516 Classical time-dependent statistical mechanics
̃b(ω) =√^1
2 π∫∞
−∞dte−iωtb(xt). (13.4.5)Now, consider the product ̃a(ω) ̃b∗(ω):
̃a(ω) ̃b∗(ω) =1
2 π∫∞
−∞dt∫∞
−∞dt′e−iωteiωt′
a(xt′)b(xt)=
1
2 π∫∞
−∞dt∫∞
−∞dt′e−iω(t−t′)
a(xt′)b(xt). (13.4.6)If a change of variabless=t−t′is made for thetintegral, we find
̃a(ω) ̃b∗(ω) =1
2 π∫∞
−∞ds∫∞
−∞dt′e−iωsa(xt′)b(xt′+s)=
1
2 π∫∞
−∞dse−iωs∫∞
−∞dt′a(xt′)b(xt′+s). (13.4.7)Multiplying both sides by eiωs
′
and integrating overωusing the fact that
∫∞−∞dωeiωs′
e−iωs= 2πδ(s−s′), (13.4.8)we obtain ∫∞
−∞dt a(xt)b(xt+s) =∫∞
−∞dωeiωs ̃a(ω) ̃b∗(ω). (13.4.9)Thus, for very largeT, we have, to a good approximation,
CAB(t) =1
T
∫∞
−∞dωeiωt ̃a(ω) ̃b∗(ω). (13.4.10)In practice, since the time interval is actually finite and time is discrete, eqn. (13.4.10)
can be evaluated using discrete fast Fourier transforms (FFTs).FFTs are nothing
more than canned routines capable of evaluating the transforms ineqn. (13.4.5) as
discrete sums of the form
̃ak=M∑− 1
j=0a(xj∆t)e−^2 πijk/M̃bk=M∑− 1
j=0b(xj∆t)e−^2 πijk/M, (13.4.11)corresponding to discrete frequencies and timesωk= 2πk/M∆tandtj=j∆t, respec-
tively, inO(MlnM) operations.
The efficiency of the Fourier transform method lies in the fact that the correlation
function can be computed by performing just three FFTs: Two FFTs are needed to