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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.

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  • $\begingroup$ Intuitively, if the product $\vec q \vec r$ is small, the exponential term is basically constant, i.e. the integral is almost proportional to the volume integral of the potential and the integral is insensitive to the variations of the potential where it is large (small $r$). If you look at it in more detail, this is very similar to the optical resolution problem, which has a naive 19th century solution (Rayleigh criterion) and a better modern one which takes the signal to noise ratio in the image (in this case $f(\theta)$) into account. $\endgroup$ – CuriousOne Jan 26 '16 at 16:11
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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$^1$:

enter image description here

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

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  • $\begingroup$ So in order to produce a sharp image intensity function (corresponding to a high resolution image) we need to include the high frequency components in the fourier transform. Making this analogy with scattering, in order to produce a sharp differential cross section, we need to have sufficient high frequency components in the fourier transform (that is have a high $\mathbf{q}$). So I guess the question that remains is why does having a 'sharp' differential cross section allow you to better probe a small object? $\endgroup$ – Watw Jan 26 '16 at 16:47
  • $\begingroup$ I guess that the best way to look at this is to suppose we have low momentum - here we get an almost uniform scattering of the particles because the scattering cross section doesn't vary greatly. On the other hand with high momentum where the flux of scattered particles is varying significantly with position we can deduce much more about the structure of the scattering body (like Rutherford deduced the existence of a nucleus from the significantly varying flux with scattering angle). Is this correct? $\endgroup$ – Watw Jan 26 '16 at 16:49
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    $\begingroup$ @Watw yes, that's exactly what's going on. We need a high momentum to get a sensitive enough data from the experiments. If you think of the incoming particle as a plane wave (Born approximation), then $\mathrm e^{-i\boldsymbol k\cdot\boldsymbol r}$ represents the incoming wave. If you want to measure details in a lenght scale $R$, then you better have $kR\stackrel{>}{\small\sim} 1$, because otherwise $\mathrm e^{-i\boldsymbol k\cdot\boldsymbol r}$ will be nearly constant over the interesting values of $r\sim R$. Therefore, if $R$ is small, you need to use a high value of $k$. $\endgroup$ – AccidentalFourierTransform Jan 26 '16 at 20:58

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