Derivation of the Fluctuation Dissipation Theorem

We previously derived the time evolution operators for both Hamiltonian and Schrodinger equations.

Time evolution operator as a Volterra series

A Heisenberg observable ${\hat{A}}_{H}$ in classical or quantum mechanics will evolve by the time evolution operator

{\hat{A}}_{H} (t_{b}) = 𝒰 (t_{a}, t_{b}) {\hat{A}}_{H} (t_{a})

where the derivative is the Liouville operator, and the time evolution operator is the operator exponentiation of the integrated Liouville operator.

\frac{d {\hat{A}}_{H} (t)}{d t} = ℒ (t) {\hat{A}}_{H}

𝒰 (t_{a}, t_{b}) = e^{\int_{t_{a}}^{t_{b}} ℒ (t^{'}) d t^{'}}

Now if we substitute back into the derivative

\frac{d}{d t} (𝒰 (t_{a}, t) {\hat{A}}_{H} (t_{a})) = ℒ (t) 𝒰 (t_{a}, t) {\hat{A}}_{H} (t_{a})

then integrate both sides

𝒰 (t_{a}, t_{b}) {\hat{A}}_{H} (t_{a}) - 𝒰 (t_{a}, t_{a}) {\hat{A}}_{H} (t_{a}) = \int_{t_{a}}^{t_{b}} ℒ (t) 𝒰 (t_{a}, t) {\hat{A}}_{H} (t_{a}) d t

replacing $𝒰 (t_{a}, t_{a})$ with 1

𝒰 (t_{a}, t_{b}) {\hat{A}}_{H} (t_{a}) = {\hat{A}}_{H} (t_{a}) + \int_{t_{a}}^{t_{b}} ℒ (t) 𝒰 (t_{a}, t) {\hat{A}}_{H} (t_{a}) d t

we get the following relation.

𝒰 (t_{a}, t_{b}) {\hat{A}}_{H} (t_{a}) = (1 + \int_{t_{a}}^{t_{b}} ℒ (t) 𝒰 (t_{a}, t) d t) {\hat{A}}_{H} (t_{a})

And since it should hold for arbitrary observable, the operators on both sides of the equation must be equal.

𝒰 (t_{a}, t_{b}) = 1 + \int_{t_{a}}^{t_{b}} ℒ (t) 𝒰 (t_{a}, t) d t

Expanding the Volterra series

If we keep substituting into itself, on iteration 0:

𝒰 (t_{a}, t_{b}) = 1

On iteration 1:

𝒰 (t_{a}, t_{b}) = 1 + \int_{t_{a}}^{t_{b}} ℒ (t) d t

Then on iteration 2:

𝒰 (t_{a}, t_{b}) = 1 + \int_{t_{a}}^{t_{b}} ℒ (t) d t + \int_{t_{a}}^{t_{b}} \int_{t_{a}}^{t_{1}} ℒ (t_{1}) ℒ (t_{2}) d t_{1} d t_{2}

Then on iteration 3:

𝒰 (t_{a}, t_{b}) = 1 + \int_{t_{a}}^{t_{b}} ℒ (t) d t + \int_{t_{a}}^{t_{b}} \int_{t_{a}}^{t_{1}} ℒ (t_{1}) ℒ (t_{2}) d t_{1} d t_{2} + \int_{t_{a}}^{t_{b}} \int_{t_{a}}^{t_{1}} \int_{t_{a}}^{t_{2}} ℒ (t_{1}) ℒ (t_{2}) ℒ (t_{3}) d t_{1} d t_{2} d t_{3}

And so on. If we assume it converges the limit should be a solution

𝒰 (t_{a}, t_{b}) = 1 + \int_{t_{a}}^{t_{b}} ℒ (t) d t + \int_{t_{a}}^{t_{b}} \int_{t_{a}}^{t_{1}} ℒ (t_{1}) ℒ (t_{2}) d t_{1} d t_{2} + \int_{t_{a}}^{t_{b}} \int_{t_{a}}^{t_{1}} \int_{t_{a}}^{t_{2}} ℒ (t_{1}) ℒ (t_{2}) ℒ (t_{3}) d t_{1} d t_{2} d t_{3} + \dots

Just for syntax we will group all the higher order terms.

