Talk:Fluctuation-dissipation theorem

message: Depending on which convention one uses for the Fourier transform, the FDT adopts different signs.
Definition (1):
:  \hat A(\omega) = \int\limits_{-\infty}^\infty \mathrm{e}^{i\omega t} A(t)\, dt 
gives
:  S_x(\omega) = \frac{2k_B T}{\omega} \, \mathrm{Im}[\hat\chi(\omega)].
Definition (2):
:  \hat A(\omega) = \int\limits_{-\infty}^\infty \mathrm{e}^{-i\omega t} A(t)\, dt 
gives
:  S_x(\omega) = - \frac{2k_B T}{\omega} \, \mathrm{Im}[\hat\chi(\omega)].
The article uses the first form of the FDT, but in its derivation a Fourier transform of form (2) is used.
This inconsistency should be corrected, but I wanted to discuss this first.
Benjamin.friedrich (talk) 22:30, 6 December 2017 (UTC)
thread_title: Potential sign error

message: I've changed the headline paragraph for the FDT entry to correct misleading text. The rest of the introductory section is problematic. Please see my Talk page. 
-- Pwfen (talk) 14:26, 25 January 2015 (UTC)
replies: text: Sure, Benjamin, I'll be glad to help you, although I'm apt to over-commit myself.  I appreciate all your work so far in shaping up this little article; most of it was just dashed off quickly while I was researching the equipartition theorem.  I meant to return to it, but I got distracted... :(  I just added some references and touched up the wording in the lead and my derivation section.  I think both derivations should go at the end, similar to what I did at Laplace-Runge-Lenz vector.  I'll need to read up on the fluctuation dissipation theorem a lot more, though, before I can really give you much help — sorry!  I learn pretty fast, though. :)  Willow (talk) 01:13, 19 December 2007 (UTC)

text: "Citation needed" / "Violations in glassy systems": "In the mid 1990s, in the study of non-equilibrium dynamics of spin glass models, a generalization of the fluctuation-dissipation theorem was discovered": I suspect you really intended one of the references you've listed in "Further reading", Cristani A, Ritort F (2003), as your citation for this point? - but I'm not familiar with it so I don't feel free to edit quite so boldly as to just put it in!!    SquisherDa (talk) 03:19, 9 April 2012 (UTC)
thread_title: Rewrite

message: I actually miss the theorem itself. Only examples, derivation and applications..--85.146.199.125 (talk) 20:22, 20 February 2008 (UTC)
thread_title: Definition of the theorem 

message: (Note: This is my first attempt at editing a Wikipedia page; I'm still unsure of syntax and etiquette. Advice and/or guidance will be very much appreciated.)
Equation (**) implies a discontinuous step-function response for  \langle x(t) \rangle - \langle x \rangle_0  at t=0, which is incorrect.  As t approaches zero from minus infinity,  \langle x(t) \rangle - \langle x \rangle_0  adiabatically approaches \  \beta f_0 A(0) .   Equation (**)  can be corrected by removing the outer parentheses on the right-hand side.   As a consequence of this change, the outer parentheses in the second term on the right-hand side in the final equation of the derivation must also be removed.  Thus the derivation beginning with Equation (**) should read: 
:  (**) -\chi(t) = \beta {\operatorname{d}A(t)\over\operatorname{d}t} 
\theta(t) . 
For stationary processes, the Wiener-Khinchin theorem states that
the power spectrum equals twice the Fourier transform of the auto-correlation
function
:  S_x(\omega) = 2 \tilde{A}(\omega). 
The last step is to Fourier transform equation (**) and to take the 
imaginary part. For this it is useful to recall that the Fourier transform
of a real symmetric function is real, while the Fourier transform of a real
antisymmetric function is purely imaginary.
We can split  {\operatorname{d}A(t)\over\operatorname{d}t} 
\theta(t)       into a symmetric and an
anti-symmetric part
:  2 {\operatorname{d}A(t)\over\operatorname{d}t} 
\theta(t)  \ = {\operatorname{d}A(t)\over\operatorname{d}t} + {\operatorname{d}A(t)\over\operatorname{d}t}{\rm sign}(t) . 
Now the fluctuation-dissipation theorem follows.
(Remark to reviewer, not to be included in the actual article: Observe that the corrected derivation above does not change the imaginary (absorptive) part of chi, but it does modify the real (dispersive) part.)
PaulH (talk) 01:58, 16 February 2011 (UTC)
replies: text: Hi Paul, welcome at Wikipedia, and please feel free to correct, improve, expand this article in whatever way. This entire field suffers from a lack of expert attention. -- Marie Poise (talk) 14:40, 16 February 2011 (UTC)
replies: text: Hey Paul, I think you have the etiquette mastered already! Using this Talk page is excellent, and it was especially kind of you to write. :)   I replied on my Talk page, but I neglected to mention there that we also need more references.  Willow (talk) 18:51, 16 February 2011 (UTC)
replies: text:  Can you provide a reference why we should have a factor of two between the power spectrum and the Fourier transform of the autocorrelation function? Maybe this is just a matter of defintion, but both this article as well as the article on the Wiener-Khinchin theorem do not have this factor of two. Benjamin.friedrich (talk) 09:44, 7 December 2017 (UTC)
thread_title:  Fix Error in Equation (**) in Derivation 

