Mori-Zwanzig formalism

Start with the Liouville operator $ℒ$ where for any observable $A_{H} (t)$

\frac{d A_{H} (t)}{d t} = ℒ (t) A_{H} (t)

and

A_{H} (t) = e^{\int_{t_{0}}^{t} ℒ (t) d t} A_{H} (t_{0})

If the Hamiltonian isn't time dependent, then neither with the Heisenberg Hamiltonian, and thus the Liouvill operator won't be time dependent. Then time evolution can just be simplified to

A_{H} (t) = e^{(t - t_{0}) ℒ (t)} A_{H} (t_{0})

Observable correlations

Defined a binary product on operators $O$ for classical systems

O (A, B) = \int_{V} d Γ ρ_{eq} (Γ) A (Γ) B (Γ)

and for quantum systems

O (A, B) = Tr (ρ_{eq} A B)

where $ρ_{eq}$ is the equilibrium distribution or density if it were a cannonical Gibbs ensemble.

Next define correlation of two observables to be

C_{A B} (t) = O (A_{H} (t_{0}), B_{H} (t_{0} + t))

Relevant variables

Let $A_{0} \dots A_{N}$ be $N$ arbitrary observables chosen as relevant. Define a matrix

G_{i j} = O (A_{i}, A_{j})

and a projection operator, assuming the inverse of the matrix exists

𝒫 = \sum_{k} \sum_{l} A_{k} G_{k l}^{- 1} O (A_{l}, \cdot)

It is trivial to see that when the projection operator is applied to one of the relevant observables it doesn't affect it

𝒫 A_{j} = \sum_{k} \sum_{l} A_{k} G_{k l}^{- 1} O (A_{l}, A_{j})

= \sum_{k} \sum_{l} A_{k} G_{k l}^{- 1} G_{l j} = A_{j}

Laplace transform

The Laplace transform of

e^{t ℒ}

gives

\int_{0}^{\infty} d t e^{- s t} e^{t ℒ}

= \int_{0}^{\infty} d t e^{t (ℒ - s)}

We know that

\frac{d}{d t} e^{t (ℒ - s)} = (ℒ - s) e^{t (ℒ - s)}

so if the inverse exists then

e^{t (ℒ - s)} = (ℒ - s)^{- 1} \frac{d}{d t} e^{t (ℒ - s)}

so then

\int_{0}^{\infty} d t e^{t (ℒ - s)} = (ℒ - s)^{- 1} \int_{0}^{\infty} d t \frac{d}{d t} e^{t (ℒ - s)}

= (ℒ - s)^{- 1} (e^{\infty (ℒ - s)} - 1)

And some justification why the infinite exponent goes to zero

= (s - ℒ)^{- 1}

Assuming some more inverses exist

= (s - ℒ)^{- 1} (s - (1 - 𝒫) ℒ) (s - (1 - 𝒫 ℒ))^{- 1}

= (s - ℒ)^{- 1} (s - ℒ) (s - (1 - 𝒫 ℒ))^{- 1} + (s - ℒ)^{- 1} (𝒫 ℒ) (s - (1 - 𝒫 ℒ))^{- 1}

= (s - (1 - 𝒫 ℒ))^{- 1} + (s - ℒ)^{- 1} (𝒫 ℒ) (s - (1 - 𝒫 ℒ))^{- 1}

TODO: inverse Laplace transform it to get

e^{t ℒ} = e^{t (1 - 𝒫) ℒ} + \int_{0}^{t} d τ e^{(t - τ) ℒ} 𝒫 ℒ e^{τ (1 - 𝒫) ℒ}

Apply it to the operator $(1 - 𝒫) ℒ A_{j}$ .

e^{t ℒ} (1 - 𝒫) ℒ A_{j} = e^{t (1 - 𝒫) ℒ} (1 - 𝒫) ℒ A_{j} + (\int_{0}^{t} d τ e^{(t - τ) ℒ} 𝒫 ℒ e^{τ (1 - 𝒫) ℒ}) (1 - 𝒫) ℒ A_{j}

Expand the left side

e^{t ℒ} ℒ A_{j} - e^{t ℒ} 𝒫 ℒ A_{j} = e^{t (1 - 𝒫) ℒ} (1 - 𝒫) ℒ A_{j} + (\int_{0}^{t} d τ e^{(t - τ) ℒ} 𝒫 ℒ e^{τ (1 - 𝒫) ℒ}) (1 - 𝒫) ℒ A_{j}

