Help with the Proof of an Operator Identity

[Opus_723]

I'm trying to come up with a proof of the operator identity typically used in the Mori projector operator formalism for Generalized Langevin Equations,

$$e^{tL} = e^{t(1-P)L}+\int_{0}^{t}dse^{(t-s)L}PLe^{s(1-P)L}$$,

where L is the Liouville operator and P is a projection operator that projects onto a finite subset of Hilbert space, abstractly defined by

$$PB = \sum_{j,k}(B,A_j)((A,A)^{-1})_{j,k}A_k$$

Where A is the set of functions to be projected onto and ( . , . ) denotes an unspecified inner product.

Unfortunately, I'm pretty rusty with operators in general, and although I can write down a "derivation," it relies on an assumption that I can't figure out a justification for. Here is what I did:

$$e^{tL} = e^{t(1-P)L}+e^{tL}-e^{t(1-P)L}$$
$$e^{tL} = e^{t(1-P)L}+e^{tL}(1-e^{-tPL})$$
$$e^{tL} = e^{t(1-P)L}+e^{tL}(e^{-0*PL}-e^{-tPL})$$
$$e^{tL} = e^{t(1-P)L}+e^{tL}(-e^{-tPL})\bigg|_0 ^t$$
$$e^{tL} = e^{t(1-P)L}+e^{tL}\int_{0}^{t}\frac{d}{ds}(-e^{-sPL})$$
$$e^{tL} = e^{t(1-P)L}+e^{tL}\int_{0}^{t}PLe^{-sPL}$$
$$e^{tL} = e^{t(1-P)L}+e^{tL}\int_{0}^{t}PLe^{-sL+s(1-P)L}$$
$$e^{tL} = e^{t(1-P)L}+\int_{0}^{t}e^{(t-s)L}PLe^{s(1-P)L}$$

Notice, however, that throughout this process I liberally assumed, in many steps, that the projection operator P and the Liouville operator L *commute*, mostly by blithely using the typical properties of the exponential as if the operators were numbers, but also in the last step more explicitly. So I see 3 options:

1) L and P are guaranteed to commute for some reason that's not obvious to me.
2) This is a bad way to prove the identity and if I did it some other way I wouldn't need them to commute, or maybe I am completely misunderstanding how to manipulate these objects.
3) This assumption is *not* guaranteed, but it *is* actually necessary for the identity to hold even though I haven't seen this assumption stated explicitly anywhere.

My guess is that either 1 or 2 is the correct answer, but I'm not able to puzzle it out. Can anyone with more experience with projection operators help me understand this identity? Thank you for your time.

EDIT: If it helps, I'm following the treatment in Ch. 8 of Nonequilibrium Statistical Mechanics by Robert Zwanzig, although this same identity seems to pop up along with an "it is easy to show that..." in every treatment I can find.

 

[vanhees71]

Let's define
$$F(t)=\exp(t L).$$
Then $F$ is defined by the differential equation
$$\mathrm{d}_t F(t)=L F(t)=F(t) L, \quad F(0)=1.$$
The only thing we need to do is to show that also the right-hand side of the equation fulfills the same differential equation and initial condition. The latter is trivial.

Now let's take the derivative wrt. $t$ of the right-hand side, which we call $\tilde{F}(t)$
$$\mathrm{d}_t \tilde{F}(t)=(1-P)L \exp[t (1-P)L]+P L \exp[t(1-P)L]+L \int_0^t \mathrm{d} s \exp[(t-s)L] P L \exp[s(1-P)L]=L \tilde{F}(t).$$
So indeed $\tilde{F}$ fulfills the differential equation and the initial condition as $F$. So $F=\tilde{F}$. Note that I've nowhere used that $P$ and $L$ maybe commute.

I hope, I've not overlooked any detail, because this looks so simple ;-).