Can you explain what happens if two possible photon paths interfere with each other but are different in their length? When does the interference pattern appear? I suggest it happens after the time for passing through the longer path passed. Is this true? I am asking this question in the light of Wheeler delayed choice tought experiment when a photon passes on both sides of a galaxy and then interferes with itself on the semireflecting mirror at the detector on Earth.

  • $\begingroup$ They will interfere as long as the path length difference is shorter than the coherence length. I'm not sure what you mean by "after the time for passing through the longer path passed". $\endgroup$
    – garyp
    Commented Jan 18, 2016 at 22:25
  • $\begingroup$ I checked this answer "There are cases when a photon cannot interfere with itself. Take an interferometer with two nearly equal paths and observe interference. As soon as a photon has a limited length itself, it can interfere with itself only if the path difference is inferior to its own length The trick is to have a superposition of waves at the same time. If wave trains of photons arrive without overlapping, there is no interference because there is no superposition" quote from Vladimir Kalitvianski here physics.stackexchange.com/questions/6234/… $\endgroup$
    – Kolyo
    Commented Jan 18, 2016 at 22:34
  • $\begingroup$ there are many interferometer setups with different length of the two paths but there is still interference at detector. How to explain them from the notion of single photon interference? $\endgroup$
    – Kolyo
    Commented Jan 18, 2016 at 22:40
  • $\begingroup$ You have to define very carefully what you mean by "a photon" (and you shouldn't be visualizing a little moving ball). At the naive level of the second quantization an excitation of the field is not confined in space (but has a precisely known wavenumber). $\endgroup$ Commented Jan 19, 2016 at 2:13
  • 1
    $\begingroup$ Photons do not have a length. The quote you provide is not a very good description of what is going on. I'm not sure what he means by "length" He might mean "coherence length", or he might mean the length of a wavepacket. A wavepacket is not a photon. Don't use Kalivianski's answer as a source. $\endgroup$
    – garyp
    Commented Jan 19, 2016 at 2:36

1 Answer 1


Think about it this way. If you had a radioactive atom it might decay and emit a photon.

And you don't know when. So there is a whole range of times for when a nearby photon detector might go off.

But there is also an issue of coherence. Which is a specific technical term, not a word people use for emotional purposes.

If you have a short coherence length then effectively after a short time the object has either decayed in that time or it hasn't. If you have a longer coherence length, then if the detectors are all far enough away then for that same time interval, it might not be the case that it decayed or didn't decay.

There are things you can do to increase the coherence, and usually it's about isolating something. Though there are limits to how much something can be isolated. For instance a spontaneous decay can sometimes be thought of as a decay stimulated by the quantum vacuum.

But don't expect that you can forget to isolate something and expect that later you can start to isolate it at the last minute and hope it can not decohere. That's not a good idea.

I don't understand very well what will happen if the path difference is significant but the coherence length is greater than the path difference.

You write as if there is some special length that is intrinsically significant. There are the lengths of the arms, the wavelengths, and the coherence length. If all of them were scaled up or down, would the physics really be different?

Does it mean that the interference from both path will produce an interference pattern

The interference pattern happens spatially. Some places will get hit more often than others when you repeat the experiment. And it will take some time from the creation of the emitter until a detector a ways away goes off.

It's not like you'll see some splotch and then see it turn into bands some time later. You see a hit or not every moment. And a pattern builds up over repeated experiments.

If you specifically want to see a correlation between the time a detector goes off and where the detector is then you have to deal not only with the emitter but the detector. You can't resolve a monochromatic beam into a perfect spot at a perfect time. If you start to focus better on when it arrives, you'll affect the wavelengths the detector reacts to, and no source is perfectly monochromatic at a fixed frequency. For instance if it is an atom decaying, there is a doppler shift from the motion of the atom and the motion and the location aren't unrelated. So the wavelength and the location of the atom are related. So the phase of the photon from different motions of the atom matters.

The coherence matters. It might be good to step back and ask yourself what you want to learn, what you want to understand, and what you want to predict. The universe never promised to be simple, so if your goal was to hear a simple story, think carefully.

There are entire textbooks on quantum optics, and they are trying to make it as simple as it can be, they don't write them in a complicated to hide understanding from lay people. They just address the issues that matter for making precision experiments. And the emitter and the detector and their details and the coherence of the interaction all matter.

  • $\begingroup$ Thank you very much for the explanation! I don't understand very well what will happen if the path difference is significant but the coherence length is greater than the path difference. Does it mean that the interference from both path will produce an intereference pattern, even if that need great time uncertainity of the photon emission from the excited atom at the start of the setup? $\endgroup$
    – Kolyo
    Commented Jan 19, 2016 at 19:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.