Operators always commute with their own exponent

ℒ e^{t ℒ} A_{j} - e^{t ℒ} 𝒫 ℒ A_{j} = e^{t (1 - 𝒫) ℒ} (1 - 𝒫) ℒ A_{j} + (\int_{0}^{t} d τ e^{(t - τ) ℒ} 𝒫 ℒ e^{τ (1 - 𝒫) ℒ}) (1 - 𝒫) ℒ A_{j}

Then expand some definitions

ℒ A_{j} (t) - e^{t ℒ} \sum_{k} \sum_{l} A_{k} G_{k l}^{- 1} O (A_{l}, ℒ A_{j}) = e^{t (1 - 𝒫) ℒ} (1 - 𝒫) ℒ A_{j} + (\int_{0}^{t} d τ e^{(t - τ) ℒ} 𝒫 ℒ e^{τ (1 - 𝒫) ℒ}) (1 - 𝒫) ℒ A_{j}

And then

\frac{\partial A_{j} (t)}{\partial t} - \sum_{k} \sum_{l} A_{k} (t) G_{k l}^{- 1} O (A_{l} (t), ℒ A_{j} (t)) = e^{t (1 - 𝒫) ℒ} (1 - 𝒫) ℒ A_{j} + (\int_{0}^{t} d τ e^{(t - τ) ℒ} 𝒫 ℒ e^{τ (1 - 𝒫) ℒ}) (1 - 𝒫) ℒ A_{j}

then expand

\frac{\partial A_{j} (t)}{\partial t} - \sum_{k} \sum_{l} A_{k} (t) G_{k l}^{- 1} O (A_{l} (t), ℒ A_{j} (t)) = e^{t (1 - 𝒫) ℒ} (1 - 𝒫) ℒ A_{j} + \int_{0}^{t} d τ \sum_{k} \sum_{l} A_{k} (t - τ) G_{k l}^{- 1} O (A_{l}, ℒ e^{τ (1 - 𝒫) ℒ} (1 - 𝒫) ℒ A_{j})

Using integration by parts and assuming the equilibrium distribution vanishes at infinity you can show that $O (A, ℒ B) = - O (ℒ A, B)$ so

\frac{d A_{j} (t)}{d t} - \sum_{k} \sum_{l} A_{k} (t) G_{k l}^{- 1} O (A_{l} (t), ℒ A_{j} (t)) = e^{t (1 - 𝒫) ℒ} (1 - 𝒫) ℒ A_{j} - \int_{0}^{t} d τ \sum_{k} \sum_{l} A_{k} (t - τ) G_{k l}^{- 1} O (ℒ A_{l}, e^{τ (1 - 𝒫) ℒ} (1 - 𝒫) ℒ A_{j})

Projection Operator Properties

Projection Operator Idempotence

We show that $𝒫 𝒫 = 𝒫$ , the projection operator is idempotent.

𝒫 (𝒫 B) = \sum_{k} \sum_{l} A_{k} (G^{- 1})_{k l} O (A_{l}, 𝒫 B)

= \sum_{k} \sum_{l} A_{k} (G^{- 1})_{k l} O (A_{l}, \sum_{m} \sum_{n} A_{m} (G^{- 1})_{m n} O (A_{n}, B))

In both the classical and quantum definition of $O$ distributes over addition

= \sum_{k} \sum_{l} A_{k} (G^{- 1})_{k l} \sum_{m} \sum_{n} O (A_{l}, A_{m} (G^{- 1})_{m n} O (A_{n}, B))

and commutes with scalar multiplication

= \sum_{k} \sum_{l} A_{k} (G^{- 1})_{k l} \sum_{m} \sum_{n} (G^{- 1})_{m n} O (A_{n}, B) O (A_{l}, A_{m})

= \sum_{k} \sum_{l} A_{k} (G^{- 1})_{k l} \sum_{m} \sum_{n} (G^{- 1})_{m n} G_{l m} O (A_{n}, B)

= \sum_{k} \sum_{l} A_{k} (G^{- 1})_{k l} \sum_{n} δ_{n l} O (A_{n}, B)

= \sum_{k} \sum_{l} A_{k} (G^{- 1})_{k l} O (A_{l}, B)

= 𝒫 B

Projection Adjointness

We also show $O (𝒫 X, Y) = O (X, 𝒫 Y)$ .

