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8.11: Appendix II- Distributions and Functionals

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Let xR 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 tn=nτ. A functional of η(t) becomes a multivariable function of the values ηnη(tn). The metric then becomes Dηndηn .

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(t1)δJ(tn)|J(t)=0=η(t1)η(tn) . The function J(t) is an arbitrary source function. We differentiate with respect to it in order to find the η-field correlators.

[Fdiscretize] Discretization of a continuous function η(t). Upon discretization, a functional Φ[η(t)] becomes an ordinary multivariable function Φ({ηj}).

Let’s compute the generating function for a class of distributions of the Gaussian form, P[η]=exp(12Γdt(τ2˙η2+η2))=exp(12Γdω2π(1+ω2τ2)|ˆη(ω)|2) . Then Fourier transforming the source function J(t), it is easy to see that Z[J]=Z[0]exp(Γ2dω2π|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(12dtdtJ(t)G(tt)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(t1)η(t2)=G(t1t2)NNη(t1)η(t2)η(t3)η(t4)=G(t1t2)G(t3t4)+G(t1t3)G(t2t4)+G(t1t4)G(t2t3) . The generalization is now easy to prove, and is known as Wick’s theorem: η(t1)η(t2n)=contractionsG(ti1ti2)G(ti2n1ti2n) , where the sum is over all distinct contractions of the sequence 1-22n into products of pairs. How many terms are there? Some simple combinatorics answers this question. Choose the index 1. There are (2n1) other time indices with which it can be contracted. Now choose another index. There are (2n3) indices with which that index can be contracted. And so on. We thus obtain C(n)\# of contractionsof 1-2-3\,2n=(2n1)(2n3)31=(2n)!2nn! .


This page titled 8.11: Appendix II- Distributions and Functionals is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Daniel Arovas.

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