Linear response 501
iL(s)f 0 (H(x)) =−
∂f 0
∂H
j(x)Fe(s). (13.2.18)
Now, suppose thatf 0 (H(x)) is given by a canonical distribution
f 0 (H(x)) =
exp[−βH(x)]
Q(N,V,T)
. (13.2.19)
Then, since∂f 0 /∂H=−βf 0 , eqn. (13.2.18) becomes
iL(s)f 0 (H(x)) =βf 0 (H(x))j(x)Fe(s). (13.2.20)
Note that eqn. (13.2.16) also holds whenf 0 (H(x)) is an isothermal-isobaric or grand
canonical distribution. When we substitute eqn. (13.2.20) into eqn.(13.2.14), we obtain
∆f(x,t) =−β
∫t
0
dse−iL^0 (t−s)f 0 (H(x))j(x)Fe(s). (13.2.21)
We now substitute eqn. (13.2.21) into eqn. (13.2.10) to express theaverage ofa(x) in
the nonequilibrium ensemble as
A(t) =〈a〉t=〈a〉−β
∫
dxa(x)
∫t
0
dse−iL^0 (t−s)f 0 (H(x))j(x)Fe(s)
=〈a〉−β
∫t
0
ds
∫
dxa(x)e−iL^0 (t−s)f 0 (H(x))j(x)Fe(s)
=〈a〉−β
∫t
0
ds
∫
dxf 0 (H(x))a(x)e−iL^0 (t−s)j(x)Fe(s). (13.2.22)
The second term in eqn. (13.2.22) involves a classical propagator exp[−iL 0 (t−s)].
However, since the propagator contains onlyiL 0 , its action generates the evolution
of the undriven system obtained by solving Hamilton’s equations in theabsence of
any perturbation. If we write the propagator as exp[iL 0 (−(t−s))], it is clear that
this propagator evolves a system backwards in time. Recall from Section 3.10 that if
the phase space vector evolves in time from an initial condition x 0 to xtaccording to
Hamilton’s equations (under the action of exp[iL 0 t]), the evolution of any phase space
functiona(x) is determined by
da
dt
=iL 0 a
a(xt) = eiL^0 ta(x 0 ). (13.2.23)
However, since it can be shown thatL 0 is a Hermitian operator, if we take the adjoint
of both sides, we obtain