O (𝒫 X, Y) = O (\sum k l A_{k} (G^{- 1})_{k l} O (A_{l}, X), Y)

move the addition and scalar multiplictation out

= \sum k l (G^{- 1})_{k l} O (A_{l}, X) O (A_{k}, Y)

then move it back in the other one

= O (\sum k l (G^{- 1})_{k l} A_{l} O (A_{k}, Y), X)

= O (\sum k l (G^{- 1})_{k l} A_{l} O (A_{k}, Y), X)

= O (𝒫 Y, X) = O (X, 𝒫 Y)

Projected force

Let the projected force be defined as

f_{j} (t) = e^{t (1 - 𝒫) ℒ} (1 - 𝒫) ℒ A_{j}

so that

\frac{d A_{j} (t)}{d t} - \sum_{k} \sum_{l} A_{k} (t) G_{k l}^{- 1} O (A_{l} (t), ℒ A_{j} (t)) = f_{j} (t) - \int_{0}^{t} d τ \sum_{k} \sum_{l} A_{k} (t - τ) G_{k l}^{- 1} O (ℒ A_{l}, f_{j} (τ))

We can show that the projection of the projected force is zero, or $𝒫 f (t) = 0$ .

𝒫 f (t) = 𝒫 e^{t (1 - 𝒫) ℒ} (1 - 𝒫) ℒ A_{j}

𝒫 (\sum_{n = 0}^{\infty} \frac{1}{n!} (t (1 - 𝒫) ℒ)^{n}) (1 - 𝒫) ℒ A_{j}

And each term goes to zero since $n = 0$ the the term vanishes and $𝒫 (1 - 𝒫) = 𝒫 - 𝒫 𝒫 = 𝒫 - 𝒫 = 0$ from idempotence, and every $n \geq 1$ has the $(1 - 𝒫)$ term in front of it too.

𝒫 f (t) = 0

and trivially

(1 - 𝒫) f (t) = f (t)

then at the derivative we can do

\frac{d A_{j} (t)}{d t} - \sum_{k} \sum_{l} A_{k} (t) G_{k l}^{- 1} O (A_{l} (t), ℒ A_{j} (t)) = f_{j} (t) - \int_{0}^{t} d τ \sum_{k} \sum_{l} A_{k} (t - τ) G_{k l}^{- 1} O (ℒ A_{l}, (1 - 𝒫) f_{j} (τ))

and using adjointness

\frac{d A_{j} (t)}{d t} - \sum_{k} \sum_{l} A_{k} (t) G_{k l}^{- 1} O (A_{l} (t), ℒ A_{j} (t)) = f_{j} (t) - \int_{0}^{t} d τ \sum_{k} \sum_{l} A_{k} (t - τ) G_{k l}^{- 1} O ((1 - 𝒫) ℒ A_{l}, f_{j} (τ))

then

\frac{d A_{j} (t)}{d t} - \sum_{k} \sum_{l} A_{k} (t) G_{k l}^{- 1} O (A_{l} (t), ℒ A_{j} (t)) = f_{j} (t) - \int_{0}^{t} d τ \sum_{k} \sum_{l} A_{k} (t - τ) G_{k l}^{- 1} O (f_{l} (0), f_{j} (τ))

And $O (X, Y) = O (X (t), Y (t))$ can be shown to be time invatiant so we can just write it without the times

\frac{d A_{j} (t)}{d t} = \sum_{k} \sum_{l} A_{k} (t) G_{k l}^{- 1} O (A_{l}, ℒ A_{j}) - \int_{0}^{t} d τ \sum_{k} \sum_{l} A_{k} (t - τ) G_{k l}^{- 1} O (f_{l} (0), f_{j} (τ)) + f_{j} (t)

And making some definition

Ω_{k j} = G_{k l}^{- 1} O (A_{l}, ℒ A_{j})

γ_{k j} (t) = \sum_{l} G_{k l}^{- 1} O (f_{l} (0), f_{j} (t))

Makes the generalized langevin equation

\frac{d A_{j} (t)}{d t} = \sum_{k} Ω_{k j} A_{k} (t) - \int_{0}^{t} d τ \sum_{k} γ_{k j} (τ) A_{k} (t - τ) + f_{j} (t)

Then you somehow are supposed to get to Langevin dynamics using some assumptions, idk what though.

\frac{d p (t)}{d t} = - \frac{\partial H}{\partial q} - γ M^{- 1} p + \sqrt{2 k_{B} T γ} \frac{d W}{d t}