I was wondering why a laser beam diverges. If all the photons are in the same direction, I would imagine that it would stay that way over a long distance. I am aware that a perfectly collimated beam with no divergence cannot be created due to diffraction, but I am looking for an explanation based on photons rather than wave physics.
[Physics] Why does a laser beam diverge
laseroptics
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There is a limit to how small you can focus an ideal single-mode laser beam. The product of the divergence half-angle $\Theta$ and the radius $w_0$ of the beam at its waist (narrowest point) is constant for any given beam. (This quantity is called the beam parameter product, and is related to the $M^2$ beam quality measure you may have heard of.) For an ideal Gaussian ("diffraction-limited") beam, it is:
$$\Theta w_0 = \lambda/\pi$$
So, to answer what I interpret as your main question:
Let's say that I have a laser beam of some given power that starts with some diameter $D_0$ at the point of emission and increases to $D_f$ at some distance $r$ away. Would this be sufficient information to imply a limit to the power per unit area (W/m^2) that could be obtained through focusing and what would that be?
The answer is no.
The parameters you have given are sufficient for calculating $\Theta$, but only if $r$ is large enough so that the points at which you measure the diameter are in each other's far field.
You would also need to know the beam radius at the waist, so you could calculate the beam parameter product. Then, to get the minimum spot size, you would need to refocus the beam so that it is maximally convergent. The absolute limit is the fictitious divergence half-angle of $\pi/2$, or 90 degrees, although in practice the theory breaks down for half-angles of more than 30 degrees (this number is from Wikipedia) since the paraxial approximation stops being valid. For an ideal beam at this impossible opening half-angle, this gives you a minimum waist radius of $2\lambda/\pi^2$. So yes, it does depend on the wavelength.
What lens characteristics and approaches would someone look for in order to do this with a laser pointer?
You need a lens with a very short focal length. This gives you the largest convergence. Note that the more convergent the beam, and the smaller the waist size, the smaller the Rayleigh range is. That is, the beam radius will get very small, but it won't stay very small, it'll get bigger very quickly as you move away from the focus. (The Rayleigh range is the distance over which the beam radius increases by $\sqrt{2}$.
In addition, thinking of a Gaussian beam as being "straight" is not quite correct. There is always a waist, always a Rayleigh range less than infinity, and always a nonzero divergence angle.
EDIT
Also, it is important to realize that there is no difference between an unfocused and a focused Gaussian beam. Refocusing a Gaussian beam with a lens just moves and resizes the waist.
The aperture size of the laser is not the same as the waist size. If the beam is more or less collimated, then the aperture will still be larger, because the waist radius is usually defined in terms of the radius at which the intensity drops to $1/e^2$ of its peak value. If the beam is cut off by an aperture at that radius, then even if it were close to diffraction-limited, it certainly wouldn't be anymore. So, apertures are always larger.
The waist is the thinnest point of the beam. Usually this point is inside the laser cavity, or outside the laser if there are focusing optics involved, which there often are. So still, the answer to your question is no. You are not missing the definition of $\lambda$; rather, you are comparing your minimum waist radius to the value of $2\lambda/\pi^2$ that I said was "impossible". I called it impossible, because to make a beam converging that strongly, you would need a lens with a focal length of zero!
Let's try a more realistic example with some numbers. Take your red laser pointer with $\lambda$ = 671 nm. Laser pointer beams are often crappy, but not so crappy as you might think, if they are single-mode. Let's assume that this particular laser pointer has an $M^2$ ("beam quality parameter", which is the beam parameter product divided by the ideal beam parameter product of $\lambda/\pi$) of 1.5. A quick Google search didn't give me typical $M^2$s of red laser pointers, but this doesn't seem to me to be too much off the mark.
Note that if you know the $M^2$ and measure the divergence of a beam, then you can calculate the waist radius. We are going to do that now. Suppose the laser pointer beam is nearly collimated: you measure a divergence of 0.3 milliradians, about 0.017 degrees. Then the waist size is
$$ w_0 = \frac{M^2 \lambda} {\pi\Theta} = \frac{1.5 \times 671 \times 10^{-9}} {\pi \times 3 \times 10^{-4}} \approx 1\,\text{mm}. $$
In this case, they probably designed the laser pointer with an aperture radius of 2 or 3 mm.
