Editing Fluctuation–dissipation theorem (section)

===Classical version===
We derive the fluctuation–dissipation theorem in the form given above, using the same notation.
Consider the following test case: the field {{math|''f''}} has been on for infinite time and is switched off at {{math|1=''t'' = 0}}

<math display="block"> f(t) = f_0 \theta(-t) , </math>
where <math> \theta(t)</math> is the [[Heaviside function]].
We can express the expectation value of {{mvar|x}} by the probability distribution {{math|''W''(''x'',0)}} and the transition probability <math> P(x',t | x,0) </math>

<math display="block"> \langle x(t) \rangle = \int dx' \int dx \, x' P(x',t|x,0) W(x,0) . </math>

The probability distribution function ''W''(''x'',0) is an equilibrium distribution and hence given by the [[Boltzmann distribution]] for the Hamiltonian <math> H(x) = H_0(x) - x f_0 </math>

<math display="block"> W(x,0)= \frac{\exp(-\beta H(x))}{\int dx' \, \exp(-\beta H(x'))} \,, </math>
where <math>\beta^{-1} = k_\text{B}T</math>.
For a weak field <math> \beta x f_0 \ll 1 </math>, we can expand the right-hand side

<math display="block"> W(x,0) \approx W_0(x) [1+\beta f_0 (x-\langle x \rangle_0)], </math>

here <math> W_0(x) </math> is the equilibrium distribution in the absence of a field. Plugging this approximation in the formula for <math> \langle x(t) \rangle </math> yields
{{NumBlk||<math display="block">\langle x(t) \rangle = \langle x \rangle_0 + \beta f_0 A(t),</math>|{{EquationRef|1}}}}
where ''A''(''t'') is the auto-correlation function of ''x'' in the absence of a field:

<math display="block"> A(t)=\langle [x(t)-\langle x \rangle_0][ x(0)-\langle x \rangle_0] \rangle_0. </math>

Note that in the absence of a field the system is invariant under time-shifts. We can rewrite <math> \langle x(t) \rangle - \langle x \rangle_0 </math> using the susceptibility of the system and hence find with the above {{EqNote|1|equation '''(1)'''}}

<math display="block"> f_0 \int_0^{\infty} d\tau \, \chi(\tau) \theta(\tau-t) = \beta f_0 A(t) </math>

Consequently,
{{NumBlk||<math display="block">-\chi(t) = \beta {dA(t) \over dt} \theta(t) . </math>|{{EquationRef|2}}}}

To make a statement about frequency dependence, it is necessary to take the Fourier transform of {{EqNote|1|equation '''(2)'''}}. By integrating by parts, it is possible to show that
<math display="block"> -\hat\chi(\omega) = i\omega\beta \int_0^\infty e^{-i\omega t} A(t)\, dt -\beta A(0).</math>

Since <math>A(t)</math> is real and symmetric, it follows that
<math display="block"> 2 \operatorname{Im}[\hat\chi(\omega)] = -\omega\beta \hat A(\omega).</math>

Finally, for [[stationary process]]es, the [[Wiener–Khinchin theorem]] states that the two-sided [[power spectrum|spectral density]] is equal to the [[Fourier transform]] of the auto-correlation function:
<math display="block"> S_x(\omega) = \hat{A}(\omega).</math>

Therefore, it follows that
<math display="block"> S_x(\omega) = -\frac{2k_\text{B} T}{\omega} \operatorname{Im}[\hat\chi(\omega)].</math>