Derivation of the Fluctuation Dissipation Theorem

Start with the Schrödinger equation for some arbitrary Hamiltonian.

i ℏ \frac{\partial}{\partial t} Ψ (t) = \hat{H} (t) Ψ (t)

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

Then we can write a closed form expression for any evolution given a point in time $t_{0}$ .

Ψ (t) = 𝒯 e^{- \frac{i}{ℏ} \int_{t_{0}}^{t} \hat{H} (t^{'}) d t^{'}} Ψ (t_{0})

The $𝒯$ is the time ordering operator, an "operator" (which as far as I know is not actually an operator) that makes sure the order of multiplication of the Hamiltonians is in order in any Volterra series expansion of the exponential.

If we define a new operator called the "time evolution operator", $\hat{U} (t, t_{0}) = 𝒯 e^{- \frac{i}{ℏ} \int_{t_{0}}^{t} \hat{H} (t^{'}) d t^{'}}$ , then we can have the below equation.

Ψ (t) = \hat{U} (t, t_{0}) Ψ (t_{0})

Schrödinger picture

If we represent $Ψ$ as a Hilbert space with conjugate symmetric inner product $⟨ \cdot, \cdot ⟩$ then any observable $\hat{A}$ (where observables are defined to be self-adjoint operators where $⟨ \hat{A} x, y ⟩ = ⟨ x, \hat{A} y ⟩$ ) can be measured by the below equation.

A (t) = ⟨ Ψ (t) | \hat{A} | Ψ (t) ⟩

This expression for getting the values of observables is called the Schrödinger picture, where the $Ψ$ 's evolve with time and the operator $\hat{A}$ stays the same.

Heisenberg picture

Another option is the Heisenberg picture where the $\hat{A} (t)$ 's evolve with time while the $Ψ$ 's stay constant. If we let ${\hat{A}}_{S}$ be the obervable in the Schrödinger picture and ${\hat{U}}_{S}$ be the same with the time evolution operator then

⟨ Ψ (t) | {\hat{A}}_{S} | Ψ (t) ⟩ = ⟨ Ψ (t_{0}) | {\hat{U}}_{S}^{†} (t, t_{0}) {\hat{A}}_{S} {\hat{U}}_{S} (t, t_{0}) | Ψ (t) ⟩ = ⟨ Ψ (t_{0}) | {\hat{A}}_{H} (t) | Ψ (t_{0}) ⟩

where ${\hat{A}}_{H} (t)$ is the observable operator in the Heisenberg picture while $Ψ (t_{0})$ stays constant.

{\hat{A}}_{H} = {\hat{U}}_{S}^{†} (t, t_{0}) {\hat{A}}_{S} {\hat{U}}_{S} (t, t_{0})

And observables are measured like this.

⟨ Ψ (t_{0}) | {\hat{A}}_{H} (t) | Ψ (t_{0}) ⟩

Interaction picture

You can also do a combination of the two. If you split the Hamiltonian into a time dependent and time independent part.

{\hat{H}}_{S} (t) = {\hat{H}}_{0} + {\hat{H}}_{1} (t)

Then you can set the state vectors to

Ψ_{I} (t) = e^{\frac{i}{ℏ} {\hat{H}}_{0} (t - t_{0})} Ψ_{S} (t)

and observables become

{\hat{A}}_{I} (t) = e^{\frac{i}{ℏ} {\hat{H}}_{0} (t - t_{0})} {\hat{A}}_{S} e^{- \frac{i}{ℏ} {\hat{H}}_{0} (t - t_{0})}

