B How do orthogonal waves interfere?

[Kuribali]

TL;DR Summary

How do two highly directional, orthogonal light beams (or any other kind of waves) with the same frequency interfere with each other?

How do two highly directional, orthogonal light beams (or any other kind of waves) with the same frequency interfere with each other?

 

[DrChinese]

Generally, they don't. Do you have a specific setup in mind?

 

[Kuribali]

Yes, this one:

Excuse me, the labeling is in german (Spiegel = mirror, Strahlteiler = beam splitter, Weg = path)
D1 registers the photon in 50% of the cases, so does D2.

The explanation is, that the light must behave as a particle and doesn't behave as a wave. I can't see the reason, why the wave function can't simply split, so that half of it travels path 1 and the other half path 2. The only reason I can image this wouldn't work is because of some strange interference when the two pathes intersect.

 

[vanhees71]

A photon is a photon and neither a classical wave nor a classical localized particle. It's much less the latter than the former in some sense.

The only way to really understand photons is relativistic quantum field theory, i.e., quantum electrodynamics.

Qualitatively one can roughly say that in many cases like the one depicted in #1 you can think about the intensity of the em. radiation calculated by classical electrodynamics, normalized to 1, gives the probability to register a photon on the one or the other detector.

Using classical em. waves (quantum-field-theoretically these are no single-photon states or any state of a fixed photon number but a socalled coherent state, and the more intense you make this state the more it behaves like a classical em. wave) you'd simply get 50% of the intensity at D1 and 50% intensity at D2. If you repeat the experiment with a single-photon source, this single photon occurs either completely at D1 or completely at D2. You can never have the situation that it occurs in both detectors, somehow split in half. QFT tells you that a single photon state in some sense is the minimal intensity of an em. wave with a given frequency you can have.

The only way to split a true single-photon state in two other photons is in non-linear optics, where you have a material where you can use high-intensity laser fields such that the usual linear-response approximation (the usual linear constitutive laws you learn about in the classical-electromagnetism lecture like $\vec{D}=\epsilon \vec{D}$ etc.). In non-linear optics it can happen that one photon out of the laser beam gets converted by the interaction with the excited medium to two photons, but both photons have lower frequencies than the original one, i.e., because of energy conservation the original frequency is the sum of the two frequencies of the new photons.

Within the linear realm of quantum optics there's no way to "split a photon" into two. A single photon is either registered at a detector or it's not registered. With a single photon in the initial state you cannot have two detectors registering the photon simultaneously.

 

[DrChinese]

There is a point to consider. The item labeled "beam splitter" is not a polarizing beam splitter. Therefore the beams are not orthogonal. A polarizing beam splitter is needed instead if that is the issue you wish to consider. Also, there are sometimes splitters placed where the beams overlap prior to being detected.

As I read it, this is a classical setup. On the other hand, it is possible to change it. Here is a similar setup that demonstrates the quantum nature of light:

http://people.whitman.edu/~beckmk/QM/grangier/Thorn_ajp.pdf

Not sure if this is relevant to your comments or not. The key point is that the detections at D1 and D2 require a single photon as an input, never 2. To be sure of that, the input is entangled and coincidence counted against its twin as a control. This guarantees a Fock state, in which the photon number is known.

 

[Kuribali]

Thank you for your replies! But the point I was trying to formulate is that the probability wave for the location of the photon could split, so it has 50% in front of D1 and 50% in front of D2. And in the moment of the detection it is "decided" where the photon is, so that it is only detected by one of the detectors. The reason I'm wondering wether this is a possible model is that it is required for a part of https://en.wikipedia.org/wiki/Wheelers_delayed-choice_experiment that the only explanation for this behaviour is that the light behaves as a particle and not as a wave.

And btw.: excuse my sometimes broken English, it's not my native language, I'm a high school freshman from germany

 

[vanhees71]

No, Wheeler's delayed-choice experiment cannot explained in any classical way. The worst idea is to think about photons in terms of bullet-like little particles which are simply massless. It's a contradiction to the fact that the only correct theory we have, which is QED, tells you that there's not even an observable for the position of a photon to begin with. The only thing that makes physical sense are detection probabilities at the place defined by the position of the detector (or by a registered interaction of a photon with the detector material in the case of a photo plate or CCD cam).

The delayed-choice experiment can be understood only with quantum theory. There's no satisfactory other model to describe it. It's the most "quantum thing" one can think of! A very astonishing version of it, which was really done in the lab, is Walborn's quantum eraser:

S. P. Walborn, M. O. Terra Cunha, S. P´adua, C. H. Monken,
Double-slit quantum eraser, Phys. Rev. A 65 (2002) 033818.
https://dx.doi.org/10.1103/PhysRevA.65.033818
https://arxiv.org/abs/quant-ph/0106078