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Electrons have spin 1/2, and as they are charged, they also have an associated magnetic moment, which can be measured by an electron beam splitting up in an inhomogeneous magnetic field or through the interaction of the electrons's magnetic moment with an external magnetic field in spectroscopic measurements.

On the other hand, a photon is neutral - how can one measure its spin if there's no magnetic moment? How do we know it has spin 1?

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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/11197/2451 and links therein. $\endgroup$ – Qmechanic Aug 12 '13 at 8:39
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    $\begingroup$ Not really a duplicate, I think. The other answer only mentions neutrinos and charged particles, so the photon question is still open. $\endgroup$ – Malabarba Aug 12 '13 at 11:29
  • $\begingroup$ Have you considered conservation of spin in interactions? $\endgroup$ – Will Aug 12 '13 at 13:20
  • $\begingroup$ Of course, there is also the fact that the Maxwell Lagrangian comes out identical to the Lagrangian for a massless spin-1 particle with U(1) gauge symmetry. $\endgroup$ – Jerry Schirmer Jan 11 '14 at 18:23
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One method is based on the conservation of angular momentum.

The electronic transition must follow the selection rule $\Delta l=\pm 1$. So the first thing to do is to choose an atom with zero total angular momentum, then let the atom absorb a photon and make a transition to $l=1$ state.

Secondly, we use the Stern-Gerlach experiment to detect the magnetic moment of this atom, which are $m=0,\pm 1$ in our case. Repeating experiments with random photons, we should see that there are mainly three bright spots on the screen: (1) non-deflected, (2) up and (3) down. The distance of outer spot from the center can also be calculated from the theory. Make sure the half-life of the excited state is long enough for the experiment.

In this way you can prove that the photon has spin=1. (That is what I thought, no reference for the actual experiment.)

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  • $\begingroup$ Does anybody have a reference for such an experiment? $\endgroup$ – yippy_yay Aug 15 '13 at 13:09
  • $\begingroup$ This can be a viable idea for atoms with the nuclear spin 0; otherwise the confusion between angular momentum of the atom and one of its electron shells will disrupt conservation arguments; see physics.stackexchange.com/questions/11197/… for details. Also, it isn’t clear whether the author refers to the total angular momentum j or to orbital angular momentum ℓ; only the former is a conserved quantity. $\endgroup$ – Incnis Mrsi Aug 20 '14 at 9:05
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Oddly, polarizing sunglasses provide a quite solid proof that photons are spin 1.

That's because if you rotate polarizers by only 90$^\circ$, you will find that you can break photons down into two mutually exclusive populations of photons. That is geometrically possible only if the particle in question is a vector boson, that is, a spin 1 particle.

In contrast, if you were worried instead about "electron glare" from oncoming spin 1/2 electron radiation (please don't try this at home, yes?), you would instead have to rotate your "electron polarizers" by 180$^\circ$ to observe and fully isolate the two distinct electron populations. Both of these polarization detector angles -- 180$^\circ$ for spin 1/2 particles, and 90$^\circ$ for spin 1 particles -- are deeply linked to the underlying symmetries of the particle spins, and so uniquely identify the spins of those particles.

However, I should also note that there is a profound strangeness in how such photon polarization states operate. You might call it the "missing state" problem.

By way of contrast, silver atoms are also spin 1, and conveniently have magnetic moments that allow "easy" separation. For example, by using three sequential Stern-Gerlach devices you can in principle divide up a population of silver atoms into six (yes, I said six, not three) populations that correspond geometrically to atoms with spin axes oriented along $\pm$X, $\pm$Y, and $\pm$Z. For a single silver atom, excruciatingly carefully use of this same arrangement can in principle create spatially isolated fractions of the wave function of the silver atom. When this done, though, you wind up with no more than three of the six "holders" being used, one along each axis. You might for example end up with 71% of the silver atom wave function in the +X Stern-Gerlach holder, 55% in -Y, and 45% in +Z. (It's a vector sum, so those percentages add as vector components.)

But where is the comparable experiment for photons? You can break them up along horizontal and vertical (or X and Y) polarizations, sure. But since you can't stop along their direction of propagation, how can you handle that aspect of the separation process?

(Hmm, odd thought: Actually, nowadays, there are a few labs that can bring photons to a full halt now by using specially tuned metal atom Bose condensates. So, have any of those folks thought to devise a way to look more closely at the "hidden photon state" issue, I wonder? Might that be related to some of the experimentation others mentioned in answers?)

(Second odd thought: Only photons in a vacuum travel at $c$. Shouldn't spin 1 photons traveling through refractive media at less than $c$ thus have some sort of explicit, accessible version of vector states along their propagation axes? After all, such slowing could I think be viewed (maybe?) as a mix of $c$ and "stopped" photon states, with the latter presumably showing explicit propagation axis states.)

Finally, at least in comparison to spin 1 atoms, photons also don't behave very nicely with respect to expressing a singular spin axis direction. For example, a single X-axis Stern-Gerlach divides silver atoms up into three groups: +X, -X, and "other" (that is, zero X spin). Later Stern-Gerlach devices can then further subdivide the "other" group into the remaining $\pm$Y and $\pm$Z populations.