Now suppose you focus your collimated beam with a 1 cm focal length positive lens, which is quite a strong lens. The beam's new waist will be at the lens's focal length. That means you can calculate the divergence half-angle: it is the smaller acute angle of a right triangle with legs 1 mm and 10 mm. So,
$$\tan\Theta = 1/10,$$
or $\Theta\approx$ 6 degrees. Applying the formula once more to calculate the waist yields a waist radius of 3.2 microns, which is quite small indeed.
A "safe" laser pointer might have a power of 1 mW. The peak intensity is equal to $2P/\pi w_0^2$, so before the lens the peak intensity is about 600 W/m^2. After the lens it is about 100000 times larger.
So, to summarize:
- yes, there is a fundamental limit to the intensity, and it does depend on the wavelength, but you cannot even come close with a real-world cheap laser pointer.
- you need to know two of any of these quantities: divergence half-angle, waist radius, Rayleigh range, beam parameter product.
- really, the minimum size and maximum intensity depend quite heavily on what optics you use and how good they are.
In one sense you are right: the only free space "perfectly collimated" optical field is the plane wave in the sense that these are the only eigenfields of Maxwell's equations, being fields which conserve their form under propagation and only undergo scaling by an eigenvalue in such propagation. Since Maxwell's equations conserve energy in free space, propagation is represented by a unitary operator and the eigenvalue is therefore always the simple phase delay $e^{i\,\vec{k}\cdot \vec{r}}$. A wavefront of finite breadth must always undergo some "scrambling" by diffraction; Fourier analysis shows that it is always a superposition of plane waves with a spread of directions, so we can actually think of diffraction through a homogeneous medium as a three step process: (1) decompose the field at one infinite plane surface into a superposition of plane wave components through a Fourier transform; (2) impart the direction-dependent phase delay each of these constituent plane ways undergoes in crossing the homogeneous medium from infinite plane boundary to a parallel infinite plane boundary and (3) reassembling electromagnetic field from its constituent plane wave components at the output boundary. It is the fact that the phase delay is direction dependent (being $\delta\vec{z} \cdot \vec{k}$ radians where $\delta\vec{z}$ is parallel displacement between the input and output infinite planes) that scrambles the wavefront, thus begetting the phenomenon we call "diffraction". So it is the propagation distance dependent interference between a field's constituent plane waves.
So in this sense the only truly "collimated" field is indeed the plane wave. However the word "collimated" is to me a kind of "laboratory" word: one "collimates" a field, i.e. makes its phase front maximally flat, through a laboratory collimation procedure such as making adjustments to field-processing lenses and suchlike whilst the wavefront is observed through a suitable "collimation detector" such as a shear plate interferometer, wavefront sensor, wavefront camera or point diffraction interferometer. So a "collimated" field in this sense is one that has been flattened by such a procedure. Optically, a "perfectly" collimated field of this kind (i.e. with perfectly flat phasefronts but finite breadth) is actually the focal plane field of an electromagnetic field with an extremely small numerical aperture, with $NA$ of the order of $\lambda / w$ where $\lambda$ is the field's wavelength and $w$ its transverse breadth. It diverges, its phasefronts take on curvature exactly as would any other focal plane field and the mathematics describing this process is precisely the same as for any focal plane field of course; it's just that, owing to the extremely small numerical aperture, the divergence is very very slow with increasing axial distance from the focal plane.
For a Gaussian collimated field, the phase front at the focal plane is flat (as for any other collimated field), the phasefronts themselves are a family of ellipsoids and the integral curves of the rays (i.e. unit normals to the phasefronts) are the family of orthogonal hyperboloids (see the Wikipedia page for Gaussian beam).
