User:WillowW/FD

Let there be two Hamiltonians, H and H+h, that differ only by a small amount from one another. More precisely, let h be much smaller than the thermal energy kBT, where kB is Boltzmann's constant and T is the temperature in Kelvin.

Consider a system that has reached equilibrium under the latter Hamiltonian H+h, so that the probability of any state is proportional to its Boltzmann factor e-β(H + h) where β=1/kBT. At time t=0, let the perturbation h be turned off; given ergodicity, the system will gradually relax to a new equilibrium, which has Boltzmann factors e-βH. The fluctuation-dissipation theorem addresses the question of how quickly the system reaches its new equilibrium.

For brevity and generality, we introduce the trace notation, which applies both to classical and quantum systems



\mathrm{Tr} X = \int d\Gamma X = \sum_{\mathrm{quantum\ states}\ n}  $$

where dΓ represents an infinitesimal volume in phase space. Let angular brackets refer to taking the average value of a variable A



\langle A \rangle = \mathrm{Tr} \left\{ A(q, p) \ f(q, p) \right\} = \frac{\mathrm{Tr} \left\{ A e^{-\beta H} \right\}}{\mathrm{Tr} \ e^{-\beta H}} $$

Let the correlation function c(t) be defined



c(t) = \langle \delta A (t) \ \delta A (0) \rangle = \lim_{T\rightarrow\infty} \frac{1}{2T} \int_{-T}^{T} dt^{\prime}\ \delta A (t^{\prime} + t) \ \delta A (t^{\prime}) $$

which may be written as



c(t) = \int d\Gamma\ \delta A(t; q, p) \ \delta A(0; q, p)\  f(q, p) $$

where δA(t; q, p) is the devation from its mean at a time t given that it began at position in phase space (q, p). Thus, the integration is over all initial positions of the system.

The mean value of A as it evolves towards its new equilibrium is given by



\bar{A}(t) = \frac{\mathrm{Tr} \left\{ A(t; p, q) e^{-\beta \left(H + h \right)} \right\}}{\mathrm{Tr} \ e^{-\beta \left(H + h \right)}} $$

Since h is much smaller than the thermal energy kBT, we may expand the numerator



\mathrm{Tr} \left\{ A(t; p, q) e^{-\beta \left(H + h \right)} \right\} = \mathrm{Tr} \left\{ A(t; p, q) e^{-\beta H} \left(1 - \beta h + \ldots \right) \right\} \approx \mathrm{Tr} \left\{ A(t; p, q) e^{-\beta H} \right\} - \mathrm{Tr} \left\{ A(t; p, q) e^{-\beta H} \beta h \right\} $$

We may likewise expand the denominator



\frac{1}{\mathrm{Tr} \ e^{-\beta \left(H + h \right)}} \approx \frac{1}{\mathrm{Tr} \left\{ e^{-\beta H } \right\} - \mathrm{Tr} \left\{ e^{-\beta H } \beta h \right\}} = \frac{1}{\mathrm{Tr} e^{-\beta H }} \cdot \frac{1}{1 - \langle \beta h \rangle} \approx \frac{1 + \langle \beta h \rangle}{\mathrm{Tr} e^{-\beta H }} $$

where we have used



\langle \beta h \rangle = \frac{\mathrm{Tr} \left\{ \beta h e^{-\beta H} \right\}}{\mathrm{Tr} \ e^{-\beta H}} $$

which is much less than one, by our assumption.

Combining the numerator and denominator, dropping quadratic and high-order terms in <βh>, and using the indifference of equilibrium to time, we obtain



\bar{A}(t) - \langle A \rangle = - \beta \langle A h \rangle + \beta \langle A \rangle \langle h \rangle $$

Let the perturbation h=-gA be proportional to the variable A with a constant -g. Then this formula becomes



\bar{A}(t) - \langle A \rangle = \beta g \left\{ \langle A(t) A(0) \rangle - \langle A \rangle^{2} \right\} = \beta g \langle \delta A(t) \delta A(0) \rangle = \beta g c(t) $$

Note that the system's relaxation is independent of A and linear in g. These results imply that perturbations will relax independently of one another; if two perturbations, g1 and g2 are applied, the net relaxation will be the sum of the individual relaxations to g1 and g2 taken separately. Such linear systems have been well-studied, and many methods developed for their solution.