message: I don't like to delete the whole section, but don't see another way. The problem is this one: In order to expand in the pertubation, h and H would have to commute. (cf. Baker-Campbell-Hausdorff formula). But if they commute, A(t) is not anymore time-dependent and there will always hold  \bar A(t) = \langle A \rangle 
I don't think one can reconcile that and thus I vote for deleting the entire section.
129.206.196.179 (talk) 10:13, 9 June 2011 (UTC)
thread_title:  Derivation in trace notation is wrong 

message: There have been edits of this page that delete text in response to old comments about missing citations. An example of this was the deletion of the following statement:
"A proof can be found by means of the LSZ reduction, an identity from quantum field theory.[citation needed]"
* There was a partial citation (to the LSZ article in Wikipedia).
* Removing this statement did not improve the quality of the article. It degraded it.
* Please invest the effort find a peer-reviewed article that is missing if that is what bothers you.
This article is not in great shape, and deletions are generally not what is required. The work I did a while back on the Intro was intended to drag the article out of the "catastrophically wrong" category (a quote from a UC Berkeley faculty member, with which I agreed completely at the time).
("Catastrophically wrong" because prior versions of the FDT article had the basis and origin of the theorem completely wrong: e.g. the text appeared to have been written by sitting down with an undergraduate exposition that assumed equilibrium conditions for ease of derivation)
Pwfen (talk) 06:54, 29 August 2018 (UTC)
thread_title:  Deletion of partially cited or uncited text from FDT page 

message: The 9 April 2019 version of the FDT page <https://en.wikipedia.org/w/index.php?title=Fluctuation-dissipation_theorem&oldid=891240986) states that "The fluctuation–dissipation theorem was originally formulated by Harry Nyquist in 1928". This is incorrect. Einstein's 1905 Brownian motion paper is the first published example of a fluctuation-dissipation relation. Pwfen (talk) 14:22, 9 April 2019 (UTC)
thread_title:  First Example of the Fluctuation-Dissipation Theorem 

message: The fluctuation-dissipation theorem does not require thermal equilibrium to hold. It requires detailed balance to hold. Please see https://en.wikipedia.org/wiki/User_talk:Pwfen for a discussion of problems with the FDT page. Pwfen (talk) 09:37, 11 April 2019 (UTC)
thread_title:  Detailed balance versus thermal equilibrium 

