Wikipedia states that a photon has a spin of value 1.

What I want to know is this: are there two possible spins for photons, i.e. +1 and -1 (like electrons, which can have +1/2 or -1/2)?

If the spin quantum number (the same as "spin"?) has to be essentially positive (to quote Wikipedia: "The conventional definition of the spin quantum number, s, is s = n/2, where n can be any non-negative integer"), then what is the term for possible +/- sign states of that spin-value? Is it called "parity"?

Would there be any difference of behavior, e.g. a Circular polarization event, such as through a solution of D-glucose?

Since through a solution of such chirally-asymmetric molecules, whole-portion of applied plane-polarized-light, light turns in a single direction (from which the compound is analyzed and given a +/- nomenclature), could I conclude that all the photons present in the applied plane-polarized-light contain only 1 type of spin out of 2 for that particular experiment (optical rotation)?


By definition of spin $S$ it is a positive integer number or zero. Not to confuse with the spin projection possible values $S_z$, which may run from $-S$ to $S$.

  • $\begingroup$ That's called "spin projection value"? thanks. $\endgroup$ – Always Confused Jun 30 '16 at 8:41
  • 1
    $\begingroup$ (And just to avoid confusion on a HNQ, $S$ can also be a half-integer like 1/2.) $\endgroup$ – Emilio Pisanty Jul 1 '16 at 0:31

The mention of the spin of elementary particle is closely related to the representations of the Poincare group, since the mathematical definition of the elementary particle is irreducible representation of the Poincare group. Our world within some approximation obeys the special relativity, and the above definition of the elementary particle just reflects this fact.

Massless particles, as the representations of the Poincare group, are characterized not by spin, but by helicity $\lambda$ - the projection of total angular momentum on the direction of motion. Formally this is because the Casimir operator, which defines the spin of representation (squared Pauli-Lubanski operator), is always zero for massless representations, and, also, the little group for massless Lorentz orbit is ISO(2) group. Physically this can be understood from the meaning of the spin - it is the total angular momentum at rest, and therefore can't be defined for massless particles moving with the speed of light.

Now let's discuss your question. In general, irreducible states with helicities $\lambda , -\lambda$ belong to different representations, and hence to different elementary particles. This is because there is no continuous transformation (of the Poincare group) which can convert $\lambda$ state to $-\lambda$ state. This is the huge difference in compare with case of non-zero mass, for which the representation with given spin $s$ includes helicities $-s,-s+1,...,s$ states initially, since there exist continuous transformation, which relates these values of helicities.

Note, however, that state with helicity $\lambda$ can be converted into the state with helicity $-\lambda$ by discrete transformations of the Lorentz group, for example, by spatial inversion transformation. So if the given theory is parity invariant, we have to introduce the particle, which corresponds to the direct sum of representations with $h = \pm\lambda$. The EM theory is an example of parity invariant theory, and corresponding interaction mediator - photon - thus has two possible helicities.


Spin 1 just means that the spin in any direction can assume values out of {-1,0,1}. The 0 is only possible for massive particles, so the photon can have spin -1 or +1. That's like clockwise and anticlockwise circular polarization


protected by Qmechanic Jun 30 '16 at 9:20

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.