But for photons, you only get one X axis group labeled "horizontal polarization." Where does the spin axis point in that case? It doesn't. The unique $\pm$X groups that are easily seen with silver atoms don't really have an analog with photons, at least none that I'm aware of. Perhaps someone else may have more insights?

So, my apologies for all the "added extras," but the original point remains: While odd in many ways, photons readily and provably give away their spin 1 natures by the way they can be broken up into two unique and isolated populations via a 90$^\circ$ rotation of a photon detector (polarizer). Beyond that, photon spins get pretty weird and not nearly so simple.

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    $\begingroup$ The sunglasses and electron glare example is wonderful. Believe it or not I was having trouble coming up with a good, simple example of how the "$SO(3)$ double cover ribbon trick" shows up in fundamental particles and this definitely IS IT- I've found that even small children in science demonstrations at my daughter's school are fascinated by the belt trick and the idea that not everything comes back to its beginning state when rotated through $360^o$ - you even get questions like "can you make fancier ribbons so they need to spin three times to get untangled?" from children as young as 7. $\endgroup$ – WetSavannaAnimal Aug 13 '13 at 23:54
  • $\begingroup$ Sx,Sy,Sz are non-commute, how can you get six? And how those six distributed? $\endgroup$ – unsym Aug 14 '13 at 1:44
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    $\begingroup$ Hmm, right or wrong, here was my thought: Using Feynman's hypothetical accelerate-then-stop S-Gs oriented along X, Y, and Z axes: (1) Apply S-G(X) to isolate {+1 X, 0 X, -1 X}; (2) Feed 0 X (only, oops?) into S-G(Y) (did I err there? maybe!) to isolate {+1 Y, 0 Y, -1 Y}; (3) Feed 0 X 0 Y into S-G(Z) to isolate {+1 Z, 0 Z, -1 Z}, with the assumption that 0 Z will be empty. So: The entire process definitely should be reversible, and you will definitely get either 6 or 7 groups... but what did I do with that process? In particular, will there be 6 or 7 groups? (Might make a nice question...) $\endgroup$ – Terry Bollinger Aug 14 '13 at 17:35
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    $\begingroup$ "That's because if you rotate polarizers by only 90∘, you will find that you can break photons down into two mutually exclusive populations of photons. That is geometrically possible only if the particle in question is a vector boson, that is, a spin 1 particle." I don't understand why this is the case. $\endgroup$ – yippy_yay Aug 15 '13 at 13:11
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    $\begingroup$ Although the polaroid argument is valid, the statement “silver atoms are spin 1” is incorrect. The spin of such object as a silver atom is undefined; see physics.stackexchange.com/questions/11197/… for details. $\endgroup$ – Incnis Mrsi Aug 20 '14 at 8:35
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In Richard Beth's Mechanical Detection and Measurement of the Angular Momentum of Light, bright light from a mercury arc lamp was circularly polarized and passed through a half-wave plate (which reverses the sense of circular polarization) attached to a torsion pendulum. A bit of clever experimental design sent the light through the half-wave plate twice so that each photon would exchange angular momentum $4\hbar$ with the fiber. Beth confirmed this prediction, and also showed it was consistent with angular momentum stored in the fields of classical electromagnetism.

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Spin of the photon is an ongoing theoretical research. For classical electromagnetic field the total angular momentum is $\vec{J}=\vec{r}\times <\vec{E}\times \vec{B}>$ (all vectors). In field theory this quantity is $\vec{J}=\vec{L}+\vec{S}$, it is gauge invariant and possible to observe. But! Nobody found a correct way to represent it in the sum of two gauge invariant operators for angular orbital momentum and spin (L and S). This concludes the theoretical knowledge - operator of photon's spin is not known.

It is possible to measure the Z-component of the angular momentum - helicity.

If you are thinking of photon's spin as a MAGNETIC quantity then you are definitely not correct (not wrong either). There is no double slit experiment for photons :) Unfortunately.

You can read further here : http://arxiv.org/abs/arXiv:1006.3876

Frankly speaking photon is not as simple as people are trying to pretend. Cracking it's nature will need more efforts.

Thank you for thinking.

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A perhaps not completely rigorous, but easy to understand derivation:

It can be shown that the magnitude of angular momentum of circularly polarized classical radiation of frequency $\omega$ and energy $E$ is given by $E/\omega$. If we now assume that radiation is quantized in packets of energy $E = \hbar \omega$, we arrive at the conclusion that the angular momentum of light is quantized in packets of $\hbar$.

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I would say this is an empirical fact. In atomic physics you don't observe optical transitions (e.g. induced by a laser) without angular momentum transfer. The change in angular momentum is always $\pm$ 1, that's what the photon can transmit. See http://en.wikipedia.org/wiki/Selection_rule The atomic states with different angular momenta are identified via their different energy levels due to the Zeeman effect in a magnetic field.

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    $\begingroup$ You've just restated the fact that the OP asked about. Say how it is known that there is always an angular momentum transfer... $\endgroup$ – dmckee Mar 21 '14 at 20:28
  • $\begingroup$ @dmckee this answer basically explains it in the same was as accepted answer. $\endgroup$ – doc Dec 11 '16 at 23:21

protected by Qmechanic Jan 11 '14 at 17:58

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