# If a polarized light wave is indistinguishable from its original self after being flipped 180°, why doesn't a photon have a spin of two?

The spin of a photon has a counterpart in classical physics, it's polarization, right?

And if you spin a polarized light wave by 180°, (or pi radians), it is now the same as before, correct?

So why isn't the spin of a photon said to be two, rather than one?

EDIT: To Michael Seifert and Anna V: I deeply appreciate the answer(s) and comment(s) on this and other questions of mine, etc. ... But....

HOW does the phase shift of 180° (or 1-pi radians) manifest itself in 'observable(s)'? How can one distinguish a '-i' photon from one that was twisted by 1-pi radians in the opposite way (now i, rather than -i)?

• Can you explain why you think that the spin has to be two if the wave is invariant under a $180^\circ$ rotation? – Cream Apr 13 at 11:42
• I think you should check your assumptions. For example, clockwise and counter-clockwise polarization states are different, but obtained from one another by a 180 degree rotation. – Eric Smith Apr 13 at 12:21
• @EricSmith He is talking about rotation around the direction of propagation. Circular polarizations are invariant under such a rotation. But they do affect linearly polarized waves different for different spin (I also understood that now). – Cream Apr 13 at 12:38
• @MichaelSeifert What do you mean with half-cycle phase shift? Both obtain a phase of $i \pi$ (i.e., a minus sign) under $180^\circ$ rotation. They behave differently under other rotations (e.g. $90^\circ$). But one does not obtain clockwise polarization from counter-clockwise polarization under $180^\circ$ rotation as you claimed. – Cream Apr 14 at 8:13
• @Cream: I didn't say that a 180° rotation affects their handedness; I said that a 180° rotation about the propagation axis is equivalent to a phase shift of π radians, i.e., a half-cycle (which is what you said.) My comment was due to my confusion about your statement that "circular polarizations are invariant under such a rotation." I read that to be saying that at 180° rotation about the axis of propagation leaves a circularly polarized wave invariant, and I didn't think that was correct. But we both seem to agree that's not the case. – Michael Seifert Apr 14 at 11:43

If you rotate a polarized classical electromagnetic wave by 180° about its axis of propagation, you don't get the same wave back; you get the same wave phase-shifted by half a cycle. You need to make a full rotation of 360° to get the same classical wave solution back, so by your logic, the photon should indeed have spin 1.

This is contrast to a gravitational wave, for which a rotation of 180° about the propagation direction yields exactly the same solution with no phase shift. This manifests on the quantum level through the graviton being a spin-2 particle.

• has anybody ever rotated a gravitional wave 180°? Or has this been observed in nature? – Michael Apr 14 at 21:10
• But how is the quantum or classical solution manifestly different when only shifted by 180° instead of the full 360°? What is the physically noticeable difference? – Kurt Hikes Apr 16 at 15:08
• @KurtHikes: The main difference (I think) is that a 180°-rotated wave will interfere destructively with the original wave. A 360°-rotated wave, of course, will not. – Michael Seifert Apr 16 at 16:59

The photons are point quantum mechanical elementary particles of spin either +1 or -1, to their direction of motion. Their mass is zero and their energy equal to $$hν$$ where $$ν$$ is the frequency of the classical wave built up by thousands at least of photons.

The way classical electromagnetic polarized waves are built up by their constituent photons is seen qualitatively in this a simple drawing of the connection of photon spin projection to circular polarization:

Photons are not electric and magnetic fields varying in space. The quantum mechanical (rightmost column) superposition of a large number of photons adds up to the polarized classical electromagnetic wave on the left column.

The mathematical way this happens is complicated and needs knowing quantum field theory, but experimentally , that individual photons add up to the classical wave can be seen in this classroom experiment of interference of classical light appearing by a superposition of a large number of photons.

• How can a right-handed shift (-i ) be distinguished in some way from a left-handed shift (i)? – Kurt Hikes Apr 16 at 15:11
• @KurtHikes Not for single photons, the wave function is not measureable. It is the collective effect seen on the left polarization that says how the photons making up the beam are polarized. – anna v Apr 16 at 15:15

I assume the photon moves in $$z$$-direction here:

The two polarizations of the photon (corresponding to its two spin states) are not linear polarizations, but clockwise and counter-clockwise circular polarizations (you obtain linear polarization as a superposition of a clockwise and counter-clockwise one). These are completely rotationally invariant. The spin points in $$z$$-direction or against it.

