In the book "Path Integrals in Quantum Mechanics" by J. Zinn-Justin the following integral is evaluated:
In the book he states:
Quite generally we consider integrals of the following form:
\begin{equation}
I=\int_{\mathbb{R}^2}dxdyf(x,y)=-i\int dzdz'f(x(z,z'),y(z,z'))
\end{equation}
makeing use of the following transformation:
\begin{equation}
z=(x+iy)/\sqrt{2}, z'=(x-iy)/\sqrt{2}
\end{equation}
Then he goes on to add the following statement:
Formal complex conjugation. Notation. We now introduce a standard, convenient but somewhat dangerous notation: we denote the variable $z'$ by $\bar{z}$. We stress that the complex variables $z$ and $\bar{z}$ remain independent integration variables and are complex conjugate only in a formal sense. Indeed, the integration contours over the variables $x$ and $y$ could be deformed in complex space and, then, these variables would also take complex values. The symbol $dzd\bar{z}$ thus corresponds to an integration over a surface of real dimension 2, embedded in $\mathbb{C}^2$.
Now we consider the following expression:
\begin{equation}
\mathcal{Z}(\mathbf{A})=\int\left(\prod_{i=1}^n \frac{\mathrm{d} z_i \mathrm{~d} \bar{z}_i}{2 i \pi}\right) \mathrm{e}^{-A(\overline{\mathbf{z}}, \mathbf{z})}
\end{equation}
Where:
\begin{equation}
A(\overline{\mathbf{z}}, \mathbf{z})=\sum_{ij}\bar{z_i}A_{ij}z_j
\end{equation}
$\mathbf{A}$ is a complex matrix with a non-vanishing determinant. The author claims that the integral can be calculated from a purely algebraic viewpoint by an asymmetric change of variables:
\begin{equation}
z_i\rightarrow z_i'=\sum_{j}A_{ij}z_j
\end{equation}
leaving the variable $\bar{z}$ unchanged. This transformation has for jacobian $1/\text{det}\mathbf{A}$. What we get is:
\begin{equation}
\mathcal{Z}(\mathbf{A})=\frac{1}{\text{det}\mathbf{A}}\int\left(\prod_{i=1}^n \frac{\mathrm{d} z_i \mathrm{~d} \bar{z}_i}{2 i \pi}\right) \mathrm{e}^{-(\overline{\mathbf{z}}, \mathbf{z})}=\frac{1}{\text{det}\mathbf{A}}\left[\int\frac{\mathrm{d}z\mathrm{d}\bar{z}}{2i\pi}\mathrm{e}^{-\bar{z}z}\right]^n=\frac{1}{\text{det}\mathbf{A}}
\end{equation}
My Questions:
$(1)$: Why can this transformation be done in the first place? $z$ and $\bar{z}$ are complex conjugates of each other, how can I ignore one when I'm transforming the other? The statement I added from his book somewhat clarifies things but still confuses me a lot.
$(2)$: I don't understand why in the last integral $\mathrm{d}z_i$ isn't replaced by $\mathrm{d}z_i'$, I think this should be the case. Then the integral:
\begin{equation}
\int\frac{\mathrm{d}z'\mathrm{d}\bar{z}}{2i\pi}\mathrm{e}^{-\bar{z}z'}
\end{equation}
is not equal to $1$ because $\bar{z}$ and $z'$ are not complex conjugates. So isn't the last equation written wrong?
Best Answer
Notation in this answer: In this answer, let $z,\bar{z}\in \mathbb{C}$ denote two independent complex variables. Let $z^{\ast}$ denote the complex conjugate of $z$.
Concerning $n=1$ dimension: When Ref. 1 talks about integration $$\int_{\mathbb{C}}\frac{\mathrm{d}z^{\ast}\wedge\mathrm{d}z}{2\pi\mathrm{i}}\ldots~:=~\int_{\mathbb{R}^2}\frac{\mathrm{dRe}z \wedge\mathrm{dIm}z}{\pi} \ldots \tag{A}$$ in the complex plane $\mathbb{C}\cong \mathbb{R}^2$, it is mainly talking about Gaussian integration (to later in Ref. 1 be applied to perturbation theory in QM). It is a well-known property of Gaussian integrals that $$\forall a,b\in\mathbb{C}:~~ \int_{\mathbb{R}^2}\frac{\mathrm{dRe}z \wedge\mathrm{dIm}z}{\pi} e^{-({\rm Re}z-a)^2-({\rm Im}z-b)^2}~=~1, \tag{B}$$ or equivalently, $$\forall a,b\in\mathbb{C}:~~ \int_{\mathbb{C}}\frac{\mathrm{d}z^{\ast}\wedge\mathrm{d}z}{2\pi\mathrm{i}} e^{-(z^{\ast}-a)(z-b)}~=~1. \tag{C}$$ This fact more or less justifies the following quote from Ref. 1:
See also e.g. this related Phys.SE post.
Concerning a positive definite $n\times n$ matrix ${\bf A}$: The Gaussian formula $$ {\cal Z}({\bf A})~=~\frac{1}{\det {\bf A}} \tag{D}$$ can be proven via diagonalization with a unitary matrix, as is done in Ref. 1. Purely algebraic substitution with an asymmetric change of variables in the integral is justified, since it leads to the very same formula. This answers OP's 1st question.
Ref. 1 argues furthermore that the rational right-hand side can be extended by analytic continuation to any invertible complex matrix (although the Gaussian integral might not be convergent).
See also e.g. this related Math.SE post.
References: