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i need to derive a formula for the photon gas correlation function $\left\langle\partial n_i\partial n_j\right\rangle $ where $$\partial n_i=n_i -\left \langle n_i \right \rangle.$$

whilst solving I saw that $$ \left \langle n_in_j \right \rangle = \frac{1}E. \frac{\partial ^2E}{\partial \beta \epsilon _i.\partial \beta \epsilon _j} =\frac{1}{E}\frac{\partial }{\partial (-\beta \epsilon _j)}(E\left \langle n_i \right \rangle)$$ but that says $\left \langle n_i \right \rangle\left \langle n_j \right \rangle i\neq j $ that means $n_i$ and $n_j$ are uncorrelated. I don't understand. How is this possible?

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Well photons do not really interact, especially when seen as a thermal gas. Intuitively you can see across the room despite all the reflection from the walls because the photons coming from across the room do not interact with other ones reflected by other walls that cross its path. If you look at field theory you find that photons interact very weakly. Lastly photons can be put in quantum states that are highly correlated (entangled) but that is not the thermal state you have in mind I don't think.

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  • $\begingroup$ yes i found it and i did understand it ... $\left \langle n_i \right \rangle$ $i=j$ since $\left \langle n_i^2 \right \rangle =\left \langle n_i \right \rangle$ therefore $g_i_j = \left \langle n_i \right \rangle \left \langle n_j \right \rangle (1-\partial _i_j) $ $\endgroup$ – relston mendonsa Jul 3 '14 at 21:43
  • $\begingroup$ good job on finding it! $\endgroup$ – Ramis Movassagh Jul 17 '14 at 22:30

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