Quantum mechanics pictures
Object	Schrödinger	Heisenberg	Interaction
State vector	$Ψ_{S} (t) = {\hat{U}}_{S} (t, t_{0}) Ψ_{S} (t_{0})$	$Ψ_{S} (t_{0})$	$Ψ_{I} (t) = e^{\frac{i}{ℏ} {\hat{H}}_{0} (t - t_{0})} Ψ_{S} (t)$
Observables	${\hat{A}}_{S}$	${\hat{A}}_{H} (t) = {\hat{U}}_{S}^{†} (t, t_{0}) {\hat{A}}_{S} (t) {\hat{U}}_{S} (t, t_{0})$	${\hat{A}}_{I} (t) = e^{\frac{i}{ℏ} {\hat{H}}_{0} (t - t_{0})} {\hat{A}}_{S} (t) e^{- \frac{i}{ℏ} {\hat{H}}_{0} (t - t_{0})}$
Measurement	$A (t) = ⟨ Ψ_{S} (t) \| {\hat{A}}_{S} \| Ψ_{S} (t) ⟩$	$A (t) = ⟨ Ψ_{S} (t_{0}) \| {\hat{A}}_{H} (t) \| Ψ_{S} (t_{0}) ⟩$	$A (t) = ⟨ Ψ_{I} (t) \| {\hat{A}}_{I} (t) \| Ψ_{I} (t) ⟩$

Deriving the Dyson Series

Starting with the interaction picture

Ψ_{I} (t) = e^{\frac{i}{ℏ} {\hat{H}}_{0} (t - t_{0})} Ψ_{S} (t)

{\hat{H}}_{S} (t) = {\hat{H}}_{0} + {\hat{H}}_{1} (t)

and take the time derivative.

\frac{\partial}{\partial t} Ψ_{I} (t) = \frac{\partial}{\partial t} e^{\frac{i}{ℏ} {\hat{H}}_{0} (t - t_{0})} Ψ_{S} (t)

Chain rule

= \frac{i}{ℏ} {\hat{H}}_{0} e^{\frac{i}{ℏ} {\hat{H}}_{0} (t - t_{0})} Ψ_{S} (t) + e^{\frac{i}{ℏ} {\hat{H}}_{0} (t - t_{0})} \frac{\partial}{\partial t} Ψ_{S} (t)

Simplify the left term.

= \frac{i}{ℏ} {\hat{H}}_{0} Ψ_{I} (t) + e^{\frac{i}{ℏ} {\hat{H}}_{0} (t - t_{0})} \frac{\partial}{\partial t} Ψ_{S} (t)

Substitute in the right term with the Schrödinger equation.

= \frac{i}{ℏ} {\hat{H}}_{0} Ψ_{I} (t) + e^{\frac{i}{ℏ} {\hat{H}}_{0} (t - t_{0})} \frac{- i}{ℏ} {\hat{H}}_{S} Ψ_{S} (t)

Factor

= \frac{i}{ℏ} ({\hat{H}}_{0} Ψ_{I} (t) - e^{\frac{i}{ℏ} {\hat{H}}_{0} (t - t_{0})} {\hat{H}}_{S} Ψ_{S} (t))

Substitute the split Hamiltonian

= \frac{i}{ℏ} ({\hat{H}}_{0} Ψ_{I} (t) - e^{\frac{i}{ℏ} {\hat{H}}_{0} (t - t_{0})} ({\hat{H}}_{0} + {\hat{H}}_{1}) Ψ_{S} (t))

An operator always commutes with its own exponent

= \frac{i}{ℏ} ({\hat{H}}_{0} Ψ_{I} (t) - {\hat{H}}_{0} e^{\frac{i}{ℏ} {\hat{H}}_{0} (t - t_{0})} Ψ_{S} (t) - e^{\frac{i}{ℏ} {\hat{H}}_{0} (t - t_{0})} {\hat{H}}_{1} Ψ_{S} (t))

= \frac{i}{ℏ} ({\hat{H}}_{0} Ψ_{I} (t) - {\hat{H}}_{0} Ψ_{I} (t) - e^{\frac{i}{ℏ} {\hat{H}}_{0} (t - t_{0})} {\hat{H}}_{1} Ψ_{S} (t))

Cancel those terms

= - \frac{i}{ℏ} e^{\frac{i}{ℏ} {\hat{H}}_{0} (t - t_{0})} {\hat{H}}_{1} Ψ_{S} (t)

And $e^{- \frac{i}{ℏ} {\hat{H}}_{0} (t - t_{0})} e^{\frac{i}{ℏ} {\hat{H}}_{0} (t - t_{0})}$ is always just the identity operator so we can put it between the Hamiltonian and the state vector.

