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)e−b†Ab+f(ϕ,ϕ†,ψ,ψ†)b+b†g(ϕ,ϕ†,ψ,ψ†)=∫D(ϕ,ψ)det(A−1)ef(...)A−1g(...)
valid when f≠g∗?
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(...)e−Ax2−Ay2+x(f+g)+iy(f−g)=πA∫D(...)e(4A)−1((f+g)2−(f−g)2)
But then this implies that Hubbard Stratonovich transformations don't need to be of squares.. so I can decouple any interaction e2fg=∫dϕdϕ†e−|ϕ|2+fϕ+ϕ†g.
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{−z†Az+f†z+z†g} = exp{f†A−1g}∫R2ndnx dny exp{−(z†−f†A−1)A(z−A−1g)} = πndet(A)exp{f†A−1g},zk ≡ xk+iyk.
Sketched proof:
The normal matrix A=U†DU 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.
There exist two complex numbers x0,y0∈C such that4 x0−iy0 = f†A−1andx0+iy0 = A−1g.
We can shift the real integration contour into the complex plane ∫Rdx∫Rdy exp{−(z†−f†A−1)A(z−A−1g)}
= ∫R+x0dx∫R+y0dy exp{−z†Az} = πA,with no new non-zero contributions arising from closing the contour, cf. Cauchy's integral theorem.◻
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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 ∫Cndnz∗dnz means ∫R2ndnx dny up to a conventional factor, cf. my Phys.SE answer here. Here zk≡xk+iyk and zk∗≡xk−iyk.
3 More generally, under a holomorphic change of variables uk+ivk≡wk=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(∂w∂z)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,g∈Cn are independent complex constants.
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