Equation (4.11) on page 146 of Sidney Coleman's book *Aspects of Symmetry* is the holomorphic path integral,
\begin{equation}
I=\int \exp(-z^{*}Az)\Pi dzdz^{*}=\frac{1}{\det(A)},
\end{equation}
where $A$ is Hermitian positive definite and $*$ is complex conjugation. As Coleman suggests, it is fairly easy to get this result by diagonalizing $A$. However, I also tried a different route, which gives a completely different answer and I cannot see my error. Here is my alternative calculation.

Put $z=(p+iq)/\sqrt{2}$ and use real integrals.
\begin{eqnarray}
I&=&\int\exp\left(-\frac{1}{2}(p-iq)A(p+iq)\right)\Pi dzdz^{*}\\
&=&\int\exp\left(-\frac{1}{2}(pAp+ipAq-iqAp+qAq\right)\Pi dzdz^{*}\\
&=&\int dq\exp\left(-\frac{1}{2}qAq\right)\int dp\exp\left(-\frac{1}{2}pAp-\frac{i}{2}p(A-A^{T})q\right).
\end{eqnarray}
The real measure is,
\begin{equation}
dp=\frac{d^{m}p}{(2\pi)^{m/2}}
\end{equation}
and an analogous measure for $dq$. The real integral over $dp$ is a case of,
\begin{equation}
J=\int dp \exp\left(-\frac{1}{2}pAp+bp\right)=\frac{\exp\left(\frac{1}{2}bA^{-1}b\right)}{\sqrt{\det{A}}}
\end{equation}
which is equation (4.9) on the same page of Coleman's book. I've checked $J$ is correct. Using $J$ in $I$,
\begin{eqnarray}
I&=&\frac{1}{\sqrt{\det{A}}}\int dq\exp\left(-\frac{1}{2}qAq\right)\exp\left(-\frac{1}{8}q(A-A^{T})^{T}A^{-1}(A-A^{T})q\right)\\
&=&\frac{1}{\sqrt{\det{A}}}\int dq\exp\left(-\frac{1}{2}q(A+\frac{1}{4}(A-A^{T})^{T}A^{-1}(A-A^{T}))q\right)\\
&=&\frac{1}{\sqrt{\det{(A)}\det{(A+\frac{1}{4}(A-A^{T})^{T}A^{-1}(A-A^{T}))}}}
\end{eqnarray}
where the last line used another application of $J$. If $A$ is Hermitian so that $A^{T}=A^{*}$ the extra term in the second determinant does not vanish and so the result is not the same as Coleman's calculation of $I$ by diagonalizing $A$. My question is , "Where is my error in the use of the real integrals to calculate $I$?"

Edit: Thanks to Qmechanic's hints, the error is because I used real integral $J$ which assumes $A$ is real, symmetric, positive definite, but $A$ is actually Hermitian.

This post imported from StackExchange Physics at 2014-12-19 09:03 (UTC), posted by SE-user Stephen Blake