# Interference of EM Waves with Orbital Angular Momentum

If you have two coherent collinear e-m beams of same frequency and polarization, but 180 degrees out of phase, they will destructively interfere.

If you introduce orbital angular momentum of L=3 (three revolutions per wavelength) to one beam and L=1 to the other beam, ceteris paribus, will they still destructively interfere in the exact same manner?

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Depends a lot what is the state of your "orbital angular momentum L": a system with well define $L$ still has $2L+1$ states (Zeeman sublevels). What is their density matrix? – Slaviks Sep 15 '11 at 18:25
Not sure about density matrix but both OAM beams could be circularized polarized. Maybe a better way to ask my question is: Is there a way to generate light with OAM such that one could differentiate between beams (of same frequency) with different revolutions per wavelength, by destructive interference, where a beam with the same OAM would cancel a beam with exact same OAM, but not cancel a similar beam with a different OAM? Thanks for any help you can give me. – user5277 Sep 15 '11 at 20:02
Sorry, I didn't understand that you are talking about the OAM of light, not of a medium regular, $L=0$, light is supposed to go through. – Slaviks Sep 15 '11 at 20:14

## 2 Answers

I don't know how you introduce L=3 to a beam of light, but I know if you give them repectively +1 and -1 for angular momentum there is no destructive interference; that is to say, the powers of the beams are simply additive.

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yea, there's no such thing as light with L=3 in free space. The electromagnetic wave equation requires that the electric field components all have the same period as the wavelength. They can't have structure at one-third the wavelength. Also, in case readers don't know, +1 and -1 angular momentum mean the same thing as "right circular polarization" and "left circular polarization" respectively (or maybe vice-versa). – Steve B Sep 17 '11 at 3:09

Free-propagating light with non-zero orbital angular momentum is similar to transverse modes of optical fiber higher than the central one. There will be no destructive interference because the transverse profiles of $L=1$ and $L=3$ light are orthogonal to each other. This should be true both clasically and quantum mechanically.

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