Graphene dispersion at Dirac points

Question

After deriving the dispersion relation of graphene:

$$ E(k) = \pm t \sqrt{3+2\cos{(k_y \sqrt{3} a)} + 4\cos{\bigg(\frac{3a}{2} k_x \bigg)} \cos{ \bigg(\frac{\sqrt{3}}{2}a k_y \bigg)} } $$

how do I see that the energies are degenerate at the Dirac point $K = (4 \pi/3a, 0, 0)$? When I plug this in, I just get: $$ E(k) = \pm t \sqrt{3+2+4}$$ which is certainly not degenerate. What am I misunderstanding here? Thanks!

Answer 1

The Wikipedia page about graphene's [electronic properties] has a different expression for the energy bands: $$ E_1(k) = \pm\,\gamma_0\, \sqrt{1+4\cos^2\frac{ak_x}{2} +4\cos\frac{ak_x}{2}\,\cos\frac{\sqrt{3}\ ak_y}{2}} \tag{1} $$ which actually does give $E(k) = \pm 0$ for $k=({\large\frac{4\pi}{3a}},0)$. If we plot the square root part of the Wikipedia result we get this: (with the green sticks marking the zeroes)

And if we plot your result $$ E_2(k) = \pm\,t\, \sqrt{3+2\cos\,\sqrt{3}\ ak_y +4\cos\frac{3ak_x}{2}\,\cos\frac{\sqrt{3}\ ak_y}{2}} \tag{2} $$ which is apparently also the result from [RevModPhys.81.109] we get the following (for $t=1$), where we see 10 instead of 6 zeroes and the hexagonal pattern is smaller and $30^{\large \circ}$ rotated:

There seems to be a different choice of basis (and scale). You can convert $(1)$ into $(2)$ by using: (if we assume $t=\gamma_0$) $$ E_2(k_x,k_y) = E_1(\pm\sqrt3\, k_y, \pm\sqrt3\, k_x) $$ and of course the degeneracy in $E_2$ is now at $(k_x,k_y)=(0,{\large \frac{4\pi}{3\sqrt3 \,a}})$

Answer 2

just adding to the answer above: you also need to be careful with conventions of the unit cell because this will affect the BZ and the position of the K points. there are 2 main conventions I'm aware of:

1. 2 vertices of the real space hexagon are on the x axis leading to 2 vertices on the ky axis

2. 2 vertices on the y axis leading to 2 vertices on the kx axis.

I believe your original dispersion relation is in the second convention and the Wikipedia page in the first one.

To see the difference yourself I suggest you try to derive the dispersion relation from a tight binding model:

• write down the tight binding hamiltonain (I suggest to do it in bra-ket notation)
• perform discrete fourier transform (this is where the different conventions will differ- the hopping vectors will be different)
• you will get an expression you can write as a vector-matrix-vector multiplication: (|k,A>,|k,B>)H (<k,A|, <k,B|)^T
• finally the energy you will get from the determinant of the Bloch hamiltonian

hope that helps clearing up the confusion

Edit: I saw new the edit of the answer above- I think we see 4 "extra" degeneracy points because the plot goes beyond the first BZ.