Talk:Coherence length

Semiconductor laser coherence length

The article currently (2006-02-14) suggests that semiconductor lasers have longer coherence lengths (100 m) than helium-neon lasers (20 cm). I don't think that's generally correct. Perhaps the article should say 100 microns, rather than 100 m. Opinions?

^I also think these statements are incorrect. Typical coherence lengths of monomode HeNe lasers are 100m to several km, while even a monomode semiconductor laser with a linewidth of about 10pm has a coherence length of about 10cm.

^This statement is technically correct, but is deceptive. A HeNe laser without frequency stabilization typically has a 20-30 cm coherence length. A frequency stabilized (monomode, as the previous comment states) HeNe can have a coherence length greater 1 km. A semiconductor laser can have a coherence length less than 1 mm or greater than 100 m. Again, the longest coherence lengths are achieved via frequency stabilization.

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"The expression above is a frequently used approximation. Due to ambiguities in the definition of spectral width of a source, however, the following definition of coherence length has been suggested[citation needed]"

Does anybody have a citation related with the expression of linewidth vs. coherence length? Some sources refer to a pi factor added in the denominator of the expression while some others do not (this article does not). I did some simple numerical simulations with a sum of planar waves with a lorentzian distribution of the amplitude this gave a visibility of 50% for a path difference around ~3x times smaller than given by the expression in this article. This seems to confirm that pi is needed for a lorentzian linewidth. WB.

Most that I know use the coefficient of 1, which is close enough for most applications. Otherwise, you have to carefully define line width, and how to measure it. Gah4 (talk) 07:55, 5 April 2018 (UTC)

Merger proposal

There is a merger template on this article and the article coherence time. I didn't put it there and the person who did has not discussed it. I will remove it in one day unless there is actually a discussion to be had. JHobbs103 (talk) 17:24, 13 June 2010 (UTC)

No objection, as long as you remove it from both articles.--Srleffler (talk) 04:11, 14 June 2010 (UTC)

Done. JHobbs103 (talk) 17:40, 16 June 2010 (UTC)

distance from source?

The article indicates that coherence length is the distance from the source. Now, if that is the way it is used then I suppose, but as I understand it is the distance along the beam, not necessarily from the source. For holography, it needs to be more than, more or less, the distance between the plate (film) and object, which might be a long way from the laser source. Gah4 (talk) 11:46, 2 February 2013 (UTC)

You're right. I fixed it.--Srleffler (talk) 19:48, 2 February 2013 (UTC)

single photon

From the formula is seems that the coherence length of a single photon is infinite, because the spectral width is zero. But interference only shows with multiple photons, so in practice the visible coherence length depends on the spectral width of all photons together (the same for coherence time). Is that a right? DParlevliet (talk) 09:26, 14 January 2014 (UTC)

No, single photons can interfere with themselves. --Srleffler (talk) 06:19, 15 January 2014 (UTC)

You are right, but you can only see, so measure the interference when you have more photons. The result is that the wave of a single photon also interferes after a very long distance DParlevliet (talk) 08:33, 15 January 2014 (UTC)

A single photon does *not* have zero spectral width, due to the nonzero uncertainty in the photon's energy. And quantum coherence effects can certainly be observed with a single photon--for example, we will never detect a photon at a double-slit null (even if we're not sure which peak it will land at). Olawlor (talk) 00:39, 22 January 2014 (UTC)

Thanks. I had a bad feeling about that paragraph, but my recollection of this material was too dim to immediately refute it.--Srleffler (talk) 02:28, 22 January 2014 (UTC)

Of course coherence effects can be observed. My edit mentioned that the coherence length was infinite, so interference of a single photon is possible over large distances. There is uncertainty during emitting a photon, and while measuring, but I don't think when it is a photon. But suppose it is, then the spectral width is very small. If I calculate the coherence length would be 7E7 meter. Still very large. DParlevliet (talk) 09:56, 22 January 2014 (UTC)

This is all linear optics, so single photons and beams act identically--a single photon can and will interfere with itself, in things as simple as an anti-reflective_coating. But this interference just plain stops working if you move beyond the coherence length--this is why a thin film of oil on water shows color interference rings, but a thicker film is dull. Folks actually do measure the frequency spectrum of individual photons using a Hong-Ou-Mandel interferometer to interfere it with itself; if each photon had zero spectral width (hence infinite coherence length), that bump would be a straight line at minimum amplitude. (Caveat reader: this isn't my field, the last course I took on it was a dozen years ago, and my copy of Mandel and Wolf is at the office) Olawlor (talk) 03:23, 25 January 2014 (UTC)

