# Are some wavelengths 'longer' than others spatially, and as such explain things such as longitudinal modes?

I'm currently learning lasers. In my research of mode-locking, I've stumbled across something called a longitudinal mode.

The allowed modes of the cavity are those where the mirror separation distance L is equal to an exact multiple of half the wavelength, λ:

L = (n λ) / 2


where n is an integer known as the mode order.

Does this mean the wavelength, the literal 450 nm length, say, has to be of that criterion? I realize increasing the wavelength will create a completely different sinusoid, and if that doesn't fit in the cavity where its 0 points are at either end, it won't lase? Because what I envision, in the first place, are photons bouncing from the gain medium, off the resonator, back through the gain medium, etc. If this is then an 'information' wave, for lack of a better word -- if its energy fits the criteria, if we took its sinusoid associated with its wavelength and extended it, if it could lase we would have 0 points at either resonator?

I guess what I'm trying to ask is, what is this wavelength in this case, and if it is what I'm thinking, then its wavelength does then have spacial significance? Is that what the wavenumber is, essentially?

What we definitely have in the laser system are photons bouncing off the gain medium to resonators and back. How on earth did this become turned into a standing wave? And how is it supposed to leak out if our abstraction we've made doesn't have nodes at either end of the mirror?

I know this is a mouthful, but I'd really appreciate any help.

Also, how does destructive interference occur if there are no 0 amplitude nodes at either end of the resonator in the first place? Having trouble visualizing that.ie, if our condition with L is not met?

• An electromagnetic wave is an oscillation of electric (and magnetic) field strength over space, so the wavelength of light is definitely a physically relevant and actual length. Is that what you're asking about? – Asher May 1 '17 at 15:38
• classical em waves are emergent from the confluence of innumerable photons. Photons are elementary particles and are described by a qm wavefunction which contains in its complex function composition the electric and magentic fields that build up the classical wave. When working with light it is easier to work with the classical em fields rather than photons , both are solutions of maxwell equations ( for photons it is quantized). The complexity of thinking in photons comes when light goes through a medium, where photons keep at velocity c but light has a smaller one. – anna v May 1 '17 at 16:09
• the path of photons is longer due to the quantum mechanical uncertainties described by their wavefunction.How the photons build the classical wave in quantum field theory can be seen here motls.blogspot.com/2011/11/… – anna v May 1 '17 at 16:12

I guess what I'm trying to ask is, what is this wavelength in this case, and if it is what I'm thinking, then its wavelength does then have spacial significance?

Wavelength is the actual distance in space between successive maxima of the field (E or M) in an EM wave.

Is that what the wavenumber is, essentially?

Wavenumber is $\frac{2\pi}{\lambda}$, so it tells you the number of cycles (measured in radians) of that wave that "fit" in a given distance.

For example, a HeNe laser typically outputs a wavelength of 632.8 nm. Let's say we have one with its output polarization in the +x direction.

That means there are 632.8 nm between locations with maximum positive E-field in the +x direction. Or between locations with maximum H-field in the -y direction.

And the corresponding wavenumber is $9.929\times10^6{\rm m}^{-1}$, meaning the phase of the wave evolves by 9,929,000 radians (~1,580,000 full cycles) as it propagates through 1 m of space.

What we definitely have in the laser system are photons bouncing off the gain medium to resonators and back. How on earth did this become turned into a standing wave?

Generally you should not think about photons when you are trying to understand how light propagates. Although photons are "particles" they don't move in straight lines like bullets. They propagate according to the wave equation, resulting in the same propagation behavior we expect from classical electromagnetics based on Maxwell's equations.

And how is it supposed to leak out if our abstraction we've made doesn't have nodes at either end of the mirror?

The assumption that the mirrors are perfect, therefore producing perfect nodes in the field pattern, is only an approximation. In fact, for there to be an output beam, the reflectivity of one mirror must be less than 1.0 exactly. But values above 0.90 are quite common.

This small change in reflectivity typically does not impact the field pattern enough to invalidate the analysis of the modes of the laser; or at least it causes no more error than other effects like thermal changes in the cavity dimension, thermal changes in the index of refraction of the medium in the cavity, etc.

• Is my thinking correct then in saying that the sinusoid has to fit, ideally, at a 0 point at either resonator? If L is fixed, then only certain EM radiation of certain spacial lengths will prevent leakage? And what is the difference between the probability curve of a photon and its EM sinusoid? – sangstar May 1 '17 at 16:17
• @SangerSteel, I can only answer the questions you asked. I don't read minds, so I don't know what you're thinking. – The Photon May 1 '17 at 16:20
• apologies, I pressed enter with intent to continue my sentence, please see edited question above ^ – sangstar May 1 '17 at 16:21
• Oh, and also, it seems unintuitive that the reflectivity must be less than 1.0 for an output beam. Isn't higher fairly preferable as long as that resonator isn't the partially reflecting one? – sangstar May 1 '17 at 16:27
• 1. I think of it as "only waves in certain modes are able to sustain oscillation". I'm not sure if the idea of "leakage" makes sense. 2. I don't know what you mean by the "probability curve" of a photon --- if you mean its (quantum mechanical) wave function, then it's essentially the same as the classical EM wavefunction. – The Photon May 1 '17 at 16:28