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).

  • 3
    $\begingroup$ Kubo uses Kramers-Kronig on p. 22 of his paper The Fluctuation-Dissipation Theorem, but only once. $\endgroup$ Mar 21, 2016 at 10:43
  • 5
    $\begingroup$ 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

1 Answer 1


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.