= - \frac{i}{ℏ} e^{\frac{i}{ℏ} {\hat{H}}_{0} (t - t_{0})} {\hat{H}}_{1} (e^{- \frac{i}{ℏ} {\hat{H}}_{0} (t - t_{0})} e^{\frac{i}{ℏ} {\hat{H}}_{0} (t - t_{0})}) Ψ_{S} (t)

Then we can move the parenthesis to this

= - \frac{i}{ℏ} (e^{\frac{i}{ℏ} {\hat{H}}_{0} (t - t_{0})} {\hat{H}}_{1} e^{- \frac{i}{ℏ} {\hat{H}}_{0} (t - t_{0})}) e^{\frac{i}{ℏ} {\hat{H}}_{0} (t - t_{0})} Ψ_{S} (t)

And simplify the state vector

= - \frac{i}{ℏ} (e^{\frac{i}{ℏ} {\hat{H}}_{0} (t - t_{0})} {\hat{H}}_{1} e^{- \frac{i}{ℏ} {\hat{H}}_{0} (t - t_{0})}) Ψ_{I} (t)

And the time dependent part of the Hamiltonian operator just transforms like any other operator in the interaction picture so

= - \frac{i}{ℏ} {\hat{H}}_{1, I} Ψ_{I} (t)

So we get something similar to the Schrödinger equation in the interaction picture.

i ℏ \frac{\partial}{\partial t} Ψ_{I} (t) = {\hat{H}}_{1, I} Ψ_{I} (t)

Expression time evolution as a Dyson series

Then if we make a time evolution operator but in the interaction picture state vectors

Ψ_{I} (t) = {\hat{U}}_{I} (t, t_{0}) Ψ_{I} (t_{0})

If we substitute that into the other equation then

\frac{\partial}{\partial t} U_{I} (t, t_{0}) Ψ_{I} (t_{0}) = - \frac{i}{ℏ} {\hat{H}}_{1, I} (t) U_{I} (t, t_{0}) Ψ_{I} (t_{0})

Then it must be that

\frac{\partial}{\partial t} U_{I} (t, t_{0}) = - \frac{i}{ℏ} {\hat{H}}_{1, I} (t) U_{I} (t, t_{0})

Then we can integrate it

U_{I} (t, t_{0}) = U_{I} (t_{0}, t_{0}) - \int_{t_{0}}^{t} \frac{i}{ℏ} {\hat{H}}_{1, I} (t^{'}) U_{I} (t^{'}, t_{0}) d t^{'}

And $U_{I} (t_{0}, t_{0})$ should just be 1 so

U_{I} (t, t_{0}) = 1 - \frac{i}{ℏ} \int_{t_{0}}^{t} {\hat{H}}_{1, I} (t^{'}) U_{I} (t^{'}, t_{0}) d t^{'}

This expression can be expanded by into a Volterra series iteratively similar to a taylor series. For the zeroth order

U_{I, 0} (t, t_{0}) = 1

then put it back in to get

U_{I, 1} (t, t_{0}) = 1 - \frac{i}{ℏ} \int_{t_{0}}^{t} {\hat{H}}_{1, I} (t^{'}) d t^{'}

and iterate again

U_{I, 2} (t, t_{0}) = 1 - \frac{i}{ℏ} \int_{t_{0}}^{t} {\hat{H}}_{1, I} (t^{'}) d t^{'} - \frac{1}{ℏ} \int_{t_{0}}^{t} \int_{t_{0}}^{t^{'}} {\hat{H}}_{1, I} (t^{''}) d t^{''} d t^{'}

