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Depending on which convention one uses for the Fourier transform, the FDT adopts different signs.
Definition (1):
gives
Definition (2):
gives
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)
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)
I started a major rewrite of this article.
What was the state before: IMO, the second part of the original version was not very clear.
What I have done: I included a section 'General form' where I give a general FDT in its simplest form; I also included a derivation.
What can be done next:
I would appreciate your help.
-- Benjamin.friedrich ( talk) 12:18, 18 December 2007 (UTC)
I actually miss the theorem itself. Only examples, derivation and applications..-- 85.146.199.125 ( talk) 20:22, 20 February 2008 (UTC)
Perhaps nitpicking, but: In "general formulation," the perturbing field f seems to be off a minus sign: dp/dt = -dH/dx = -dH0/dx - f(t) as written. So then f is *minus* the applied force. This should be changed to be consistent with the definition of susceptibility in that wikipedia article. Dstrozzi ( talk) 00:02, 27 March 2017 (UTC)
The old definition is not correct:
Rather and are swapped. It should be:
such that only depends on the perturbation before . 76.199.99.43 ( talk) 10:41, 21 April 2009 (UTC)
(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 at t=0, which is incorrect. As t approaches zero from minus infinity, adiabatically approaches . 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:
For stationary processes, the Wiener-Khinchin theorem states that the power spectrum equals twice the Fourier transform of the auto-correlation function
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 into a symmetric and an anti-symmetric part
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)
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
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)
Correction: Now I think, it's right. But there is a huge lack of explanation. I hope, I'll get back to inserting it.
129.206.91.26 ( talk) 16:36, 9 June 2011 (UTC)
Another correction: I checked it, it's definitely wrong. I hope, I'll find the time to point out, why.
129.206.91.26 ( talk) 17:49, 9 June 2011 (UTC)
So, here it comes: The Taylor-expansion in numerator and denominator need the following approximation to hold (which is far from obvious for non-commuting H and h, check Baker-Campbell-Hausdorff formula):
and in the denominator:
While the second approximation seems to hold (I checked some matrices with oracle), the first approximation is simply wrong. Demanding A to be small is probably a solution, but it's highly unphysical to expect some observable to be small compared to the Hamiltonian. Another possibility is to demand and to commute. But then the time evolution of our equilibrium state is trivial and the FDT states: .
For these reasons I will now delete the whole "Derivation in trace notation" - section. Hope, your ok with that. — Preceding unsigned comment added by 129.206.196.143 ( talk) 16:10, 12 June 2011 (UTC)
I am trying to comprehend why the field is always taken to be time-independent. I understand why it is turned off at time zero, but there should be a way to derive this without requiring time independence of the field. 129.255.226.50 ( talk) 18:06, 19 August 2013 (UTC)
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]"
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)
The 9 April 2019 version of the FDT page < https://en.wikipedia.org/?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)
The fluctuation-dissipation theorem does not require thermal equilibrium to hold. It requires detailed balance to hold. Please see /info/en/?search=User_talk:Pwfen for a discussion of problems with the FDT page. Pwfen ( talk) 09:37, 11 April 2019 (UTC)
There are few things that do not look 100% fine in that section and probably need some adjustment.
I am ok with the initial discussion but I would refer to and just drop this strange . Then I am ok with Kubo formula, this is a very good starting point
if we take Kubo for granted then the final result can be derived just by a simple Fourier transform
so
then the following part is very correct: "In the canonical ensemble, the second term can be re-expressed as
where in the second equality we re-positioned using the cyclic property of trace (in this step we have also assumed that the operator is bosonic, i.e. does not introduce a sign change under permutation). Next, in the third equality, we inserted next to the trace and interpreted as a time evolution operator with imaginary time interval ."... and if we plug this in the previous equation we get
and finally, with minor reshuffling of the factors, to
where : note here we have a , not . The makes sense but only because the same calculation for yields
without the +1. So the cited formula with is just the symmetric part of power spectral density . In quantum noise is not symmetric and, consistently, the autocorrelation contains an imaginary part. This is something that is not particularly trivial/intuitive and might deserve a separate discussion... possible reference/examples here [1]
Wikirod76 ( talk) 00:57, 9 January 2021 (UTC)
The parenthetical statement that "in this step we have also assumed that the operator is bosonic, i.e. does not introduce a sign change under permutation" appears unnecessary. The cyclic property of the trace is general, and doesn't require this assumption.
WinterSyzygy ( talk) 03:33, 10 June 2021 (UTC)
References
{{
cite journal}}
: CS1 maint: multiple names: authors list (
link)
In the General Formulation section, there is the sentence "The observable will fluctuate around its mean value with fluctuations characterized by a power spectrum ." Using the definition of the Fourier transform given below, has units of , which do no match with the fluctuation dissipation theorem as stated. On the power spectrum page, is defined with a , which resolves the issue.
Proposed change: "The observable will fluctuate around its mean value with fluctuations characterized by an auto correlation function . Here, the fluctuations are assumed to be stationary (independent of )." Skell013 ( talk) 16:41, 21 May 2024 (UTC)
This
level-5 vital article is rated Start-class on Wikipedia's
content assessment scale. It is of interest to the following WikiProjects: | |||||||||||
|
Depending on which convention one uses for the Fourier transform, the FDT adopts different signs.
