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

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

The scalar field $\phi$ is defined on a two-dimensional curved sphere. The mass-like term $\frac{1}{\lambda}M^2(x) \phi^2$ depends explicitly on $x$ and I'm interested in limit $\lambda\to 0$. Formally the result is the functional determinant $\log Z=\log \operatorname{det} \Big(-\Delta+\frac{1}{\lambda} M^2(x)\Big)$ and I'm interested the small $\lambda$ 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 $\lambda$ expansion here does not seem to reduce to the conventional large mass expansion. Moreover, the heat kernel coefficients have the form like $a_2=\int d^2x\sqrt{g}\Big(\frac16R-\frac{1}{\lambda}M^2(x)\Big)$ while I naively expect that the leading order in small $\lambda$ limit should be

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

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.

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

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

The scalar field $\phi$ is defined on a two-dimensional curved sphere. The mass-like term $\frac{1}{\lambda}M^2(x) \phi^2$ depends explicitly on $x$ and I'm interested in limit $\lambda\to 0$. Formally the result is the functional determinant $\log Z=\log \operatorname{det} \Big(-\Delta+\frac{1}{\lambda} M^2(x)\Big)$ and I'm interested the small $\lambda$ 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 $\lambda$ expansion here does not seem to reduce to the conventional large mass expansion. Moreover, the heat kernel coefficients have the form like $a_2=\int d^2x\sqrt{g}\Big(\frac16R-\frac{1}{\lambda}M^2(x)\Big)$ while I naively expect that the leading order in small $\lambda$ limit should be

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

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.

Maybe this comment will help to connect the question with the existing literature.

I've learned (obvious in retrospect) fact that such functional determinants can be related to the Schrodinger operators. In this case $\frac{1}{\lambda}M^2(x)$ plays the role of the potential. The small $\lambda$ limit then should be equivalent to $\hbar\to0$ and allow to use the WKB approximation. In 1d one could probably use it in combination with the Gelfand-Yaglom theorem to compute the determinant, but whether it is useful in 2d I'm not sure.

I need to compute a functional determinant of the following form $$\operatorname{det} \Big(-\Delta+m^2+V(x)\Big)$$ The underlying space is a curved two-sphere. Operator $\Delta=\nabla_\mu\nabla^ {\mu}$ is the Laplacian acting on functions (the covariant derivative is the standard Riemannian covariant derivative), $V(x)$ is a potential term and I am interested in $m\to\infty$ limit.

In other words this is a determinant representing the one-loop effective action of a scalar field $\phi$ of mass $m$ in a quadratic potential $\phi^2 V(x)$ on a curved sphere and I would like to get the large $m$ expansion.

This must be a pretty basic stuff, but I'm not sure how to do it or where to look it up. Any pointers are welcome. Below I give my attempt to compute it.

A standard approach seems to be via the heat kernel method which asserts $$\log \det(D)=-\int_{\Lambda^{-2}}^\infty \frac{dt}{t}e^{-tm^2}\operatorname{Tr} e^{-t(-\Delta+V(x))}$$ and $\Lambda$ is a regularization parameter that needs to be sent to infinity. And as far as I can tell the heat kernel small $t$ expansion in this case is given by (I use formulas (2.21) and (4.26), (4.27) from this review $$\operatorname{Tr} e^{-t(-\Delta+V(x))}=\frac{1}{t}a_0+a_2+t a_4+O(t^2)$$ with $$a_0=\frac{1}{4\pi}\int d^2x\sqrt{g}=\frac{Area}{4\pi}\\ a_2=\frac{1}{4\pi}\int d^2x \sqrt{g}\left(-V(x)+\frac16 R(x)\right)=-\frac{Area*Average(V)}{4\pi}+\frac13\\ a_4=\dots$$ Substituting this expansion into the integral and evaluating it gives (I'm not sure about all the constants, but the spirit must be correct) $$\log\det(-\Delta+m^2+V(x))=-m^2a_0\left(\frac{\Lambda^2}{m^2}+\log \frac{m^2}{\Lambda^2}+\gamma-1\right)+a_2(\log\frac{m^2}{\Lambda^2}+\gamma)-\frac{a_4}{m^2}+O(\frac{m^2}{\Lambda^2}\log\frac{m^2}{\Lambda^2})$$ where $\gamma=-\int_0^\infty \log(t)e^{-t}\approx 0.58$ is Euler's constant.

