8.11: Appendix II- Distributions and Functionals
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Let x∈R be a random variable, and P(x) a probability distribution for x. The average of any function ϕ(x) is then ⟨ϕ(x)⟩=∞∫−∞dxP(x)ϕ(x)/∞∫−∞dxP(x) .
Let η(t) be a random function of t, with η(t)∈R, and let P[η(t)] be the probability distribution functional for η(t). Then if Φ[η(t)] is a functional of η(t), the average of Φ is given by ∫DηP[η(t)]Φ[η(t)]/∫DηP[η(t)] The expression ∫DηP[η]Φ[η] is a functional integral. A functional integral is a continuum limit of a multivariable integral. Suppose η(t) were defined on a set of t values t∗n=nτ. A functional of η(t) becomes a multivariable function of the values η∗n≡η(t∗n). The metric then becomes Dη⟶∏ndη∗n .
In fact, for our purposes we will not need to know any details about the functional measure Dη; we will finesse this delicate issue18. Consider the generating functional, Z[J(t)]=∫DηP[η]exp(∞∫−∞dtJ(t)η(t)) . It is clear that 1Z[J]δnZ[J]δJ(t∗1)⋯δJ(t∗n)|†J(t)=0=⟨η(t∗1)⋯η(t∗n)⟩ . The function J(t) is an arbitrary source function. We differentiate with respect to it in order to find the η-field correlators.
Let’s compute the generating function for a class of distributions of the Gaussian form, P[η]=exp(−12Γ∞∫−∞dt(τ2˙η2+η2))=exp(−12Γ∞∫−∞dω2π(1+ω2τ2)|ˆη(ω)|2) . Then Fourier transforming the source function J(t), it is easy to see that Z[J]=Z[0]⋅exp(Γ2∞∫−∞dω2π|ˆJ(ω)|21+ω2τ2) . Note that with η(t)∈R and J(t)∈R we have η∗(ω)=η(−ω) and J∗(ω)=J(−ω). Transforming back to real time, we have Z[J]=Z[0]⋅exp(12∞∫−∞dt∞∫−∞dt′J(t)G(t−t′)J(t′)) , where G(s)=Γ2τe−|s|/τ,ˆG(ω)=Γ1+ω2τ2 is the Green’s function, in real and Fourier space. Note that ∞∫−∞dsG(s)=ˆG(0)=Γ . We can now compute ⟨η(t∗1)η(t∗2)⟩=G(t∗1−t∗2)N∑N⟨η(t∗1)η(t∗2)η(t∗3)η(t∗4)⟩=G(t∗1−t∗2)G(t∗3−t∗4)+G(t∗1−t∗3)G(t∗2−t∗4)+G(t∗1−t∗4)G(t∗2−t∗3) . The generalization is now easy to prove, and is known as Wick’s theorem: ⟨η(t∗1)⋯η(t∗2n)⟩=∑contractionsG(t∗i∗1−t∗i∗2)⋯G(t∗i∗2n−1−t∗i∗2n) , where the sum is over all distinct contractions of the sequence 1-2⋯2n into products of pairs. How many terms are there? Some simple combinatorics answers this question. Choose the index 1. There are (2n−1) other time indices with which it can be contracted. Now choose another index. There are (2n−3) indices with which that index can be contracted. And so on. We thus obtain C(n)≡\# of contractionsof 1-2-3\,⋯2n=(2n−1)(2n−3)⋯3⋅1=(2n)!2nn! .