Mori-Zwanzig formalism

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

dAH(t)dt=(t)AH(t)

and

AH(t)=et0t(t)dtAH(t0)

This generalizes to both classical mechanics where 𝔸=2n, a function over the positions and momentum, or quantum mechanics where 𝔸=(n×n) 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(tt0)AH(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 𝒫𝔸𝔸.

Weird Identity

Let

V(t)=etet(1𝒫)

then

dV(t)dt=etet(1𝒫)+et(1𝒫)et(1𝒫)

and operators commute with their exponent

=et(+(1𝒫))et(1𝒫)
=et𝒫et(1𝒫)

And fundamental theorem of calculus

V(t)V(0)=0tdτeτ𝒫eτ(1𝒫)
etet(1𝒫)1=0tdτeτ𝒫eτ(1𝒫)

Then left multiply

et(1𝒫)et=0tdτe(tτ)𝒫eτ(1𝒫)

So

et=et(1𝒫)+0tdτe(tτ)𝒫eτ(1𝒫)

We have

dA(t)dt=etA=et(𝒫+1𝒫)A=et𝒫A+et(1𝒫)A

Then use the operator idendity on the right one

=et𝒫A+(et(1𝒫)+0tdτe(tτ)𝒫eτ(1𝒫))(1𝒫)A

more stuff

=et𝒫A+et(1𝒫)(1𝒫)A+0tdτe(tτ)𝒫eτ(1𝒫)(1𝒫)A

Projected force

Let the projected force be defined as

fA(t)=et(1𝒫)(1𝒫)A

Then gives

dA(t)dt=et𝒫A+0tdτe(tτ)𝒫fA(τ)+fA(t)

Note that we didn't make any assumptions about the projection operator 𝒫 or the Liouville operator , 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)dΓρeq(Γ)A(Γ)δ(g(Γ)g(Γ))

Where ρeq(Γ)=1ZeH(Γ)kBT is the cannonical equilibrium distribution, and g2nm 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

𝒫g(Γ)=1Z(Γ)dΓρeq(Γ)(g(Γ))δ(g(Γ)g(Γ))

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

=1Z(Γ)dΓg(Γ)(ρeq(Γ)δ(g(Γ)g(Γ)))
=1Z(Γ)dΓg(Γ)(ρeq(Γ)δ(g(Γ)g(Γ))+ρeq(Γ)δ(g(Γ)g(Γ)))

And peq=0 so

=1Z(Γ)dΓg(Γ)ρeq(Γ)δ(g(Γ)g(Γ))
=1Z(Γ)dΓg(Γ)ρeq(Γ)[δ(g(Γ)g()),H](Γ)
=kBT1Z(Γ)dΓg(Γ)[δ(g(Γ)g()),ρeq](Γ)

Then integrate by parts again

=kBT1Z(Γ)dΓρeq[g(Γ),δ(g(Γ)g())](Γ)

Chain rule

=kBT1Z(Γ)dΓρeq](Γ)kδ(g(Γ)g(Γ))ck[gk,g](Γ)
=kBT1Z(Γ)kg(Γ)kdΓρeq(Γ)δ(g(Γ)g(Γ))[gk,g](Γ)
=kBT1Z(Γ)kg(Γ)kdΓρeq(Γ)δ(g(Γ)g(Γ))[gk,g](Γ)
=kBT1Z(Γ)kg(Γ)k((𝒫[gk,g])Z(g(Γ)))(g(Γ))
=kBT1Z(Γ)kg(Γ)k((𝒫[gk,g])Z(g(Γ)))(g(Γ))
=kBT1Z(Γ)kg(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)=1ZeF(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(tτ)𝒫fg(τ)

Looking at just

𝒫fg(τ)(Γ)=dΓpeq(Γ)δ(g(Γ)g(Γ))fg(τ)(Γ)
=dΓpeq(Γ)fg(τ)(Γ)δ(g(Γ)g(Γ))
=dΓpeq(Γ)fg(τ)(Γ)kδ(g(Γ)g(Γ))ckgk(Γ)
=kgk(Γ)dΓpeq(Γ)fg(τ)(Γ)δ(g(Γ)g(Γ))gk(Γ)

Next we show that

𝒫g=0
=dΓpeq(Γ)δ(g(Γ)g(Γ))g(Γ)