Is scattering amplitude a real function?

Question

I am reading Zetilli's text on Quantum Mechanics. In Chapter 11 (Scattering Theory) the authors argue that a particle is scattered through a target has a wave function of the form

$$ \psi = \phi_{inc} + \phi_{sc} \approx e^{i\vec{k}_0\cdot\vec{r}} + f(\theta,\varphi) \frac{e^{i\vec{k}\cdot\vec{r}}}{r},$$

where $\vec{k}_0$ is a the wave vector of the incident particle, and $\vec{k}$ the wave vector of the scattered one. They also develop the probability density currents of the incoming and scattered particles, which both have the form

$$\vec{J} = \frac{i\hbar}{2\mu}(\phi\nabla\phi^* - \phi^*\nabla\phi).\tag{1}$$

Here $\mu$ is the mass of the particle (of a ficticious particle because the problem is written as a 2-body problem). Now, for the incoming (eq. 11.37) it can be correctly shown that

$$J_{inc} = \frac{\hbar}{\mu}k_0.$$

They also claim that $$ J_{sc} = \frac{\hbar k}{\mu r^2} |f(\theta,\varphi)|^2.\tag{2}$$

However, I don't see why this is true since equation (1) would mean that I have to take derivatives with respect to the angular coordinates. When I do this I obtain: $$\vec{J}_{sc} = \frac{i\hbar}{2\mu}\left[\frac{f}{r}\left\{ \nabla \left(\frac{f^*}{r}\right) -i\vec{k}\frac{f^*}{r}\right\} - \frac{f^*}{r}\left\{ \nabla \left(\frac{f}{r}\right) +i\vec{k}\frac{f}{r}\right\}\right]$$

$$= \frac{i\hbar}{2\mu}\left[\frac{f}{r}\nabla \left(\frac{f^*}{r}\right)-\frac{f^*}{r}\nabla \left(\frac{f}{r}\right)-2i\vec{k}\frac{|f|^2}{r^2}\right].$$

If $f$ is real, then then (2) would be true. If not, I don't see how the equality (2) holds. I'm grateful if any of you points out any mistake I'm making. Thanks.

Answer 1

The scattering amplitude $f(\theta, \varphi)$ is in general not a real-valued function. I also don't see anything wrong with the calculation you posted. But, given the nabla operator in spherical coordinates $$ \nabla = \hat{r} \partial_r + \hat{\theta} \frac{1}{r} \partial_\theta + \hat{\phi} \frac{1}{r\sin{\theta}} \partial_\varphi$$ you can see that all your extra terms will behave like $\sim r^{-3}$ (since $f$ does not depend on $r$). In scattering theory, one is typically interested in the behavior far away from the scattering center, which is why I'm pretty sure the text you got this from only considered the leading order in $r$ without mentioning it explicitly. A more correct way to write this would be $$J_\mathrm{sc}=\frac{\hbar k}{\mu r^2}|f(\theta, \varphi)|^2 + \mathcal{O}(r^{-3})$$