12
$\begingroup$

It is often repeated that "the spin observable is purely quantum and has no classical counterpart". What is actually meant by that? I see no principle difference between the set of spin observables and the set of position-momentum observables in that respect (of being purely quantum): both sets have non-commuting observables. The only interesting difference I see is that spin, having a discrete spectrum, can nonetheless transform continously in time between two orthogonally spin states thanks to the principle of superposition, whereas a classical coordinate (of phase space) must have a continuum of possible values for the value to evolve continously in time (So in this respect spin cannot be classical but position-momentum could ; but I don't think this is the answer since also energy is discrete for bound systems)

$\endgroup$

5 Answers 5

10
$\begingroup$

Spin is not less classical than position. It shows up in any relativistic field theory, whether it's quantized or not.


The accepted answer by Adam says

because [spin] can't be larger than a given number of the order of $s\hbar$, where $s$ is typically smaller than two, there is no way you can construct a classical limit out of the spin: it is a purely quantum quantity, always of the order of $\hbar$.

That makes no more sense than saying that energy is a purely quantum property of waves because $E=\hbar ω$, so $E\to 0$ in the $\hbar\to 0$ limit.

The energy of an electromagnetic wave is $n\hbar ω$, not $\hbar ω$. There is no upper limit on the energy; only the spacing of adjacent energy levels is constrained. That constraint disappears classically, so no relation like $E=n\hbar ω$ is needed, but energy still exists.

Likewise, the quantum restriction $|S|\le n\hbar s$ (where $|S|$ is the magnitude of a component of the spin) doesn't impose any maximum on the spin angular momentum of the field, and spin angular momentum survives in the classical limit, just without the quantization.

Spin angular momentum is not completely unconstrained in the classical case, because by substituting $n\hbar=E/ω$ into $|S|\le n\hbar s$ you get $|S|/E \le s/ω$, a constraint that is independent of the particle count and $\hbar$. And that constraint is obeyed by Maxwell's equations. This answer calculates the angular momentum of a plane wave and finds it to be $$S=\frac{1}{\omega}\int \left(|\mathbf{F}_+|^2-|\mathbf{F}_-|^2\right)d^{3}\mathbf{r}$$ where $\mathbf{F}_\pm$ are the right/left chiral parts of the field. The energy is $$E=\int \left(|\mathbf{F}_+|^2+|\mathbf{F}_-|^2\right)d^{3}\mathbf{r}$$ from which you can see that $|S|/E \le 1/ω$, with the extreme being attained when $\mathbf{F}_+$ or $\mathbf{F}_-$ is zero, i.e., when the wave is circularly polarized.

A similar calculation in linearized general relativity will give you $|S|/E \le 2/ω$. GR is a spin-2 classical field theory.

Dirac's equation is a spin-½ classical field equation. It has nothing that would justify calling it quantum: no quantization of particle number, no fermionic statistics, no violation of Bell's inequality. Dirac intended it to replace Schrödinger's equation, but that didn't happen. Instead, the modern home of the Dirac field is in the QFT Lagrangian, alongside the Maxwell field, and quantization is introduced separately through the path integral.

Dirac put a factor of $\hbar$ in the equation to make it look more like Schrödinger's equation, but it has no physical significance. You can eliminate it along with $m$ by substituting $mc^2/\hbar=ω$. You can also eliminate the factor of $i$ (which he probably also added to mimic Schrödinger's equation) by absorbing it into the $γ$ matrices, but there's really no need to do that, since there's nothing wrong with using complex numbers in classical wave mechanics.

There are no fields in nature for which the Dirac field is a useful classical limit, because there is no classical limit of fermionic statistics (as far as I know), but it is a classical field that has the mathematical property of being spin-½.

The Klein-Gordon equation is a spin-0 classical field equation. You can eliminate the factor of $\hbar^2$ by the same substitution. It may have practical uses in condensed-matter physics (I don't know).


People sometimes point to the discrete-valuedness of spin (e.g. the Stern–Gerlach experiment) as evidence that spin is nonclassical, but that argument is nonsensical since orbital angular momentum is also quantized in that way, and is obviously not purely quantum.

Spin is not discrete even in quantum mechanics. An electron's spin can lie anywhere on the Bloch sphere, and all points on the Bloch sphere are equivalent under $SO(3)$ spatial symmetry: there are no distinguished polar points of which other directions are mere superpositions.