𝒰 (t_{a}, t_{b}) = 1 + \int_{t_{a}}^{t_{b}} ℒ (t) d t + 𝒪 (ℒ^{2})

Time evolution of probability

Classical

If we have a probability distribution $ρ (t, \vec{q}, \vec{p})$ over the phase space we can find the expected value of an observable by

𝔼 [A] (t) = \int_{V} A (\vec{q}, \vec{p}) ρ (t, \vec{q}, \vec{p}) d \vec{q} d \vec{p}

And the probability distribution flows according the the continuity equation

\frac{d ρ}{d t} (t, \vec{x} (t)) = - \nabla_{x} \cdot (\frac{d \vec{x}}{d t} ρ)

where the state vector $\vec{x} = ⟨ q_{1}, q_{2}, \dots, p_{1}, p_{2}, \dots ⟩$ so we can split it up and say the below.

\frac{d ρ}{d t} (t, \vec{q} (t), \vec{p} (t)) = - \nabla_{q} \cdot (\frac{d \vec{q}}{d t} ρ) - \nabla_{p} \cdot (\frac{d \vec{p}}{d t} ρ)

Use the chain rule

= - \sum_{k} (\frac{\partial ρ}{\partial q_{k}} \frac{d q_{k}}{d t} + ρ \frac{\partial}{\partial q_{k}} \frac{d q_{k}}{d t}) - \sum_{k} (\frac{\partial ρ}{\partial p_{k}} \frac{d p_{k}}{d t} + ρ \frac{\partial}{\partial p_{k}} \frac{d p_{k}}{d t})

then substitue in Hamiltons equations.

= - \sum_{k} (\frac{\partial ρ}{\partial q_{k}} \frac{\partial H}{\partial p_{k}} + ρ \frac{\partial^{2} H}{\partial q_{k} \partial p_{k}}) - \sum_{k} (- \frac{\partial ρ}{\partial p_{k}} \frac{\partial H}{\partial q_{k}} - ρ \frac{\partial^{2} H}{\partial p_{k} \partial q_{k}})

Cancel those two terms.

= - \sum_{k} (\frac{\partial ρ}{\partial q_{k}} \frac{\partial H}{\partial p_{k}} - \frac{\partial ρ}{\partial p_{k}} \frac{\partial H}{\partial q_{k}})

It's just the Possion bracket

= {H, ρ}

Define an operator that does this

= 𝒦 (t) ρ (t)

Quantum

If we have a density matrix $ρ (t)$ over the phase space we can find the expected value of an observable by

𝔼 [A] (t) = Tr (ρ (t) \hat{A})

where the density matrix is defined below.

ρ (t) = \sum_{k} p_{k} | ψ_{k} (t) ⟩ ⟨ ψ_{k} (t) |

If we take the derivative of the density matrix we use the chain rule

\frac{d ρ (t)}{d t} = \sum_{k} p_{k} \frac{d}{d t} (| ψ_{k} (t) ⟩ ⟨ ψ_{k} (t) |) = \sum_{k} p_{k} (| \frac{d ψ_{k} (t)}{d t} ⟩ ⟨ ψ_{k} (t) | + | ψ_{k} (t) ⟩ ⟨ \frac{d ψ_{k} (t)}{d t} |)

and substituting in Schrodingers equation we get

= \sum_{k} p_{k} (- \frac{i}{ℏ} \hat{H} (t) | ψ_{k} (t) ⟩ ⟨ ψ_{k} (t) | + | ψ_{k} (t) ⟩ ⟨ ψ_{k} (t) | \frac{i}{ℏ} {\hat{H}}^{†} (t))

and the Hamiltonian is its on Hermetian so

= - \frac{i}{ℏ} (\hat{H} (t) ρ (t) = ρ (t) \hat{H} (t))

It's just the commutator bracket

= - \frac{i}{ℏ} [\hat{H} (t), ρ (t)]

Define an operator that does this

= 𝒦 (t) ρ (t)

Since the derivative of the probability is

\frac{d ρ}{d t} = 𝒦 (t) ρ (t)

Make a time evolution operator as the exponentiation

ρ (t_{b}) = e^{\int_{t_{a}}^{t_{b}} 𝒦 (t^{'}) d t^{'}} ρ (t_{a}) = 𝒫 (t_{a}, t_{b}) ρ (t_{a})

Interation picture

In the interaction picture we split the Hamiltonian into time independent and a time dependent part.

H = H_{0} + H_{1} (t)

Classical

Define an interaction picture version.