This generailizes to

U_{I, N} (t, t_{0}) = 1 + \sum_{n = 1}^{N} (- \frac{i}{ℏ})^{n} \int_{t_{0}}^{t} \dots \int_{t^{n - 2}}^{t^{n - 1}} {\hat{H}}_{1, I} (t^{'}) \dots {\hat{H}}_{1, I} (t^{n}) d t^{n} \dots d t

And the convergence is the actual solution

U_{I} (t, t_{0}) = 1 + \lim_{N \to \infty} \sum_{n = 1}^{N} (- \frac{i}{ℏ})^{n} \int_{t_{0}}^{t} \dots \int_{t^{n - 2}}^{t^{n - 1}} {\hat{H}}_{1, I} (t^{'}) \dots {\hat{H}}_{1, I} (t^{n}) d t^{n} \dots d t

Which is equal to the matrix exponentiation with the time ordering operator from before

U_{I} (t, t_{0}) = 𝒯 e^{- \frac{i}{ℏ} \int_{t_{0}}^{t} {\hat{H}}_{1, I} (t^{'}) d t^{'}}

Evolution of the density matrix.

A statistical mechanical ensemble is represented by a density matrix defined as below

ρ = \sum_{i} p_{i} | Ψ_{i} ⟩ ⟨ Ψ_{i} |

The expected value of an observable in an ensemble can be measured by

𝔼 [A] = \sum_{i} p_{i} ⟨ Ψ_{i} | \hat{A} | Ψ_{i} ⟩ = Tr (\hat{A} \sum_{i} p_{i} | Ψ_{i} ⟩ ⟨ Ψ_{i} |) = Tr (\hat{A} ρ)

We can substitute the state vectors to write the density matrix like

ρ (t) = \sum_{i} p_{i} U (t, t_{0}) | Ψ_{i} (t_{0}) ⟩ ⟨ Ψ_{i} (t_{0}) | U^{†} (t, t_{0}) = U (t, t_{0}) ρ (t_{0}) U^{†} (t, t_{0})

Then if we replace the time evolution operator with the Dyson series

= (1 + \sum_{n = 1}^{\infty} (- \frac{i}{ℏ})^{n} \int_{t_{0}}^{t} \dots \int_{t^{n - 2}}^{t^{n - 1}} \hat{H} (t^{'}) \dots \hat{H} (t^{n}) d t^{n} \dots d t) ρ (t_{0}) (1 + \sum_{n = 1}^{\infty} (- \frac{i}{ℏ})^{n} \int_{t_{0}}^{t} \dots \int_{t^{n - 2}}^{t^{n - 1}} \hat{H} (t^{'}) \dots \hat{H} (t^{n}) d t^{n} \dots d t)^{†}

Multiply the density matrix inside the left sum

= (ρ (t_{0}) + \sum_{n = 1}^{\infty} (- \frac{i}{ℏ})^{n} \int_{t_{0}}^{t} \dots \int_{t^{n - 2}}^{t^{n - 1}} \hat{H} (t^{'}) \dots \hat{H} (t^{n}) ρ (t_{0}) d t^{n} \dots d t) (1 + \sum_{n = 1}^{\infty} (- \frac{i}{ℏ})^{n} \int_{t_{0}}^{t} \dots \int_{t^{n - 2}}^{t^{n - 1}} \hat{H} (t^{'}) \dots \hat{H} (t^{n}) d t^{n} \dots d t)^{†}

After expanding yields a first order term with many higher order. Collect the higher order terms together.

