# Second-order correlation function

1. Oct 30, 2012

### svenvbins

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 $g^{(2)}(\tau)=\frac{<I(t) I(t+\tau)>}{<I(t)><I(t+\tau)>}$

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

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
2. Oct 30, 2012

### George Jones

Staff Emeritus
Taking time averages gets rid of the $t$ dependence.

Suppose $f\left(t\right) = I\left(t\right)$ and $g\left(t\right) = I\left(t + 1\right)$. What do you have to do to the graph of $f\left(t\right)$ to get the graph of $g\left(t\right)$?

3. Oct 30, 2012

### svenvbins

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 :)

4. Oct 30, 2012

### George Jones

Staff Emeritus
How about left? :)

5. Oct 30, 2012

### svenvbins

Ugh, yeah. That was really stupid of me :P