message: There are few things that do not look 100% fine in that section and probably need some adjustment.
* The present strange formula \text{Im}\left[\chi(t)\right]=(\chi(t)-\chi(-t))/2i comes from equation (2) in the cited paper... where it is stated that \chi^d(t)=(\chi(t)-\chi(-t))/2i is the "dissipative part" of the response function. I kind of understand what all this means... since the Fourier transform of \chi^d(t) is indeed \text{Im}\left[\chi(\omega)\right]. However, this looks a bit like a misleading formula to me and anyway calling this \text{Im}\left[\chi(t)\right] is really weird since \chi(t) is a real causal funcion and strictly speaking it does not have any imaginary part.
* There seems to be a sign inconsistency between :\chi(t-t')=i\theta(t-t')\langle [\hat{x}(t),\hat{x}(t')] \rangle and \text{Im}\left[\chi(t)\right]=-\frac{1}{2}\left[\langle \hat{x}(t) \hat{x}(0)\rangle - \langle \hat{x}(0) \hat{x}(t)\rangle\right]
* The final formula with the "+1/2" is just the symmetric part of S_x(\omega)... indeed S_x(\omega) = 2\hbar[n_{BE}(\omega)+1]\chi(\omega)... plus 1, not plus 1/2.
* sometimes \hbar is dropped, other times it's not... I am ok with \hbar=1 but then it should be used all the times.
=== Suggestions ===
I am ok with the initial discussion but I would refer to \text{Im}\left[\chi(\omega)\right] and just drop this strange \text{Im}\left[\chi(t)\right]. Then I am ok with Kubo formula, this is a very good starting point
:\chi(t) = \frac{i\theta(t)}{\hbar}\langle\left[x(t),x(0)\right]\rangle
if we take Kubo for granted then the final result can be derived just by a simple Fourier transform 
:\chi(\omega) = \frac{i}{\hbar}\int_0^{\infty}\langle \left[\hat{x}(t),\hat{x}(0)\right] \rangle e^{i\omega t}dt
so
:\text{Im}\left[\chi(\omega)\right] = \frac{\chi(\omega)-\chi^*(\omega)}{2i}=
\frac{1}{2\hbar}\int_0^{\infty}\langle \left[\hat{x}(t),\hat{x}(0)\right] \rangle e^{i\omega t}dt
+\frac{1}{2\hbar}\int_0^{\infty}\langle\left[\hat{x}(0),\hat{x}(t)\right]\rangle e^{-i\omega t}dt = \frac{1}{2\hbar}\int_{-\infty}^{+\infty}\langle\left[\hat{x}(t),\hat{x}(0)\right]\rangle e^{i\omega t}dt
then the following part is very correct: "In the canonical ensemble, the second term can be re-expressed as
:\langle \hat{x}(0) \hat{x}(t)\rangle=\text{Tr } e^{-\beta \hat{H}}\hat{x}(0)\hat{x}(t)=\text{Tr } \hat{x}(t) e^{-\beta \hat{H}}\hat{x}(0)=\text{Tr } e^{-\beta \hat{H}}\underbrace{e^{\beta \hat{H}}\hat{x}(t) e^{-\beta \hat{H}}}_{\hat{x}(t-i\hbar\beta)}\hat{x}(0)=\langle \hat{x}(t-i\hbar\beta) \hat{x}(0)\rangle
where in the second equality we re-positioned \hat{x}(t) using the cyclic property of trace (in this step we have also assumed that the operator \hat{x} is bosonic, i.e. does not introduce a sign change under permutation). Next, in the third equality, we inserted e^{-\beta \hat{H}}e^{\beta \hat{H}} next to the trace and interpreted e^{-\beta\hat{H}} as a time evolution operator e^{-\frac{i}{\hbar}\hat{H}\Delta t}  with imaginary time interval \Delta t=-i\hbar\beta."... and if we plug this in the previous equation we get
:\text{Im}\left[\chi(\omega)\right] = \frac{1}{2\hbar}\int_{-\infty}^{+\infty}\langle\hat{x}(t)\hat{x}(0)-\hat{x}(t-i\hbar\beta)\hat{x}(0)\rangle e^{i\omega t}dt= \frac{1-e^{-\beta\hbar\omega}}{2\hbar}\int_{-\infty}^{+\infty}\langle\hat{x}(t)\hat{x}(0)\rangle e^{i\omega t}dt=
 \frac{1-e^{-\beta\hbar\omega}}{2\hbar}S_{x}(\omega)
and finally, with minor reshuffling of the factors, to
:S_{x}(\omega)=2\hbar\left[n_{\rm BE}(\omega)+1\right]\text{Im}\left[\chi(\omega)\right]
where n_{\rm BE}(\omega)=1/\left(e^{\beta\hbar\omega}-1\right): note here we have a +1, not +1/2. The 1/2 makes sense but only because the same calculation for -\omega yields 
:S_{x}(-\omega)=2\hbar\left[n_{\rm BE}(\omega)\right]\text{Im}\left[\chi(\omega)\right]=e^{-\beta\hbar\omega}S_{x}(\omega)
without the +1. So the cited formula with 1/2 is just the symmetric part of power spectral density (S_x(\omega)+S_x(-\omega))/2. In quantum noise S_x(\omega) is not symmetric and, consistently, the autocorrelation \langle x(t)x(0)\rangle contains an imaginary part. This is something that is not particularly trivial/intuitive and might deserve a separate discussion... possible reference/examples here
Wikirod76 (talk) 00:57, 9 January 2021 (UTC)
thread_title:  Problems with quantum derivation of FDT 

