Scattering Theory In non-relativistic quantum mechanical scattering theory you can derive an expression for the differential scattering cross section under the first order Born approximation as $$\frac{d\sigma}{d\Omega}=|f(\theta)|^2$$ where $$f(\theta)=-\frac{m}{2 \pi {\hbar}^2}\int_{all space}e^{i \mathbf{q} \cdot \mathbf{r}}V(\mathbf{r})d^3\mathbf{r}$$ where $\mathbf{q}=\mathbf{k}-\mathbf{k}'$ is the difference between the incoming and detected wavevector and $V(\mathbf{r})$ is the potential under consideration. This expression is simply the fourier transform of the potential with respect to the variable $\mathbf{q}$.
My notes then state that this implies that in order to probe a small object you need a high $\mathbf{p}=\hbar\mathbf{k}$. Does anybody see how this follows from the above results? Thankyou.
 A: As what enters into the formula is $\boldsymbol q$ instead of $\boldsymbol k$, I'd say we need a high $\boldsymbol q$ (which, of course, implies a high $\boldsymbol k$, because of conservation of energy/momentum). For example, if $\boldsymbol k$ is very high, but $\boldsymbol q$ is not, this means that there was barely no scattering, which means you didn't actually measure anything. This means that what you actually need is a high $\boldsymbol q$.
Now, why would we need a high $\boldsymbol q$ in order to measure small objects? well, the answer is fairly simple: because of the properties of the Fourier Transform.
It is well known that the low frequencies (read, low $\boldsymbol q$) of the Fourier Transform encode the coarse properties of an image, and the high frequencies encode the details$^1$:

In the end, it all boils down to the uncertainty principle $\Delta x\Delta k\ge 1$, which is actually a property of the Fourier Transform!

$^1$ see, for example, http://www.robots.ox.ac.uk/~az/lectures/ia/lect2.pdf
