Even "incoherent" light as the one of a light bulb has some coherency, and would interfer in the double-slit experiment (even if more blurry because the different wavelengths don't trigger the same superimposable pattern).

The spatial coherence is due to the fact that even for a single emitted photon it's the same wave that reach the 2 slits. Now I'm more puzzled by the time coherency: does it correspond to the "length of the photon", i.e. the delay between the beginning and the end of the transcient photonic emission by an electronic layer? Then, how much is it? Is it constant, varying on some factor, stochastic?

The experimental counterpart: I modify the double slit experiment so that light path passing through one of the two slits is elongated using mirrors. Will the interference pattern disappear once the extra length (i.e. delay) difference between the 2 light pathes get too long? Which length / duration would it be?

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    $\begingroup$ You need to stop thinking about photons as microscopic constituents of electromagnetic waves. Quantum mechanics is not a resurrection of the corpuscular theory of light. Whether a light bulb is spatially coherent or not depends on how far the observer is away from it. Even a luminous object the size of a star will show near perfect interference in a telescope. It takes hundreds of meters of base length in a stellar interferometer to resolve that spatially coherent light as actually incoherent (by which we can measure the stellar diameter). $\endgroup$
    – CuriousOne
    Commented Feb 13, 2016 at 1:33
  • $\begingroup$ I perfectly know that. My question is about time coherence. And you don't tell either what the experiment with time delay of one light path would give. $\endgroup$ Commented Feb 13, 2016 at 1:38
  • $\begingroup$ The same thing applies to temporal coherence, except that in that case it's not the distance that matters but how long your measurement is. If you measure white light for no longer than $10^{-15}s$, it will show coherence. Coherence depends on your experiment, not just on the light. That's why we are specifying coherence length and coherence time. If you want a coherence time of $10^{-12}s$, then you have to limit the relative bandwidth of the light to approx. $10^{-3}$. $\endgroup$
    – CuriousOne
    Commented Feb 13, 2016 at 1:42
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    $\begingroup$ You are way over-thinking this. Interference is all a trivial consequence of harmonic waves being described by $e^{i( \omega t - \vec k \vec x)}$ and Fourier transforms. $\endgroup$
    – CuriousOne
    Commented Feb 13, 2016 at 3:14
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    $\begingroup$ This is tautological, especially concerning the phase (missing in your formula). This one might varies in space and time. Sometime it's interesting to understand where continuous quantities emerge from ( plus there exist transitional states; fields are not always stationary). $\endgroup$ Commented Apr 27, 2016 at 13:16

2 Answers 2


The spatial coherence is due to the fact that even for a single emitted photon it's the same wave that reach the 2 slits.

I'm noot too sure what you mean by that. Spatial coherence has nothing to do with photons, it comes from the apparent size of the source as seen by the observer. Every source you might want to use in an interference experiment (a spectral lamp, a star, a laser) has a finite size. To increase contrast you want to use a source that is as much point-like as possible to avoid superposition of interference patterns coming from different incoherent points of the source. This is why, as @CuriousOne pointed out, stars produce very coherent interference patterns : their angular size is very small from an observer sitting on the surface of the Earth.

Regarding the laser, you do not need to worry about the size of the beam because its quantum nature makes it spatially coherent.

Time coherence is a different story. There are two equivalent ways to picture it (linked by Fourier transform). Take the case of a pointlike source that emits some radiation at frequency $\omega$. Because of thermal excitations, the phase of emission $\phi (t)$ will jump randomply on a typical timescale $\tau$. Two rays superposing on the interference pattern will be coherent if their respective delay is less than $\tau$, which defines the coherence time. It is linked to the coherence lenght of the source by a factor of $c$, the speed of light.

In the frequency domain, this random change of phase $\phi(t)$ causes a broadening of the peak of emission at frequency $\omega$, which becomes a lorentzian peak centered at $\omega$ with some width $\Delta \omega$. The quantity

$\tau = \frac{1}{\Delta \omega}$

defines in this case the coherence time of the source, which is again linked to the coherence lenght by $l_c = c \tau$. Every source you can think of has a finite coherence lenght, even the most stable lasers (whose coherence lenght can still reach hundreds of kilometers). To compare, regular spectral lamps used in school labs $l_c$ of order of a few milimeters to a few centimeters, and a light bulb is more of order of a few microns.

  • $\begingroup$ thaks for your answer. What kind of derivations would give an estimation of $\tau$ ? $\endgroup$ Commented Apr 27, 2016 at 9:37
  • $\begingroup$ It is mainly an experimental quantity, measured with an interference experiment or a spectrometer. $\endgroup$
    – Dimitri
    Commented Apr 27, 2016 at 10:11
  • $\begingroup$ Yep, but I'm interested in the microphysics causing it. Including though experiments. $\endgroup$ Commented Apr 27, 2016 at 13:12

As for your last question, a similar experiment has been done though it doesn't involve a double-slit. It's called the Michelson experiment, and using mirrors it tests the interference pattern created by light when the light-waves are combined with time-delayed versions of themselves. By changing the distance of one of the mirrors, the time-delay can be controlled. A certain amount of delay fades the interference, and a certain amount more makes the interference completely disappear.

Also, the spatial coherence you mention needed to see interference in the double-slit experiment, unless you're using a source of light that's spatially coherent enough, is provided by diffraction. For example, a single-slit is often placed at a distance in front of the double-slit. The effect makes the light-waves travel to the two slits in-phase with each other (assuming monochrome light), which is precisely what's needed for the interference.

EDIT: "The effect makes the light-waves travel to the two slits in-phase with each other (assuming monochrome light), which is precisely what's need for the interference." I should add that this doesn't mean, however, that the light-waves will hit the surface with the double-slit at the same time. It just means that the waves will travel to the double-slit oriented crest-to-crest and trough-to-trough. That's what's needed to see the patters of destructive interference.


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