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In classic books of quantum optics visibility of interference fringes (for example in Michelson interferometer) is equal to the modulus of the degree of first order coherence $g^{(1)}$, when the intensity of the two beams are equal. Here I am focusing on temporal coherence.

It is known that the spectrum of the radiation is related to the first-order coherence by a Fourier transform (Wiener-Khinchin theorem). For example, single mode light has a spectrum in frequency like Dirac's delta function and the first-order coherence is an imaginary exponential, so the modulus is one. Single mode light is first order coherent. Now, it is seems that saying that a radiation is first order coherent means saying that the visibility of the fringe is unity. But take for example coherent states in multimode, or classically a sum of a certain number of deterministic waves with different frequencies. They are first-order coherent, see for example Loudon, Quantum theory of light.

This means that there is a fixed relationship on the phase of the fields, and this is connected to the ability to form clear fringe pattern. But I have a doubt. Taking a sum of waves with different frequencies, when i change the path difference of a Michelson interferometer the fringes formed by every component in frequency shift in a different way, so for a certain path difference it's no more possible to see fringes, even if the waves are deterministic (in a sense that there are not random change of phase).

Taking incoherent sources we know that there are random changes in phase of the electric field. Because of the Wenier-Kinchin theorem the coherence time is inversely proportional to the bandwidth of the spectrum.

In conclusion it appear to me that first order coherence doesn't imply always $100\%$ of visibility, because even if a beam is in a multimode coherent states, different frequencies shift the pattern in different ways and a certain point fringes are not clear enough.

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2 Answers 2

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The issue seems to be about the definition of coherence. The OP states that multimode light could be perfectly coherent. In the context of temporal coherence the term multimode means multiple frequencies. However, as explained by the OP, a light source with multiple frequencies can produce a reduced fringe visibility.

Well, the definition of first order coherence is based on fringe visibility. Therefore, multiple frequencies do not produce a perfectly coherent optical field. In fact, the coherence length of an optical source is inversely proportional to width of the frequency spectrum of the source. If the spectrum is a Dirac delta function, the coherence length would be infinite, which means the source is perfectly coherent. However, physical light sources always have a spectrum with a finite width. Therefore, the coherence length of such sources are finite. It means that multimode light in this context is not perfectly coherent.

For a full treatment of optical (first order) coherence, you can consult the book by Mandel and Wolf "Optical coherence and quantum optics." Emil Wolf is known for having developed a comprehensive theory of optical coherence.


Based on the comments, it may be useful to say a bit more about coherent states and whether one can say that a coherent state is always perfectly coherent. Firstly, a coherent state is nominally an eigenstate of the annihilation operator. If we incorporate the additional degrees of freedom, we have $$ \hat{a}_s(\mathbf{k}) |\alpha\rangle = |\alpha\rangle \alpha_s(\mathbf{k}) , $$ where the eigenvalue function $\alpha_s(\mathbf{k})$ is a complex valued function with a spin index $s$ and wave vector dependence $\mathbf{k}$ taken over from the annihilation operator. Each eigenvalue function is associated with a unique eigenstate, which is the coherent state. One can turn the annihilation operator into the annihilating part of a field operator by a Fourier transform and then the eigenvalue function will become a function of space and time. Hence $$ \hat{E}(\mathbf{x},t) |\alpha\rangle = |\alpha\rangle \mathbf{E}(\mathbf{x},t) , $$ where the eigenvalue function $\mathbf{E}(\mathbf{x},t)$ become an electric field that parameterized the coherent state. For the sake of this discussion, let's focus only on time. So we consider $$ \hat{E}(t) |\alpha\rangle = |\alpha\rangle \mathbf{E}(t) . $$