𝒦 (t) = {H (t), \cdot}

= {H_{0} (t) + H_{1} (t), \cdot}

The Possion bracket distributes over addition.

= {H_{0}, \cdot} + {H_{1} (t), \cdot}

Then define 2 new operators for each term

= 𝒦_{0} + 𝒦_{1} (t)

Quantum

Define an interaction picture Liouville operator.

𝒦 (t) = - \frac{i}{ℏ} [H (t), \cdot]

= - \frac{i}{ℏ} [H_{0} + H_{1} (t), \cdot]

The commutator bracket distributes over addition.

= - \frac{i}{ℏ} [H_{0}, \cdot] - \frac{i}{ℏ} [H_{1} (t), \cdot]

Then define 2 new operators for each term

= 𝒦_{0} + 𝒦_{1} (t)

𝒫_{0} (t_{a}, t_{b}) = e^{(t_{b} - t_{a}) 𝒦_{0}}

𝒫_{1} (t_{a}, t_{b}) = e^{\int_{t_{a}}^{t_{b}} 𝒦_{1} (t^{'}) d t^{'}}

𝒦_{1, I} (t) = e^{- (t - t_{a}) 𝒦_{0}} 𝒦_{1} (t) e^{(t - t_{a}) ℒ_{0}}

𝒫_{1, I} (t) = e^{\int_{t_{a}}^{t_{b}} 𝒦_{1, I} (t^{'}) d t^{'}}

and can prove with showing starting condition is the same and derivative is the same

𝒫 (t_{a}, t_{b}) = 𝒫_{0} (t_{a}, t_{b}) 𝒫_{1, I} (t_{a}, t_{b})

We can do a Volterra series it as

𝒫_{1, I} (t_{a}, t_{b}) = 1 + \int_{t_{a}}^{t_{b}} 𝒦_{1, I} (t) 𝒫_{1, I} (t_{a}, t)) d t

and expand it

= 1 + \int_{t_{a}}^{t_{b}} 𝒦_{1, I} (t) d t + 𝒪 (𝒦_{1, I}^{2})

Then

𝒫 (t_{a}, t_{b}) = 𝒫_{0} (t_{a}, t_{b}) + \int_{t_{a}}^{t_{b}} 𝒫_{0} (t_{a}, t_{b}) 𝒦_{1, I} (t) d t + 𝒪 (𝒦_{1, I}^{2})

= 𝒫_{0} (t_{a}, t_{b}) + \int_{t_{a}}^{t_{b}} e^{(t_{b} - t_{a}) 𝒦_{0}} e^{- (t - t_{a}) 𝒦_{0}} 𝒦_{1} (t) e^{(t - t_{a}) 𝒦_{0}} d t + 𝒪 (𝒦_{1, I}^{2})

= 𝒫_{0} (t_{a}, t_{b}) + \int_{t_{a}}^{t_{b}} e^{(t_{b} - t) 𝒦_{0}} 𝒦_{1} (t) e^{(t - t_{a}) 𝒦_{0}} d t + 𝒪 (𝒦_{1, I}^{2})

= 𝒫_{0} (t_{a}, t_{b}) + \int_{t_{a}}^{t_{b}} 𝒫_{0} (t, t_{b}) 𝒦_{1} (t) 𝒫_{0} (t_{a}, t) d t + 𝒪 (𝒦_{1, I}^{2})

Linear response

Consider a closed system where the Hamiltonian is time independent, and is then perturbed by a force along some observable

H (t) = H_{0} - h (t) \hat{B}

Here

H_{1} (t) = - h (t) \hat{B}

Classical

Here $𝒦_{1} (t)$ is

𝒦_{1} (t) = {- h (t) \hat{B}, \cdot} = - h (t) {\hat{B}, \cdot}

So let's define

𝒦_{B} = {\hat{B}, \cdot}

So that

𝒦_{1} (t) = - h (t) 𝒦_{B}

Quantum

Here $ℒ_{1} (t)$ is

𝒦_{1} (t) = - \frac{i}{ℏ} [- h (t) \hat{B}, \cdot] = - h (t) (- \frac{i}{ℏ} [\hat{B}, \cdot])

So let's define

𝒦_{B} (t) = - \frac{i}{ℏ} [\hat{B} (t), \cdot]

So that

𝒦_{1} (t) = - h (t) 𝒦_{B} (t)

