   PHY.K02UF Molecular and Solid State Physics

## Two dimensional periodic functions

A periodic function consists of a primitive unit cell that is repeated on a Bravais lattice. There are five Bravais lattices in two dimensions. Oblique Rectangular Centered rectangular Square Hexagonal

A Bravais lattice can be specified by giving two primitive lattice vectors, $\vec{a}_1$ and $\vec{a}_2$. The lattice parameters (or lattice constants) in two dimensions are the lengths of the primitive lattice vectors $a = |\vec{a}_1|$, $b=|\vec{a}_2|$, and the angle $\gamma$ between the primitive lattice vectors, $\cos\gamma = \vec{a}_1\cdot\vec{a}_2/(ab)$. For convenience $\vec{a}_1$ can be chosen to point in the $x-$direction,

$$\vec{a}_1 = a\,\hat{x},\qquad\vec{a}_2 = b\cos\gamma\,\hat{x} + b\sin\gamma\,\hat{y}.$$

The primitive lattice vectors in reciprocal space can then be calculated using the conditions,

$$\vec{a}_i\cdot\vec{b}_j=2\pi\delta_{ij}.$$

The primitive lattice vectors in reciprocal space are,

$$\vec{b}_1 = \frac{2\pi}{a}\,\hat{k_x}-\frac{2\pi\cos\gamma}{a\sin\gamma}\,\hat{k_y},\qquad\vec{b}_2 = \frac{2\pi}{b\sin\gamma}\,\hat{k_y}.$$

A Fourier series can be used to describe a periodic function in two dimensions. The expression for a Fourier series is,

$$f(\vec{r})=\sum \limits_{\vec{G}} f_{\vec{G}}\exp \left(i\vec{G}\cdot\vec{r}\right).$$

The reciprocal lattice vectors are $\vec{G}_{\nu_1,\nu_2}=\nu_1\vec{b}_1+\nu_2\vec{b}_2$, where $\nu_1$ and $\nu_2$ are integers. The Fourier coefficients can be specified with the same integers $f_{\nu_1,\nu_2}=f_{\vec{G}_{\nu_1,\nu_2}}$

A two-dimensional periodic function is plotted below. The form can be used to adjust the ratio $b/a$ and the angle $\gamma$. It is possible to specify four Fourier coefficients $f_{\nu_1,\nu_2}$. The Fourier coefficients are set such that $f_{\vec{G}} = f^*_{\vec{G}}$ so that the periodic function is real. All other Fourier coefficients are set to zero.

 $b/a=$ $\gamma=$ °

$\vec{a}_1=1\,\hat{x}$   $\vec{a}_2=$ $\hat{x} + ($ $) \hat{y}$

$\vec{b}_1=$ $\hat{k}_x + ($ $)\hat{k}_y$   $\vec{b}_2=$ $\hat{k}_y$

 $f_{00}=$ $f_{01}=f^*_{0-1}=$ $+i$ () $f_{10}=f^*_{-10}=$ $+i$ () $f_{11}=f^*_{-1-1}=$ $+i$ () $f_{1-1}=f^*_{-11}=$ $+i$ ()

Determining the Fourier coefficients

If the periodic function $f(\vec{r})$ is known, the Fourier coefficients $f_{\vec{G}}$ can be determined by multiplying both sides of a Fourier series by $\exp (-i\vec{G}'\cdot \vec{r})$ and integrating over a primitive unit cell.

$$\int\limits_{\text{unit cell}}f(\vec{r})\exp (-i\vec{G}'\cdot \vec{r})d\vec{r} = \sum\limits_{\vec{G}}\int\limits_{\text{unit cell}}f_{\vec{G}}\exp (-i\vec{G}'\cdot \vec{r})\exp (i\vec{G}\cdot \vec{r})d\vec{r}$$

On the right-hand side, only the term where $\vec{G} = \vec{G}'$ contributes and the integral evaluates to $f_{\vec{G}}$ times the volume $V_{\text{uc}}$ of the primitive unit cell.

$$f_{\vec{G}} = \frac{1}{V_{\text{uc}}}\int\limits_{\text{unit cell}}f(\vec{r})\exp (-i\vec{G}\cdot \vec{r})d\vec{r}$$

Example 1: Checkerboard pattern

Consider a two-dimensional function that has a constant value $C$ on the black squares of a checkerboard and has a value 0 on the white squares of a checkerboard. The lattice constant $a$ is from the center of a black square to the center of a neighboring black square so the sides of a black square are $\sqrt{2}a/2$. This is a two-dimensional analog to the square wave problem that was solved in one dimension. This function can be expressed as a Fourier series,

$$f(\vec{r})=\sum \limits_{\vec{G}} f_{\vec{G}}\exp \left(i\vec{G}\cdot\vec{r}\right).$$

