## 21 October 2013

### Coherence 3 - The Michelson interferometer

The Michelson interferometer (shown below) is an excellent setup for illustrating the concept of longitudinal coherence.  We will simplify the formalism of the previous post down to three essential ingredients:
• The beam consists of wave pulses of length l (corresponding to ξl in post 2).
• They are "split in half"; each half goes through one of the arms.
• The "twin" pulses arrive at the detector with a shift ΔL.
 The image on the left is from Aaron Kennard. The light beam is split in two: beam 1 is reflected towards the mirror M1 and beam 2 continues towards M2. After reflection in the mirrors, the two sub-beams reach the beam splitter again; 1 goes through and 2 is reflected, so they recombine before reaching the detector, but with a path difference ΔL (twice the difference in arm length).

Intuitively, since the pulses are emitted at random times, we would estimate that "each pulse only interferes with its twin", so that the visibility of the interference fringes goes to zero for  ΔL > l (when the twins no longer superpose).

Let us now try to give a rigorous expression to the above reasoning. Each pulse (of duration $$\Delta t = l /c$$) is described by the function $$f(t)$$ (with a Fourier transform $$F(\nu)$$), which in practice is an oscillation modulated by an envelope of width $$\Delta t$$, as in the example below (note that $$\Delta \nu \simeq 1/ \Delta t$$, a general property of the Fourier transform):
The full signal is:
\label{pulses} V(t) = \sum_{n} f(t-t_n) where pulse $$n$$ is emitted at time $$t_n$$ and the $$\left \lbrace t_n \right \rbrace$$ sequence is completely random (e.g. individual atomic emission events in a thermal source).

The signal that reaches the detector is of the form: $$V_{\text{det}}(t)=V_1(t)+V_2(t)=V(t)+V(t-\tau)$$, with $$\tau = \Delta L /c$$.

We are interested in the intensity averaged over a long time T:
\label{interf}
I(\tau) = \left \langle \left | V_{\mbox{det}}(t)\right | ^2 \right \rangle _{T}=
\frac{1}{2T} \int_{-T}^{T} \text{d} t \, \left [ \left | V(t)\right | ^2 + \left | V(t-\tau)\right | ^2 +
2 V(t)\tilde{V}(t-\tau) \right ]
To evaluate the third term, which is exactly the coherence function (see post 1) we apply the Wiener-Khinchin theorem (W.-K.): $\frac{1}{2T} \int_{-T}^{T} \text{d} t \, V(t)\tilde{V}(t-\tau) = \Gamma _V(\tau) \stackrel{\mbox{W.-K.}}{=} \frac{1}{2T} \int_{-\infty}^{\infty} \text{d} \nu \, \left | V(\nu)\right | ^2 \text{e}^{2i\pi \nu \tau}$
Let us now take the Fourier transform of (\ref{pulses}):
$V(t)=\sum_{n} f(t-t_n) \Longrightarrow V(\nu) = F(\nu) \sum_{n=1}^{N} \text{e}^{2i\pi \nu t_n} \Longrightarrow \tilde{V}(\nu) = \tilde{F}(\nu) \sum_{m=1}^{N} \text{e}^{-2i\pi \nu t_m}$$\left | V(\nu)\right | ^2 = \left | F(\nu)\right | ^2 \sum_{n,m=1}^{N} \text{e}^{2i\pi \nu (t_n - t_m)} = N \left | F(\nu)\right |^2 + \left | F(\nu)\right | ^2 \times \overbrace{\sum_{n \neq m} \text{e}^{2i\pi \nu (t_n - t_m)}}^0$ Finally, from (\ref{interf}): \label{GVT} \Gamma _V(\tau) = \underbrace{\frac{N}{2T}}_{n/2} \int_{-\infty}^{\infty} \text{d} \nu \, \text{e}^{2i\pi \nu \tau} \left | F(\nu)\right | ^2 =
\frac{n}{2}\int_{-\infty}^{\infty} \text{d} \nu \, \cos (2\pi \nu \tau) \left | F(\nu)\right | ^2 Remarkably, in only a few lines we managed to obtain the full coherence function $$\Gamma _V(\tau)$$ in terms of the power spectrum of the pulses $$F(\nu)$$! Let us now show from Equation (\ref{GVT}) that $$\Gamma _V(\tau)$$ does indeed vanish for $$\Delta L > l \Leftrightarrow \tau > \Delta t$$.

We will do this graphically, using the Figure above as support. When the lobes of the cosine function are wider than the spectral width $$\Delta \nu$$ of the pulses (i.e. when $$\tau \Delta \nu \ll 1$$) the integral oscillates as a function of the shift $$\tau$$. When, on the other hand, $$\tau \Delta \nu > 1$$ and the cosine oscillates many times over the width $$\Delta \nu$$, the integral is practically zero, since the positive and negative contributions cancel out, and this independently of the shift. Defining the visibility criterion as $$\tau \Delta \nu = 1$$ and recalling that $$\Delta \nu \simeq 1/ \Delta t$$ yields simply $$\tau = \Delta t$$.

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