A thicker oil film is dull because it is absorbing photons or to much/fast variations in phase. Look to Feynmans example of glass, where interference happens even with thick glass. In the Hong-Ou-Mandel interferometer the photon does not interfere with itself, but with another photon (beam). So you can use it to measure the difference in wave length. The article also states that if both photons are identical, there will be a zero dip (but only when they differ 180 degree phase shift!).DParlevliet (talk) 16:40, 26 January 2014 (UTC)

Hg

I moved the statement about an Hg line. (I suspect 546nm, but didn't check the reference.) It seems that the way to make it longer is to use isotopic pure ¹⁹⁸
₈₀Hg
. We could also add some other sources and their coherence length. Gah4 (talk) 07:58, 5 April 2018 (UTC)

It seems to be usual to give line widths in wavenumbers, that is, delta reciprocal wavelength. Ignoring small constant factors, that will then be the reciprocal of the coherence length. One source seems to give 0.03 cm^-1 for natural Hg, which would be about 30cm coherence length. Gah4 (talk) 20:28, 5 April 2018 (UTC)

Coherence length formula for a Gaussian source

I copied the discussion below from my talk page. --Srleffler (talk) 03:59, 10 April 2022 (UTC)

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Hi Srleffler, I see that on the Coherence Length wiki page, you revert changes to the coherence length equation for a Gaussian source that remove the square root. I looked at the paper you referenced, and I believe the paper is incorrect. In the paper, they say that the coherence length is the FWHM of the modulus of the complex temporal coherence function (Eq. 2 in the paper), but actually the coherence length should be the HWHM (half width half maximum), which is the FWHM/2. The coherence length being the HWHM agrees with the explanation that the coherence length can be thought of as the path offset at which the fringe visibility drops to 50% (when measured with a Michelson interferometer), and is also why the path offset is +-L. This is why Equation 8 in the paper is off by a factor of 2 from the "correct" answer, i.e., they report the FWHM number incorrectly as the coherence length, when the coherence length should actually be the HWHM. As for why Equation 12 looks "correct" in the paper but actually refers to a Lorentzian source, again they are off by a factor of 2 and Equation 12 should also be divided by 2. Thus, the conclusion should be that given some bandwidth, the coherence length of a Lorentzian source would be half the coherence length of a Gaussian source.

We can derive the correct answer (which again is Equation 8 in the paper divided by 2) using the fact that the fringe visibility is related to the Fourier transform of the source spectrum (in frequency). Starting with a Gaussian source that has a FWHM of Δν and some center frequency ν0, we first find the standard deviation of the Gaussian, σ, using the fact that FWHM = 2*sqrt{2 ln(2)}*σ, then we can write the full description of the Gaussian in frequency space. If we compute the Fourier transform of the Gaussian in frequency space using a Fourier kernel of e^{-i2*pi*ντ), where τ = l/c (τ is time, which corresponds to length divided by the speed of light), we get another Gaussian in length space with some standard deviation σ', which we can find in terms of σ. The coherence length is the HWHM of the Gaussian in length space, and when you plug in for σ' in terms of σ, and σ in terms of Δν, then use the fact that c/Δv = λ^2/Δλ, you find that there is no square root. We can sort of see this now, since the FWHM and the HWHM equations both have sqrt{2 ln(2)}, and when you do the math the two square roots are multiplied together and thus gets rid of the square root. — Preceding unsigned comment added by 2603:7080:6E01:72AE:9922:7A36:ACAC:8CC0 (talk) 19:07, 8 April 2022 (UTC)

Thanks for explaining. Looking back over the history, I see that the equation has been flipped back and forth between the two forms at least five times over the last nine years. It appears that there are two different forms of this equation in the published literature, and users are editing the page to match whichever book or paper they happen to have.

We can't derive the correct answer. One of Wikipedia's foundational policies is No original research. We only report what has been published elsewhere, ideally in reliable secondary sources.

The current article cites a paper by Akcay, which you assert is incorrrect. Interestingly, the article originally had the form you prefer and cited a book by Dreyler. An editor back in 2015 asserted that that book was incorrect, and changed both the formula and the reference. Between a book and a paper I would normally take the book as the more reliable source but given the number of times this has been flipped I think the article probably needs to address both to prevent continual churn on this formula. I'll take a look and see what I can do.--Srleffler (talk) 03:59, 10 April 2022 (UTC)

I documented both opinions for now, with references for each. Perhaps someone who has read both can clearly explain the assumptions that yield each of the expressions. I doubt that either is incorrect. Too many editors have tried to switch the formula for this to be just an error by an author of one paper, or one book.--Srleffler (talk) 04:26, 10 April 2022 (UTC)