# Problem evaluating a holomorphic path integral [closed]

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.

## closed as off-topic by ACuriousMind♦, Kyle Kanos, Brandon Enright, Pranav Hosangadi, PraharDec 9 '14 at 2:39

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## 1 Answer

Hints:

1. Real Gaussian integral: $$\tag{1} \int_{\mathbb{R}^n} \! d^n x ~e^{- x^t M x} ~=~ \sqrt{\frac{(2\pi)^n}{\det (M+M^t)}}.$$

2. Determinant of block matrices: $$\tag{2} \det\begin{pmatrix} A & B \\ C & D \end{pmatrix} ~=~\det(A) \det(D-CA^{-1}B)~=~ \det(D) \det(A-BD^{-1}C).$$