So we get

𝒫 (t_{a}, t_{b}) = 𝒫_{0} (t_{a}, t_{b}) + \int_{t_{a}}^{t_{b}} 𝒫_{0} (t, t_{b}) (- h (t) 𝒦_{B}) 𝒫_{0} (t_{a}, t) + 𝒪 (h^{2})

And we find that

ρ (t_{b}) - ρ_{0} (t_{b}) = \int_{t_{a}}^{t_{b}} 𝒫_{0} (t, t_{b}) (- h (t) 𝒦_{B}) 𝒫_{0} (t_{a}, t) ρ (t_{a}) + 𝒪 (h^{2})

Expected value of observable at equilibium

Assume that at time $t_{a}$ there was no perturbation and the system is in thermal equilibrium. For all $t \leq t_{a}$ the probability distribution is at some equilibrium distribution $ρ (t) = ρ_{0}$ and there was no force $h (t) = 0$ .

Classical

If we have a probability distribution $ρ (t, \vec{q}, \vec{p})$ over the phase space we can find the expected value of an observable by

𝔼 [A] (t) = \int_{V} A (\vec{q}, \vec{p}) ρ (t, \vec{q}, \vec{p}) d \vec{q} d \vec{p}

The difference in expected values is

𝔼 [A] (t_{b}) - 𝔼_{0} [A] (t_{b}) = \int_{V} ρ (t_{b}) A d \vec{q} d \vec{p} - \int_{V} ρ_{0} (t_{b}) A d \vec{q} d \vec{p}

= \int_{V} (ρ (t_{b}) - ρ_{0} (t_{b})) A d \vec{q} d \vec{p}

Then put in the Volterra series

= \int_{V} (\int_{t_{a}}^{t_{b}} 𝒫_{0} (t, t_{b}) (- h (t) 𝒦_{B}) 𝒫_{0} (t_{a}, t) ρ (t_{a}) d t) A d \vec{q} d \vec{p} + 𝒪 (h^{2})

and move the integrals around

= \int_{t_{a}}^{t_{b}} h (t) \int_{V} - (𝒫_{0} (t, t_{b}) 𝒦_{B} 𝒫_{0} (t_{a}, t) ρ (t_{a})) A d \vec{q} d \vec{p} d t + 𝒪 (h^{2})

Call the factor with first order of $h$ the linear response function

χ_{A B} (t - t_{b}) = - \int_{V} (𝒫_{0} (t, t_{b}) 𝒦_{B} 𝒫_{0} (t_{a}, t) ρ (t_{a})) A d \vec{q} d \vec{p}

put in the equilbrium distribution $ρ (t_{a}) = ρ_{0} (\vec{q}, \vec{p}) = \frac{e^{- \frac{H_{0} (\vec{q}, \vec{p})}{k_{B} T}}}{Z}$

χ_{A B} (t - t_{b}) = - \int_{V} (𝒫_{0} (t, t_{b}) 𝒦_{B} 𝒫_{0} (t_{a}, t) ρ_{0}) A d \vec{q} d \vec{p}

And by definition the distribution shouldn't evolve under the unberturbed Hamiltonian (you can prove this).

χ_{A B} (t - t_{b}) = - \int_{V} (𝒫_{0} (t, t_{b}) 𝒦_{B} ρ_{0}) A d \vec{q} d \vec{p}

so its like

= - \int_{V} (𝒫_{0} (t, t_{b}) \sum_{i} \frac{\partial B}{\partial q_{i}} \frac{\partial ρ_{0}}{\partial p_{i}} - \frac{\partial B}{\partial p_{i}} \frac{\partial ρ_{0})}{\partial q_{i}}) A d \vec{p} d \vec{q}

= \frac{1}{k_{B} T} \int_{V} 𝒫_{0} (t, t_{b}) (\frac{e^{- \frac{H_{0} (\vec{q}, \vec{p})}{k_{B} T}}}{Z} (\sum_{i} \frac{\partial B}{\partial q_{i}} \frac{\partial H_{0}}{\partial p_{i}} - \frac{\partial B}{\partial p_{i}} \frac{\partial H_{0}}{\partial q_{i}})) A d \vec{p} d \vec{q}

= \frac{1}{k_{B} T} \int_{V} 𝒫_{0} (t, t_{b}) (ρ_{0} ℒ_{0} B)) A d \vec{p} d \vec{q}

where $ℒ_{0} = {\cdot, H_{0}}$ is a Liouville operator for only the equilibrium Hamiltonian. The time evolution operator distributes over multiplication.