= ρ (t_{0}) - \frac{i}{ℏ} \int_{t_{0}}^{t} \hat{H} (t^{'}) ρ (t_{0}) d t^{'} + ρ (t_{0}) (- \frac{i}{ℏ} \int_{t_{0}}^{t} \hat{H} (t^{'}) d t^{'})^{†} + 𝒪 ({\hat{H}}^{2})

= ρ (t_{0}) - \frac{i}{ℏ} \int_{t_{0}}^{t} \hat{H} (t^{'}) ρ (t_{0}) d t^{'} + ρ (t_{0}) (- \frac{i}{ℏ})^{†} \int_{t_{0}}^{t} {\hat{H}}^{†} (t^{'}) d t^{'} + 𝒪 ({\hat{H}}^{2})

Remember since the Hamiltonian is self-adjoint that $\hat{H} = {\hat{H}}^{†}$ .

= ρ (t_{0}) - \frac{i}{ℏ} \int_{t_{0}}^{t} \hat{H} (t^{'}) ρ (t_{0}) d t^{'} + ρ (t_{0}) \frac{i}{ℏ} \int_{t_{0}}^{t} \hat{H} (t^{'}) d t^{'} + 𝒪 ({\hat{H}}^{2})

= ρ (t_{0}) - \frac{i}{ℏ} \int_{t_{0}}^{t} \hat{H} (t^{'}) ρ (t_{0}) - ρ (t_{0}) \hat{H} (t^{'}) d t^{'} + 𝒪 ({\hat{H}}^{2})

Deriving the linear response function

From the dyson series of the time evolution operator:

U_{I} (t, t_{0}) = 1 - \frac{i}{ℏ} \int_{t_{0}}^{t} {\hat{H}}_{1, I} (t^{'}) d t^{'} + 𝒪 ({\hat{H}}_{1, I}^{2})

Then

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

subsitute

= Tr (ρ (t_{0}) \hat{A} - \frac{i}{ℏ} \int_{t_{0}}^{t} (\hat{H} (t^{'}) ρ (t_{0}) - ρ (t_{0}) \hat{H} (t^{'})) \hat{A} d t^{'} + 𝒪 ({\hat{H}}^{2}) \hat{A} - ρ (t_{0}) \hat{A})

= Tr (- \frac{i}{ℏ} \int_{t_{0}}^{t} (\hat{H} (t^{'}) ρ (t_{0}) - ρ (t_{0}) \hat{H} (t^{'})) \hat{A} d t^{'}) + Tr (𝒪 ({\hat{H}}^{2}) \hat{A})

= - \frac{i}{ℏ} \int_{t_{0}}^{t} Tr ((\hat{H} (t^{'}) ρ (t_{0}) - ρ (t_{0}) \hat{H} (t^{'})) \hat{A}) d t^{'} + Tr (𝒪 ({\hat{H}}^{2}) \hat{A})

= - \frac{i}{ℏ} \int_{t_{0}}^{t} Tr (\hat{H} (t^{'}) ρ (t_{0}) \hat{A}) - Tr (ρ (t_{0}) \hat{H} (t^{'}) \hat{A}) d t^{'} + Tr (𝒪 ({\hat{H}}^{2}) \hat{A})

And traces are have the circular property, so $Tr (A B C) = Tr (B C A) = Tr (C A B)$ .

= - \frac{i}{ℏ} \int_{t_{0}}^{t} Tr (ρ (t_{0}) \hat{A} \hat{H} (t^{'})) - Tr (ρ (t_{0}) \hat{H} (t^{'}) \hat{A}) d t^{'} + Tr (𝒪 ({\hat{H}}^{2}) \hat{A})

= - \frac{i}{ℏ} \int_{t_{0}}^{t} Tr (ρ (t_{0}) (\hat{A} \hat{H} (t^{'}) - \hat{H} (t^{'}) \hat{A})) d t^{'} + Tr (𝒪 ({\hat{H}}^{2}) \hat{A})

So a final expression of

𝔼 [A (t)] - 𝔼 [A (t_{0})] = - \frac{i}{ℏ} \int_{t_{0}}^{t} 𝔼_{0} [\hat{A} \hat{H} (t^{'}) - \hat{H} (t^{'}) \hat{A}] d t^{'} + Tr (𝒪 ({\hat{H}}^{2}) \hat{A})

