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Homework Help: First-order coherence degree

  1. Jul 19, 2012 #1
    1. The problem statement, all variables and given/known data
    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]

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

    3. The attempt at a solution
    Here is my attempt at the first part if [itex]\varphi[/itex] is kept fixed. Lets 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?
  2. jcsd
  3. Jul 19, 2012 #2
    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]
    Last edited by a moderator: Mar 1, 2014
  4. Mar 1, 2014 #3
    Hi. did you find the answer?I am also interested to know
  5. Feb 26, 2016 #4
    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,
    E = E_0 \left ( e^{-i \omega_1 t} + e^{-i (\omega_2 t - \varphi)} \right).
    Determine the first-order temporal ##(z = 0)## coherence function for a randomly varying [itex]\varphi[/itex],
    g^{(1)}(t,\tau) = \frac{\langle E^*(t) E(t +\tau) \rangle }{\langle |E(t)|^2 \rangle }.
    Let's start by looking at the numerator,
    \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]

    Independent components containing the random variable [itex]\varphi[/itex] will vanish in the averaging. Thus we have,
    \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}.
    Let's now look at the denominator,
    \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)}.

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