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I am trying to understand why single mode laser diodes have lower divergence after collimation than multimode laser diodes. From reading the following quote from another discussion group, something must be missing in my understanding. I thought the smaller the emitter the less divergence but the following seems to indicate the opposite:
Due to Heisenberg uncertainty principle ΔxΔp≳ℏ2 , 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 ∼λ/d=532nm/100μm∼0.3 ∘ .
•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 Δx small and thus Δp high, and this is what leads to high divergence. Divergence angle would be ∼λ/d=640nm/1μm∼40 ∘ . Actual angle would depend on which transverse direction you select, because active zone is ∼10× 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.