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Second-order correlation function

  1. Oct 30, 2012 #1
    Hi all,

    I'm studying for my Quantum Optics exam and still have problems with the second-order correlation function.

    The question concerned is question 3b, 3c etc which can be found here: http://www.arago.utwente.nl/comms/sotn/tentamendatabase/optics/qo/351500_Quantum_Optics_2010-11-03.pdf [Broken]

    They want to know [itex]g^{(2)}(\tau)=\frac{<I(t) I(t+\tau)>}{<I(t)><I(t+\tau)>}[/itex]

    For [itex]\tau=0[/itex], this simplifies to [itex]\frac{<I(t)^2>}{<I(t)>^2}[/itex]
    For [itex]\tau=1[/itex], it is [itex]\frac{I(t)I(t+1)>}{<I(t)><I(t+1)>}[/itex]

    Now, I'd say that in the second case, the answer depends on time, since (take t=1) I(1) is not equal to I(2).

    If anyone can help me out (especially understanding what exactly is going on) I'd be grateful!

    Sven
     
    Last edited by a moderator: May 6, 2017
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  3. Oct 30, 2012 #2

    George Jones

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    Taking time averages gets rid of the [itex]t[/itex] dependence.

    Suppose [itex]f\left(t\right) = I\left(t\right)[/itex] and [itex]g\left(t\right) = I\left(t + 1\right) [/itex]. What do you have to do to the graph of [itex]f\left(t\right)[/itex] to get the graph of [itex]g\left(t\right)[/itex]?
     
  4. Oct 30, 2012 #3
    Ah, yeah. I was a bit confused about the time average, since in some (other) cases you average over very short times (much smaller than the 1 second in this assignment, thus not deleting the time dependence :P)

    To get g(t) from f(t) you'd have to move the whole graph one unit to the right.

    Thanks, I think with that tip I know what should be done :)
     
  5. Oct 30, 2012 #4

    George Jones

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    How about left? :)
     
  6. Oct 30, 2012 #5
    Ugh, yeah. That was really stupid of me :P
     
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