Angular Momentum of a Photon Why is it that the angular momentum of a photon is $\hbar$, irrespective of its energy? I encountered such a claim in a text about Raman spectroscopy. Is there an explanation for this using basic basic quantum mechanics? As I am studying this as part of a chemistry course. Any help is much appreciated!
 A: (sorry, I couldn't write this in the comment section) Have you met the postulates of quantum mechanics?
Here is a summary of them
http://vergil.chemistry.gatech.edu/notes/quantrev/node20.html
Postulate 3 says if an observable has associate a (hermitian) operator, the only values we would observe for one photon the spin-angular momentum are the eigenvalues of the equation
$\hat{S_z} | \hspace{2mm}{\psi}> = \pm \hbar |\hspace{2mm} \psi >$
where $\hat{S_z}$ is the projection of spin-angular momentum in along the $z$-axis.
More generally with the Total angular momentum (i.e. spin-momentum coupling) $\hat{J} = \hat{L} + \hat{S}$ It turns out for photons in circularly polarised light the eigenvalues are the same. 
It does require a little more computation, using properties of the operators and matrix mechanics, but in general it is not the case that a photon has an orbital angular momentum of $\pm \hbar$. in fact, photons can be plane waves, circularly polarised waves, even elliptically polarised etc. And with each of these modes, that in general represents different angular momenta; a good visual example can be found here
https://en.wikipedia.org/wiki/Orbital_angular_momentum_of_light
Note again the presence of that postulate we saw earlier: the eigenvalues of the operator $\hat{L_z}$ turn out to be $m\hbar$ where $m$ can take values $-l,-(l-1)...,(l-1),l$ in steps of one, and $l$ is the operator of the angular momentum operator $\hat{L}$.
