Mori-Zwanzig formalism

Start with the Liouville operator ℒ where for any Heisenberg observable AH(t)∈𝔸

dAH(t)dt=ℒ(t)AH(t)

and

AH(t)=e∫t0tℒ(t)dtAH(t0)

This generalizes to both classical mechanics where 𝔸=ℝ2n→ℝ, a function over the positions and momentum, or quantum mechanics where 𝔸=(ℝn×ℝn)→ℂ is a self Hermetian matrix. If the Hamiltonian isn't time dependent, then neither is the Heisenberg Hamiltonian, and thus the Liouville operator won't be time dependent. Then time evolution can just be simplified to

AH(t)=e(t−t0)ℒAH(t0)

Projection Operator

We want to split the observables into "relevant" and "irrelevant" groups, then split equations of motion of the relevant observables into a part that only depends on the relevant observables and a part that depends on the irrelevant observables. We assume, with minimal assumptions, arbitrary projection operator 𝒫∈𝔸→𝔸.

Weird Identity

Let

V(t)=e−tℒet(1−𝒫)ℒ

then

dV(t)dt=−ℒe−tℒet(1−𝒫)ℒ+e−tℒ(1−𝒫)ℒet(1−𝒫)ℒ

and operators commute with their exponent

=e−tℒ(−ℒ+(1−𝒫)ℒ)et(1−𝒫)ℒ
=−e−tℒ𝒫ℒet(1−𝒫)ℒ

And fundamental theorem of calculus

V(t)−V(0)=−∫0tdτe−τℒ𝒫ℒeτ(1−𝒫)ℒ
e−tℒet(1−𝒫)ℒ−1=−∫0tdτe−τℒ𝒫ℒeτ(1−𝒫)ℒ

Then left multiply

et(1−𝒫)ℒ−etℒ=−∫0tdτe(t−τ)ℒ𝒫ℒeτ(1−𝒫)ℒ

So

etℒ=et(1−𝒫)ℒ+∫0tdτe(t−τ)ℒ𝒫ℒeτ(1−𝒫)ℒ

We have

dA(t)dt=etℒℒA=etℒ(𝒫+1−𝒫)ℒA=etℒ𝒫ℒA+etℒ(1−𝒫)ℒA

Then use the operator idendity on the right one

=etℒ𝒫ℒA+(et(1−𝒫)ℒ+∫0tdτe(t−τ)ℒ𝒫ℒeτ(1−𝒫)ℒ)(1−𝒫)ℒA

more stuff

=etℒ𝒫ℒA+et(1−𝒫)ℒ(1−𝒫)ℒA+∫0tdτe(t−τ)ℒ𝒫ℒeτ(1−𝒫)ℒ(1−𝒫)ℒA

Projected force

Let the projected force be defined as

fA(t)=et(1−𝒫)ℒ(1−𝒫)ℒA

Then gives

dA(t)dt=etℒ𝒫ℒA+∫0tdτe(t−τ)ℒ𝒫ℒfA(τ)+fA(t)

Note that we didn't make any assumptions about the projection operator 𝒫 or the Liouville operator ℒ, so it should work for any projection operator on any time evolution.


Zwanzig Projection Operator

The Zwanzig projection operator can be defined by

𝒫A(Γ)=1Z(Γ)∫dΓ′ρeq(Γ′)A(Γ′)δ(g(Γ)−g(Γ′))

Where ρeq(Γ)=1Ze−H(Γ)kBT is the cannonical equilibrium distribution, and g∈ℝ2n→ℝm is some collection of observables. Let these be called the "relevant" observables. We want to find the time evolution of these relevant observables.

First Term

𝒫ℒg(Γ)=1Z(Γ)∫dΓ′ρeq(Γ′)(ℒg(Γ′))δ(g(Γ)−g(Γ′))

we can do integration by parts and assume peq goes to zero at infinity

=1Z(Γ)∫dΓ′g(Γ′)ℒ(ρeq(Γ′)δ(g(Γ)−g(Γ′)))
=1Z(Γ)∫dΓ′g(Γ′)(ℒρeq(Γ′)δ(g(Γ)−g(Γ′))+ρeq(Γ′)ℒδ(g(Γ)−g(Γ′)))

And ℒpeq=0 so

=1Z(Γ)∫dΓ′g(Γ′)ρeq(Γ′)ℒδ(g(Γ)−g(Γ′))
=1Z(Γ)∫dΓ′g(Γ′)ρeq(Γ′)[δ(g(Γ)−g(⋅)),H](Γ′)
=−kBT1Z(Γ)∫dΓ′g(Γ′)[δ(g(Γ)−g(⋅)),ρeq](Γ′)

Then integrate by parts again

=−kBT1Z(Γ)∫dΓ′ρeq[g(Γ′),δ(g(Γ)−g(⋅))](Γ′)

Chain rule

=−kBT1Z(Γ)∫dΓ′ρeq](Γ′)∑k∂δ(g(Γ)−g(Γ′))∂ck[gk,g](Γ′)
=−kBT1Z(Γ)∑k∂∂g(Γ)k∫dΓ′ρeq(Γ′)δ(g(Γ)−g(Γ′))[gk,g](Γ′)
=−kBT1Z(Γ)∑k∂∂g(Γ)k∫dΓ′ρeq(Γ′)δ(g(Γ)−g(Γ′))[gk,g](Γ′)
=−kBT1Z(Γ)∑k∂∂g(Γ)k((𝒫[gk,g])Z(g(Γ)))(g(Γ))
=−kBT1Z(Γ)∑k∂∂g(Γ)k((𝒫[gk,g])Z(g(Γ)))(g(Γ))
=−kBT1Z(Γ)∑k∂∂g(Γ)k((𝒫[gk,g])Z(g(Γ)))(g(Γ))
=−kBT1Z(Γ)∑k((∂∂g(Γ)k𝒫[gk,g])Z(g(Γ))+(𝒫[gk,g])∂∂g(Γ)kZ(g(Γ)))(g(Γ))

Let

Z(Q)=1Ze−F(Q)kBT
=−kBT1Z(Γ)∑k((∂∂g(Γ)k𝒫[gk,g])Z(g(Γ))+(𝒫[gk,g])Z(g(Γ))∂F(g(Γ))∂g(Γ)k(−1kBT))(g(Γ))
=∑k(−kBT(∂∂g(Γ)k𝒫[gk,g])+(𝒫[gk,g])∂F(g(Γ))∂g(Γ)k)(g(Γ))

Second and Third Term

We probably can't get an exact expression for the second term but we can relate it to the third term.

∫0tdτe(t−τ)ℒ𝒫ℒfg(τ)

Looking at just

𝒫ℒfg(τ)(Γ)=∫dΓ′peq(Γ′)δ(g(Γ)−g(Γ′))ℒfg(τ)(Γ′)
=−∫dΓ′peq(Γ′)fg(τ)(Γ′)ℒδ(g(Γ)−g(Γ′))
=∫dΓ′peq(Γ′)fg(τ)(Γ′)∑k∂δ(g(Γ)−g(Γ′))∂ckℒgk(Γ′)
=∑k∂∂gk(Γ)∫dΓ′peq(Γ′)fg(τ)(Γ′)δ(g(Γ)−g(Γ′))ℒgk(Γ′)

Next we show that

𝒫ℒg=0
=∫dΓ′peq(Γ′)δ(g(Γ)−g(Γ′))ℒg(Γ′)