Is the spin of a photon necessarily parallel or antiparallel to its momentum?

[Zarathustra0]

I ask only because I thought that if a particle moves at the speed of light, its spin angular momentum must be along the axis of its movement (parallel or antiparallel). That said, if one were to measure the component of a photon's spin angular momentum along its direction of movement, one would inevitably get hbar, 0, or -hbar. The magnitude of the photon's spin angular momentum must be root-2 * hbar, but since this is greater than any possible value of the component in the direction of motion, musn't there be another non-zero component?

 

[Bill_K]

No, you're trying to apply formulas that only apply to particles with mass, like S² = S(S+1). The photon does not have a state with spin component 0, just +1 or -1.

 

[Zarathustra0]

OK, thanks--to make sure I understand, then, the magnitude of a photon's spin angular momentum is in fact always hbar?

 

[tom.stoer]

The "problem" exists even for the massless photon b/c S₃= +1 or -1 but S² = 2.

The solution is that even if the values for S₁ and S₂ are undefined b/c S₃ has a sharp value +1 or-1, these two values are not zero. So the modulus of S² is "distributed in S₃ and in the directions perpendicular to the 3-axis".

Caveat: it does not really makes sense to try to find the right words b/c spin is intrinsically quantum. What one can savely say is that the formulas predict the experimentally results correctly.

 

[Zarathustra0]

Oh, OK, in that case is it not true that the spin angular momentum of the photon must be along its direction of motion? I thought I had been led to believe that that was the case.

 

[tom.stoer]

Of course the spin eigenstate in z-direction (better: helicity) means either +1 or -1, never 0. But the state is not an eigenstate to the other two operators.

S_z|+1> = |+1>
S²|+1> = 2|+1>
<+1|S_x² + S_y²|+1> = <+1|S² - S_z²|+1> = 2-1 = 1

... strange but true ...

 

[Bill_K]

I guess I'm not understanding this either, Tom. I think you're also quoting SO(3) results where they do not apply. S_x² + S_y² + S_z² with eigenvalues s(s+1) is the Casimir operator for SO(3). And as I'm sure you know, the little group for a photon is not SO(3) but the two-dimensional Poincare group, in which [s₁, s₂] = 0. For a photon state with helicity n, s₁|n> = 0 and s₂|n> = 0, the expectation of s₁² + s₂² is zero, and the eigenvalue of the Casimir operator is just n².

 

[tom.stoer]

Bill is right; my idea to use SO(3) for photons is an oversimplification, misleading - or simply wrong :-(

The formula is correct, but does not apply to massless spin-1 vector bosons.