This means for a "pure" state (with only one polarization), rotation around the $$z$$-axis will not change the polarization of the photon. It will only change its phase: $$|\uparrow\rangle \to e^{-i \theta} |\uparrow\rangle,~ |\downarrow\rangle \to e^{+i \theta} |\downarrow\rangle.$$ As you see, this depends on the spin of the photon ($$+1$$ in and $$-1$$ against direction of the $$z$$-axis). If the photon had spin-$$2$$, we would have a factor of 2 in the exponent.
For a superposition, however, the respective "pure" states can have a relative phase and will change: $$|\uparrow\rangle + |\downarrow\rangle \to e^{-i \theta} |\uparrow\rangle + e^{+i \theta} |\downarrow\rangle$$ For a $$180^\circ$$ rotation ($$\theta = i\pi$$), the mixed state will pick up an overall minus sign, but the linear polarization will be the same: $$|\uparrow\rangle + |\downarrow\rangle \to -(|\uparrow\rangle + |\downarrow\rangle) \sim |\uparrow\rangle + |\downarrow\rangle$$ For $$360^\circ$$ we are back to at the starting point.
Only for $$90^\circ$$ degree ($$\theta = i\pi/2$$), we notice a real change: $$|\uparrow\rangle + |\downarrow\rangle \to -i(|\uparrow\rangle - |\downarrow\rangle )$$ (a relative minus sign) and the polarization changes. So spin-1 is exactly what corresponds to the photon.

Your question is posing a characterization that's grounded in a misunderstanding and does not fit the situation at all. What you described is not what spin is, nor how it is characterized. And, in fact, some of the replies are also grounded in a misunderstanding and mischaracterization.

First, photons don't have spin. They have helicity. Spin is an attribute only of tardions (i.e. the class corresponding to positive rest-mass systems, sometimes also called bradyons). The description is applicable both classically and quantum theoretically; in both cases the captured by the symplectic geometry of the system. The key property is that tardions with spin 0 have 3 complementary pairs of symplectic coordinates, while those with non-zero spin have 4. For quantum systems, the 4th is quantized customarily as the "m" quantum number, and is complementary to the azimuthal angle; while the others are described by the Heisenberg conjugates of position $$𝐫$$ and linear momentum $$𝐩$$.

The most important property is that there is a decomposition of the total angular momentum for such a system into an orbital angular momentum - corresponding to the motion of its center of mass with respect to a fixed point (usually denoted by $$𝐋$$, and defined by the vector product $$𝐋 = 𝐫×𝐩$$) - and the internal angular momentum (denoted $$𝐒$$) which is the angular momentum of the system with reference to its own center of mass. The total angular momentum is $$𝐉 = 𝐋 + 𝐒$$.

For composite systems, $$𝐒$$ is the total of the orbital angular momenta of the system's components about the system's center of mass plus the total internal angular momenta of each of its components. For elementary systems, it does not decompose any further, is referred to as its "spin", and is not the result of anything "in" the system circulating or "spinning" at all. It's just simply the intrinsic angular momentum of the system.

All of this applies to the descriptions of systems in both classical and quantum theory, though its retro-applicability to classical theory was not recognized until around the middle of the 20th century.

The situation is very different for luxons. They may or may not be helical. The most general case, for non-helical luxons, has 4 complementary pairs of symplectic coordinates and sometimes also characterized as systems with "continuous spin", which however is not spin in the above sense at all and involves nothing like a spin-orbit decomposition.

The more restricted subclass of luxons are those with "helical" angular momentum: those in which the axis for the intrinsic angular momentum $$𝐒$$ is collinear with the linear momentum $$𝐩$$. The photon fall into this subclass (as does the Higgs, by the way). For such systems, the ratio of the two is fixed and is an invariant property of the system. Only its orientation can vary: the axis for $$𝐒$$ aligns with $$+𝐩$$, or it aligns with $$-𝐩$$ ... if it is non-zero.

The helical class reduces further to the case of 0 helicity (e.g. the Higgs) and non-zero helicity (the photon). Both classes have only three complementary pairs of symplectic coordinates; but only the former has a semblance of a "spin-orbit" decomposition - namely 0 helcity, with angular momentum that is 100% orbital: $$𝐉 = 𝐋 = 𝐫×𝐩$$ and "$$𝐒 = 𝟬$$". So, sometimes they are referred to as "spin 0", though incorrectly.

Helical luxons with non-zero helicity are visualized as left and right circular polarization; the polarization states of a photon are one and the same as its helical states - which you've seen vividly illustrated here, in other replies.

But there is no "m" quantum number, since there isn't a 4th complementary pair in the first place; and no spin-orbit decomposition at all. In fact, its symplectic geometry can be described in the usual way by the three complementary pairs of position $$𝐫$$ and momentum coordinates $$𝐩$$, except that the coordinates are singular in much the same way that spherical coordinates are for spherical geometry; and for these systems, with a suitable definition for $$𝐫$$, complementary to $$𝐩$$ - after choosing a unit vector $$𝐧$$, you can write $$𝐉 = 𝐫×𝐩 + σp²c/E 𝐧×𝐩×𝐧/|𝐧×𝐩|²$$ where its helicity is $$σ$$, $$c$$ is light-speed and $$E$$ is its energy.

(The situation is analogous to what happens with the symplectic description, discussed in Lecture Notes in Physics 188 (https://doi.org/10.1007/3-540-12724-0_1), for magnetic monopoles, except the roles of $$𝐩$$ and $$𝐫$$ are somewhat reversed).

There's a fairly well-known "no position operator" theorem that's often cited (and mis-applied) as a folklore result, which forbids the existence of a single coordinate chart for its symplectic geometry ... but this does not rule out a position-operator, per se. The situation, here too, is analogous to the role that the "no hair" theorem plays in ruling out a single coordinate chart for spherical geometry.