4
$\begingroup$

In my physics classes, we described spin as a quantum property of electrons---that is, their angular momentum. Photons also have angular momentum, and when the angular momentum of photons is aligned, we say that we have polarized light. But I have never seen the angular momentum of photons described as photonic spin or a photon's spin. Why is that? Is it wrong to describe the photon's angular momentum as its spin?

$\endgroup$
3
$\begingroup$

No it is not wrong. The intrinsic angular momentum of a photon is spin. But note that photons can also have orbital angular momentum (and so can electrons).

Orbital angular momentum is defined as $$ \hat{L} = \hat{\bf x} \times \hat{\bf p} $$ where $\hat{\bf x}$ is the position operator and $\hat{\bf p}$ is the linear momentum operator. Just as in classical mechanics, the amount of such angular momentum which any given entity has depends on the choice of the origin of coordinates, as well as on the entity's own motion. It is a quantity useful to physics because it is conserved in isolated systems in the absence of spin-orbit interaction (also true in classical mechanics). Finally, the total angular momentum $$ \hat{\bf J} \equiv \hat{\bf L} + \hat{\bf S} $$ is conserved in isolated systems (and this also holds in classical mechanics). Spin-orbit interaction can move angular momentum between the two forms, without changing $\bf J$.

One way to define spin is to say that it is that part of the total angular momentum which is not orbital angular momentum. Another way to define spin is simply to assert at the outset that one is dealing with a certain kind of quantum field, described by scalar quantities (spin zero) or the simplest spinor quantities (spin half) or vector quantities (spin one) etc. In either case the conservation of total angular momentum, as opposed to just orbital angular momentum, is connected to symmetries under rotation. But that requires a more advanced mathematical level than I think this question and answer is looking for.

P.S. The word 'orbital' in 'orbital angular momentum' does not necessarily mean any particle need be in an orbit about something. A particle moving in a straight line can and does have orbital angular momentum about any point not on that line. I mention this as an antidote to some nonsense statements that I saw in other answers.

P.P.S. It is the photon's orbital angular momentum that ensures conservation of angular momentum in atomic transitions when the atom's angular momentum changes by more than one through the emission of a single photon; such transitions are not dipole transitions (they are quadrupole or octapole etc.)

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ What is the difference between the orbital angular momentum and regular angular momentum? How would a photon have both? $\endgroup$ – vy32 Jan 5 at 20:51
  • $\begingroup$ @vy32: Orbital angular momentum is the angular momentum it has because it's moving through space. Intrinsic angular momentum is the angular momentum that is built into the particle, even when it's not moving through space. The spin 1/2 of an electron is intrinsic angular momentum. $\endgroup$ – user4552 Jan 5 at 21:15
  • $\begingroup$ @BenCrowell, thank you for your comment. But photons are always moving through space at the speed of c. Photons only have 6 variables: x,y,z, azimuth & elevation (ie: direction of motion), frequency f (or wavelength λ), and angular momentum (spin). So what would Orbital angular momentum be? $\endgroup$ – vy32 Jan 6 at 1:54
  • $\begingroup$ @vy32: Not sure what point you're making in your comment. Its orbital momentum is $r\times p$, as with any angular momentum. $\endgroup$ – user4552 Jan 6 at 19:16
  • $\begingroup$ @BenCrowell, what is $r$ ? $\endgroup$ – vy32 Jan 8 at 17:52
2
$\begingroup$

Light can have orbital and intrinsic angular momentum. It is the latter that is related to polarisation and spin. The problem is that there is no gauge invariant expression for it. In the standard, gauge theory of light all angular momentum is described as orbital momentum!

This paradox is at the root of your question. Light has spin but the present theory cannot not explicitly describe it. Either you have gauge invariance and no spin, or spin and no gauge invariance. Present physics takes the first option. I published a paper showing that you can also take the second. See https://arxiv.org/abs/physics/0106078.

