Diffraction limited divergence angle I was reading an article and came across the following paragraph:
"A thin lens of focal length f is used to collimate the light emerging from an optical fibre. The fibre has small core diameter and numerical aperture NA. It achieves a diffraction limited divergence angle θ for the collimated beam."
I am wondering how can there be a "diffraction limited divergence angle θ"? If the lens is placed at exactly focal length f away from the end of the optical fibre, shouldn't all the light be properly collimated with θ=0?
 A: Divergence $\theta=0$ is not possible -- from a quantum mechanical perspective, it would violate the Heisenberg uncertainty principle for position and momentum of a photon. "Collimated" is not really a technical term; it is a qualitative description of a beam with a relatively small divergence angle.
The fact that the divergence is diffraction-limited means that the lens does not cause aberrations, so that the product of the beam width and beam divergence is as small as possible. This can be true regardless of how big the beam divergence is, and instead has to do with the beam quality. A diffraction-limited beam (e.g., a Gaussian) is one for which this product is equal to the smallest value allowed by the uncertainty principle.
A: In classical electromagnetism the uncertainty principle does not apply as it represents the limit of $h \rightarrow 0$.
For cases where the paraxial approximation applies the minimum beam waist is $2 w_0=\frac{4 \lambda f}{\pi d}$ for focal length $f$ and collimating lens diameter $d$.
http://www.ophiropt.com/laser-measurement-instruments/beam-profilers/knowledge-center/tutorial/what-is-m2
A: You can actually show that no collimated beam exists using ray optics and some radiometry. 
Perfect collimation would require that the chief ray angle be zero and the marginal ray height be nonzero angle be zero. If your chief ray angle is zero, that means your beam came from a source of zero size. 
From radiometry we know that would require a source of infinite radiant exitance. Outside that exception, there would be no energy in the perfectly collimated beam.
As others have mentioned, even if that source existed, you would then be defeated by wave optics which would disallow a perfectly collimated beam anyway. 
Practically speaking, however, you can generally pick beam diameters sufficiently large to produce "zero" divergence as far as your application is concerned. 
Also, a diffraction limited divergence is in contrast to a worse performing beam. If you have a non-Gaussian beam (i.e. M^2 > 1) the beam will diverge more. 
EDIT:
It's actually more simple than that. I originally used the wrong definition of what a perfectly collimated beam would be. If you wanted a completely collimated beam (all rays traveling parallel to the optical axis), that would mean your marginal and chief ray angles would be zero, which would make the value of your Lagrange invariant zero, which means you'd have zero light.
