Complex Gaussian integral with different source terms

Question

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

$$\int D({\phi,\psi,b}) e^{-b^\dagger A b +f(\phi, \phi^\dagger,\psi, \psi^\dagger )b +b^\dagger g(\phi, \phi^\dagger,\psi, \psi^\dagger )} = \int D(\phi,\psi) \det(A^{-1}) e^{f(...) A^{-1} g(...)} $$

valid when $f \ne 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

$$\int D(...) e^{- Ax^2 - A y^2 +x(f + g) + i y(f-g)} = \frac{\pi}{A}\int D(...) e^{(4A)^{-1}((f+g)^2 - (f-g)^2)}$$ $$=\frac{\pi}{A}\int D(...) e^{A^{-1} fg}.$$

But then this implies that Hubbard-Stratonovich transformations don't need to be of squares.. so I can decouple any interaction $$e^{2fg} = \int d \phi d\phi^\dagger e^{-|\phi|^2 +f\phi + \phi^\dagger g}.$$ This can't be right?

Answer 1

Theorem: Given a normal$^1$ $n\times n$ matrix $A$ where ${\rm Re}(A)>0$ is positive definite, then the complex Gaussian integral is convergent with value$^2$ $$\begin{align} I&~:=~\int_{\mathbb{R}^{2n}} \! d^nx ~d^ny~ \exp\left\{-z^{\dagger}Az +f^{\dagger}z +z^{\dagger}g\right\}\cr &~=~\exp\left\{f^{\dagger}A^{-1}g\right\}\int_{\mathbb{R}^{2n}} \! d^nx ~d^ny~ \exp\left\{-(z^{\dagger}-f^{\dagger}A^{-1})A(z-A^{-1}g)\right\}\cr &~=~\frac{\pi^n}{\det(A)}\exp\left\{f^{\dagger}A^{-1}g\right\}, \qquad z^k~\equiv~ x^k+iy^k.\end{align}$$

Sketched proof:

1. The normal matrix $A=U^{\dagger}DU$ can be diagonalized with a unitary transformation $U$. Here $D$ is a diagonal matrix with ${\rm Re}(D)>0$. Next change integration variables$^3$ $w=Uz$. 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 $x_0,y_0\in\mathbb{C}$ such that$^4$ $$ x_0-iy_0~=~f^{\dagger}A^{-1}\qquad\text{and}\qquad x_0+iy_0~=~A^{-1}g.$$

3. We can shift the real integration contour into the complex plane $$\int_{\mathbb{R}} \! dx \int_{\mathbb{R}} \! dy~ \exp\left\{-(z^{\dagger}-f^{\dagger}A^{-1})A(z-A^{-1}g)\right\}$$ $$~=~\int_{\mathbb{R}+x_0} \! dx \int_{\mathbb{R}+y_0} \! dy~ \exp\left\{-z^{\dagger}Az\right\}~=~\frac{\pi}{A},$$ with no new non-zero contributions arising from closing the contour, cf. Cauchy's integral theorem.$\Box$

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$^1$ The Gaussian integral is also convergent for a pertinent class of non-normal matrices $A$, cf. my other Phys.SE answer, but in this answer we only consider normal matrices for simplicity.

$^2$ Recall that the notation $\int_{\mathbb{C}^n}d^nz^{\ast} d^nz$ means $\int_{\mathbb{R}^{2n}} \! d^nx ~d^ny$ up to a conventional factor, cf. my Phys.SE answer here. Here $z^k \equiv x^k+iy^k$ and $z^{k\ast} \equiv x^k-iy^k$.

$^3$ More generally, under a holomorphic change of variables $u^k+iv^k\equiv w^k=f^k(z)$, the absolute value of the Jacobian determinant in the formula for integration by substitution is $$ |\det\left(\frac{\partial (u,v)}{\partial (x,y)} \right)_{2n\times 2n}|~=~ |\det\left(\frac{\partial w}{\partial z} \right)_{n\times n}|^2. $$

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

Answer 2

Theorem: Given a $2n\times 2n$ complex symmetric matrix $S$ where $${\rm Re}(J^TSJ)~>~0$$ is positive definite, then the complex Gaussian integral is convergent with value $$ \begin{align} I~:=~&\int_{\mathbb{C}^n} \! \left[\prod_{k=1}^n\frac{\mathrm{d}z^{k\ast} \wedge \mathrm{d}z^k}{2\pi i}\right]~ \exp\left\{ -\frac{1}{2}Z^TSZ +B^TZ\right\}\cr ~=~&\sqrt{\frac{(-1)^n}{\det(S)}}\exp\left\{\frac{1}{2}B^TS^{-1}B \right\}. \end{align} $$ Here we have defined $$ \begin{align} Z~:=~&(z^1, \ldots, z^n, z^{\ast 1}, \ldots, z^{\ast n})^T, \cr B~:=~&( b^{\ast 1}, \ldots, b^{\ast n},b^1, \ldots, b^n)^T, \cr J~:=~&\begin{pmatrix} 1 & i \cr 1 & -i \end{pmatrix} \otimes \mathbb{1}_{n\times n}. \end{align}$$ Note that $b^{\ast j}$ is not necessarily the complex conjugate of $b^j$, it could be an independent complex number; however $z^{\ast j}$ is assumed to be the complex conjugate of $z^j$.

Sketched proof:

1. Make a coordinate transformation into real and imaginary parts $$\begin{align} z^k~\equiv~& x^k+iy^k, \cr z^{k\ast}~\equiv~& x^k-iy^k, \cr Z~=~&JX, \cr X~:=~&(x^1, \ldots, x^n, y^1, \ldots, y^n)^T. \cr \end{align} $$ Then the integral becomes $$ \begin{align} I~=~&\int_{\mathbb{R}^{2n}} \! \left[\prod_{k=1}^n\frac{\mathrm{d}x^k \wedge \mathrm{d}y^k}{\pi}\right]~ \exp\left\{ -\frac{1}{2}X^TJ^TSJX +B^TJX\right\}\cr ~=~&\sqrt{\frac{2^n}{\det(J^TSJ)}}\exp\left\{\frac{1}{2}B^TJ( J^TSJ)^{-1}J^TB \right\}. \end{align} $$ In the second and last equality we used the complex Gaussian formula from this Math.SE post.

2. Next use that $$\begin{align} JJ^T~=~&\begin{pmatrix} 0 & 2 \cr 2 & 0 \end{pmatrix} \otimes \mathbb{1}_{n\times n},\cr \det(JJ^T)~=~&(-2)^n, \end{align} $$ to derive the theorem. $\Box$