- #1
Azelketh
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Homework Statement
Hi, I am trying to work through exercise 2.1 on page 37 of Microcavities (by alexy kavokin, jeremy baumberg, guillaume malpuech and fabrice laussy)
the problem is to prove
[tex]
| g^{(1)}(\tau) | = | cos( \frac{1}{2}(\omega_1 - \omega_2)\tau) ) |
[/tex]
where:
[tex]
g^{(1)}(\tau)=\frac{\langle E^{\ast}(t)E(t+\tau)\rangle}{\langle |E(t)|^2 \rangle}
[/tex]
and
[tex]
E(t)=E_0(t)\exp^{i[k_1z-\omega_1t]}+E_0(t)\exp^{i[k_2z-\omega_2t+\varphi]}
[/tex]
where
[tex]
\varphi [/tex] varies randomly between measurements
how do you deal mathmatically with [tex] \varphi [/tex] varying??
Also more simply above the exercise the text states a simpler apparently 'trivial' result using the same formula for [tex] g^{(1)}(\tau) [/tex] that the sine wave of
[tex]
E(t)=E_0(t)\exp^{i[\omega t - kz + \varphi]}
[/tex]
by direct application of the formula for [tex] g^{(1)}(\tau) [/tex] yields:
[tex]
g^{(1)}(\tau)= \exp^{i\omega \tau}
[/tex]
i cannot show even this 'trivial' application, i find that:
[tex]
\langle E^{\ast}(t)E(t+\tau)\rangle = \langle E_0\exp^{-i\omega \tau}\rangle
[/tex]
and
[tex]
\langle |E(t)|^2 \rangle = \langle |E_0^2 \exp^{2i(\omega t -kx + \varphi)} \rangle
[/tex]
How does that evaluate to
[tex]
g^{(1)}(\tau)= \exp^{-i\omega \tau}
[/tex] ??
If anyone can give me any pointers( or show me the complete workings of this XD ) then it would much appreciated. Thanks for reading my post.
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