What happens to light at points of intersection? [duplicate]

Background Knowledge

Researching on light led me to wave-particle duality and its properties regarding photons. All the sources I have seen agree that the frequency of the wave nature defines how we perceive light. And the amount of photons (maybe particle nature, it's sort of unclear) defines the light's intensity or brightness. There is also little to no interactions between photons. But I've come across a conundrum of when photons interact.

Question

If I point two lasers that emits different frequencies with the same intensity at a single spot at the same time, what would be observed?

Edit: More specifically what happens to the photons when they intersect?

Constraints

When answering, keep in mind the following:

• The photons are traveling in a vacuum between the source and the detector.

• The observation should describe the configuration of the photons, not the color representation, at the point of intersection.

(When I say the "configuration of the photons" I mean what happens to the wave and particle nature of the photons, like if interference occurs)

marked as duplicate by Conifold, Kyle Kanos, Jon Custer, David Hammen, YashasJul 15 '17 at 7:08

• @Conifold My question asks what happens to the photons when they intersect but maybe my phrasing could have some similarities to the question you linked to. – Qubic Lens Jul 12 '17 at 23:48
• The answers there answer your question as well. What happens is nothing interesting, you get two waves superimposed. Or in terms of photons, they do not "bump" into each other as classical intuitions suggest, that would be scattering. – Conifold Jul 12 '17 at 23:52
• @Conifold Now reading it again, you are right that it also answers my question. Do you suggest I delete this question or leave it on the site as a sign post? – Qubic Lens Jul 13 '17 at 0:22
• Leave it, it will probably be marked as duplicate eventually. – Conifold Jul 13 '17 at 0:26

I will answer this, as the chosen answer is not doing so.

If I point two lasers that emits different frequencies with the same intensity at a single spot at the same time, what would be observed?

One has to keep clearly in mind that classical electromagnetic beams emerge from zillions of photons ( which are quantum mechanical entities) in a straightforward but mathematically complicated way.

In quantum mechanics everything is explained by wavefunctions which are the solutions of the corresponding to the particle quantum equations. For photons it is a quantized form of Maxwell equations. . Wavefunctions can be superposed, and the complex conjugate square of the superposed functions which gives the probabilities of detecting the photons, will show interference.

Lasing is a quantum mechanical phenomenon.

This is a very instructive video that shows the interference between laser beams, yes there is interference. The whole system, laser source and setup is in a quantum mechanical state, the power source taking part in the appearance and disappearance of the laser beams.

One has to keep clear in mind that superposition is not interaction in quantum mechanics. The two photon beams are not interacting , the whole system is superposing the photon wavefunctions in specific (x,y,z,t) so that when measured the interference pattern appears, similar to the double slit single photon patterns.

The above video uses the same frequency because interference effects are much more prominent for same frequency waves. But this answer here is relevant for differing frequencies , where for waves in general there will be interference patterns depending on the frequencies. For details have a look at this link. Keep in mind that at the photon level the mathematics of the sine and cosine functions are the same but they refer to probability distributions, not the photons themselves.

Edit: More specifically what happens to the photons when they intersect?

There exist higher order diagrams where photon-photon interactions are non zero. This is shown here,

where the solid line represent virtual particles of the whole gamut of charged elementary particles, the larger their mass the more depressed the diagrams, so usually only the electron loop is shown.

If the energy of the interaction of the two photons in the center of mass is below the pair production energy threshold the photons scatter elastically and get out of the beam line.

The frequency of the photons has to be high for pair production to materialize.

After the electron pair production energy threshold, at least 0.5 MeV per photon, particles may be produced at the intersection of two laser beams. There exist proposals for a gamma gamma colliders.

Photon beams can be made so energetic and so intense that when brought into collision with each other they can produce copious amounts of elementary particles.

Energetic refers to frequency, E=h*nu. Intense to the number of photons per cm^2.

Classical electrodynamics is linear, meaning that to light waves do not interact directly with each other. The electric field measured at the target will be the sum of that of each field, so something like $$\mathbf E = \vec{\epsilon}_1 \sin(\omega_1 t) + \vec{\epsilon}_2 \sin(\omega_2 t + \alpha)$$ for linear polarizations $\vec{\epsilon}_i$, frequencies $\omega_i$, and some phase shift between the lasers $\alpha$.1 From the electric field you can calculate the observed intensity, et cetera. This is the end of the story in classical electrodynamics, and as long as your laser intensity is much less than $10^{28}\, \text{W/cm}^2$ or so, you don't need to care about quantum electrodynamics. For comparison, the total power of the sun is $\approx 4 \cdot 10^{23}\, \text{W}$.

In quantum electrodynamics, on the other hand, particles can be created and annihilated. Two photons can produce an electron-positron pair;t The resulting particles can annihilate through the inverse process, leading to an effective $\gamma\gamma \to \gamma\gamma$ process^2:

However, for these processes to be of any importance, there has to be enough energy in the laser fields, because of the non-zero electron mass. Because of Heisenberg's principle an electron cannot be more localized than the Compton length $\hbar/(mc)$. Roughly speaking, the work done by the electric field over this length must be comparable to $mc^2$, so $q E_0 \hbar/(mc) \sim mc^2$ where $E_0$ is the peak electric field. This gives $E_0 \sim 1.4\cdot10^{18}, \text{V/m}$.

The intensity is proportional to $E^2$, and this works out to needing laser intensities of the order $10^{28}\, \text{W/cm}^2$. This is several orders of magnitude beyond the world's most powerful laser systems, which are the size of small houses.

You can get around this limit if instead of two lasers you consider photons propagating through a strong static background magnetic field. Here "strong" means $\sim E_0/c \approx 10^{10} \,\text{T}$. The world's strongest lab magnetic fields are maybe $10\, \text{T}$, so they are enormously weak, a billion times weaker, compared to this. But neutron stars can come close, and then one expects to see photon splitting $\gamma \to \gamma + \gamma$. Also, the photons polarized parallel to or perpendicular to the magnetic field propagate at different speeds, so the magnetic field can act as a polarizer. This was recently observed but the effect is small because even neutron stars are below the quantum limit.

The canonical reference for these things is Probing the Quantum Vacuum by Dittrich and Gies.

1 It is rather simple to extend this to the more general case of elliptical polarizations but that would clutter the notation correspondingly.

2 Ignore anyone telling you things like "space is full of virtual particles that appear and vanish all the time". That is not an accurate description, and it should be regarded as, to be frank, bullshit.

• there should be a distinction between intensity of a classical beam and the frequency of the photons. Pair production cannot occur for frequencies bellow .5 mev energy , no matter how intense the beam – anna v Jul 13 '17 at 4:49

Because of the different frequencies the phase difference of the two lasers at the single spot will change all the time. Meaning on average you will not see any interferences.

• What's do you mean by "change all time" or is it "change all the time"? – Qubic Lens Jul 12 '17 at 23:50
• Yes I was missing a "the" there, thank you. – Nikl Jul 13 '17 at 6:28

What happens to the photons depends on the surface material (atoms and molecular arrangements) and whether or not the photons are absorbed or not. Some or all of the photon frequencies may or may not be absorbed. If the photons are absorbed they will be converted to heat energy or a new photon will be re-emitted. The frequencies of the new photons are determined by everything mentioned above.