An optical fibre is different again; here the matter of the fibre interacts with the electromagnetic field and there are many non-plane eigenfields (indeed all, even the radiation fields are non-plane). Again these fields are the ones that undergo only a simple scaling in propagation through the translationally invariant waveguide, and again, if the waveguide is made of lossless material, the scaling is a simple phase delay. The full tale is rather long winded: begin at chapter 11 of Snyder and Love, "Optical Waveguide Theory" (Chapman 1983) and keep reading! However you can get an intuitive feel for what's going on by looking in more detail into the Beam Propagation Method. The finite difference finding of a solution of Maxwell's equations is equivalent to solving a Master-like equation in the Lie group $U(N)$ for the square matrix $T \in U(N)$ which maps the input field to the output field with $N$ being of the order of the number of transverse grid points in the simulation (that's an awfully big Lie group!) depending on whether a vector scaler simulation is done:
$${\rm d}_z T(z) = K(z) T(z)$$
where $K(z) \in \mathfrak{u}(N)$ wanders around in the Lie algebra $\mathfrak{u}(N)$ of $N\times N$ skew-Hermitian matrices to represent the local action of diffraction and either focusing or de-focusing by the waveguide. A short, translationally invariant, length of waveguide has a transfer matrix of the form $\exp((D + L) z)$, where $D$ represents the diffraction through the "mean homogeneous" medium and $L$ the diffraction-free lensing of the waveguide's local profile. The beam propagation method implements "operator splitting", which is another way of saying splitting apart the effect of diffraction and lensing through the Trotter product formula of Lie theory:
$$\lim\limits_{m\to\infty}\left(\exp\left(D\,\frac{z}{m}\right)\,\exp\left(L\,\frac{z}{m}\right)\right)^m = \exp((D + L) z)$$
So the beam propagation method amounts to nothing more than a succession of diffractions through thin slivvers of homogeneous medium followed by the diffraction free lensing imparted by the equivalent local refractive index profile of each slivver. You can, with a bit of a stretch, imagine a step index refractive index profile to be a badly pixellated version of the smooth GRIN lens: the wave near the centre is delayed more by the higher refractive index of the core than the wave near the waveguide edges, so it tends to cancel out the divergence begotten of the pure diffraction of the foregoing slivver. It's not too hard to imagine that for certain special field shapes this lensing, although arising from a "bad" lens, exactly cancels the diffraction and one has a waveguide mode whose phasefront stays perfectly flat and whose transverse profile perfectly conserved as it propagates through the waveguide.
Best Answer
Due to Heisenberg uncertainty principle $\Delta x\Delta p\gtrsim \frac{\hbar}2$, one can't really make a quantum have zero momentum in any direction. So you can't say that photons go in the same direction - this is just a simplified description of laser operation. In reality, the thinner the beam, the higher the divergence.
Compare e.g. a DPSS laser (e.g. green laser pointer) with a diode laser (e.g. a red laser pointer).
In a DPSS laser the active material will have diameter of order of hundreds of micrometer, and the exiting beam will start from even smaller diameter for various reasons. The divergence is quite small: if you remove the collimating lens, your light image from a green laser pointer will be several centimeters after the light goes several meters. Divergence angle would be $\sim \lambda/d=532\text{nm}/100\operatorname{\mu m}\sim0.3^\circ$.
If you try doing the same with a red laser pointer, you'll see that its light diverges quite a lot: after going several centimeters in direction of propagation, it'll already give image of several centimeters. The reason for this is that active zone of diode laser has diameter of order of several micrometers. This makes output beam quite thin, making $\Delta x$ small and thus $\Delta p$ high, and this is what leads to high divergence. Divergence angle would be $\sim \lambda/d=640\text{nm}/1\operatorname{\mu m}\sim 40^\circ$. Actual angle would depend on which transverse direction you select, because active zone is $\sim 10\times$ longer in one direction than in another.
In general, the thicker your starting laser beam, the more collimated it is, so if you manage to make a (visible wavelength) laser with beam starting at 1cm thickness, you'll have almost perfectly collimated laser beam.