How Does First-Order Coherence Degree Relate to Signal Visibility?

[Sigurdsson]

Homework Statement

Hi guys, appreciate all the help I can get. This has been bugging me for 24 hours now. I'm starting to think I'm missing something in the question.

We are exploring first-order coherence degree. That is, exploring the coherence of two separate signals (wave packets) by using the equation
[tex]g^{(1)}(\tau, t) = \frac{\langle E^\ast (t) E(t + \tau) \rangle}{\langle |E(t)|^2 \rangle }[/tex]
If you are familiar with the Michelson interferometer, then you should be familiar with the equation of fringe visibility
[tex] V = \frac{I_{\text{max}} - I_{\text{min}} }{I_{\text{max}} + I_{\text{min}} } [/tex]
Which equates to
[tex] V = |g^{(1)}(\tau)| [/tex]

So here is the question. We are given the field of two different signals
[tex]\frac{E(x,t)}{E_0} = e^{i(k_1 z - \omega_1 t)} + e^{i(k_2 z - \omega_2 t + \varphi)}[/tex]
with the common amplitude [itex]E_0[/itex] and dephasing difference [itex]\varphi[/itex]. The goal is to show that if [itex]\varphi[/itex] is kept fixed we get
[tex] V = |g^{(1)}(\tau)| = 1 [/tex]

and if it varies randomly between measurements, the averaging should yield
[tex] V = |g^{(1)}(\tau)| = \left|\cos{\left( \frac{1}{2}(\omega_1 - \omega_2) \tau \right) } \right| [/tex]

Homework Equations

This is exercise 2.1 in the book Microcavities by Alexey V. Kavokin.

The Attempt at a Solution

Here is my attempt at the first part if [itex]\varphi[/itex] is kept fixed. Let's put [itex]a = k_1 z - \omega_1 t[/itex] and [itex]b = k_2 z - \omega_2 t + \varphi[/itex] for simplicity's sake. Then we have

[tex] \frac{\langle E^\ast (t) E(t + \tau) \rangle}{\langle |E(t)|^2 \rangle } = \frac{ \langle \left( e^{-ia} + e^{-ib} \right) \left( e^{ia} e^{-i\omega_1 \tau} + e^{ib} e^{-i\omega_2 \tau} \right) \rangle }{ \langle 2 + e^{-i(a -b)} + e^{i(a-b)} \rangle } = \frac{\langle (e^{i(a-b)} + 1) e^{-i\omega_1 \tau} + (e^{-i(a-b)} + 1) e^{-i\omega_2 \tau} \rangle}{\langle 2 + e^{-i(a -b)} + e^{i(a-b)} \rangle} [/tex]

This doesn't look like unity to me. The problem is (I think) is that I'm not sure how to apply the method of "averaging" in this example. Any tips?

 

[Sigurdsson]

Getting a little closer, I just realized that the average value of cosine over 2[itex]\pi[/itex] is zero. Meaning...

[tex] \frac{\langle E^\ast (t) E(t + \tau) \rangle}{\langle |E(t)|^2 \rangle } [/tex]
[tex] = \frac{ \langle \left( e^{-ia} + e^{-ib} \right) \left( e^{ia} e^{-i\omega_1 \tau} + e^{ib} e^{-i\omega_2 \tau} \right) \rangle }{ \langle 2 + e^{-i(a -b)} + e^{i(a-b)} \rangle } [/tex]
[tex]= \frac{\langle (e^{i(a-b)} + 1) e^{-i\omega_1 \tau} + (e^{-i(a-b)} + 1) e^{-i\omega_2 \tau} \rangle}{\langle 2 + 2 \cos{(a-b)} \rangle}[/tex]
[tex] = \frac{\langle (e^{i(a-b)} + 1) e^{-i\omega_1 \tau} + (e^{-i(a-b)} + 1) e^{-i\omega_2 \tau} \rangle}{2}[/tex]

 

[hamid112]

Hi. did you find the answer?I am also interested to know

 

