I am using Born Approximation to calculate this problem and I've already got the solutions to both of the problems. What confused me is the definition of the unpolarized differential cross section: $$\langle\frac{d\sigma}{d\Omega}\rangle=\frac{\int{\frac{d\sigma}{d\Omega}}d\Omega'}{\int{d\Omega'}}$$Where $\theta'$ is the angle between $\vec{q}=\vec{k}-\vec{k'} $ and $\vec{d}$. k, k' stand for the initial state and outgoing state respectively. How should I understand this definition? I have a feeling that we should integrate over all possible relative position of $\vec{q}$ and $\vec{d}$. But why this is the correct way? What I expect is that this answer fit with what people observe in the lab for unpolarized scattering. Can someone explain this definition from the perspective of experimental physics? enter image description here enter image description here


1 Answer 1


The idea behind the definition of the unpolarized differential scattering cross section is simply to average over all orientations of the $\mathbf{d}$ vector, assuming an isotropic orientational distribution function. One way to do this is to define polar coordinates with $\mathbf{q}$ lying along the $z$ axis, write $\mathbf{q}\cdot\mathbf{d}=qd\cos\theta$ and average $d\sigma/d\Omega$ over $\theta$ and $\phi$, exactly as you have written in your very first equation: $$ \left(\frac{d\sigma}{d\Omega}\right)_{\text{unpol}}= \frac{1}{4\pi} \int_0^{2\pi} d\phi \int_0^\pi d\theta \, \frac{d\sigma}{d\Omega} $$ Starting with the expression in part (a), this leads to answer (b), except that, as written, it contains a typographical error: $\sin 2qk$ should be $\sin 2qd$.

Experimentally, the reasoning will be that you can't actually specify the orientations of the diatomic molecules in the experiment under consideration. Perhaps the sample is disordered in some way. To a good approximation the molecules may be independent of each other, so you only need to consider the scattering from individual molecules, but the result needs to be averaged over this isotropic distribution, and will end up being independent of the direction of $\mathbf{q}$. (If the molecules were all aligned in the same direction in the laboratory frame of reference, then the scattering would be dependent on the angle of $\mathbf{q}$ relative to that direction).

I have to say that I find the term "unpolarized" a bit confusing, because it gives the impression that we are discussing something that is seen for an unpolarized beam of neutrons, rather than a polarized beam. The polarization in this case would relate to the neutron spin. I don't believe that this is the case, from the context given in your question. Polarized incident neutrons, or indeed an analysis of the polarization of the outgoing beam for some cases, even when the incoming beam is unpolarized, can be useful, in understanding magnetic solids, for instance. But, as I say, this doesn't seem to be what's being discussed here.

I won't provide any more details, because once you have understood the underlying idea, completing the derivation is a homework-like exercise.

  • $\begingroup$ Thank you. I have a difficulty in understanding the formula you give. It seems plausible but I try to understand it like this: from an fixed angle of scattering (angle between $\vec{k'} $ and $\vec{k}$), if there are lots of molecules randomly distributed(I only consider single scattering here), each scattering has its own contribution to the average differential cross-section. How should I proceed like this? $\endgroup$ Jan 8, 2019 at 12:16
  • $\begingroup$ if we assume the total number of molecules is N, I guess finally N cancels out? $\endgroup$ Jan 8, 2019 at 12:19
  • $\begingroup$ Derivations of this kind start from the differential scattering cross section from a single target nucleus. Then one can extend the calculation, to discuss a single diatomic molecule, as stated in this problem. If there are $N$ such molecules in the target region, one has to make some assumption: the simplest is that one can neglect correlations between them and treat them independently; then the differential scattering cross section will just be proportional to $N$. I believe those are the assumptions being made here. $\endgroup$
    – user197851
    Jan 8, 2019 at 12:32

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