Condider a linear force applied to a constant Hamiltonian across one of the directions of an observable starting at time $t_{0}$ , so

{\hat{H}}_{S} (t) = {\hat{H}}_{0} - F (t) {\hat{B}}_{S} (t)

Then if we use the interaction picture and set

H_{1} = - F (t) {\hat{B}}_{S} (t)

Then

H_{1, I} (t) = e^{\frac{i}{ℏ} {\hat{H}}_{1} (t)} {\hat{H}}_{1} (t) e^{- \frac{i}{ℏ} {\hat{H}}_{0} (t - t_{0})}

= - e^{\frac{i}{ℏ} {\hat{H}}_{0} (t - t_{0})} (F (t) {\hat{B}}_{S} (t)) e^{- \frac{i}{ℏ} {\hat{H}}_{0} (t - t_{0})}

And $F$ is just a scalar so you can move the operators

= - F (t) (e^{\frac{i}{ℏ} {\hat{H}}_{0} (t - t_{0})} {\hat{B}}_{S} (t) e^{- \frac{i}{ℏ} {\hat{H}}_{0} (t - t_{0})})

= - F (t) {\hat{B}}_{I} (t)

Now we start with the difference in expected value of an operator

𝔼 [A (t)] - 𝔼 [A (t_{0})] = - \frac{i}{ℏ} \int_{t_{0}}^{t} 𝔼_{0} [{\hat{A}}_{I} F (t^{'}) {\hat{B}}_{I} (t^{'}) - F (t^{'}) {\hat{B}}_{I} (t^{'}) {\hat{A}}_{I}] d t^{'} + Tr (𝒪 ({\hat{H}}^{2}) \hat{A})

𝔼 [A (t)] - 𝔼 [A (t_{0})] = - \frac{i}{ℏ} \int_{t_{0}}^{t} F (t^{'}) 𝔼_{0} [{\hat{A}}_{I} {\hat{B}}_{I} (t^{'}) - {\hat{B}}_{I} (t^{'}) {\hat{A}}_{I}] d t^{'} + Tr (𝒪 ({\hat{H}}^{2}) \hat{A})

We define this term to be the "linear response function" of $A$ to a perturbation along $B$

𝔼 [A (t)] - 𝔼 [A (t_{0})] = - \frac{i}{ℏ} \int_{t_{0}}^{t} F (t^{'}) 𝔼_{0} [\hat{A} \hat{B} (t^{'}) - \hat{B} (t^{'}) \hat{A}] d t^{'} + Tr (𝒪 ({\hat{B}}^{2}) \hat{A})

𝔼 [A (t)] - 𝔼 [A (t_{0})] = - \frac{i}{ℏ} \int_{t_{0}}^{t} F (t^{'}) χ_{A B} (t - t^{'}) d t^{'} + Tr (𝒪 ({\hat{B}}^{2}) \hat{A})

where

χ (t - t^{'}) = - \frac{i}{ℏ} 𝔼_{0} [\hat{A} \hat{B} (t^{'}) - \hat{B} (t^{'}) \hat{A}]

We can take the fourier transform of it

χ (ω) = \int_{- \infty}^{\infty} e^{i ω t} χ (t) d t = - \frac{i}{ℏ} \int_{- \infty}^{\infty} e^{i ω t} 𝔼_{0} [\hat{A} \hat{B} (t^{'}) - \hat{B} (t^{'}) \hat{A}] d t

And the linear response function must be zero at negative times so the future can't affect the past so the integral can start at zero.