[Sigurdsson]

Sorry for the freakishly late reply. I revisited the book Microcavities again and remembered that I never finished the problem here. Here is my take on it,

The solution is as follows:
Let's consider the following electric field composed of two plane waves,
[tex]
E = E_0 \left ( e^{-i \omega_1 t} + e^{-i (\omega_2 t - \varphi)} \right).
[/tex]
Determine the first-order temporal ##(z = 0)## coherence function for a randomly varying [itex]\varphi[/itex],
[tex]
g^{(1)}(t,\tau) = \frac{\langle E^*(t) E(t +\tau) \rangle }{\langle |E(t)|^2 \rangle }.
[/tex]
Let's start by looking at the numerator,
[tex]
\begin{align} \notag

E^*(t) E(t+\tau) & = E_0^2 \left( e^{i \omega_1 t} + e^{i (\omega_2 t - \varphi)} \right) \left( e^{-i \omega_1 (t+\tau)} + e^{-i (\omega_2 (t+\tau) - \varphi)} \right) \\ \notag

& = E_0^2 \left[ e^{- i \omega_1 \tau} + e^{-i \omega_2 \tau} + e^{i(\omega_1 - \omega_2) t} e^{- i \omega_2 \tau} e^{i \varphi} + e^{-i(\omega_1 - \omega_2) t} e^{- i \omega_1 \tau} e^{-i \varphi} \right] \\ \notag

& = E_0^2 e^{-i( \omega_1 + \omega_2)\tau/2} \left[ e^{-i(\omega_1 - \omega_2) \tau/2} \left( 1 + e^{-i (\omega_1 - \omega_2) t} e^{-i \varphi} \right) + e^{i (\omega_1 - \omega_2)\tau/2} \left( 1 + e^{i(\omega_1 - \omega_2)t } e^{i \varphi} \right) \right]

\end{align}
[/tex]
Independent components containing the random variable [itex]\varphi[/itex] will vanish in the averaging. Thus we have,
[tex]
\langle E^*(t) E(t + \tau) \rangle = E_0^2 e^{-i( \omega_1 + \omega_2)\tau/2} \left[ e^{-i(\omega_1 - \omega_2) \tau/2} + e^{i (\omega_1 - \omega_2)\tau/2} \right] = 2E_0^2 \cos{ \left( \frac{\omega_1 - \omega_2}{2} \tau \right)} e^{-i( \omega_1 + \omega_2)\tau/2}.
[/tex]
Let's now look at the denominator,
[tex]
\begin{align} \notag

|E(t)|^2 & = E_0^2 \left[ 2 + e^{-i(\omega_1 - \omega_2)t} e^{-i \varphi} + e^{i(\omega_1 - \omega_2)t} e^{i \varphi}\right] \\ \notag

& = E_0^2 \left[ 2 + 2 \cos{ \left( (\omega_1 - \omega_2) t + \varphi \right) } \right] \\ \notag

& = 4 E_0^2 \cos^2{ \left( \frac{\omega_1 - \omega_2 }{2} t + \frac{\varphi}{2}\right)}.

\end{align}
[/tex]
The average of the cosine will be [itex]1/2[/itex]. Thus in the end we have,
[tex]
g^{(1)}(t,\tau) = \frac{\langle E^*(t) E(t +\tau) \rangle }{\langle |E(t)|^2 \rangle } = \frac{2E_0^2 \cos{ \left( \frac{\omega_1 - \omega_2}{2} \tau \right)} e^{-i( \omega_1 + \omega_2)\tau/2}}{2 E_0^2} = \cos{ \left( \frac{\omega_1 - \omega_2}{2} \tau \right)} e^{-i( \omega_1 + \omega_2)\tau/2}.
[/tex]
And thus,
[tex]
|g^{(1)}(t,\tau)| = \left| \cos{ \left( \frac{\omega_1 - \omega_2}{2} \tau \right)} \right|.
[/tex]
Which is what was supposed to be shown.