How can a photon have a spin of $1$ when it has no mass? Since the spin is intrinsic angular momentum, if $m = 0$, then the spin must be zero.


A classical picture of spin angular momentum treats particles like tops, and says that their spin is given by some form of $I\,\omega$, where $I$ is derivable from the objects' geometry and mass, and $\omega$ is its angular velocity.

For quantum mechanics, this picture does not work, at least when describing spin angular momentum. Even for the electron, it if you try and work out some boundaries on what $I$ can be from experiments about the size of the electron, you get implausible values for the angular velocity.

It is better to just think of spin as an intrinsic property of quantum particles that relates to the particle's symmetery under rotation. (spin 0 particles do not alter under rotation, spin 1 particles rotate to themselves after being rotated through a full circle, spin 1/2 particles rotate to themselves after being rotated through 720 degrees, etc.). Within the formalism of quantum mechanics, there are ways of relating these concepts clearly and showing why rotations should give rise to this sort of property. But it is not best to think of the photon spinning like a top.

  • $\begingroup$ Can we view spin as related to the number of rotational symmetries of the mathematical object used to represent the particle in its field? $\endgroup$ – sakurashinken Mar 7 at 22:12

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