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  Functional determinant of a massive scalar field with variable mass

+ 1 like - 0 dislike

I hope to compute a functional integral $Z=\int \mathcal{D}\phi\,\, e^{-S[\phi]}$ with an action

$$S[\phi]=\int d^2x \sqrt{g}\Big((\nabla \phi)^2+\frac{1}{g}M^2(x) \phi^2\Big)$$

The scalar field $\phi$ is defined on a two-dimensional curved sphere.. The mass-like term $\frac{1}{g}M^2(x) \phi^2$ depends explicitly on $x$ and I'm interested in limit $g\to 0$. Formally the result is the functional determinant $\log Z=\log \operatorname{det} \Big(-\Delta+\frac{1}{g} M^2(x)\Big)$ and I'm interested the small $g$ expansion as a functional of $M^2(x)$.

I'm not very familiar with functional determinants but I've tried to apply the heat kernel method here without much success. The small $g$ expansion here does not seem to reduce to the conventional large mass expansion. Moreover the heat kernel coefficients have the form like $a_2=\int d^2x\sqrt{g}\Big(\frac16R-\frac{1}{g}M^2(x)\Big)$ while I naively expect that the leading order in small $g$ limit should be

$$\log \operatorname{det} \Big(-\Delta+\frac{1}{g} M^2(x)\Big)\sim \log \operatorname{det} \Big(\frac{1}{g} M^2(x)\Big)=\operatorname{Tr}\log \Big(\frac{1}{g} M^2(x)\Big)\sim \\\int d^2z \sqrt{g} \log\Big(\frac{1}{g} M^2(x)\Big)$$

Where the last line is my guess for what the functional trace of a function (diff operator of order 0) should be. Heat kernel method does not seem to produce logarithms like that.

Any comments and pointers to the literature are welcome.

asked Jun 29, 2020 in Theoretical Physics by Weather Report (240 points) [ no revision ]

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