message: I am ok with the initial discussion but I would refer to \text{Im}\left[\chi(\omega)\right] and just drop this strange \text{Im}\left[\chi(t)\right]. Then I am ok with Kubo formula, this is a very good starting point
:\chi(t) = \frac{i\theta(t)}{\hbar}\langle\left[x(t),x(0)\right]\rangle
if we take Kubo for granted then the final result can be derived just by a simple Fourier transform 
:\chi(\omega) = \frac{i}{\hbar}\int_0^{\infty}\langle \left[\hat{x}(t),\hat{x}(0)\right] \rangle e^{i\omega t}dt
so
:\text{Im}\left[\chi(\omega)\right] = \frac{\chi(\omega)-\chi^*(\omega)}{2i}=
\frac{1}{2\hbar}\int_0^{\infty}\langle \left[\hat{x}(t),\hat{x}(0)\right] \rangle e^{i\omega t}dt
+\frac{1}{2\hbar}\int_0^{\infty}\langle\left[\hat{x}(0),\hat{x}(t)\right]\rangle e^{-i\omega t}dt = \frac{1}{2\hbar}\int_{-\infty}^{+\infty}\langle\left[\hat{x}(t),\hat{x}(0)\right]\rangle e^{i\omega t}dt
then the following part is very correct: "In the canonical ensemble, the second term can be re-expressed as
:\langle \hat{x}(0) \hat{x}(t)\rangle=\text{Tr } e^{-\beta \hat{H}}\hat{x}(0)\hat{x}(t)=\text{Tr } \hat{x}(t) e^{-\beta \hat{H}}\hat{x}(0)=\text{Tr } e^{-\beta \hat{H}}\underbrace{e^{\beta \hat{H}}\hat{x}(t) e^{-\beta \hat{H}}}_{\hat{x}(t-i\hbar\beta)}\hat{x}(0)=\langle \hat{x}(t-i\hbar\beta) \hat{x}(0)\rangle
where in the second equality we re-positioned \hat{x}(t) using the cyclic property of trace (in this step we have also assumed that the operator \hat{x} is bosonic, i.e. does not introduce a sign change under permutation). Next, in the third equality, we inserted e^{-\beta \hat{H}}e^{\beta \hat{H}} next to the trace and interpreted e^{-\beta\hat{H}} as a time evolution operator e^{-\frac{i}{\hbar}\hat{H}\Delta t}  with imaginary time interval \Delta t=-i\hbar\beta."... and if we plug this in the previous equation we get
:\text{Im}\left[\chi(\omega)\right] = \frac{1}{2\hbar}\int_{-\infty}^{+\infty}\langle\hat{x}(t)\hat{x}(0)-\hat{x}(t-i\hbar\beta)\hat{x}(0)\rangle e^{i\omega t}dt= \frac{1-e^{-\beta\hbar\omega}}{2\hbar}\int_{-\infty}^{+\infty}\langle\hat{x}(t)\hat{x}(0)\rangle e^{i\omega t}dt=
 \frac{1-e^{-\beta\hbar\omega}}{2\hbar}S_{x}(\omega)
and finally, with minor reshuffling of the factors, to
:S_{x}(\omega)=2\hbar\left[n_{\rm BE}(\omega)+1\right]\text{Im}\left[\chi(\omega)\right]
where n_{\rm BE}(\omega)=1/\left(e^{\beta\hbar\omega}-1\right): note here we have a +1, not +1/2. The 1/2 makes sense but only because the same calculation for -\omega yields 
:S_{x}(-\omega)=2\hbar\left[n_{\rm BE}(\omega)\right]\text{Im}\left[\chi(\omega)\right]=e^{-\beta\hbar\omega}S_{x}(\omega)
without the +1. So the cited formula with 1/2 is just the symmetric part of power spectral density (S_x(\omega)+S_x(-\omega))/2. In quantum noise S_x(\omega) is not symmetric and, consistently, the autocorrelation \langle x(t)x(0)\rangle contains an imaginary part. This is something that is not particularly trivial/intuitive and might deserve a separate discussion... possible reference/examples here
Wikirod76 (talk) 00:57, 9 January 2021 (UTC)
thread_title:  Suggestions 

message: In the General Formulation section, there is the sentence "The observable x(t) will fluctuate around its mean value \langle x\rangle_0
with fluctuations characterized by a power spectrum S_x(\omega) = \langle \hat{x}(\omega)\hat{x}^*(\omega) \rangle."  Using the definition of the Fourier transform given below, S_x(\omega) has units of [x]^2[t]^2, which do no match with the fluctuation dissipation theorem as stated.  On the power spectrum page, S_x(\omega) is defined with a 1/T, which resolves the issue.
Proposed change:
"The observable x(t) will fluctuate around its mean value \langle x\rangle_0
with fluctuations characterized by an auto correlation function S_x(\omega) = \int_{-\infty}^\infty dt e^{-i\omega t}\langle (\hat{x}(t_0-t)-\langle x\rangle_0)(\hat{x}(t_0)-\langle x\rangle_0) \rangle. Here, the fluctuations are assumed to be stationary (independent of t_0)." Skell013 (talk) 16:41, 21 May 2024 (UTC)
thread_title:  Misleading definition of power spectrum