. However, such phenomenon must disappear as the speed of light $c$ goes to infinity, but there is no $c$ in the spin value of the elementary particles,

so why do the spins appear in these irreps if they are purely quantum mechanical? can it be that they are both quantum and relativistic phenomena? if so, where is $c$?

After some thought, I've made some new observations. Wigner's classification of irreps of the Poincaré group doesn't specify the exact value of spin. It only tells us units (quanta) of spins (0,1/2,1,...). The exact values are determined by quantum mechanics. The Poincaré symmetry group(the symmetry of our universe spacetime) provides the boundary and the geometry. For a moment, let's consider the particle in a box, Schrödinger equation (quantum mechanics) gives us the exact values of the energy, but what determines the quanta is the shape of the box and the boundary (the symmetry). I am not sure if this reasoning is correct at all and I still don't get why wouldn't $c$ appear in the spin values.

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    $\begingroup$ Just because spin is a phenomenon that shows up in relativistic quantum mechanics does not necessarily mean that is must vanish in the classical limits. But our ability to measure its more tiny effects on a macroscopic scale might be too limited. Still, the Pauli exclusion principle is so fundamental to our description of matter that we essentially cannot get away with arguing spin “vanishes” in any limit if we wish to know how matter is put together. $\endgroup$ Commented Jun 17 at 13:17
  • $\begingroup$ It shows up in irreps of the Poincare group. I fail to see which part of this group is quantum mechanical. $\endgroup$
    – Saeed
    Commented Jun 17 at 13:21
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    $\begingroup$ I am certainly no expert so hopefully somebody corrects me if I’m wrong: if I’m not mistaken, half integer spin is projective reps of the Poincaré group, and we can only use projective reps in quantum theory. Meanwhile, integer spin, like that of the photon, is sort of a “classical” phenomenon (ie polarization). It’s of course not fully classical, but perhaps behaves a bit more (eg when there are many photons) than half integer spin. $\endgroup$
    – Joe
    Commented Jun 17 at 13:40
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    $\begingroup$ Intrinsic spin is a quantum and relativistic phenomenon: it naturally appears in the context of Galilean relativity by studying the projective representations of the central extension of the Galilei group in a projective Hilbert space. More details here: physics.stackexchange.com/questions/348279/… $\endgroup$
    – DanielC
    Commented Jun 27 at 21:12

5 Answers 5

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SPIN ORIGIN

Spin is a purely relativistic property. It comes in fact from the representation theory of the Lorentz group (the relativistic symmetries group).

In classical mechanics, you have representations which are of "integer" spin, so you have spin-0 fields, spin-1 fields, spin-2 fields and so on. The EM field has intrinsic spin 1, for example.

What is a purely quantum effect is the fact that fields with half-integer spins are allowed. The motivation for that is the fact that in QM phase is not observable, and this gives rise to half-integer spins (I can expand if needed). The complete story is that in quantum mechanics you have a different type of representation theory, so the symmetry group you are interested in is not the Lorentz group (which has integer spin) but its universal cover, the Spin group, which has ALSO half-integer spins. The Spin group is the quantum version of the relativistic symmetry.

So to recap: spin is a property which is purely due to the relativistic symmetry of the universe. If you keep this symmetry classical it gives you integer intrinsic spin, while if you go quantum this symmetry allows half-integer intrinsic spins.

THE LIGHT SPEED

Where does $c$ and $\hbar$ come into play?

$c$ enters the game by defining the symmetry group. In $D=4$ the fact that $c$ is finite gives you symmetry groups $SO(1,3)$ or $SO(3,1)$, which are the Lorentz "classical" groups (this is a bit incorrect but it's just to simplify). If you allow $c\rightarrow \infty$ then the symmetry group changes to $SO(1,3)\rightarrow Gal(3)$, which is just the Galileian group in 3 spatial dimensions. Here is where spin disappears.

As already stated, the $\hbar$ enters the game when introducing half integers, so that the group becomes $SO(1,3) \rightarrow Spin(1,3)$ and the Spin group is the "quantum version" of the Lorentz group (it takes into account the phase mentioned earlier).

In principle, you can have a quantum version of the Galileian group, where something different should appear, but I'm not very prepared on that matter, so I can't tell you more.

To recap: $c$ defines the symmetry group, which allows or not spin to exist, while $\hbar$ extends the group at the quantum level and allows half integers to exist (not sure for the Galileian version though).