= \frac{1}{k_{B} T} \int_{V} (𝒫_{0} (t, t_{b}) ρ_{0}) (𝒫_{0} (t, t_{b}) ℒ_{0} B) A d \vec{p} d \vec{q}

= \frac{1}{k_{B} T} \int_{V} ρ_{0} 𝒫_{0} (t, t_{b}) (ℒ_{0} B) A d \vec{p} d \vec{q}

And since $𝒦_{0} (t) = - ℒ_{0} (t)$ , and operators commute with their own exponent:

= \frac{1}{k_{B} T} \int_{V} ρ_{0} (ℒ_{0} e^{- (t_{b} - t) ℒ_{0}} B) A d \vec{p} d \vec{q}

= \frac{1}{k_{B} T} \int_{V} ρ_{0} ({\dot{B}}_{0} (t - t_{b})) A d \vec{p} d \vec{q}

Where $\dot{B}$ is time shifted to measure $t - t_{b}$ in the future. And this is just the expected value at equilibrium

= \frac{1}{k_{B} T} 𝔼_{0} [\dot{B} (t_{a} + t - t_{b}) A (t_{a})]

χ_{A B} (t - t_{b}) = \frac{1}{k_{B} T} \frac{d}{d t} 𝔼_{0} [B (t_{a} + t - t_{b}) A (t_{a})]

χ_{A B} (t) = \frac{1}{k_{B} T} \frac{d}{d t} 𝔼_{0} [B (t_{a} + t) A (t_{a})]

Quantum

So the expected value is

𝔼 [A] (t) = Tr (ρ (t) \hat{A})

And the difference is

𝔼 [A] (t_{b}) - 𝔼_{0} [A] (t_{b}) = Tr (ρ (t_{b}) \hat{A}) - Tr (ρ_{0} (t_{b}) \hat{A})

= Tr ((ρ (t_{b}) - ρ_{0} (t_{b})) \hat{A})

Put the volterra series in there

= Tr (\int_{t_{a}}^{t_{b}} (𝒫_{0} (t, t_{b}) (- h (t) 𝒦_{B}) 𝒫_{0} (t_{a}, t) ρ (t_{a})) \hat{A} d t) + 𝒪 (h^{2})

Move some stuff around

= - \int_{t_{a}}^{t_{b}} h (t) Tr (𝒫_{0} (t, t_{b}) 𝒦_{B} (t) 𝒫_{0} (t_{a}, t) ρ (t_{a})) \hat{A}) d t + 𝒪 (h^{2})

Call this first order term the linear response function

χ_{A B} (t - t_{b}) = - Tr ((𝒫_{0} (t, t_{b}) 𝒦_{B} (t) 𝒫_{0} (t_{a}, t) ρ (t_{a})) \hat{A})

Assume equilibrium at $t_{a}$ and put in the equilbium density matrix $ρ_{0} = \frac{e^{- \frac{{\hat{H}}_{0}}{k_{B} T}}}{Z}$ .

= - Tr ((𝒫_{0} (t, t_{b}) 𝒦_{B} (t) 𝒫_{0} (t_{a}, t) ρ_{0}) \hat{A})

And you can check that it doesn't change in time from the equilbrium Hamiltonian

= - Tr ((𝒫_{0} (t, t_{b}) 𝒦_{B} (t) ρ_{0}) \hat{A})

Then expand

= Tr (\frac{i}{ℏ} (𝒫_{0} (t, t_{b}) (ρ_{0} B - B ρ_{0})) \hat{A})

and because $𝒦 (t) = - ℒ (t)$

= Tr (\frac{i}{ℏ} (ρ_{0} B (t - t_{b}) - B (t - t_{b}) ρ_{0}) \hat{A})

split

= \frac{i}{ℏ} (Tr (e^{- \frac{{\hat{H}}_{0}}{k_{B} T}} B (t - t_{b}) \hat{A}) - Tr (B (t - t_{b}) e^{- \frac{{\hat{H}}_{0}}{k_{B} T}} \hat{A}))

and remember the cycle property of traces, and cycle the second one

= \frac{i}{ℏ} (Tr (e^{- \frac{{\hat{H}}_{0}}{k_{B} T}} B (t - t_{b}) \hat{A}) - Tr (e^{- \frac{{\hat{H}}_{0}}{k_{B} T}} \hat{A} B (t - t_{b})))

