Linear response 499can use a perturbative approach to solve the Liouville equation. When the external
perturbation is small, we assume that the solutionf(x,t) can be written in the form
f(x,t) =f 0 (H(x)) + ∆f(x,t), (13.2.7)whereH(x) =H(q,p) is the Hamiltonian andf 0 (H(x)) is the equilibrium phase space
distribution function generated by the corresponding unperturbed system (Ci=Di=
0). Eqns. (13.2.1) could be coupled to a heat bath, barostat, or particle reservoir, so
thatf 0 (H(x)) can be any of the equilibrium ensemble distributions introduced thus
far. All that is required off 0 (H(x)) is that it satisfy the equilibrium Liouville equation
iL 0 f 0 (H(x)) = 0, (13.2.8)whereiL 0 is the unperturbed Liouville operatoriL 0 ={...,H}. We will assume that
f(x,t) is normalized, so that ∫
dxf(x,t) = 1. (13.2.9)Using the ansatz in eqn. (13.2.7) gives the expression for the ensemble average of any
functiona(x):
〈a〉t=∫
dxa(x)f(x,t)=
∫
dxa(x)f 0 (H(x)) +∫
dxa(x)∆f(x,t)=〈a〉+∫
dxa(x)∆f(x,t)=A(t), (13.2.10)where〈a〉is the average ofa(x) in the unperturbed ensemble described byf 0 (H(x)),
and the notationA(t) =〈a〉tindicates an average in the nonequilibrium ensemble
corresponding to the time-dependent propertyA(t). Note that if we assume the system
to be in equilibrium att= 0, then〈a〉 0 =〈a〉.
Since eqns. (13.2.1) are of the form ̇x = ̇x 0 + ∆ ̇x(t), the Liouville operator can be
written as
iL= ̇x·∇x= ( ̇x 0 + ∆ ̇x(t))·∇x=iL 0 +i∆L(t). (13.2.11)
Thus, the Liouville equation becomes
∂
∂t
(f 0 (H(x)) + ∆f(x,t)) + (iL 0 +i∆L(t))(f 0 (H(x)) + ∆f(x,t)) = 0. (13.2.12)Assuming that the driving force terms in eqns. (13.2.1) constitute asmall perturba-
tion, we neglect the second order termi∆L∆f(x,t). This approximation constitutes
a linearization of the Liouville equation, which is the basis oflinear response theory.
Thus, the results we derive by neglecting second-order terms will only be valid within