= - \frac{i}{ℏ} \int_{0}^{\infty} e^{i ω t} 𝔼_{0} [\hat{A} \hat{B} (t^{'}) - \hat{B} (t^{'}) \hat{A}] d t

To find the imaginary part of any value we can do

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

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

= - \frac{1}{2 ℏ} (\int_{0}^{\infty} e^{i ω t} 𝔼_{0} [\hat{A} (t_{0}) \hat{B} (t^{'}) - \hat{B} (t^{'}) \hat{A} (t_{0})] d t - \int_{0}^{\infty} e^{- i ω t} 𝔼 [{\hat{B}}^{†} (t^{'}) {\hat{A}}^{†} (t_{0}) - {\hat{A}}^{†} (t_{0}) {\hat{B}}^{†} (t^{'})] d t)

And once agan observables are by definition their own hermetion conjugate

= - \frac{1}{2 ℏ} (\int_{0}^{\infty} e^{i ω t} 𝔼_{0} [\hat{A} (t_{0}) \hat{B} (t^{'}) - \hat{B} (t^{'}) \hat{A} (t_{0})] d t - \int_{0}^{\infty} e^{- i ω t} 𝔼 [\hat{B} (t^{'}) \hat{A} (t_{0}) - \hat{A} (t_{0}) \hat{B} (t^{'})] d t)

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

And if we make $A = B$

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

And assuming the autocorrelation should be the same under time reversal

𝔼 [\hat{A} (t_{0}) \hat{A} (t^{'}) - \hat{A} (t^{'}) \hat{A} (t_{0})] = 𝔼 [\hat{A} (t_{0}) \hat{A} (- t^{'}) + \hat{A} (- t^{'}) \hat{A} (t_{0})]

Then

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

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

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

Quantum cannonical ensemble

At equilibrium the cannonical ensemble should maximize the Von Neumann entropy. This is given by the quantum Boltzmann distribution, if it has a constant Harmiltonian. Assume that the Hamilitoian is constnat before time $t_{0}$

ρ (t_{0}) = \frac{e^{- β {\hat{H}}_{0}}}{Tr (e^{- β {\hat{H}}_{0}})} = \frac{e^{- β {\hat{H}}_{0}}}{Z}

So with a constant Hamiltonian at equilibrium we can write

𝔼_{0} [\hat{A} (t_{0}) \hat{A} (t^{'})] = Tr (ρ (t_{0}) (\hat{A} (t_{0}) \hat{A} (t^{'}))) = \frac{1}{Z} Tr (e^{- β {\hat{H}}_{0}} (\hat{A} (t_{0}) \hat{A} (t^{'})))

cycle the trace

= \frac{1}{Z} Tr ((\hat{A} (t^{'}) e^{- β {\hat{H}}_{0}} \hat{A} (t_{0})))

= \frac{1}{Z} Tr ((e^{- β {\hat{H}}_{0}} e^{β {\hat{H}}_{0}} \hat{A} (t^{'}) e^{- β {\hat{H}}_{0}} \hat{A} (t_{0})))

Imaginary time

This is equivelent to moving in imaginary time

= \frac{1}{Z} Tr ((e^{- β {\hat{H}}_{0}} \hat{A} (t^{'} - i ℏ β) \hat{A} (t_{0})))

= 𝔼_{0} [\hat{A} (t^{'} - i ℏ β) \hat{A} (t_{0})]

Taking the fourier transform

\int_{- \infty}^{\infty} e^{i ω t} 𝔼_{0} [\hat{A} (t^{'} - i ℏ β) \hat{A} (t_{0})] d t = e^{- β ℏ ω} \int_{- \infty}^{\infty} e^{i ω t} 𝔼_{0} [\hat{A} (t^{'}) \hat{A} (t_{0})] d t

So this gives the formula

\int_{- \infty}^{\infty} e^{i ω t} 𝔼_{0} [\hat{A} (t_{0}) \hat{A} (t^{'})] d t = e^{- β ℏ ω} \int_{- \infty}^{\infty} e^{i ω t} 𝔼_{0} [\hat{A} (t^{'}) \hat{A} (t_{0})] d t

then substitute

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

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

So this means

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

Where $S (ω)$ is the Fourier transform of the time shifted autocorrelation of the observable.

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)]