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Short version

After integrating over all possible outgoing angles, the total cross-section of coherent elastic scattering from a fixed target of characteristic length $L$ scales like $L^4$. Does this mean, given a beam of sufficiently small angular dispersion and detectors capable of sufficiently fine angular resolution, there are coherent effects over macroscopic distances arbitrarily longer than the wavelength of the beam?

Long Version

Consider a target of $N$ identical atoms located at positions $\vec{x}_n$ (where $n = 1,\ldots, N$) which is bombarded by a flux of incoming particles, which we'll call neutrinos. Set $\hbar = 1$ and let $\vec{k} = 2\pi/\lambda$ be the momentum of incoming neutrino, where $\lambda$ is the de Broglie wavelength. Let $f(\theta)$ be the scattering amplitude for a neutrino on a single free atom so the differential cross section is \begin{align} \frac{d \sigma_0}{d\Omega} = \vert f(\theta) \vert^2 \end{align} for one atom. As is befitting neutrinos, assume that $f(\theta)$ is very tiny so that the Born approximation is applicable. In particular, we can ignore multiple scattering events.

For many scattering processes the cross section of the target $\sigma_\mathrm{T}$is just $N$ times the cross section of each atom: $N \sigma_0$. However, for elastic scattering of extremely low energy neutrinos from the relatively heavy nuclei in atoms, the very long wavelength of the neutrinos means the various nuclei contribute coherently to the cross section. When $\lambda \gg L$, where $L$ is the size of the target, one has $\sigma_\mathrm{T} = N^2 \sigma_0$ rather than $N \sigma_0$ [1]. For general $k$, the total cross section is [2] \begin{align} \frac{d \sigma_\mathrm{T}}{d\Omega} = \left\vert \sum_{n=1}^N f(\theta) \, e^{-i \vec{x}_n \cdot (\vec{k}-\vec{l})} \right\vert^2 \end{align} where $\vec{l}$ is the momentum of the outgoing neutrino [3] and $\theta$ is the angle between $\vec{k}$ and $\vec{l}$. Because it's elastic, $|\vec{l}|=|\vec{k}|=k$.

A property of equation (1) is that for $L \gg \lambda$ there is a large suppression of scattering in most directions because the phase in the exponential tends to cancel for the different atoms in the sum. The exception is when $\vec{l}$ is very close to $\vec{k}$ (i.e. low momentum transfer, very slight scattering), because then the phase of the exponent varies very slowly from atom to atom. This means that for large targets, the vast majority of the scattering is in the forward direction.

Now restrict to the case $A \lesssim \lambda \lesssim L$, where $A$ is the typical atomic spacing. What's initially confusing about this is that if we ask for the total cross section by integrating over $\hat{l}$, we find for large $L$ that [4] \begin{align} \sigma_\mathrm{T} = \int_\Omega d \hat{l} \frac{d \sigma_\mathrm{T}}{d\Omega} \sim \frac{N^2}{L^2 k^2} = \rho^{2/3} \lambda^2 N^{4/3} \end{align} where $\rho = N/L^3$ is the number density of the atoms. This means that, for fixed density and fixed neutrino momentum, the total cross section grows faster that than the number of atoms in the target---even over distance scales much larger than the neutrino wavelength. In this sense the coherent scattering effects is non-local over large distances.

The story I have been told is that this is resolved by incorporating the realities of a real-world detector. For any traditional experiment, there is always a minimum forward acceptance angle $\theta_0$ that it can detect. Particles which are scattered at smaller angles are indistinguishable from unscattered particles in the beam. Indeed, if we let $\tilde{\sigma}_\mathrm{T}$ be the detectable cross section scattered at angles greater than $\theta_0$ for any fixed $\theta_0>0$, we find \begin{align} \tilde{\sigma}_\mathrm{T} = \int_{\theta > \theta_0} d \hat{l} \frac{d \sigma_\mathrm{T}}{d\Omega} \sim \frac{N^2}{L^4 k^4} = \rho^{4/3} \lambda^4 N^{2/3}. \end{align} This accords with our intuition. Growing like $N^{2/3}$ is the same as growing like $L^2$, i.e. there is complete cancellation in the bulk of the target, and the only significant scattering is from the surface (which scales like $L^2$).

Is this all there is to say? Can we potentially see scattering of particles (with atomic-scale wavelength) which demonstrates coherent contributions from target atoms separated by meters? Are there any other limiting factors besides a finite mean free path of the incoming particle (which breaks the Born approximation) and the angular resolution of the detector?

Form of answer

For positive answer, I would accept (a) anything that points to a reputable source (textbook or journal article) which explicitly discusses the possibility of coherent effects over arbitrarily large distances or (b) an argument which significantly improves on the one I have made above. For negative answers, any conclusive argument would suffice.

Bonus

It has been argued to me that the Born approximation is invalid for small momentum transfer $\vec{k}-\vec{l}$, because the approximation requires the energy associated with this transfer to be much larger than the potential (which cannot be the case for $\vec{k}-\vec{l}=0$). This seems to explicitly conflict with textbooks on the Born approximation which state things like

For any potential $V$ there is a $\bar{\lambda} > 0$ such that the Born series (9.5) converges to the correct on-shell $T$ matrix for all values $\vec{p}$ and $\vec{p}'$, provided $|\lambda| < \bar{\lambda}$.

["Scattering Theory: The Quantum Theory of Nonrelativistic Collisions" (1972) by John R. Taylor]

(Here, $\lambda$ is the coefficient of the expansion.)

Is there any validity to this objection?


[1] For macroscopic $N$, this can be a stupendous boost. This was responsible for some (misplaced) optimism in the '70s and '80s that relic neutrinos might be detectable.

[2] This form also appears in many less exotic places than neutrino physics, like neutron and X-ray scattering.

[3] $d\Omega$ is the differential over the possible directions $\hat{l}$.

[4] This behavior is independent of the details of the geometry. The $1/3$'s come from the integrals over 3 spatial dimensions.

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I don't want to write a really answer as I'm kinda shaky on this stuff, but I believe that the matter effect in electron neutrino mixing is a very similar effect (though in a different space). –  dmckee Apr 22 '13 at 19:10
    
@dmckee, Could you recommend an accessible reference which has a point of closest approach? –  Jess Riedel Apr 23 '13 at 2:21
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I would expect that the answer would be yes except the amplitude would fall to be undetectable, mainly based on the double slit experiments which show macroscopic coherence of individual particles. –  anna v Jun 17 '13 at 4:15

1 Answer 1

I am not sure that I get the full point of your question, but I'll answer anyway: the limiting factor is the coherence length of the incoming radiation. In the visible range this is easily obtained with lasers, and the resulting coherent interference patterns are known as holograms.

In the X-ray regime, intrinsically coherent sources have only recently become available (free-electron lasers), but with synchrotron facilities one can already obtain transverse coherence lengths of several microns (search for "coherent x-ray imaging").

If this coherent length is very large (e.g. for slow neutrinos [1]), one is still limited by the domain size in the target: your exposition assumes that the crystal is perfect, but the phases scattered from two domains not in registry with one another are shifted by an arbitrary phase and we are back to incoherent superposition.

The detector should also be able to detect one particle at a time, otherwise we are back to paragraph one and the transverse coherence length of the incoming beam as a whole.

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