Definition (1):
gives
Definition (2):
gives
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)
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)
I started a major rewrite of this article.
What was the state before: IMO, the second part of the original version was not very clear.
What I have done: I included a section 'General form' where I give a general FDT in its simplest form; I also included a derivation.
What can be done next:
I would appreciate your help.
-- Benjamin.friedrich ( talk) 12:18, 18 December 2007 (UTC)
I actually miss the theorem itself. Only examples, derivation and applications..-- 85.146.199.125 ( talk) 20:22, 20 February 2008 (UTC)
Perhaps nitpicking, but: In "general formulation," the perturbing field f seems to be off a minus sign: dp/dt = -dH/dx = -dH0/dx - f(t) as written. So then f is *minus* the applied force. This should be changed to be consistent with the definition of susceptibility in that wikipedia article. Dstrozzi ( talk) 00:02, 27 March 2017 (UTC)
The old definition is not correct:
Rather and are swapped. It should be:
such that only depends on the perturbation before . 76.199.99.43 ( talk) 10:41, 21 April 2009 (UTC)
(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 at t=0, which is incorrect. As t approaches zero from minus infinity, adiabatically approaches . 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:
For stationary processes, the Wiener-Khinchin theorem states that the power spectrum equals twice the Fourier transform of the auto-correlation function
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 into a symmetric and an anti-symmetric part
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)
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
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)
Correction: Now I think, it's right. But there is a huge lack of explanation. I hope, I'll get back to inserting it.
129.206.91.26 ( talk) 16:36, 9 June 2011 (UTC)
Another correction: I checked it, it's definitely wrong. I hope, I'll find the time to point out, why.
129.206.91.26 ( talk) 17:49, 9 June 2011 (UTC)
So, here it comes: The Taylor-expansion in numerator and denominator need the following approximation to hold (which is far from obvious for non-commuting H and h, check Baker-Campbell-Hausdorff formula):
and in the denominator:
While the second approximation seems to hold (I checked some matrices with oracle), the first approximation is simply wrong. Demanding A to be small is probably a solution, but it's highly unphysical to expect some observable to be small compared to the Hamiltonian. Another possibility is to demand and to commute. But then the time evolution of our equilibrium state is trivial and the FDT states: .
For these reasons I will now delete the whole "Derivation in trace notation" - section. Hope, your ok with that. — Preceding unsigned comment added by 129.206.196.143 ( talk) 16:10, 12 June 2011 (UTC)
I am trying to comprehend why the field is always taken to be time-independent. I understand why it is turned off at time zero, but there should be a way to derive this without requiring time independence of the field. 129.255.226.50 ( talk) 18:06, 19 August 2013 (UTC)
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]"
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)
The 9 April 2019 version of the FDT page < https://en.wikipedia.org/?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)
The fluctuation-dissipation theorem does not require thermal equilibrium to hold. It requires detailed balance to hold. Please see /info/en/?search=User_talk:Pwfen for a discussion of problems with the FDT page. Pwfen ( talk) 09:37, 11 April 2019 (UTC)
There are few things that do not look 100% fine in that section and probably need some adjustment.
I am ok with the initial discussion but I would refer to and just drop this strange . Then I am ok with Kubo formula, this is a very good starting point
if we take Kubo for granted then the final result can be derived just by a simple Fourier transform
so
then the following part is very correct: "In the canonical ensemble, the second term can be re-expressed as
where in the second equality we re-positioned using the cyclic property of trace (in this step we have also assumed that the operator is bosonic, i.e. does not introduce a sign change under permutation). Next, in the third equality, we inserted next to the trace and interpreted as a time evolution operator with imaginary time interval ."... and if we plug this in the previous equation we get
and finally, with minor reshuffling of the factors, to
where : note here we have a , not . The makes sense but only because the same calculation for yields
without the +1. So the cited formula with is just the symmetric part of power spectral density . In quantum noise is not symmetric and, consistently, the autocorrelation contains an imaginary part. This is something that is not particularly trivial/intuitive and might deserve a separate discussion... possible reference/examples here [1]
Wikirod76 ( talk) 00:57, 9 January 2021 (UTC)
The parenthetical statement that "in this step we have also assumed that the operator is bosonic, i.e. does not introduce a sign change under permutation" appears unnecessary. The cyclic property of the trace is general, and doesn't require this assumption.
WinterSyzygy ( talk) 03:33, 10 June 2021 (UTC)
References
{{
cite journal}}
: CS1 maint: multiple names: authors list (
link)
In the General Formulation section, there is the sentence "The observable will fluctuate around its mean value with fluctuations characterized by a power spectrum ." Using the definition of the Fourier transform given below, has units of , which do no match with the fluctuation dissipation theorem as stated. On the power spectrum page, is defined with a , which resolves the issue.
Proposed change: "The observable will fluctuate around its mean value with fluctuations characterized by an auto correlation function . Here, the fluctuations are assumed to be stationary (independent of )." Skell013 ( talk) 16:41, 21 May 2024 (UTC)