There are several things that puzzle me about this result

1. There seems to be a tension between $\Lambda\to\infty$ (removing regularization) and $m\to \infty$ limit as the expansion parameter is effectively $\frac{m^2}{\Lambda^2}$. Can I really trust it to give the large $m$ expansion?
2. Is it true that the determinant I'm computing only depends on the potential $V(x)$ through its average value $\int d^2x \sqrt{g} V(x)$ in the lowest orders of large $m$ expansion (orders that do not vanish at $m\to\infty$)? This is no inconsistency, but I expect a different result.
3. I have little intuition about the functional determinants but it seems to me that $ \log\det m^2$ should be proportional to $\log m^2$ and this should give the leading order in the above expansion. The actual leading order seems to be higher $\sim m^2$ or $\sim m^2\log \frac{m^2}{\Lambda^2}$.

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

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

The scalar field $\phi$ is defined on a two-dimensional curved sphere. The mass-like term $\frac{1}{\lambda}M^2(x) \phi^2$ depends explicitly on $x$ and I'm interested in limit $\lambda\to 0$. Formally the result is the functional determinant $\log Z=\log \operatorname{det} \Big(-\Delta+\frac{1}{\lambda} M^2(x)\Big)$ and I'm interested the small $\lambda$ 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 $\lambda$ expansion here does not seem to reduce to the conventional large mass expansion. Moreover, the heat kernel coefficients have the form like $a_2=\int d^2x\sqrt{g}\Big(\frac16R-\frac{1}{\lambda}M^2(x)\Big)$ while I naively expect that the leading order in small $\lambda$ limit should be

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

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.

I need to compute a functional determinant of the following form $$\operatorname{det} \Big(-\Delta+m^2+V(x)\Big)$$ The underlying space is a curved two-sphere. Operator $\Delta=\nabla_\mu\nabla^ {\mu}$ is the Laplacian acting on functions (the covariant derivative is the standard Riemannian covariant derivative), $V(x)$ is a potential term and I am interested in $m\to\infty$ limit.

In other words this is a determinant representing the one-loop effective action of a scalar field of mass $m$ in a potential $V(x)$ on a curved sphere and I would like to get the large $m$ expansion.

This must be a pretty basic stuff, but I'm not sure how to do it or where to look it up. Any pointers are welcome. Below I give my attempt to compute it.

A standard approach seems to be via the heat kernel method which asserts $$\log \det(D)=-\int_{\Lambda^{-2}}^\infty \frac{dt}{t}e^{-tm^2}\operatorname{Tr} e^{-t(-\Delta+V(x))}$$ and $\Lambda$ is a regularization parameter that needs to be sent to infinity. And as far as I can tell the heat kernel small $t$ expansion in this case is given by (I use formulas (2.21) and (4.26), (4.27) from this review $$\operatorname{Tr} e^{-t(-\Delta+V(x))}=\frac{1}{t}a_0+a_2+t a_4+O(t^2)$$ with $$a_0=\frac{1}{4\pi}\int d^2x\sqrt{g}=\frac{Area}{4\pi}\\ a_2=\frac{1}{4\pi}\int d^2x \sqrt{g}\left(-V(x)+\frac16 R(x)\right)=-\frac{Area*Average(V)}{4\pi}+\frac13\\ a_4=\dots$$ Substituting this expansion into the integral and evaluating it gives (I'm not sure about all the constants, but the spirit must be correct) $$\log\det(-\Delta+m^2+V(x))=-m^2a_0\left(\frac{\Lambda^2}{m^2}+\log \frac{m^2}{\Lambda^2}+\gamma-1\right)+a_2(\log\frac{m^2}{\Lambda^2}+\gamma)-\frac{a_4}{m^2}+O(\frac{m^2}{\Lambda^2}\log\frac{m^2}{\Lambda^2})$$ where $\gamma=-\int_0^\infty \log(t)e^{-t}\approx 0.58$ is Euler's constant.