= - \frac{i}{ℏ} Tr (e^{- \frac{{\hat{H}}_{0}}{k_{B} T}} [A, B (t - t_{b})])

χ_{A B} (t - t_{b}) = - \frac{i}{ℏ} Tr (e^{- \frac{{\hat{H}}_{0}}{k_{B} T}} [A (t_{a}), B (t_{a} + t - t_{b})])

χ_{A B} (t) = - \frac{i}{ℏ} Tr (e^{- \frac{{\hat{H}}_{0}}{k_{B} T}} [A (t_{a}), B (t_{a} + t)])

χ_{A B} (t) = - \frac{i}{ℏ} 𝔼_{0} [[A (t_{a}), B (t_{a} + t)]]

This is the Kubo formula

The time evolution operator

𝒰 (t_{a}, t_{b}) \hat{A} = e^{(t_{b} - t_{a}) \frac{i}{ℏ} {\hat{H}}_{0}} \hat{A} e^{(t_{b} - t_{a}) \frac{- i}{ℏ} {\hat{H}}_{0}}

Next we consider imaginary time

= \frac{i}{ℏ} (Tr (e^{- \frac{{\hat{H}}_{0}}{k_{B} T}} B (t) \hat{A}) - Tr (e^{- \frac{{\hat{H}}_{0}}{k_{B} T}} \hat{A} B (t)))

= \frac{i}{ℏ} (Tr (e^{- \frac{{\hat{H}}_{0}}{k_{B} T}} B (t) e^{\frac{{\hat{H}}_{0}}{k_{B} T}} e^{- \frac{{\hat{H}}_{0}}{k_{B} T}} \hat{A}) - Tr (e^{- \frac{{\hat{H}}_{0}}{k_{B} T}} \hat{A} B (t)))

= \frac{i}{ℏ} (Tr (B (t - \frac{i ℏ}{k_{B} T}) e^{- \frac{{\hat{H}}_{0}}{k_{B} T}} \hat{A}) - Tr (e^{- \frac{{\hat{H}}_{0}}{k_{B} T}} \hat{A} B (t)))

Then cycle the trace again

χ_{A B} (t) = \frac{i}{ℏ} (Tr (e^{- \frac{{\hat{H}}_{0}}{k_{B} T}} \hat{A} B (t - \frac{i ℏ}{k_{B} T})) - Tr (e^{- \frac{{\hat{H}}_{0}}{k_{B} T}} \hat{A} B (t)))

fourier transform (linear response must be zero for negative times so the integral can start at zero)

χ_{A B} (ω) = \int_{0}^{\infty} e^{- i ω t} \frac{i}{ℏ} (Tr (e^{- \frac{{\hat{H}}_{0}}{k_{B} T}} \hat{A} B (t - \frac{i ℏ}{k_{B} T})) - Tr (e^{- \frac{{\hat{H}}_{0}}{k_{B} T}} \hat{A} B (t))) d t

χ_{A B} (ω) = \frac{i}{ℏ} (\int_{0}^{\infty} e^{- i ω t} Tr (e^{- \frac{{\hat{H}}_{0}}{k_{B} T}} \hat{A} B (t - \frac{i ℏ}{k_{B} T})) d t - \int_{0}^{\infty} e^{- i ω t} Tr (e^{- \frac{{\hat{H}}_{0}}{k_{B} T}} \hat{A} B (t)) d t)

And we can remove the phase shift in the first term

= \frac{i}{ℏ} (e^{\frac{ℏ ω}{k_{B} T}} \int_{0}^{\infty} e^{- i ω t} Tr (e^{- \frac{{\hat{H}}_{0}}{k_{B} T}} \hat{A} B (t)) d t - \int_{0}^{\infty} e^{- i ω t} Tr (e^{- \frac{{\hat{H}}_{0}}{k_{B} T}} \hat{A} B (t)) d t)

= \frac{i}{ℏ} (e^{\frac{ℏ ω}{k_{B} T}} - 1) \int_{0}^{\infty} e^{- i ω t} Tr (e^{- \frac{{\hat{H}}_{0}}{k_{B} T}} \hat{A} B (t)) d t

= \frac{i}{ℏ} (e^{\frac{ℏ ω}{k_{B} T}} - 1) \int_{0}^{\infty} e^{- i ω t} 𝔼_{0} [\hat{A} B (t)] d t