If we now ask whether the coherent state is coherent according to the definition of first order coherence, as determined by the observation of fringe visibility, then the obvious way to answer this question is to perform an experiment. Such an experiment would involve an interferometer, such as a Mach-Zender, consisting of two beam splitters that respectively separate the state into two paths and then recombines them again. Prior to the second beam splitter, a relative phase is introduced between the two paths. Without going through the detailed calculations, I can hopefully convince you what the interferometer does to the coherent state is to convert its parameter function into a superposition of the parameter function and a shifted version of that parameter function. In other words, $$ \mathbf{E}(t) \rightarrow \frac{1}{\sqrt{2}} \left[ \mathbf{E}(t)+\mathbf{E}(t+\tau) \right] . $$ With a slight abuse of notation, we may represent the coherent state as $$ |\alpha_{out}\rangle = \left| \mathbf{E}(t)+\mathbf{E}(t+\tau) \right\rangle , $$ where I'm discarding the normalization to simplify the expression.

Next we need to measure the intensity of the output to see the fringes. For this purpose, we use a number operator, which would give us the expectation value for the number of photons, which is proportional to the intensity. Hence $$ \langle \hat{n} \rangle = \langle \alpha_{out}|\hat{n} |\alpha_{out}\rangle = |\mathbf{E}(t)+\mathbf{E}(t+\tau)|^2 . $$ So, we see that the final result is precisely the same expression that we would expect to find in classical optics, from which it then follows that the visibility of the fringes would depend on the coherence properties of the electric field, which is also the parameter function of the coherent state. In general, it would not be perfectly coherent, as for instance in the case of an electric field with a finite bandwidth. I am not sure what Loudon is referring to, but it is clear that a coherent state does not necessarily represent a state that is perfectly coherent according to first order coherence theory.

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  • $\begingroup$ Let's take a coherent state, the eigenstate of positive frequency electric field operator, and this coherent state is an excitation of differents mode with k pointing on the same direction, so it's composed of multiple frequencies, but it's a parallele beam. If you compute the $g^{(1)} (z_1,t_1;z_2,t_2)$ the modulus is 1 for every space-time points, then is first order temporal coherent, other than spatial coherent. See for example chapter 6 of Loudon, Quantum theory of light. $\endgroup$
    – Mark_Bell
    Nov 21, 2021 at 10:29
  • $\begingroup$ But this seems to contradict the fact that bandwidth is inversely proportional to the coherence time. For other kinds of light, this is perfectly fine for me. But here you have multimode light perfectly coherent at first order. What am i missing? $\endgroup$
    – Mark_Bell
    Nov 21, 2021 at 10:29
  • $\begingroup$ Sorry, I don't have access to Loudon. Are you saying that a coherent state must always be perfectly coherent? That is evidently not the case. $\endgroup$ Nov 21, 2021 at 12:27
  • $\begingroup$ Yes. Coherent state is an eigeinstate of positive frequency electric field: $E^{(+)}|\alpha> = \epsilon |\alpha>$. Computing $g^{1}(z_{1}, t_{1}; z_{2}, t_{2}) = \frac{<E^{(-)}(z_{1}, t_{1}) E^{(+)}(z_{2}, t_{2})>}{(<E^{(-)}(z_{1}, t_{1})E^{(+)}(z_{1}, t_{1})><E^{(-)}(z_{2}, t_{2})E^{(+)}(z_{2}, t_{2})>)^{\frac{1}{2}}} = \frac{<\epsilon_{1}^{*} \epsilon_{2}>}{(<\epsilon_{1}^{*} \epsilon_{1}><\epsilon_{2}^{*} \epsilon_{2}>)^{\frac{1}{2}}} = ({\frac{\epsilon_{1}^{*}\epsilon_{2}}{\epsilon_{2}\epsilon_{1}^{*}})^{\frac{1}{2}}} = e^{i\theta}$. Then $|g^{(1)}| = 1$, so it's perfect coherent. $\endgroup$
    – Mark_Bell
    Nov 23, 2021 at 19:02
  • $\begingroup$ Not sure I quite understand your notation. However, I think I can see where you are going with this. If I have some time, I'll try to add something to my answer about the way it should work with coherent states. $\endgroup$ Nov 24, 2021 at 4:28
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This is one of those questions that fits into a pretty large category of confusions caused by Loudon's lack of clarity. I love his book, I just wished he said more things.

