Simple square lattice

• 💼 Case: square lattice, single atom unit cell with atomic orbital $\phi(\vec r)$ of isolated orbital energy $E_0$, were we assume tunnelling occurs only between nearest neighbours with amplitude $-t$

• The square lattice is generated by two vectors $\vec a_1=a(1,0)$ and $\vec a_2 =a(0,1)$

  Each atom connected to neighbours by vectors $\pm \vec a_1,\pm \vec a_2$

• We label unit cell using $l_x,l_y$ for columns and rows respectively. Where $l_x=0,1,\ldots,N_x-1$, $l_y=0,1,\ldots,N_y-1$, $N_xN_y=N$ and $N$ is the total number of unit cells

  🗒️ Note: Assume periodic boundary conditions.

• Hopping around $(l_x,l_y)$ can be to $(l_x\pm 1,l_y)$ and $(l_x,l_y \pm 1)$

• We can write the tight binding wave function

  $$ \psi_{\vec k} (\vec r)=\frac{1}{\sqrt{N}} \sum^{N_x-1}{l_x=0} \sum^{N_y-1}{l_y=0}c_{l_x,l_y}\phi(\vec r- \vec r_{l_x,l_y}) $$

• Since the position of $(l_x,l_y)$ is $\vec r_{l_x,l_y}=l_x\vec a_1+l_y\vec a_2=(l_xa,l_ya)$ equation for $E$ becomes

  $$ (E_0-E)c_{l_x,l_y}-t(c_{l_x+1,l_y}+c_{l_x-1,l_y}+c_{l_x,l_y+1} +c_{l_x,l_y-1})=0 $$

  where we sub in $c_{l_x,l_y}=\tilde c_{\vec k}e^{i\vec k \cdot \vec r_{l_x,l_y}}$ where $\vec k =(k_x,k_y)$ is a $\rm 2D$ vector we get

  $$ \begin{aligned} &\;E_{\vec k}=E_0 -t(e^{ik_x a}+e^{-ik_x a}+e^{ik_y a}+e^{-ik_y a}) \\ &\boxed{E_{\vec k}=E_0-2t[\cos(k_xa)+\cos(k_ya)]} \end{aligned} $$

💎 Conclusion:

• The energy is periodic in both $k_x$ and $k_y$ with period $2\pi /a$
• The Brillouin zone is $-\pi/a\le k_x \le \pi /a$ and $-\pi/a\le k_y \le \pi /a$ ie it is the minimum region in $\vec k$ -space for periodicity

• We can find $k_x$ and $k_y$ by applying the boundary condition $c_{N_x,l_y}=c_{0,l_y}$ and $c_{l_x,N_y}=c_{l_x,0}$

  $$ \left \{ \begin{aligned} e^{ik_xN_xa} =1 \\ e^{ik_y N_y a}=1 \end{aligned}\right. \qquad \Rightarrow \qquad \left \{ \begin{aligned} k_x=\frac{2\pi n_x}{N_x a} \quad n_x=0,1,\ldots,N_x-1 \\ k_y=\frac{2\pi n_y}{N_y a} \quad n_y=0,1,\ldots , N_y-1

  \end{aligned} \right . $$

  Therefore in the Brillouin zone the number of wave vector $\vec k$ is $N_x\times N_y = N$

💎 Conclusion:

• There are as many $\vec k$ points in the Brillouin zone as numbers of unit cells of the crystal
• The number of state in each band is $2N$ where $2$ is the spin degeneracy

(left) plot of $E_k$

• Minimum at $\vec k_\text {min}=(0,0)$, centre
• Maximum at $\vec k _\text{max} = (\pm \pi/a,\pm\pi/a)$, four corners

🗒️ Note: four corners correspond to a single $\vec k$ point due to the periodicity in $\vec k$ space

• $E_{\vec k_\text{min}}=E_0-4t$
• $E_{\vec k _\text{max}}=E_0 +4t$

💎 Conclusion: total bandwidth: $\Delta E\equiv E_{\vec k \text{max}}-E{\vec k _\text{min}}=8t$

(right) plot of band structures

• On the top left (the red arrows) are the path taken in $\rm 3D$
• The plot is the values of $E/t$ at those points of $k$ space
• We see $\Gamma$ is the lowest point, $M$ is the highest and $X$ is a saddle point

🗒️ Note:

• $\Gamma,X,M$ are common names for the minimum, side and corner respectively

• The symmetry of crystal allows us to know the value of other points (ie 90 degree rotations)

• Expand to see symmetries

Fermi surfaces, electrons and holes