# Gaussian Integrals : Functional determinant expressed as a trace

Be $A_{ij}$ a symmetric matrix. Then I can easily write $$\int \exp\left(-\frac{1}{2}\sum_{i,j}x_i A_{ij} x_j+\sum_{i} B_i x_i\right)\; d^nx= \sqrt{(2\pi)^n}\exp\left\{-\frac{1}{2}\mathrm{Tr}\log A\right\}\exp\left\{\frac{1}{2}\vec{B}^{T}A^{-1}\vec{B}\right\}.$$ and this formula holds both in the finite and inifinite dimensional case.

If I consider a functional integration, and a differential operator $\hat A$, the result is usually expressed as $$\int \exp\left( - \frac 1 2 \varphi \hat A \varphi +J \varphi \right) D\varphi \; \propto \; \exp \left( {1\over 2} \int d^4x \; d^4y J\left ( x \right ) A^{-1}\left ( x - y \right ) J\left( y \right ) \right)$$ or in some cases as $$\int \exp\left( - \frac 1 2 \varphi \hat A \varphi +J \varphi \right) D\varphi \; =\\\mathcal{N}\,\exp\left\{-\frac{1}{2}\mathrm{Tr}\log \hat A \right\}\, \exp \left( {1\over 2} \int dx \; dx^\prime J\left ( x \right ) A^{-1}\left ( x - x^\prime \right ) J\left( x^\prime \right ) \right)$$ where $\varphi \hat A \varphi = \int dx dx^\prime \varphi(x) A(x,x^\prime) \varphi(x^\prime)$ and $J\varphi = \int dx \varphi(x) J(x)$.

A more complicated situation I have to face now would be $$\int D\psi D\varphi \exp\left\{ - \frac 1 2 \left[ \varphi \hat A \varphi + \psi \hat B \psi + \varphi \psi \hat C \psi \varphi \right] \right\}$$ where $\varphi \psi \hat C \psi \varphi=\int dx dx^\prime \varphi(x) \psi(x) C(x,x^\prime) \psi(x^\prime) \varphi(x^\prime)$ where $\psi$ is a Grassman-odd field. Anyway, the problem should be the same in the case the $\psi$ was a simple Grassman-even field.

My idea in this case would be rewriting it as $$\int D\psi \exp\left\{ - \frac 1 2 \psi \hat B \psi \right\} \int D\varphi \exp\left\{ - \frac 1 2 \varphi \left[ \hat A + \psi \hat C \psi \right]\varphi \right\}$$ in such a way to obtain something similar to $$\int D\psi \exp\left\{ - \frac 1 2 \psi \hat B \psi \right\} \exp \left\{ - \frac{1}{2} \mathrm{Tr} \log \left[ \hat A + \psi \hat C \psi \right]\right\}$$ but I don't know how to treat the Trace in this case. Would this procedure be acceptable, and how should I deal with the trace operator?

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Comment to the question(v1): antisymmetric matrix? Are $\varphi$ and $\psi$ Grassmann-even variables? –  Qmechanic Dec 8 '12 at 14:24
Symmetric matrix, sorry, I wrote it wrong. In the full calculation I have to do, $\varphi$ is a real field and $\psi$ is a grassman-odd variable. –  Juan Sebastian Totero Dec 8 '12 at 14:29
This is acceptable, and people refer to it as 'integrating out' something (in this case the field $\varphi$). There's no easy way to deal with the new $\text{Tr log}$ in the integral; you can try expanding it around $\psi = 0$ and see that it introduces an infinite number of vertices. I recall that Weinberg I chapter 9 is a good reference for this. –  Vibert Dec 8 '12 at 15:33
@Vibert : Actualy expanding the $\log$ was my idea. Provided idk how to manage an expansion over "abstract operators", I would like to project the whole $\mathrm{Tr} \log [..]$ over $x$ or $k$, and then expand the projected logarithm. But how exactly does a $\mathrm{Tr}$ projection work? –  Juan Sebastian Totero Dec 8 '12 at 15:50
In field theory, the trace should be regularized. –  Qmechanic Dec 9 '12 at 12:15