The Gentle Hum of Polarization
In 1940 Landau published a one-page paper "On the polarisation of electrons by scattering" resolving a seeming contradiction between experiment and theory. The goal of this article is to learn how to solve this dilemma on your own by adopting the proper thinking methods. But the way we will approach this is through a story.
Let’s take a novel approach in *approaching* this paper. Imagine you are a working physicist in the 1940s and a client comes in and hands you a purported dilemma. He tells you, befuddled, that his experiments scatter electrons off a gold foil but no polarization is visible, hence, something must be wrong with the theory (Mott formula). Now, of course, your subconscious immediately screams “no, there must be something wrong in the experiment, not theory,” since you know that the likelihood of an experiment ruling out an established theory is low, a more frequent event is the other way around. This does not mean you are correct in this instance, but you already are somewhat biased in favor of what the actual problem might be.
You sit back in your chair and ruminate a bit, part of your automatized knowledge contains solving the cross section for the Yukawa potential, and also determining the polarisation after scattering when initial polarisation is non zero. You know that the Mott experiment is dependent on Rutherford’s formula, it suffices to know that the cross section is proportional to (sin (theta/2))^-4. But the details are hazy so you need to refresh your mind.
The basics of scattering
The fundamental relation that everything builds on is the cross section \( d\sigma = \lvert f(\theta) \rvert^2 \, d\Omega \), where \(f\) is the scattering amplitude, and currently just a number.
We determine this quantity with the Born approximation \( f(\theta) = -\frac{m}{2\pi \hbar^2} \int U(\mathbf{r}) e^{i \mathbf{q} \cdot \mathbf{r}} \, dV \) where q is the momentum transfer with a magnitude of \( q = 2k \sin\left(\frac{\theta}{2}\right) \).
Now unfortunately this integral diverges, and we have to restrict it. One way we do this is by imposing a Yukawa potential, which is another word for “screened potential” if you agree with me that at large distances the nucleus will be screened by the electron cloud.
It looks like:
and the FT of this \( \int U(\mathbf{r}) e^{i \mathbf{q} \cdot \mathbf{r}} \, dV = \frac{4\pi Z e^2}{q^2 + a^{-2}} \) so
The expression for the diff cross section now becomes
Ah, finally you arrive at the most basic result. Ok. You quickly notice that the angle matters greatly. If \( \theta \) is small then it is prop to \( 1/\theta^4 \), and it blows up. So infinite scattering at zero angle, well not quite, but physically this means tiny deflection since large angles are rare. Now this alone should be something to flag.
Scattering amplitude as a spin operator
Since this is a Mott experiment and you know Mott wanted to introduce spin, we need to make some changes. When the electron passes through the nucleus, since it is a spin ½ particle it will couple to a magnetic field (feeling this through its rest frame). Now because of this, \( f \) really becomes a 2x2 matrix, a spin operator, not a scalar. This changes how we compute the cross section. We have to take the trace.
The polarization of the beam is
where \( \nu = \mathbf{n} \times \mathbf{n}' \) and A and B are complex functions of the angle. Importantly the polarization is perpendicular to the scattering plane, its magnitude depends on the real part of AB*. You might know that B is coming from the spin-orbit interaction, which goes of \( 1/r^3 \), ie, close to the nucleus means a large scattering angle, or rather polarization requires a large angle.
At this point it would be reasonable to ask yourself, what are the angles the electrons are deflected at in the experiment? Because the client told you that the electrons do end up having a measurable angle. You get the inkling that they may be scattering many times such that deflections get accumulated. But can you prove this somehow?
Deriving the probability of deflection
Typically in nuclear physics the story goes … let’s consider a thin slab of the material (Au in this case), with thickness \( dz \). The prob of deflection into \( d\Omega \) at angle \( \theta \) is \( N \cdot dz \cdot d\sigma(\theta) \). However, quickly realise that if we take the expectation value of \( \theta \) we should get zero since the direction is randomized. But if i look at the expectation value of squared \( \theta \) i won’t have to deal with the direction problem or averaging to zero issue.
Integrate over the gold foil \( \langle \theta^2 \rangle = N \ell \int \theta^2 \, d\sigma \), and with Rutherford \( d\Omega = 2\pi \sin\theta \, d\theta \). one gets \( \langle \theta^2 \rangle \)
We have a problem with divergence here so let’s deal with it. At large angles we have a cutoff \( \langle \theta^2 \rangle^{1/2} \)
Hence,
Your instinct was indeed correct. The experiment wasn’t doing large angle scattering, Mott’s formula is correct .......
I wish to point out that the negative result of experiments seems to follow from a much simpler reason, namely from the fact that the scattered electrons observed have got their deflection not in a single act of scattering but as a result of multiple scattering.
Concretization
You call up your client and ask him what the size of the thinnest foil was in the experiment, he tells you \(\ell = 7 \times 10^{-6}\,\text{cm}\). \( \text{Gold, } E = 100\,\text{keV}, \quad \ell = 7 \times 10^{-6}\,\text{cm}. \quad \text{We get:} \)