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  • $\begingroup$ I suspect this is what Joe (in the comments) was talking about when he mentioned the projective reps. So based on this reasoning, integer spins are purely relativistic phenomenon, but half-integer spins are both quantum-mechanical and relativistic phenomenon? $\endgroup$
    – Saeed
    Commented Jun 17 at 23:54
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    $\begingroup$ Yep. You cannot have half integer spin without relativity AND quantum mechanics. Both are needed for half spins $\endgroup$
    – LolloBoldo
    Commented Jun 17 at 23:55
  • $\begingroup$ Following this argument a ferromagnet is a purely relativistic phenomenon. I have my doubts, @Saeed. $\endgroup$
    – my2cts
    Commented Jun 18 at 8:36
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    $\begingroup$ Yes this is what I was alluding to in my comment :) One question for @LolloBoldo . Even without relativity, we have SO(3) symmetry, which itself has integer spin representations. Why do we need relativity for spin, then? It seems like integer spin can come from either classical non-relativistic or relativistic. And half integer can also come from either, as long as you add in quantum. I guess non-relativistic vs relativistic wouldn’t be the same type of spin, though, and fundamentally of course we have spin coming from relativity (Poincaré group). $\endgroup$
    – Joe
    Commented Jun 18 at 14:38
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    $\begingroup$ @LolloBoldo Perhaps Joe was meaning just the $SO(3)$ subgroup of rotations when we have three spatial dimensions; I feel like I'm personally fine with spin as being related to the double-cover of $SO(3)$ rather than the double-cover of the Poincare group. However, I think the latter is important for understanding the spin-statistics theorem and the classification of particles in Lorentz-covariant field theory. $\endgroup$
    – user196574
    Commented Jun 19 at 1:38
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There is nothing quantum about spin.

When the great mathematician Élie Cartan first introduced the concept of spinor in 1913, it's a purely classical concept of geometry. Quantum mechanics has not been fully developed yet back in 1913. The classical spinors have NOTHING to do with quantum mechanics. The rotation property of a spinor associated with the Poincare group is a purely classical spinor behavior, unlike the other quantum properties of a spinor when the spinor field is second-quantized in QED.

The spinor field $\Psi$ endowed with half integer intrinsic spin is NOT a quantum wave function and does NOT belong to the Hilbert space $\cal H$. To quote Weinberg in his Lectures on QM: "there is one topic I was not sorry to skip: the relativistic wave eq. of Dirac. It seems to me that the way this is usually presented in books on QM is profoundly mistaken".

Let's take a look at the specific definition of spin: $$ \vec{S} = \frac{\hbar}{2} \vec{\sigma} $$ where $\sigma_i$ are Pauli matrices. Folks see the Planck constant $\hbar$ in the above definition and automatically associate spin with quantum mechanics.

But that is a false impression. Let's look at the spin rotation $$ U(\vec{\theta}) = e^{i\vec{\theta} \cdot \vec{S}/\hbar}= e^{\frac{i}{2}\vec{\theta} \cdot \vec{\sigma}} $$ As you can see, the $\hbar$ in the definition of spin is canceled out by the denominator $\hbar$ in spin rotation, $\hbar$ has just simply disappeared!

The Planck constant $\hbar$ was manually put into the definition of spin due to historical reasons. It is redundant and unnecessary, since spin per se (without QFT second quantization) is purely a classical concept.

For more details, see a related answer here.


Added note:

The other answers/comments erroneously stated that:

  • "Spin is a purely relativistic property"
  • "Dirac equation, which is both relativistic and quantum mechanical"
  • "Spinors do not exist outside of quantum mechanics since they belong to projective vector spaces"

All the above are FALSE statements, given that:

  • Spin is NOT a purely relativistic property. One can surely define the Spin group $Spin(3) \sim SU(2)$ in 3-D non-relativistic Euclidean space.
  • Dirac equation is relativistic but NOT quantum mechanical. If you are wondering, the spinor $\Psi$ per se is NOT a quantum wave function. There is an addition step to "second quantize" $\Psi$ to make it quantum in the context of QFT, where $\Psi$ is turn into an amalgam of quantum creation and annihilation operators. However, before taking this additional step of "second quantization", there is nothing quantum about $\Psi$.
  • Pure quantum states are described by the elements of the projective space of a complex Hilbert space $\cal H$. However, as we stated in the previous bullet point, the spinor $\Psi$ is NOT a quantum wave function, hence $\Psi$ (including its spin characteristics) does NOT belong to the complex Hilbert space $\cal H$! Therefore, any attempt using "the projective space of a complex Hilbert space $\cal H$" to justify the quantumness of spin/$\Psi$ is misguided.