What is discrete is quantum measurement. You can't just measure a system, you have to measure with respect to some orthogonal Hilbert-space basis, and if the system has finitely many degrees of freedom then you will get one of the finitely many basis directions as a result.

So the only relevant discreteness of spin is that there are only finitely many spin degrees of freedom (at a point). There's nothing quantum about that. The classical electromagnetic and gravitational fields have finitely many degrees of freedom at a point, but they are in all respects completely continuous fields. The degrees of freedom are just the components in some basis of a continuously varying geometrical quantity.

$\endgroup$
4
$\begingroup$

It is a problem of definition: what does one mean when one says that something is classical. One could say: is classical something that is not an operator, and therefore an Ising spin is classical, thus spins are as classical as momentum. This is a fine thing in statistical physics, but that is not what people were thinking as classical in the early days of quantum mechanics.

What people usually mean by classical, is that there exists a well defined classical limit. For example, states such that position and momentum are well defined (i.e. $\langle\hat X\hat P\rangle\simeq\langle\hat P\hat X\rangle$). For orbital angular momentum this would correspond to $\langle \hat L\rangle\gg \hbar$, or at least that the precision you have on the measurement of $\langle \hat L\rangle$ cannot differentiate between $0$ and $\hbar$.

And now you see that you have a problem with the spin: because it can't be larger than a given number of the order of $s\hbar$, where $s$ is typically smaller than two, there is no way you can construct a classical limit out of the spin: it is a purely quantum quantity, always of the order of $\hbar$.

$\endgroup$
4
  • $\begingroup$ OK, that might be it, although I always thought that what they mean with "no classical counterpart" has something to do with the spinor nature of spin, and not with a classical limit. $\endgroup$
    – Lior
    Commented May 7, 2014 at 18:04
  • $\begingroup$ @Lior: I could agree with that: it is important that we think of the classical limit as that of a corpuscul (that is, a classical point like object, which is different from a "quantum particle") and not a wave (as for photon, where the two spin orientation are still classical). This has to do with the peculiarities of the classical limit of fermions (that become corpuscul), where spin does not have a classical counterpart. $\endgroup$
    – Adam
    Commented May 7, 2014 at 18:13
  • $\begingroup$ so if s were large enough for a certain particle, then it would have a classical counterpart? what would it be and why is the large s limit needed? $\endgroup$
    – Lior
    Commented May 7, 2014 at 18:17
  • $\begingroup$ @Lior: Even for composite particles (large nucleus for example), $s$ is still quite small (people talk about large spins for $s=9/2$, which is still a very small angular momentum), so I don't think that this is a limit worth considering. $\endgroup$
    – Adam
    Commented May 7, 2014 at 19:44
1
$\begingroup$

Spin and angular momentum

The spin observable is purely quantum and has no classical counterpart

What would be a classical counterpart of the spin ? We know that the spin is an angular momentum, a quantity associated to a system that rotates :

$$ \mathbf{\mathcal L} = \mathbf r \times m\mathbf v $$ where $\mathbf r$ denotes the position of the particle, $\mathbf v$ its velocity and $m$ its mass.

An attempt to find a classical counterpart

I propose this classical definition of spin :

An electron is a small sphere that carries mass and rotates around itself at a certain velocity. This rotation thus creates an angular momentum called spin.

Let's consider one electron of radius $R$ and mass $m_e$. Classical physics and measurements gives us an order of magnitude for the mass of the electron, which is $m_e \approx 10^{-31}\mathrm{kg}$. The order of magnitude for the spin is the Planck's constant $\hbar \approx 10^{-34} \mathrm{J.s}$.

I suppose there is $\alpha$ such as $R \approx 10^{-\alpha} \mathrm{m}$, I don't know $\alpha$ yet, we are going to find it's value. An electron is necessarily smaller than any atom, so we know that $\alpha >10$.

With the definition of angular momentum we can link the order of magnitude of the velocity (at which the electron rotates) with $\alpha$ :

$$ v \approx \frac{\hbar}{R m_e} \approx 10^{\alpha-3} $$ with $\alpha >10$ we have then $v > 10^7 \mathrm{m/s}$.

Of course $R= 10^{-10} \mathrm m$ is a crazy value for the radius of an electron and any smaller value for $R$ would lead to the associated rotation velocity to be greater than the speed of light $c = 3.10^8 \mathrm{m/s}$. Thus the definition I earlier gave for the spin of an electron is utterly false (you have to accept that the velocity of light is the greatest one, but that's another story).