There are several things that puzzle me about this result

1. There seems to be a tension between $\Lambda\to\infty$ (removing regularization) and $m\to \infty$ limit as the expansion parameter is effectively $\frac{m^2}{\Lambda^2}$. Can I really trust it to give the large $m$ expansion?
2. Is it true that the determinant I'm computing only depends on the potential $V(x)$ through its average value $\int d^2x \sqrt{g} V(x)$ in the lowest orders of large $m$ expansion (orders that do not vanish at $m\to\infty$)? This is no inconsistency, but I expect a different result.
3. I have little intuition about the functional determinants but it seems to me that $ \log\det m^2$ should be proportional to $\log m^2$ and this should give the leading order in the above expansion. The actual leading order seems to be higher $\sim m^2$ or $\sim m^2\log \frac{m^2}{\Lambda^2}$.

I need to compute a functional determinant of the following form $$\operatorname{det} \Big(-\Delta+m^2+V(x)\Big)$$ The underlying space is a curved two-sphere. Operator $\Delta=\nabla_\mu\nabla^ {\mu}$ is the Laplacian acting on functions (the covariant derivative is the standard Riemannian covariant derivative), $V(x)$ is a potential term and I am interested in $m\to\infty$ limit.

In other words this is a determinant representing the one-loop effective action of a scalar field $\phi$ of mass $m$ in a quadratic potential $\phi^2 V(x)$ on a curved sphere and I would like to get the large $m$ expansion.

This must be a pretty basic stuff, but I'm not sure how to do it or where to look it up. Any pointers are welcome. Below I give my attempt to compute it.

A standard approach seems to be via the heat kernel method which asserts $$\log \det(D)=-\int_{\Lambda^{-2}}^\infty \frac{dt}{t}e^{-tm^2}\operatorname{Tr} e^{-t(-\Delta+V(x))}$$ and $\Lambda$ is a regularization parameter that needs to be sent to infinity. And as far as I can tell the heat kernel small $t$ expansion in this case is given by (I use formulas (2.21) and (4.26), (4.27) from this review $$\operatorname{Tr} e^{-t(-\Delta+V(x))}=\frac{1}{t}a_0+a_2+t a_4+O(t^2)$$ with $$a_0=\frac{1}{4\pi}\int d^2x\sqrt{g}=\frac{Area}{4\pi}\\ a_2=\frac{1}{4\pi}\int d^2x \sqrt{g}\left(-V(x)+\frac16 R(x)\right)=-\frac{Area*Average(V)}{4\pi}+\frac13\\ a_4=\dots$$ Substituting this expansion into the integral and evaluating it gives (I'm not sure about all the constants, but the spirit must be correct) $$\log\det(-\Delta+m^2+V(x))=-m^2a_0\left(\frac{\Lambda^2}{m^2}+\log \frac{m^2}{\Lambda^2}+\gamma-1\right)+a_2(\log\frac{m^2}{\Lambda^2}+\gamma)-\frac{a_4}{m^2}+O(\frac{m^2}{\Lambda^2}\log\frac{m^2}{\Lambda^2})$$ where $\gamma=-\int_0^\infty \log(t)e^{-t}\approx 0.58$ is Euler's constant.

There are several things that puzzle me about this result

1. There seems to be a tension between $\Lambda\to\infty$ (removing regularization) and $m\to \infty$ limit as the expansion parameter is effectively $\frac{m^2}{\Lambda^2}$. Can I really trust it to give the large $m$ expansion?
2. Is it true that the determinant I'm computing only depends on the potential $V(x)$ through its average value $\int d^2x \sqrt{g} V(x)$ in the lowest orders of large $m$ expansion (orders that do not vanish at $m\to\infty$)? This is no inconsistency, but I expect a different result.
3. I have little intuition about the functional determinants but it seems to me that $ \log\det m^2$ should be proportional to $\log m^2$ and this should give the leading order in the above expansion. The actual leading order seems to be higher $\sim m^2$ or $\sim m^2\log \frac{m^2}{\Lambda^2}$.