Valley degree of freedom in graphene I know that in the energy band structure of graphene there are six points where the valence and conductance band touch (At $E=0$), called Dirac points.
Only 2 of these points are inequivalent, K and K'. If I expand the hamiltonian around those points I get the 2D Dirac Hamiltonian (hence the name "Dirac points").
What I don't understand is why you can treat the valleys as a discrete degree of freedom (like spin). It looks like the electron can only be either in the valley corresponding to K, or the one corresponding to K'. But momentum is continuous so in principle the electron could also be not exactly at the center of one valley, or somewhere in the middle between the two. How do you explain that?
 A: It’s true that if a electron can have any energy, then it could be in a high energy state near, say, the $\Gamma$ point, and all the discussion of the properties around the $K$ points is moot. But generally we have the understanding that energies are low, and the electron experienced the linear dispersion around $K$. One of the properties of $K$ is that the electron spin is “locked” in one way or the other depending on if it’s at $K$ or $K’$ (clockwise vs. counterclockwise rather than up or down). This spin locking is the same for all energies below which trigonal warping becomes significant (trigonal warping is the deviation of the circular cross section of the band structure at low energy into a triangular shape at higher energies). Because of this, all electrons near the $K$ point have the same character in this regard, as do all near $K’$, regardless of the precise momentum value. As long as the energy remains low, the electron will remain in one of these two states, and it behaves like a quantized (either $K$-like or $K’$-like) degree of freedom.
The thing to keep in mind with graphene is that it is a solid state system with many many interacting particles. The we say “electron” as a shorthand, when we actually mean quasiparticle. Electrons don’t really have zero mass, for example, the quasiparticles just act like they do when they have low energy near the $K$ points.
A: Only the valleys (K and K') are near the Fermi energy, so these are the points where the electrons go when they get excited from valence band. Notice that usually the graphene are half filled, i.e. the electrons are exactly filled up to the Dirac point energy, which in typical tight-binding calculation is just 0 energy.(You are correct that momentum is continuous, but all other momentum states are well above the Fermi energy so electrons can hardly be excited to there.)
Besides, these two points are well separated in momentum space, so it is hard to change a electron in K valley to K' valley via scattering (hard to supply such a large amount of momentum), so we can treat K and K' as discrete degress of freedom.