Conclusion

This is a way to understand that spin doesn't have a classical counterpart. It is impossible to think of spin as the angular momentum of something rotating at a given velocity, since it leads to velocities greater than $c$. Whereas position for example has obvious classical counterparts.

One way to think properly of spin is to see it as an internal degree of freedom of the particle, it is a new quantum number and it is needed to completely define the state of the particle.

Many other things make spin purely quantum. The fact there is only two values for the spin projection is hard to picture in classical physics even if it can transform continuously thanks to the superposition principle. I also think of spin and angular momenta additions. Quantum angular momenta don't add like classical ones.

$\endgroup$
10
  • 1
    $\begingroup$ I can also say that classically spin is an internal degree of freedom, which means that the argument above is irrelevant. $\endgroup$
    – Lior
    Commented May 7, 2014 at 15:27
  • 1
    $\begingroup$ I don't see why I can't add a spin degree of freedom and remain classical. instead of using 6 numbers for the state, use 9 (where I added 3 spins). I mean, this will be the suitable classical counterpart of spin. It might be needless and unhelpful (although in Ising models it is used) , but it is not less classical than position, I think. $\endgroup$
    – Lior
    Commented May 7, 2014 at 15:46
  • 1
    $\begingroup$ @Lior I agree with rob. You can decide to add any internal degree of freedom you want but if it can be expressed with the others DOF then it is not a degree of freedom. Adding it doesn't give additional information about the system and it is thus irrelevant. $\endgroup$
    – ChocoPouce
    Commented May 7, 2014 at 17:38
  • 1
    $\begingroup$ I define a system (a particle) with 9 independent degrees of freedom, three of which are spin, and I define a Hamiltonian (or Lagrangian or whatever) to depend on these extra three degrees of freedom. Thus I get a classical particle with spin as an internal degree of freedom. $\endgroup$
    – Lior
    Commented May 7, 2014 at 17:46
  • 1
    $\begingroup$ a classical electron, maybe? its spin would dictate the trajectory in the elctromagnetic field, for example, which is a better theory than a spinless electron in an electromagnetic field. It's not a perfect theory, of course, as it is classical and not quantum, but that can be said also regarding position and momentum. $\endgroup$
    – Lior
    Commented May 7, 2014 at 18:09
0
$\begingroup$

There is a description of spin in classical mechanics, see the answer here. However as explained in that answer this description cannot arise from motion of particles in real space, constrained or not. If you believe that classical mechanics should deal only with positions and momenta of particles, this classical spin is excluded. However in the quantum theory, even if you believe that the wavefunction in space is a complete description, the linear structure of quantum mechanics and Wigner's theorem mean that you are not allowed to exclude spin. Of course one could imagine a universe where all particles are spin 0. I don't know of a theoretical reason to exclude this, but it's certainly not the universe we live in.

$\endgroup$
2
  • $\begingroup$ could you please give a reference to the statement "the linear structure of quantum mechanics [...] mean that you are not allowed to exclude spin"? $\endgroup$
    – Lior
    Commented May 7, 2014 at 17:51
  • $\begingroup$ Wigner's theorem says that Lorentz invariance (special relativity) has to be implemented by a linear operator on states. You can enumerate all such implementations (representations of the Lorentz group). Most of them have non-zero spin, and there's nothing that excludes them. $\endgroup$ Commented May 13, 2014 at 1:40
-2
$\begingroup$

I think the most important difference is that the components of spin do not commute with each other. This is the key feature that makes spin weirder than position. The components of the position operator, $X$,$Y$,$Z$ commute with each other.

$$\left[X_i,X_j\right]=0$$

This means that the position of a particle can be localized to an arbitrarily small volume. A huge momentum uncertainty comes along with doing so, but it is at least possible in principle.

Different components of spin have a nonzero commutator.

$$\left[S_i,S_j\right]=i\hbar\epsilon_{ijk}S_k$$

This means that it is not possible to specify multiple components of spin at once. This leads to non-classical effects. The Stern-Gerlach experiment is a good demonstration of what makes spin inherently quantum.

$\endgroup$
1
  • 1
    $\begingroup$ But orbital angular momentum does have a classical analogue, and also has non-commuting components. This was addressed in the question. $\endgroup$
    – rob
    Commented May 7, 2014 at 15:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.