Imaginary part

Classical

Now take the fourier transform

χ_{A B} (ω) = \frac{1}{k_{B} T} \int_{- \infty}^{\infty} e^{- i ω t} \frac{d}{d t} 𝔼_{0} [B (t_{a} + t) A (t_{a})] d t

It shouldn't work for negative values so

= \frac{1}{k_{B} T} \int_{0}^{\infty} e^{- i ω t} \frac{d}{d t} 𝔼_{0} [B (t_{a} + t) A (t_{a})] d t

integrate by parts

\frac{1}{k_{B} T} (e^{- i ω \infty} 𝔼_{0} [B (t_{a} + \infty) A (t_{a})] - e^{0} 𝔼_{0} [B (t_{a}) A (t_{a})]) + \frac{1}{k_{B} T} \int_{0}^{\infty} (\frac{d}{d t} e^{- i ω t}) 𝔼_{0} [B (t_{a} + t) A (t_{a})] d t

In the limit to infinity they should become uncorrelated

- \frac{1}{k_{B} T} 𝔼_{0} [B (t_{a}) A (t_{a})]) + \frac{1}{k_{B} T} \int_{0}^{\infty} (\frac{d}{d t} e^{- i ω t}) 𝔼_{0} [B (t_{a} + t) A (t_{a})] d t

- \frac{1}{k_{B} T} 𝔼_{0} [B (t_{a}) A (t_{a})]) - \frac{i ω}{k_{B} T} \int_{0}^{\infty} e^{- i ω t} 𝔼_{0} [B (t_{a} + t) A (t_{a})] d t

taking just the imaginary part

Im [χ_{A B} (ω)] = - \frac{ω}{k_{B} T} Re [\int_{0}^{\infty} e^{- i ω t} 𝔼_{0} [B (t_{a} + t) A (t_{a})] d t]

Taking just the real part of anything is

Re [f] = \frac{f + f^{*}}{2}

So doing that gives

= - \frac{ω}{2 k_{B} T} (\int_{0}^{\infty} e^{- i ω t} 𝔼_{0} [B (t_{a} + t) A (t_{a})] d t + \int_{0}^{\infty} e^{i ω t} 𝔼_{0} [B (t_{a} + t) A (t_{a})] d t)

= - \frac{ω}{2 k_{B} T} (\int_{0}^{\infty} e^{- i ω t} 𝔼_{0} [B (t_{a} + t) A (t_{a})] d t + \int_{- \infty}^{0} e^{- i ω t} 𝔼_{0} [B (t_{a} - t) A (t_{a})] d t)

And if $A = B$ then

= - \frac{ω}{2 k_{B} T} (\int_{0}^{\infty} e^{- i ω t} 𝔼_{0} [A (t_{a} + t) A (t_{a})] d t + \int_{- \infty}^{0} e^{- i ω t} 𝔼_{0} [A (t_{a} - t) A (t_{a})] d t)

And at equlibrium it's time invariant so you can shift both the offsets in the right integral by $t$ .

= - \frac{ω}{2 k_{B} T} (\int_{0}^{\infty} e^{i ω t} 𝔼_{0} [A (t_{a} + t) A (t_{a})] d t + \int_{- \infty}^{0} e^{i ω t} 𝔼_{0} [A (t_{a}) A (t_{a} + t)] d t)

= - \frac{ω}{2 k_{B} T} \int_{- \infty}^{\infty} e^{i ω t} 𝔼_{0} [A (t_{a} + t) A (t_{a})] d t

So we get a final formulat relating the spectral density to the linear respones function.

\int_{- \infty}^{\infty} e^{i ω t} 𝔼_{0} [A (t_{a} + t) A (t_{a})] d t = - \frac{2 k_{B} T}{ω} Im [χ_{A A} (ω)]

Quantum

Taking just the imaginary part of anything is

Im [f] = \frac{f - f^{*}}{2 i}

Im [χ_{A B} (ω)] = \frac{1}{2 ℏ} (e^{\frac{ℏ ω}{k_{B} T}} - 1) ((\int_{0}^{\infty} e^{- i ω t} 𝔼_{0} [\hat{A} B (t)] d t) - (- \int_{0}^{\infty} e^{i ω t} 𝔼_{0} [B^{†} (t) {\hat{A}}^{†}] d t))