So Eq. 3.5.2 is where we get this idea that visibility of interference fringes corresponds to $|g^{(1)}(\tau)|$. But this comes from a simple generalization of the thing that's actually always true, Eq. 3.3.9, which states that the fringes have a term proportional to $1+\mathbb{Re}[g^{(1)}(\tau)]$. And to be very specific, in the example given to produce Eq. 3.5.2, the light has a first-order correlation function $\exp(-i\omega_0\tau - \gamma|\tau|)$ which contains the oscillation $e^{-i\omega_0\tau}$ which produces a single oscillation, so that a well-defined visibility can be obtained by taking the magnitude of this oscillation, the $e^{-\gamma |\tau|}$ term.

The generalizations Loudon made in these sections were alright for single-mode light. But multimode light is going to have some problems; you can work out for yourself that the first-order coherence of a more general multimode field will not have a single oscillation like the example in Chapter 3, and therefore a single visibility will not be well-defined.

You can indeed to exactly what you said in your question. It's just that the generalizations made in the single-mode case regarding visibility corresponding to $|g^{(1)}(\tau)|$ are no longer really valid.

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  • $\begingroup$ This answer is on the right track, but I feel still confused. I have two points in mind: 1) when you say single oscillation you don't mean single mode, right? Because the example $e^{(i\omega_0 \tau- \gamma |\tau|)} $ refer to multimode, Lorentzian light. The $\omega_0$ is the central frequency. Maybe with "single oscillation" you mean single peak in the spectrum? $\endgroup$
    – Mark_Bell
    Nov 23, 2021 at 10:45
  • $\begingroup$ 2) My confusion arise also indeed from these two problems 3.2 and 3.3 of Loudon. You wrote the result of problem 3.3, but you start with the question of problem 3.2. In problem 3.2 all is fixed, there is no randomness and, according to Loudon you have to prove that is coherent: $|g^{(1)} (\tau)| = 1$, while 3.3 ask for the answer when both amplitudes and phases are random and you obtain the cosine term. First when I try to compute the answer of 3.2 I obtain the answer for 3.3, but that's my problem, maybe I did some mistakes. $\endgroup$
    – Mark_Bell
    Nov 23, 2021 at 10:50
  • $\begingroup$ But taking the right answer from Loudon you have where my confusion arise. A sum of two waves with two frequencies gives a result from first order coherence that imply perfect coherence, modulus of $g^{(1)}$ is equal to 1. But is multimode. So for summarize my question: perfect coherence can arise even for multimode light, but there is no simple relationship with the fringes visibility. I hope I have explain correctly what I mean. $\endgroup$
    – Mark_Bell
    Nov 23, 2021 at 10:53
  • $\begingroup$ @Mark_Bell indeed when I say "single oscillation," what I mean is NOT single mode, but rather the fact that $g^{(1)}$ itself oscillates with a single frequency, referring here to the $e^{-i\omega_0 t}$ term. If you have for example a sum of frequencies as in Problem 3.2, the first-order coherence no longer oscillates with a single frequency. So indeed it's exactly as you described in your last comment: multimode light can be first-order coherent, but there's no simple relationship to the fringe visibility anymore. $\endgroup$ Nov 23, 2021 at 13:47
  • $\begingroup$ @Mark_Bell you're correct that I missed that difference between Problems 3.2 and 3.3. I'll probably edit that out of the answer. The point I was trying to make, without properly doing the work, was that if you work out $g^{(1)}$ for a general multi-mode field, you won't get a single $e^{i\omega t}$ term anymore like you do for a single-mode field or for some special cases like thermal light. $\endgroup$ Nov 23, 2021 at 13:49

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