Scattering Amplitude for Many Particles (Born Approximation)

[Rubiss]

Homework Statement

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

Homework 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.

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?

 

[jonnynuke]

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

 

[Rubiss]

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

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.

 

[jonnynuke]

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.

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?

 

[paranormal]

Your calculations seem to be correct. However, the discrepancy in the denominator may be due to the fact that the particles are not evenly spaced in the square configuration. In the cube configuration, the particles are evenly spaced in all three dimensions, hence the factor of 2 in the denominator.

For the two atoms a distance d apart, your proposed scattering amplitude formula seems to be correct. However, it may be more accurate to write it as:

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

This takes into account the fact that the atoms are separated in all three dimensions, not just the x-axis.

It is also worth noting that the Born approximation is only valid for small scattering angles, and may not accurately describe the scattering for larger angles or when the particles are in close proximity. Further analysis and calculations may be necessary to accurately describe the scattering for these cases.