Coherence Length

From Qwiki

Wikipedia gives a reasonable introduction to the idea of coherence of waves, so I'm not going to get into it. Here, what I will instead do is formally derive the relationship between the Spectral Density of a wave and its coherence time or, equivalently, coherence length. The result turns out to be a proof of a special case of the Wiener-Khintchine theorem, in which the process involved is deterministic.

Self-interfering waves find their way into all sorts of experiments in physics labs. In particular, the Michelson interferometer is the classic example of a beam interfering with itself, and it and related configurations have a large number of uses. But a wave must satisfy several criteria in order to produce an interference pattern with itself, and temporal coherence is an important one of these that turns out to be a direct result of the wave's spectral density.

Suppose, as in the case of the Michelson interferometer, that we pick off a beam at two points and superimpose the results on a detector. Further suppose that the beam has a spectral density $S (ω)$ . The power incident on the detector, as a function of $ω$ , is given by

$P(\omega) = S(\omega) \left[\cos(\omega t)+\cos(\omega t + \varphi(\omega))\right]^2$

where we have arbitrarily set the phase of one of the beams to zero at all frequencies. This is OK, because all that really matters is the relative phase of the two beams and this is how we will define $\varphi(\omega)$ in a few lines. After some algebra we have

$P(\omega) = S(\omega)(1+\cos(\varphi))\left\{1 + \cos\left[2 \omega t + 2 \tan^{-1}\left(\frac{\sin(\varphi)}{1 + \cos(\varphi)}\right)\right]\right\},$

and since our detector will typically have bandwidth that is much smaller than $ω$ (this is certainly the case with optical detectors) the ugly cosine term goes away:

$P(\omega) = S(\omega)(1 + \cos(\varphi)).$ At a single frequency, this result gives us the familiar sinusoidal interference fringes that we have all come to know and love. Furthermore, for a beam composed of some distribution of frequencies, we see that larger interference fringes will be produced when $\varphi$ varies little over this frequency distribution. One might guess that this requires a narrow distribution. One would be correct, as we will soon see.

We must ask, where do we get the relative phase $\varphi$ ? It simply comes from the time that the beam requires to travel the distance $x$ between the two pick-off points:

$\varphi = \frac{x \omega}{c}$

where $c$ is the speed of the wave (assumed to be frequency-independent for simplicity). Substituting this for $\varphi$ and integrating over $ω$ , we get the total power seen by the detector:

$P = \int_0^\infty S(\omega) \left[1+\cos\left(\frac{x \omega}{c}\right)\right]d\omega$ $= 1 + \Re \int_0^\infty S(\omega) \exp\left(\frac{x \omega i}{c}\right)d\omega.$

So we have that the power in the interference fringes is the inverse Fourier transform of the spectral density! This gives us plenty of mathematical insight into the phenomenon that we have already described: narrow-band sources give interference fringes over a large range of $x$ and conversely, wide-band sources give fringes over a narrow region.

Just so that we actually calculate something here, let's see what this looks like for a wave with Gaussian spectral density (because that's easy to integrate), with mean $ω 0$ and variance $σ 2$ :

$P = 1 + \cos\left(\frac{x \omega_0}{c}\right)\exp\left(-\frac{x^2 \sigma^2}{2 c^2}\right).$

In the limit $\sigma \rightarrow 0$ we get the usual sinusoidal fringes without damping characteristic of a laser source. In the opposite (white light) limit $\sigma \rightarrow \infty$ we don't see any interference at all. In general, we will see interference over a spatial region of width (defined as 1/e) approximately