Wednesday, 8 August 2018

quantum field theory - Multivariable Dirac Delta and Faddeev-Popov Determinant


From this mathstack page and in particular Qmechanic's answer:





  1. There exists an n-dimensional generalization δn(f(x)) = f(x(0))=0x(0)1|detf(x)x|δn(xx(0))
    of the substitution formula for the Dirac delta distribution under pertinent assumptions, such as e.g., that the function f:ΩRnRn has isolated zeros. Here the sum on the rhs. of eq. (1) extends to all the zeros x(0) of the function f.



Also from this page on the Faddeev-Popov procedure they say:



For ordinary functions, a property of the Dirac delta function gives: δ(xx0)=|df(x)dx|x=x0δ(f(x)) assuming f(x) has only one zero at x=x0 and is differentiable there. Integrating both sides gives :1=|df(x)dx|x=x0dxδ(f(x))

. Extending over n variables, suppose f(xi)=0 for some xi0. Then, replacing δ(xx0) with niδi(xixi0) :1=(i|f(xi)xi|)(idxi)δ(f(xi))
. Recognizing the first factor as the determinant of the diagonal matrix f(xi)xiδij (no summation implied), we can generalize to the functional version of the identity: :1=det|δGδΩ|G=0DΩδ[Ga(ϕΩ)]
, where ΔF[ϕ]det|δFδg|F=0 is the Faddeev-Popov determinant.



What I don't understand is that it seems their function f seems to be f:ΩRnR. How does the generalized Dirac formula (1) work in this case? I don't really understad their notation in:


1=(i|f(xi)xi|)(idxi)δ(f(xi))


What does f(xi)xi

mean here?




Answer



The notation fixi

means the diagonal elements of the matrix: Jij=fixj
where fi is the component of the vector f(x).




I found this very confusing a few weeks ago so. Here is the proof I wrote up for the identity based on the response I received to an earlier question of mine here:


Recall that if f(x) has one zero at x0 then, dx|df(x)dx|x=x0δ(f(x))=1

We want to generalize this to instead of having f(x) we have, g(a) for vectors of arbitrary size. To do this consider the Taylor expansion of g around its root (we assume it only has one root, a0): gi(a)=0gi(a0)+jgiaj|a0(aja0,j)+...
We want to insert this into a delta function, δ(n)(g(a)). This will only be nonzero near a=a0. Thus we have, δ(g(a))=iδ(gi(a))=iδ(jJij(aja0,j))
where Jij is the Jacobian matrix defined by Jijgiaj|a0. We have, δ(g(a))=δ(jJ1j(aja0,j))δ(jJ2j(aja0,j))...
We now use the identity, δ(α(aa0))=δ(aa0)|α|
We choose to isolate each delta function in the equation above for a different aj: δ(g(a))=δ(a1a0,1)|J1,1|δ(a2a0,2)|J2,2|...
If we take the Jacobian matrix to be greater then zero then we have the product: (J1,1J2,2..)1=1detJ
where we have used the fact that the determinant of J is independent of a unitary transformation. So we finally have, (idai)δ(n)(g(a))det(giaj)=1
where it is understood that the Jacobian matrix is evaluated at the root of g.


We write the continuum generalization of this equation as, Dα(x)δ(G(Aα))det(δG(Aα)δα)=1


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