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What is the direction of angular momentum vector of a photon?

by jartsa
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jartsa
#1
Feb10-12, 08:50 AM
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Let's say momentum vector of a photon points to direction D. What are the possible directions that this photon's angular momentum vector can point to?

The spin angular momentum is the angular momentum that I'm interested of.
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jtbell
#2
Feb10-12, 09:13 AM
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A photon's intrinsic angular momentum ("spin") is either parallel or antiparallel to its momentum. This is a consequence of its mass being zero.
lugita15
#3
Feb10-12, 02:17 PM
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Quote Quote by jtbell View Post
A photon's intrinsic angular momentum ("spin") is either parallel or antiparallel to its momentum. This is a consequence of its mass being zero.
How do you reconcile this with the fact that angular momentum operators along perpendicular directions do not commute, and thus satisfy an uncertainty relation?

Bill_K
#4
Feb10-12, 03:25 PM
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What is the direction of angular momentum vector of a photon?

This is incorrect. The members of the Lorentz group that define a particle's angular momentum form the "little group", i.e. the subgroup that leaves its 4-momentum invariant. For a particle with mass the little group is SO(3), and as you say the three generators of this group do not commute. For a massless particle the little group is the Euclidean group ISO(2), i.e. rotations and translations in two dimensions. One generator of this group is the rotation with axis along the direction of motion. The eigenvalues of this operator are Sz, the particle's helicity, namely 1 for a photon. The other two generators are null rotations in the two transverse directions. They commute with each other, and for photons their effect is simply a gauge transformation.
lugita15
#5
Feb10-12, 10:04 PM
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Quote Quote by Bill_K View Post
The members of the Lorentz group that define a particle's angular momentum form the "little group", i.e. the subgroup that leaves its 4-momentum invariant.
Why exactly are the symmetry operations that leave the 4-momentum invariant the ones used to define angular momentum? I thought that the whole reason why angular momentum is connected with rotation is that in nonrelativistic classical mechanics angular momentum is the Noether charge of SO(3)-invariance, so in nonrelativistic quantum mechanics the angular momentum operators are identified with the generators of SO(3). Where does the "little group" come into play in all this? Why would it be of paramount importance to keep the 4-momentum invariant under the exponential of the angular momentum operators?
Ben Niehoff
#6
Feb10-12, 10:33 PM
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Particles in 3-space are points. The symmetries that leave a point invariant are SO(3).

Particles in 4-dimensional spacetime are lines. Massive particles are represented by timelike lines. The symmetries that leave a timelike line invariant are SO(3). This is the "little group" of a massive particle.

Massless particles are represented by null lines. The symmetries that leave a null line invariant are ISO(2). This is the "little group" of a massless particle.

The reason we care about the 4-momentum is because this vector is tangent to the particle's worldline.

If it were possible to have a particle that appears for one instant and then disappears, so that it is pointlike in spacetime, then its angular momentum would be in a representation of the full SO(3,1).


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