More Added note:

This is to answer @tparker's question: "So you’re saying that spin is dimensionless in a classical context"?

Yes, spin is dimensionless in a classical context. Let's take a look at the Pauli equation: $$ -q\frac{\hbar\vec{\sigma} \cdot \vec{B}}{2m}\Psi=i \hbar\frac{\partial}{\partial t}\Psi $$ where for simplicity sake we assumed no space-dependence for the spinor field $\Psi$.

Pauli equation in the above format gives you the FALSE impression that the spin is of dimension $\hbar$. But do you notice that $\hbar$ appears on BOTH sides of the Pauli equation? We can simmplly divide both sides by $\hbar$, and arrive at $$ -q\frac{\vec{\sigma} \cdot \vec{B}}{2m}\Psi=i \frac{\partial}{\partial t}\Psi $$ Now we can clearly see that the above time evolution of the spinor field $\Psi$ is actually NOT dependent on the Planck constant $\hbar$.

For historical reasons, folks erroneously thought that the spinor field $\Psi$ is a quantum wave function and manually multiply both sides by $\hbar$ to make the Pauli equation resemble the Schrödinger equation. But in reality spinor field $\Psi$ is NOT a quantum wave function and the multiplication by $\hbar$ is unnecessary and misleading.

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    $\begingroup$ Spinors do not exist outside of quantum mechanics since they belong to projective vector spaces, and classical mechanics is not based on projective vector spaces, so "rotation properties of spinors in classical mechanics" is not even a thing, its just not defined $\endgroup$
    – LolloBoldo
    Commented Jun 18 at 14:07
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    $\begingroup$ Élie Cartan would tell you that the spinors he use to describe spin geometry will never come up as observable quantities in classical physics, since the bundle in which spinors live (the bundles associated to spin bundles) are complex vector spaces which does not exist in classical mechanics. The Dirac spinors you are referring to exist in geometry but are not observable in our world unless you consider as associated bundles vector spaces with the projective property, namely quantum vector spaces. Good luck with building a spin-$\frac{1}{2}$ observable in classical mechanics lol $\endgroup$
    – LolloBoldo
    Commented Jun 18 at 17:30
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    $\begingroup$ Majorana spinors are not real, they are CONSTRAINED Dirac spinors. Majorana spinors leave in the same spinor bundle as the Dirac ones, but they are constrained objects, they are not real. To see why this is case, consider the fact that the reality condition is not a Lorentz-invariant condition, so you cannot consider the Majorana spinors as a "real Dirac spinor", you need to impose extra conditions on them to have the components satisfy some properties, all of this has nothing to due with spin geometry. If you think that half spins are not quantum, give a non-quantum observable with half spin. $\endgroup$
    – LolloBoldo
    Commented Jun 18 at 19:16
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    $\begingroup$ @printf Can't I have an object with no quantum mechanical amplitude significance that nevertheless transforms under a projective representation? $\endgroup$
    – user196574
    Commented Jun 19 at 3:56
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    $\begingroup$ So you’re saying that spin is dimensionless in a classical context, and only gains an association with angular momentum in a quantum context? $\endgroup$
    – tparker
    Commented Jun 26 at 14:13
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The classical electromagnetic field has spin $S=1$. In my paper A theory of electromagnetism with uniquely defined potential and covariant conserved spin I proved this and defined the electromagnetic spin operators. This proves that spin is not an exclusive quantum phenomenon but can also be macroscopically manifest.

Non-integer spin is macroscopically manifest in ferromagnets and well described by the macroscopic Maxwell equations, whereas non-relativistic quantum mechanics is adequate to describe ferromagnetism microscopically.

"Quantum phenomena disappear as the Planck constant goes to zero and since the intrinsic spins are proportional to the Planck constant, it is a quantum phenomenon." Many molecular, atomic or subatomic phenomena are 'quantum' in this sense. I would include $e$ in this context. Yet they lead to non-quantum phenomena when large quantities of particles are involved.

In conclusion, spin is neither an exclusively relativistic nor quantum mechanical phenomenon.

