Thursday, 30 November 2017

quantum field theory - Complex Gaussian integral with different source terms


Do the source terms multiplying a complex field and its conjugate need to be conjugates for the Gaussian identity to hold? E.g. is



D(ϕ,ψ,b)ebAb+f(ϕ,ϕ,ψ,ψ)b+bg(ϕ,ϕ,ψ,ψ)=D(ϕ,ψ)det(A1)ef(...)A1g(...)


valid when fg?


If I change to real and imaginary coordinates in the b it seems fine, but I'm worried that I'm screwing up the measure in D(...) without realizing it.


Edit:


Let's say A is a c-number. To do the integral I can write b=x+iy etc. Then the integral is


D(...)eAx2Ay2+x(f+g)+iy(fg)=πAD(...)e(4A)1((f+g)2(fg)2)

=πAD(...)eA1fg.


But then this implies that Hubbard Stratonovich transformations don't need to be of squares.. so I can decouple any interaction e2fg=dϕdϕe|ϕ|2+fϕ+ϕg.

This can't be right?



Answer




Theorem: Given a normal1 n×n matrix A where Re(A)>0 is positive definite, then the complex Gaussian integral is2 I := R2ndnx dny exp{zAz+fz+zg} = exp{fA1g}R2ndnx dny exp{(zfA1)A(zA1g)} = πndet(A)exp{fA1g},zk  xk+iyk.




Sketched proof:




  1. The normal matrix A=UDU can be diagonalized with a unitary transformation U. Here D is a diagonal matrix with Re(D)>0. Next change integration variables3 w=Uz. The absolute value of the Jacobian determinant is 1. So it is enough to consider the case n=1, which we will do from now on.




  2. There exist two complex numbers x0,y0C such that4 x0iy0 = fA1andx0+iy0 = A1g.





  3. We can shift the real integration contour into the complex plane RdxRdy exp{(zfA1)A(zA1g)}

     = R+x0dxR+y0dy exp{zAz} = πA,
    with no new non-zero contributions arising from closing the contour, cf. Cauchy's integral theorem.




--


1 The Gaussian integral is presumably also convergent for a pertinent class of non-normal matrices A, but in this answer we consider only normal matrices for simplicity.


2 Recall that the notation Cndnzdnz means R2ndnx dny up to a conventional factor, cf. my Phys.SE answer here. Here zkxk+iyk and zkxkiyk.


3 More generally, under a holomorphic change of variables uk+ivkwk=fk(z), the absolute value of the Jacobian determinant in the formula for integration by substitution is |det((u,v)(x,y))2n×2n| = |det(wz)n×n|2.


4 The underlying philosophy in point 2 is similar to my Phys.SE answer here: One can in a certain sense treat z and z as independent variables! And therefore it is possible to consider OP's case where f,gCn are independent complex constants.


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