The interference function

In the theory section, we derived the following formula for the interference function:

$$ S(\mathbf q) = 1 + \frac{1}{N}\sum_{i\neq j} \int dP_{i,j}(\mathbf{R_1}, \mathbf{R_2})\, e^{i\mathbf q\cdot (\mathbf{R_1}-\mathbf{R_2})}$$

If we further assume that the probability density function depends only on the difference $\mathbf{R_1}-\mathbf{R_2}$ and we take $\rho_S = N/S$ to be the mean surface density of the particles, we can write this as:

\begin{align} S(\mathbf q) &= 1 + \frac{1}{N}\sum_{i\neq j} \int dP_{i,j}(\mathbf{R_1} - \mathbf{R_2})d^3\mathbf{R_1}\, \rho_S e^{i\mathbf q\cdot (\mathbf{R_1}-\mathbf{R_2})} \\ &= 1 + \int dP(\mathbf R)\, e^{i\mathbf q\cdot \mathbf R} \end{align}

where $P(\mathbf R)$ denotes the probability density of finding any particle at a distance $\mathbf R$ from a given particle.

If we write: $$dP(\mathbf R) = \rho_S g(\mathbf R) d^3\mathbf R$$

the interference function becomes: $$S(\mathbf q) = 1 + \rho_S \int d^3\mathbf R\, g(\mathbf R)\, e^{i\mathbf q\cdot \mathbf R}$$

Infinite lattices

In a infinite one-dimensional lattice with lattice constant $a$, we have

\begin{align} S(\mathbf q) &= 1 + \int dP(\mathbf R)\, e^{i\mathbf q\cdot \mathbf R} \\ &=\frac{2\pi}{a} \sum_{n=-\infty}^\infty \delta(q - na_*) \end{align}

with $a_* = 2\pi/a$ the reciprocal lattice constant.

Finite lattices

For a finite lattice with $N$ points, we have: \begin{align} S(\mathbf q) &= \frac{1}{N} \left| \sum_{n=0}^{N-1} e^{iqna} \right|^2 \\ &=\frac{1}{N} \frac{\sin^2(Nqa/2)}{\sin^2(qa/2)} \end{align}

Infinite lattice with decay function

If we take $$dP(R) + \delta(R) = \sum_{n=-\infty}^{\infty} f(na) \delta(R-na)$$ with $f(x)$ a decay function that encodes the loss of coherence along long distances, the interference function is:

\begin{align} S(q) &= F(q) \otimes \frac{1}{a}\sum_{n=-\infty}^\infty \delta(q - na_*) \\ &= \frac{1}{a}\sum_{n=-\infty}^\infty F(q-na_*) \end{align}

where $\otimes$ denotes the convolution operator.

We see that the interference function consists of a smooth function $F(q)$ at each reciprocal lattice point.

The paracrystal interference function

Suppose we have a particle at the origin and that the probability density for its first neighbor to the right is $p(x)$ with:

\begin{align} \int dx\, p(x) &= 1\\ \int dx\, x p(x) &= D \end{align}

where $D$ is the mean distance from the origin.

If every particle's position is determined by the previous position and the same probability density, the total probability density for all particles to the right of the origin becomes:

$$S^+(x) = \sum_{n=1}^\infty \otimes^n p(x)$$

The total density is then:

$$S_{total} = \delta(x) + S^+(x) + S^+(-x)$$

If the Fourier transform of $p(x)$ is $\mathcal P(q)$, the Fourier transform of $S^+(x)$ is:

\begin{align} \mathcal F\left[ S^+(x) \right] &= \sum_{n=1}^\infty \mathcal P^n(q) \\ &= \frac{\mathcal P(q)}{1-\mathcal P(q)} \end{align}

Finally, the total interference function is:

$$\mathcal F\left[ S_{total}(x) \right] = \mathcal{Re}\left\lbrace \frac{1+\mathcal P(q)}{1-\mathcal P(q)}\right\rbrace$$

Two dimensional lattices

All the previously mentioned one dimensional lattices have a two dimensional counterpart in BornAgain. These are all implemented as a convolution of two non-parallel one dimensional lattices. In reciprocal space, this amounts to a product of two one dimensional interference functions.