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I do not know how to discuss this for general spin, but I will try to give a concrete answer for spin 1/2.

In non-relativistic quantum mechanics spin is usually discussed as some independent observable $S$. So spin can be described as a purely non-relativistic quantum phenomenon. However, its introduction may seem ad-hoc and attached, as the contribution to the wave function turns out to be a multiplication of some spin part to the "normal" spatial wave function.

If we consider the example of spin 1/2, there is the Dirac equation, which is both relativistic and quantum mechanical. The Dirac equation works with Spinors $\Psi$ which have the desired relativistic transformation properties.

Now it is possible to take the non-relativistic limit of the Dirac equation, see e.g. https://physics.stackexchange.com/a/516986/406978 for a derivation. In that derivation we can see, how the Spinor $\Psi$ is decomposed into $u$ and $v$. And in the end of the derivation we get the Pauli equation, which is the non-relativistic Schrödinger equation for a spin 1/2 particle in a magnetic field. The $\frac{\bar h}{2} \vec{\sigma} $ can be identified with the non-relativistic spin observable and the $u$ can be identified with the non-relativistic wave function, which will turn out to be the spatial wave function times a non-relativistic spin-part.

So when going from relativistic to non-relativistic the spin does not "disappear" for large $c$, but the Spinor is somewhat decomposed.

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  • $\begingroup$ "Dirac equation, which is both relativistic and quantum mechanical": not true. Dirac equation is relativistic but NOT quantum mechanical. If you are wondering, the spinor $\Psi$ per se is NOT a quantum wave function. There is an addition step to "second quantize" $\Psi$ to make it quantum in the context of QFT, where $\Psi$ is turn into an amalgam of quantum OPERATORS. However, before taking this additional step of "second quantization", there is nothing quantum about $\Psi$. $\endgroup$
    – MadMax
    Commented Jun 17 at 17:41
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    $\begingroup$ Based on this reasoning, spin is purely a quantum mechanical phenomenon, but it still doesn't address the issue of why it appears in the irreps of Poincare group which is a purely relativistic symmetry. $\endgroup$
    – Saeed
    Commented Jun 17 at 18:12
  • $\begingroup$ @MadMax: I agree that $\Psi$ is not a quantized field and the Dirac equation is not "quantum mechanical" in the sense of second quantization. What I mean is that you, can get e.g. the Pauli equation, or the fine structure corrections from it. If you have a suggestion to put a better phrase than "quantum mechanical" to describe this, I am glad to edit my answer. What do you think of "quantum mechanical in the sense of first quantization"? $\endgroup$
    – Maik H.
    Commented Jun 17 at 19:31
  • $\begingroup$ @Saeed I think of it more like this: if you introduce spin directly to non-relativistic quantum mechanics it is done in an ad-hoc way. A way to show the connection between the way spin appears in non-relativistic quantum mechanics and the Spinor is given in the mentioned derivation. $\endgroup$
    – Maik H.
    Commented Jun 17 at 20:32
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There is nothing that distinguishes the spin from the magnetic dipole

Spin was introduced into physics when it was realized that, in addition to its interaction with electric fields, the electron also has the property of interacting with a magnetic field. While it is attracted or repelled by an electric field, it is aligned in a magnetic field or deflected while moving in the magnetic field. It was given the replacement name “spin” for the intrinsic magnetic dipole because the analogy to the gyroscopic effect seemed to exist. Later, people rowed back when they realized that a rotation of the electron made no sense.

The second case, when the concept of spin is used, is the calculation of the total angular momentum of the atom. Again, it is the magnetic dipoles that determine the mutual orientations of the subatomic particles and thus the structure of the atom. Replace the spin with magnetic dipoles and the internal cohesion of atoms and molecules will be easier to visualize and calculate.

If your actual question relates to where spin is to be taken into account, the answer is in any theory that makes use of it. And that includes both the relativistic and the QM theory, as you yourself have stated.

If you ask where the spin was introduced, the story was as follows:

The spectral lines of atoms are split into subtler lines under the influence of a magnetic field. UHLENBECK and GOUDSMITH in 1925 proposed to explain this by a self-rotation of the electron.

Uhlenbeck and Goudsmit treated spin as a consequence of classical rotation, while PAULI shortly afterwards emphasised that spin is a non-classical and intrinsic property.

(The title was deliberately chosen because I liked MadMax's answer so much.)

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