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Is there any connection between fluctuation dissipation theorem and Kramers-Kronig relations? They are often described together under linear response theory but I do not see any exact connection (like one being special case of another).

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    $\begingroup$ Kubo uses Kramers-Kronig on p. 22 of his paper The Fluctuation-Dissipation Theorem, but only once. $\endgroup$ Mar 21, 2016 at 10:43
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    $\begingroup$ Fluctuations and dissipations are the real and imaginary parts of the response function, and are related to each other via Kramers-Kronig relations. The general mathematical statement behind everything is the Hilbert transform. See these key-words (Kramers-Kronig / fluctuation-dissipation / Hilbert transform) on Wikipedia, and answer your question yourself :-) $\endgroup$
    – FraSchelle
    Mar 22, 2016 at 21:28

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Kramers-Kronig relations can be understood as requirement that $\operatorname{Re} \chi(\omega)$ is an even function, while $\operatorname{Im} \chi(\omega)$ is an odd function - see this answer and the linked Wikipedia therein.

The general Fluctuation-Dissipation theorem usually written as proportionality between fluctuation energy spectrum and imaginary part of fourier of responce function: $$S_x(\omega) = \frac{2kT}{\omega}\operatorname{Im} \chi(\omega)$$ While the lhs of the above expression $S_x(\omega)=\langle x(\omega)x^*(\omega)\rangle$ is easily interpreted as "fluctuation" part. It is not at all obvious that $\operatorname{Im} \chi(\omega)$ is related to "dissipation". The standard derivation below uses the the Kramers-Kronig symmetry requirements and links $\operatorname{Im} \chi(\omega)$ to the dissipation of the system.

Assuming that we have the observable $x(t)$, the stochastic external force $f(t)$ and the response function $\chi(t)$ are linked via: $$ x(t) = \int_{-\infty}^{\infty}d\tau\; \chi(t-\tau)f(\tau)$$

We want to compute dissipated power $f(t)\dot x(t)$ under periodic force $f(t) = A \cos\Omega t$. To do that we first simplify the expression for $\dot{x}(t)$. Taking the derivative gives us factor $-i\omega$ and integration over $\tau$ gives a double-$\delta$ fourier transform of the cosine:

$$\dot{x}(t) = \int \frac{d\omega}{2\pi}(-i\omega) e^{-i\omega t}\chi(\omega) \pi A[\delta(\omega - \Omega) + \delta(\omega+\Omega)]$$

Integrating out the $\delta$s we'll get: $$\dot{x}(t) =-iA\frac\Omega2\left[\chi(\Omega)e^{-i\Omega t} - \chi(-\Omega)e^{i\Omega t} \right]$$

Now we multiply by $f(t)$ and average over one period:

$$\langle W\rangle = \frac{\Omega}{2\pi}\int_0^{2\pi/\Omega}dt\;f(t)\dot{x}(t) = -\frac{i A^2\Omega}{4} \left[\chi(\Omega) - \chi(-\Omega)\right] = \frac12 A^2\Omega\operatorname{Im}\chi(\Omega) $$ In the last equality we've used that real part of $\chi(\omega)$ is even, while imaginary is odd.
This links $\operatorname{Im} \chi(\omega)$ to dissipation using Kramers-Kronig relations.

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  • $\begingroup$ What I understand from your answer is that there is no deeper connection between FD theorem and KK relations apart from the fact that $\chi(\omega)$ appears in both the relations and it represents dissipation. $\endgroup$ Apr 24, 2021 at 19:10
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    $\begingroup$ @mithusengupta123 I would say that KK relations are just mathematical equalities for a Fourier transform of certain class of functions. So KK don't really say anything about what $Im \chi$ represents. You can show that $Im \chi$ represents dissipation using KK relations (as done in the answer). $\endgroup$
    – Kostya
    Apr 24, 2021 at 19:53
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    $\begingroup$ I liked your answer. Particularly the second part. But I think the oddness of the imaginary part of $\chi(\omega)$ follows directly from the defintion of $\chi(\omega)$ which is the Fourier transform of the response function, and reality of theresponse function. One need not use KK relations to show that. I would like to know your comment on this. $\endgroup$ Apr 24, 2021 at 20:02

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