And observables are Hermetian

= \frac{1}{2 ℏ} (e^{\frac{ℏ ω}{k_{B} T}} - 1) ((\int_{0}^{\infty} e^{- i ω t} 𝔼_{0} [\hat{A} B (t)] d t) + (\int_{0}^{\infty} e^{i ω t} 𝔼_{0} [B (t) \hat{A}] d t))

Let A=B

Im [χ_{A A} (ω)] = \frac{1}{2 ℏ} (e^{\frac{ℏ ω}{k_{B} T}} - 1) ((\int_{0}^{\infty} e^{- i ω t} 𝔼_{0} [\hat{A} A (t)] d t) + (\int_{0}^{\infty} e^{i ω t} 𝔼_{0} [A (t) \hat{A}] d t))

= \frac{1}{2 ℏ} (e^{\frac{ℏ ω}{k_{B} T}} - 1) ((\int_{0}^{\infty} e^{- i ω t} 𝔼_{0} [\hat{A} A (t)] d t) + (\int_{- \infty}^{0} e^{- i ω t} 𝔼_{0} [A (- t) \hat{A}] d t))

And at equilibrium shifting the time shouldn't matter $𝔼_{0} [A (t_{a} - t) A (t_{a})] = 𝔼_{0} [A (t_{a}) A (t_{a} + t)]$

= \frac{1}{2 ℏ} (e^{\frac{ℏ ω}{k_{B} T}} - 1) ((\int_{0}^{\infty} e^{- i ω t} 𝔼_{0} [\hat{A} A (t)] d t) + (\int_{- \infty}^{0} e^{- i ω t} 𝔼_{0} [A \hat{A} (t)] d t))

Then combined makes

\frac{1}{2 ℏ} (e^{\frac{ℏ ω}{k_{B} T}} - 1) \int_{- \infty}^{\infty} e^{- i ω t} 𝔼_{0} [\hat{A} A (t)] d t

And we get

\int_{- \infty}^{\infty} e^{- i ω t} 𝔼_{0} [\hat{A} A (t)] d t = \frac{2 ℏ}{e^{\frac{ℏ ω}{k_{B} T}} - 1} Im [χ_{A} (ω)]

Classical limit

If we negative the frequency

S^{'} (ω) = S (- ω) = \int_{- \infty}^{\infty} e^{- i ω t} 𝔼_{0} [\hat{A} (t^{'}) \hat{A} (t_{0})] d t = 2 ℏ (\frac{1}{e^{β ℏ ω} - 1} + 1) Im [χ (- ω)]

And since $χ$ is a real function the Imaginary component is an odd function

= - 2 ℏ (\frac{1}{e^{- β ℏ ω} - 1} + 1) Im [χ (ω)]

= 2 ℏ (\frac{e^{β ℏ ω}}{e^{β ℏ ω} - 1} - 1) Im [χ (ω)]

= 2 ℏ \frac{1}{e^{β ℏ ω} - 1} Im [χ (ω)]

So in the quantum version unlike the classical version $S (ω) \neq S (- ω)$ , the spectral density isn't the same when frequency is negated. To fix this we can average the two.

\frac{S (ω) + S (- ω)}{2} = 2 ℏ (\frac{1}{e^{β ℏ ω} - 1} + \frac{1}{2}) Im [χ (ω)]

= ℏ (\frac{2}{e^{β ℏ ω} - 1} + 1) Im [χ (ω)]

= ℏ \frac{e^{β ℏ ω} + 1}{e^{β ℏ ω} - 1} Im [χ (ω)]

= ℏ \frac{e^{\frac{β ℏ ω}{2}} + e^{- \frac{β ℏ ω}{2}}}{e^{\frac{β ℏ ω}{2}} - e^{- \frac{β ℏ ω}{2}}} Im [χ (ω)]

And thats just the formula for hyperbolic cotangent

= ℏ \coth (\frac{β ℏ ω}{2}) Im [χ (ω)]

for large temepratrues

\lim_{ℏ \to 0} ℏ \coth (\frac{β ℏ ω}{2}) = \frac{2}{β ω}

\lim_{ℏ \to 0} \frac{S (ω) + S (- ω)}{2} = \frac{2}{β ω} Im [χ (ω)]

And this recovers the classical version in the classical limit

\lim_{ℏ \to 0} \frac{S (ω) + S (- ω)}{2} = \frac{2 k_{B} T}{ω} Im [χ (ω)]