Expanding the Graphene Hamiltonian near Dirac points upto second order term I was trying to solve the Graphene Hamiltonian near the Dirac points upto the second order term for the nearest neighborhood points.
So expanding the function near the Dirac Point, we get$$g(K+q)=\frac{3ta}{2}e^{i\pi/3}(q_x-iq_y)+\frac{3ta^2}{8}e^{i\pi/3}(q_x-iq_y)^2$$
Now the Hamiltonian should be $$
\begin{pmatrix}
0 & g(K+q)\\
g^*(K+q) &0
\end{pmatrix}
$$
And the energy form will be
$E=\pm |g(K+q)|$
Now for solving only for the first order term we get the massless Dirac equation as $E=\pm\hbar v|q|$, which I got.
While looking for the so called trigonal warping, I need the second term as $$=\mp\frac{3ta^2}{8}sin(3\theta_p)|q|^2   $$ where $\theta_p=\arctan(\frac{q_x}{q_y})$
Now I have tried solving for this term a lot of times using lot of ways but I can't get around that $sin(3\theta_p)$ term.
I got $$E =\sqrt{\frac{9t^2a^2}{4}(q_x^2+q_y^2)+ \frac{9t^2a^4}{64}(q_x^4+q_y^4)+ \frac{9t^2a^3}{16} (q_x^3+q_xq_y^2)  } $$ Now how to take a square root of this thing and get that $sin(3\theta_p)$ term.
Please help me!!
Also any information about this trigonal warping would be also useful. Thank You!!
 A: Something seems to be off with the energy term here... Your last term goes as $\cos\theta$.
Let's try again :)
$$
H = \begin{pmatrix}
0&-t f_\mathbf{q}
\\
-tf_\mathbf{q}^*&0
\end{pmatrix}\,,
$$
where $f_\mathbf{q} = 1 + e^{i\mathbf{q}\cdot\mathbf{d}_1}+ e^{i\mathbf{q}\cdot\mathbf{d}_2}$ and $\mathbf{d}_1$ and $\mathbf{d}_2$ are the basis vectors. If the Dirac point is at $\mathbf{q} = \mathbf{K}$, for momenta close to it, we write
$$
f_\mathbf{q} = 1 + e^{i\left(\mathbf{q}+\mathbf{K}\right)\cdot\mathbf{d}_1}+ e^{i\left(\mathbf{q}+\mathbf{K}\right)\cdot\mathbf{d}_2}
 = 1 
+ 
e^{i\mathbf{k}\cdot\mathbf{d}_1}
e^{i\mathbf{K}\cdot\mathbf{d}_1}
+ 
e^{i\mathbf{k}\cdot\mathbf{d}_2}
e^{i\mathbf{K}\cdot\mathbf{d}_2}\,.
$$
For small $k$, we get
$$
f_\mathbf{q} \approx 1 
+ 
(1+i\mathbf{k}\cdot\mathbf{d}_1 - \frac{1}{2}\left(\mathbf{k}\cdot\mathbf{d}_1\right)^2)
e^{i\mathbf{K}\cdot\mathbf{d}_1}
+ 
(1+i\mathbf{k}\cdot\mathbf{d}_2 - \frac{1}{2}\left(\mathbf{k}\cdot\mathbf{d}_2\right)^2)
e^{i\mathbf{K}\cdot\mathbf{d}_2}
\\
=
1 
+ 
e^{i\mathbf{K}\cdot\mathbf{d}_1}
+ 
e^{i\mathbf{K}\cdot\mathbf{d}_2}
+
(i\mathbf{k}\cdot\mathbf{d}_1 )
e^{i\mathbf{K}\cdot\mathbf{d}_1}
+ 
(i\mathbf{k}\cdot\mathbf{d}_2)
e^{i\mathbf{K}\cdot\mathbf{d}_2}
 - \frac{1}{2}\left(\mathbf{k}\cdot\mathbf{d}_1\right)^2
e^{i\mathbf{K}\cdot\mathbf{d}_1}
- \frac{1}{2}\left(\mathbf{k}\cdot\mathbf{d}_2\right)^2
e^{i\mathbf{K}\cdot\mathbf{d}_2}
\,.
$$
Using $\mathbf{K} = 4\pi(0, 1)/(3\sqrt{3}a)$ and $\mathbf{d}_{1/2}=a(3,\pm\sqrt{3})/2$, we have $e^{i\mathbf{K}\cdot\mathbf{d}_1} = e^{ 2\pi i/3}$ and $e^{i\mathbf{K}\cdot\mathbf{d}_2} = e^{- 2\pi i/3}$:
$$
f_\mathbf{q} 
=
i\mathbf{k}\cdot(\mathbf{d}_1 
e^{ 2\pi i/3}
+ 
\mathbf{d}_2
e^{- 2\pi i/3})
 - \frac{1}{2}\left(\mathbf{k}\cdot\mathbf{d}_1\right)^2
e^{2\pi i/3}
- \frac{1}{2}\left(\mathbf{k}\cdot\mathbf{d}_2\right)^2
e^{- 2\pi i/3}
\\
=
-\frac{3a}{2}ik e^{-i \theta}
 - \frac{1}{2}\left(\mathbf{k}\cdot\mathbf{d}_1\right)^2
e^{ 2\pi i/3}
- \frac{1}{2}\left(\mathbf{k}\cdot\mathbf{d}_2\right)^2
e^{-2\pi i/3}
\,,
$$
where $\mathbf{k} = k(\cos\theta,\sin\theta)$.
We know that the dispersion is $\pm t^2\sqrt{f_\mathbf{q}f_\mathbf{q}^*}$, so let's deal with the square root. For small $k$ (keeping to 3d order), we have
$$
f_\mathbf{q}f_\mathbf{q}^* \approx \frac{9a^2}{4}k^2 
-\frac{3a}{2}ik e^{-i \theta}
\left[
 - \frac{1}{2}\left(\mathbf{k}\cdot\mathbf{d}_1\right)^2
e^{ -2\pi i/3}
- \frac{1}{2}\left(\mathbf{k}\cdot\mathbf{d}_2\right)^2
e^{2\pi i/3}
\right]
+\frac{3a}{2}ik e^{i \theta}
\left[
 - \frac{1}{2}\left(\mathbf{k}\cdot\mathbf{d}_1\right)^2
e^{ 2\pi i/3}
- \frac{1}{2}\left(\mathbf{k}\cdot\mathbf{d}_2\right)^2
e^{-2\pi i/3}
\right]
\\
=\frac{9}{4}a^2k^2+\frac{9}{8}a^3k^3\sin(3\theta)\,.
$$
You can check this in, say, Mathematica.
From this,
$$
\sqrt{f_\mathbf{q}f_\mathbf{q}^*} =
\sqrt{\frac{9}{4}a^2k^2+\frac{9}{8}a^3k^3\sin(3\theta)}
=
\frac{3}{2}ak\sqrt{1+\frac{1}{2}a^2k^2\sin(3\theta)}
\approx
\frac{3}{2}ak\left(1+\frac{1}{4}a^2k^2\sin(3\theta)\right)\,,
$$
as required.