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ This is not a helpful response to the OP. The OP is a beginner who simply wants to know about basic terminology. The answer to their question is simply that the photon has an intrinsic spin of 1. $\endgroup$ – user4552 Jan 5 at 16:38
  • $\begingroup$ @BenCrowell I strongly disagree. We should not hide paradoxes because this too may confuse students. It is better and more respectful of their intellect to be open and honest. $\endgroup$ – my2cts Jan 5 at 17:28
  • 1
    $\begingroup$ We should not hide paradoxes because this too may confuse students. It is better and more respectful of their intellect to be open and honest. You haven't established that there is a paradox. Extraordinary claims require extraordinary proof. I haven't seen any evidence that anyone but you believes that this is some earthshaking foundational issue that has been universally ignored. $\endgroup$ – user4552 Jan 5 at 21:18
  • $\begingroup$ Now a short statement of the paradox: Either you say ${\mathbf {J}}=\epsilon _{0}\int {\mathbf {r}}\times \left({\mathbf {E}}\times {\mathbf {B}}\right)d^{{3}}{\mathbf {r}}$ , gauge invariance but no spin or you say ${\displaystyle \mathbf {J} =\epsilon _{0}\int \left(\mathbf {E} \times \mathbf {A} \right)d^{3}\mathbf {r} +\epsilon _{0}\sum _{i=x,y,z}\int \left({E^{i}}\left(\mathbf {r} \times \mathbf {\nabla } \right)A^{i}\right)d^{3}\mathbf {r} =\mathbf {S} +\mathbf {L}}$, spin but no gauge invariance. $\endgroup$ – my2cts Jan 5 at 21:49
  • $\begingroup$ Source of the expressions: en.wikipedia.org/wiki/… $\endgroup$ – my2cts Jan 5 at 21:50
2
$\begingroup$

Spin is the (usually) finite-dimensional and unitary representation of the Lorentz group in which the state transforms (unless you want to study continuous spin) and for massive particles the "little group" approach well expounded be Weinberg leads to irreducible representations labelled by $su(2)$ variables $j$ and $m_{j}$.

For massless particles the analyses is slightly different as the little group is $\mathcal{E}(2)$, the Euclidean group in 2 dimensions. This admits continuous spin representations and demanding that this is avoided leaves only trivial, one-dimensional representations labelled by the helicity, $h$. (In a theory with parity, you get $\pm h$ states connected).

Another way of saying it is that spin and helicity are different for massive particles (helicity being the projection of spin along the 3-direction of movement) whereas for massless particles, travelling at $c$, the spin and helicity coincide.

In my opinion it is perfectly fine to say a photon has spin 1, as we know that for the massless particle spin is helicity. The only danger is that a massive spin 1 triplet has a longitudinal (m=0) state that does not exist for the massless photon. This is the difference between massive and massless. See also the Pauli-Lubanski vector. Massless objects have velocity four vectors that are null and as such travel at the speed of light.

| cite | improve this answer | |
$\endgroup$
  • 4
    $\begingroup$ The level of this answer is wildly wrong for the OP. $\endgroup$ – user4552 Jan 4 at 23:28
  • $\begingroup$ The final paragraph gives a simple and accessible summary putting it in different words. It is also the correct answer. $\endgroup$ – lux Jan 4 at 23:33
  • 2
    $\begingroup$ The level of the answer is not wildly wrong for the OP! But it would be great if @lux would edit the answer and say if it is wrong or acceptable to describe the photon's angular momentum as spin. And since lux brings up the concept of massless particles, it would be great to know if massless particles must necessarily travel at c or if they can travel at other speeds as well. $\endgroup$ – vy32 Jan 4 at 23:53
  • $\begingroup$ Fair enough - thanks for confirming the utility of the answer. I'm editing now $\endgroup$ – lux Jan 5 at 2:01
  • $\begingroup$ I think the only thing that's missing here is that the helicity is effectively the projection of spin (=intrinsic angular momentum) onto momentum, and so the pure $\pm h$ corresponding to (anti-)parallel spins with the momentum. For extensive discussion of the "missing" $S_z = 0$ state, see physics.stackexchange.com/q/46643/50583 and its answers. $\endgroup$ – ACuriousMind Jan 6 at 18:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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