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I'm trying to relate a known quantum correlation function $C_j(t-t') = \left<\hat{B}_j(t)\hat{B}_j(t')\right>$ (which is not real!) of a (time dependent, but this is not super relevant) quantum operator $\hat{B}_j(t)$ to the mean squared of this operator: $\left<\hat{B}_j(t)^2\right>$, given that I know the exact form of $C_j(t-t')$.

Now, to start with I am not sure if what I want to do is possible with the information I have, or more precisely, with the information I think could be enough. If parts are missing, it is very possible we can still figure them out; I will simply have to provide you with much more (messy) context.

What I have is the following relationship:

$C_j(t-t') = \int_{0}^{\infty}{d\omega S(\omega)\left(\coth(\beta \hbar \omega /2)\cos(\omega (t-t')) - i \sin(\omega (t-t'))\right)}$ where $S(\omega)$ is the spectral density. It appears to me that from this, you cannot get what I want, or am I mistaken? Could one perhaps change the t's to t' and t'', and integrate them up to t?

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    $\begingroup$ It's simpler than you think, just set $t = t'$. This makes sense because your state is stationary (otherwise $C(t,t')$ would depend on both $t$ and $t'$ separately, rather than their difference), so the result for $\langle B_j(t)^2\rangle$ should be $t$-independent. $\endgroup$ Apr 18 '16 at 15:34
  • $\begingroup$ It really does work like that? Hm. I did think about this originally. It is simply a bit confusing because another source (a different paper) cites the result to be $(1 + \Coth)S(\omega)$, while here I do not get the 1 + term. So I will have to dive into what is different between the two. Thanks in any case! $\endgroup$
    – user129412
    Apr 19 '16 at 7:52

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