We choose the origin to be in the middle of one of the black squares. The Wigner-Seitz cell contains one black square so we integrate over the square to determine the Fourier coefficients,

$$f_{\vec{G}} = \frac{C}{a^2}\int\limits_{-\sqrt{2}a/4}^{\sqrt{2}a/4}\int\limits_{-\sqrt{2}a/4}^{\sqrt{2}a/4}\exp (-i\vec{G}\cdot \vec{r})dxdy.$$

By rewritting the exponential factor,

$$f_{\vec{G}} = \frac{C}{a^2}\int\limits_{-\sqrt{2}a/4}^{\sqrt{2}a/4}\int\limits_{-\sqrt{2}a/4}^{\sqrt{2}a/4}\exp (-iG_xx)\exp (-iG_yy)dxdy,$$

the integrals can easily be performed,

$$f_{\vec{G}} = \frac{C}{a^2}\frac{\left(\exp (-iG_x\sqrt{2}a/4)-\exp (iG_x\sqrt{2}a/4)\right)\left(\exp (-iG_y\sqrt{2}a/4)-\exp (iG_y\sqrt{2}a/4)\right)}{-G_xG_y}.$$

Using Euler's formula,

$$f_{\vec{G}} = \frac{C}{a^2}\frac{\left(\cos (-G_x\sqrt{2}a/4)+i\sin (-G_x\sqrt{2}a/4)-\cos (G_x\sqrt{2}a/4)-i\sin (G_x\sqrt{2}a/4)\right)\left(\cos (-G_y\sqrt{2}a/4)+i\sin (-G_y\sqrt{2}a/4)-\cos (G_y\sqrt{2}a/4)-i\sin (G_y\sqrt{2}a/4)\right)}{-G_xG_y}.$$

The cosine terms cancel out and the final result is,

$$f_{\vec{G}} = \frac{4C}{a^2}\frac{\sin (G_x\sqrt{2}a/4)\sin (G_y\sqrt{2}a/4)}{G_xG_y}.$$

Example 2: circles on a 2-d Bravais lattice

Consider a periodic function defined by non-overlapping circles arranged on a 2-D Bravais lattice. A function $f$ is defined such that it has a constant value $C$ inside the circles and is zero outside the circles. As long as the circles do not overlap, a primitive unit cell can be defined so that the circle at the origin lies entirely within the primitive unit cell. Since the function $f$ is zero outside circle, we just need to integrate over the circle. The Fourier coefficients $f_{\vec{G}}$ are given by,

$$f_{\vec{G}} = \frac{C}{\pi R^2}\int\limits_{\text{circle}}\exp (-i\vec{G}\cdot \vec{r})d^2r = \frac{1}{\pi R^2}\int\limits_{0}^R\int\limits_{-\pi}^{\pi}\left(\cos (|G|r\cos\theta)- i\sin(|G|r\cos\theta)\right)rd\theta dr$$

Here $R$ is the radius of the circle. Performing the integral over $\theta$.

$$f_{\vec{G}} = \frac{2C}{R^2}\int\limits_{0}^RrJ_0(|G|r)dr$$

Here $J_0$ is the zeroth order Bessel function. Integrating over $r$ yields,

$$f_{\vec{G}} = \frac{2 C J_1(|G|R)}{R|G|}$$

Here $J_1$ is the first order Bessel function. As long as the circles do not overlap, the Fourier series for circles repeated on any 2-D Bravais lattice is,

$$f(\vec{r})=\frac{2 C }{R}\sum \limits_{\vec{G}} \frac{J_1(|G|R)}{|G|}\cos \left(i\vec{G}\cdot\vec{r}\right).$$

The sine terms cancel out since $f(\vec{r})$ is an even function.

Example 3: parallelograms repeated on a 2-D Bravais lattice

Consider a periodic function defined by non-overapping parallelograms arranged on a 2-D Bravais lattice. A function f is defined such that it has a constant value C inside the parallelograms and is zero outside the parallelograms. As long as the parallelograms do not overlap, a primitive unit cell can be defined so that the parallelogram at the origin lies entirely within the primitive unit cell. Since the function f is zero outside parallelogram, we just need to integrate over the parallelogram. The Fourier coefficients fG are given by, Here A is the area of the primitive unit cell. b and h are shown in the drawing. Performing the integral over y, Integrating over x yields, As long as the parallelograms do not overlap, the Fourier series for parallelograms repeated on any 2-D Bravais lattice is, This includes the results for squares and rectangles as special cases (tan(0) = 0). e.g. rectangle: 