# Scattering Amplitude for Many Particles (Born Approximation)

1. Apr 1, 2013

### Rubiss

1. The problem statement, all variables and given/known data

Given that the scattering amplitude off of a single atom is $$f_{1}(\vec{q}),$$ find the scattering amplitude for 1) four atoms each placed in the corner of a square of length a, and 2) two atoms a distance d apart

2. Relevant equations

The total scattering amplitude can be written as

$$f_{total}(\vec{q})=f_{1}(\vec{q})\sum_{i=1}^{n}e^{-i(\vec{q}\cdot \vec{r}_{i})}$$

where $$\vec{r}_{i}$$ is a vector that points from the origin to a particle.

3. The attempt at a solution

For the square of side a, I place the origin at the center and assume the particles are in the xy plane. That means particle 1 (in quadrant 1) would be at the location (a/2,a/2), particle 2 would be at (-a/2,a/2), particle 3 would be at (-a/2,-a/2), and particle 4 would be at (a/2,-a/2). The sum is then

$$\sum_{i=1}^{4}e^{-i(\vec{q}\cdot \vec{r}_{i})} = e^{-i(q_{x}(\frac{a}{2})+q_{y}(\frac{a}{2}))}+e^{-i(q_{x}(\frac{-a}{2})+q_{y}(\frac{a}{2}))}+e^{-i(q_{x}(\frac{-a}{2})+q_{y}(\frac{-a}{2}))}+e^{-i(q_{x}(\frac{a}{2})+q_{y}(\frac{-a}{2}))} =4\cos(\frac{a}{2}q_{x})\cos(\frac{a}{2}q_{y})$$

The problem is that according to the solution, there should be √2 in the denominator instead of 2. Can anyone see where I am going wrong?

I also know that for a cube of length a with a particle at each corner, the scattering amplitude is

$$f_{total}(\vec{q})=8\cos(\frac{a}{2}q_{x})\cos(\frac{a}{2}q_{y}) \cos(\frac{a}{2}q_{z})$$

where now 2 in the denominator is correct. So, if there is any pattern, I would expect the scattering amplitude for the the 2 atoms a distance d apart (on the x axis) to look something like

$$f_{total}(\vec{q})=2\cos(\frac{d}{2^{\frac{1}{4}}}q_{x})$$

If this is correct, I can't get this to fall out of the math. Can anyone help?

2. Apr 2, 2013

### jonnynuke

The 1/√2 comes from the dot product. Remember that a$\bullet$r = |a|*|r|*Cos(θ).

3. Apr 2, 2013

### Rubiss

I'm well aware of that. I calculated the dot product above in component form: I resolved q into x and y components and the vector r into its components for eacf particle. Particle 1 had a x component of a/2 and a y component of a/2, particle 2 etc. Calculating the dot product your way does not allow one to resolve q into its components, as it is in the supposed answer.

4. Apr 3, 2013

### jonnynuke

You're right, and actually I believe the answer should have a factor of 2, not square root of 2. Solving the two atom problem gives me a factor of 2 in the denominator and not square or quartic root of 2. Perhaps the solutions are incorrect?