# Why does a laser beam stay collimated?

I am looking for a simple way of explaining the collimation of a laser beam. The typical discussion of the two slit experiment of quantum theory relies heavily on the Huygens principle. Its application to a laser beam would tend to predict spreading. From a purely electromagnetic field point of view, how can one visualize what happens at the edge of the beam?

• There is another thread relating to this question - physics.stackexchange.com/q/79417 Commented Feb 2, 2016 at 15:38
• And it is spreading, just not as much as an incoherent source. Commented Feb 2, 2016 at 15:51
• And asking about the 'edges' of an electromagnetic field is not a very useful thing to do - it is all the solution to a wavefunction. Commented Feb 2, 2016 at 15:51
• But if you are interested in trying to figure out what is physically going on its a good place to start. Commented Feb 2, 2016 at 22:25

Just as Jon Custer wrote in his comment, even a perfectly collimated laser beam with a planar wavefront will diverge. The way it happens is determined by the Huygens principle, and depends on the beam profile:

When the light intensity is abruptly cut by a sharp flat obstacle, the light will indeed diffract in almost all angles. A razor blade cutting a laser beam, even though viewed from the geometric shadow, will still have a bright rim. (In wave optics, no absolute shadows exist!)

When the intensity of the beam is modulated smoothly, it will diffract in small angles only. Intuitively it can be imagined that at the edge of a beam passing through a "soft aperture", the Huygens elementary source with slightly higher amplitude than its neighbour forces the resulting wavefront to be bent away by a minute angle, outwards from the beam axis. It is still all about the superposition of spherical waves.

Edit: To illustrate this, I employed my open-source simulation scripts, using the excellent MEEP Maxwell-equations solver, to a classical edge-diffraction experiment. In the following three animations, I computed for you a wave diffraction on a sharp edge (left), and on "soft apertures" with characteristic transition width of 50 % and 100 % of the full image width.

• Could you add a sketch or some images, please. Commented Feb 2, 2016 at 17:59
• @HolgerFiedler for some images and animations see e.g. wikipedia page on diffraction Commented Feb 2, 2016 at 19:24
• Consider the "perfectly collimated" beam in your first paragraph. I am trying to visualize what happens at the edges. Or is that a mere mathematical abstraction of a beam whose edges smoothly go to zero as one proceeds further from the center of the beam?
– user59591
Commented Feb 2, 2016 at 19:49
• @Heaviside Maybe this is not the most elegant physical approach, but you may envisage that, in compliance with Huygens principle, the center of the beam radiates into the surrounding area at circumference of the beam. The wave radiating diagonally is however retarded in time. Since the center of the beam is more intense, it partially imparts this phase retardation also to the resulting superposed wave. Its wavefront is then curved and the beam diffracts. -- This is the best way how I can envisage it without starting talking about spatial frequency and Fourier transform. Commented Feb 2, 2016 at 20:03
• Only one additional issue: In discussions of relativity, shouldn't one restrict oneself to spherical wavefronts in derivations? It seems to me that the usual practice of assigning a single velocity (albeit of speed $c$) to a light "beam" is incorrect. A real "beam" consists of many photons going in slightly different directions.
– user59591
Commented Feb 3, 2016 at 3:30

A properly collimated laser beam is called a Gaussian Beam whose transverse magnetic and electric field amplitude profiles are given by the Gaussian function.

The Gaussian beam is a transverse electromagnetic (TEM) mode. The mathematical expression for the electric field amplitude is a solution to the paraxial Helmholtz equation:

The width of such laser beam is described by:

The relationship between beam width and divergence is a fundamental characteristic of diffraction, and of the Fourier transform which describes Fraunhofer diffraction.

Here is how a Gaussian beam1 looks:

1. Excerpt from the following article: Quantitative and Qualitative Study of Gaussian Beam Visualization Techniques

• Does a Gaussian beam have the least angular spread, in some sense? Commented Feb 2, 2016 at 20:19
• @DanielSank Yes, such a beam can be focused into the "smallest" spot. Commented Feb 2, 2016 at 20:22
• Any chance I can get you to define "smallest spot"? :) Commented Feb 2, 2016 at 20:30
• @dominecf surely this is related to the fact that the Gaussian minimizes the product of the variance of the function and the variance of its Fourier transform? Commented Feb 3, 2016 at 18:12
• @DanielSank Exactly as you write. It is also tightly connected with the concept of "minimum wavepacket" in quantum mechanics. In fact, focusing a beam onto the focal plane of a lens is equivalent to Fourier-transforming the beam profile. More generally, at any point between an ideal lens and the focal plane, the beam is determined by the fractional Fourier transform of its original profile. Commented Feb 3, 2016 at 19:13