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I have been trying to get my hands around a scattering problem all day but I can't wrap my head around the idea. It's a scattering problem with First Born Approximation and the potential is a Dirac delta sitting at the origin. The standard procedure would be to just use the formula from Griffith's book

$$ f(\theta) = \frac{-2m}{\hbar^2 \kappa}\int_0^\infty a\delta(r)r\sin (\kappa r)dr $$

where $\kappa = 2k \sin({\theta/2})$. The first problem is of course the fact that Dirac delta function lacks support at $r=0$ so I have been trying to switch to cartesian coordinates where it has support at the origin. But I keep getting zero as result, so either the my answer is right and that makes the problem really boring to start with, or either I'm missing something.

I have also been trying to understand the physics around the problem. If the Dirac delta was evaluated at some radius a the problem would have made more sense since that would have been the scattering from a hard sphere, but I can't understand what a scatter from a single infinitely small point would mean?

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    $\begingroup$ 1. Recall that a Dirac delta at the origin is $\frac{\delta(r)}{4\pi r^2}$ and 2. Did you try to do the calculation with a Dirac delta at some finite radius? did the result make sense? what if you take this radius to zero? $\endgroup$ – AccidentalFourierTransform Mar 3 '17 at 22:03
  • $\begingroup$ Yeah I did not forgot the scaling factors in my calculations. I first wrote it as $4 \pi r^2 \delta(\mathbf{r}) = \delta(r)$ and then changed it to cartesian coords, but that just added another zero since $\int \delta(x)\delta(y)\delta(z)(x^2+y^2+z^2) dV = 0$. If I solve this for a finit radius, say $R$, I get (if my calculations are right) $\frac{-2\pi a \sin(2k\sin(\theta / 2)R)}{\hbar^2 2 k \sin(\theta / 2)}$. Not sure what to do with this to be honest. $\endgroup$ – Claessie Mar 3 '17 at 22:16
  • $\begingroup$ If i let R go to zero in the expression above the answer just just tend to zero. $\endgroup$ – Claessie Mar 3 '17 at 22:22
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The general formula for the first-order approximation of $f(\kappa)$ is $$ f(\kappa)=-\frac{m}{2\pi}\int\mathrm d\boldsymbol r\ \mathrm e^{-i\boldsymbol\kappa\cdot\boldsymbol r} V(\boldsymbol r)\tag{A} $$

In the particular case $$ V(\boldsymbol r)=a\delta(\boldsymbol r) $$ we have $$ f(\kappa)\overset{(\mathrm A)}=-\frac{ma}{2\pi} $$

If the potential is spherically symmetric, then $(\mathrm A)$ becomes $$ f(\kappa)=-\frac{2m}{\kappa}\int_0^\infty \mathrm dr\ rV(r)\sin (\kappa r)\tag{B} $$

Recall that the Dirac delta centred at the origin, in spherical coordinates, reads $$ V(r)=\frac{a}{4\pi r^2}\delta(r) $$ and therefore $$ f(\kappa)\overset{(\mathrm B)}=-\frac{ma}{2\pi}\int_0^\infty \mathrm dr\ \delta(r)\frac{\sin (\kappa r)}{\kappa r} $$ which agrees with the previous result, if we take $$ \left.\frac{\sin x}{x}\right|_{x\to0}=1 $$

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  • $\begingroup$ Oh I see! So it was the problem that was stated wrong from the beginning. It was actually two younger students at my school trying to solve this problem in the library and they had some expressions on the board with the integral $f(\theta) = \int_0^\infty dr r V(r) \sin(\kappa r)$ and then under it $V(r) = a \delta(r)$. I just figure that it was the problem they were trying to solve but they must have missed the scaling factor you pointed out here, so the problem was already doomed when I tried to solve it for myself. $\endgroup$ – Claessie Mar 3 '17 at 22:56
  • $\begingroup$ There is one thing I still don't understand though. Are you really allowed to use the Dirac delta at r = 0, like you did here on the end? Isn't that something that is not allowed in general, since the Dirac delta is a distribution and not a function, I'm thinking about the lack of compact support and all mathematics that comes with it? $\endgroup$ – Claessie Mar 3 '17 at 23:05
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    $\begingroup$ @Belgianhorse that depends. Mathematicians will complain, but as long as the result is consistent, we can say that we are allowed to do whatever we want to. The Dirac delta is not a function, so all our manipulations were formal (ie., unjustified but consistent). If you want to do things properly (and I encourage you to do it), you can change the Delta for a well-behaved function, e.g. a Gaussian, such that $\delta_\epsilon(r)\to \delta(r)$ as $\epsilon\to 0$. Do this and check for yourself that you arrive at the same result, thus justifying our formal manipulations. $\endgroup$ – AccidentalFourierTransform Mar 3 '17 at 23:21
  • $\begingroup$ Ok, I hear you. Thanks a lot for the explanations! $\endgroup$ – Claessie Mar 3 '17 at 23:25

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