# One-loop effective action for scalar field on the curved background in large potential

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.

• It seems that you can rewrite original action as $\lambda(...)$ and then find field configuration which saturates the integral, in other words try to use $\lambda^{-1}$ expansion and method of steepest descent. Commented Jul 2, 2020 at 12:52
• @ArtemAlexandrov I'm not sure what you mean. The action is already Gaussian so the saddle point approximation just gives back the original problem (the saddle point action is zero in this case). Commented Jul 2, 2020 at 13:09
• If $\tilde{g}_{\mu\nu}=M^{2}g_{\mu\nu}$, then the action becomes the same as the action of a massive particle with mass $1/\lambda$ in a space with metric $\tilde{g}_{\mu\nu}$. Commented Jul 5, 2020 at 12:06
• @atarasenko This is an interesting observation. However, to my understanding the measure in the path integral $\mathcal{D}\phi$ in fact depends on $g$. So although this redefinition works at the level of classical action it does not work for the full partition function. Evaluating the Jacobian of the change is probably equivalent to the original problem. The answer I get by other method also does not seem invariant under $g_{\mu\nu}\to \Omega(x) g_{\mu\nu}, M^2\to \Omega(x)^{-1}M^2$. Commented Jul 6, 2020 at 16:01

## 1 Answer

As suggested by Michael Stone in a removed answer there are hints on how to do this in a known heat kernel user's manual. It goes under the name of modified large mass expansion there.

I will set $$\lambda=1$$ and consider $$M^2$$ to be large. Standard heat kernel method begins with the following representation for the determinant $$W=\log Z=-\frac12 \log \det\left(-\Delta+M^2\right)=\frac12 \int^\infty_{\Lambda^{-2}}\frac{dt}{t}\operatorname{Tr} e^{t(\Delta-M^2)}\quad (1)$$ and here the cut-off regularization is used.

The trace can be understood as $$\operatorname{Tr} e^{t(\Delta-M^2)}=\sum_n \int d^2z \sqrt{g}\psi_n e^{t(\Delta-M^2)} \psi_n$$ with some orthonormal basis of functions $$\psi_n$$. One can use factorization $$e^{t(\Delta-M^2)}=e^{t \Delta}e^{-tM^2}(1+O(t))\qquad (2)$$ where the higher-order terms arise because $$\Delta$$ and $$M^2$$ do not commute. However, since small $$t$$ expansion corresponds to the desirable large $$M$$ expansion they are not needed to get the leading order. Using the standard small-t expansion of the heat kernel on the diagonal $$\langle x| e^{t\Delta}|x\rangle=\frac{1}{4\pi t}\Big(1+O(t)\Big)$$ and substituting factorization (2) in equation (1) and first integrating over $$t$$ one gets $$W=\frac{1}{8\pi}\int d^2z \sqrt{g} \int_{\Lambda^{-2}}^\infty \frac{dt}{t^2} e^{-tM^2}\left(1+O(t)\right)=\\\frac{1}{8\pi}\int d^2z \sqrt{g}\Big(\Lambda^2+M^2\log\frac{M^2}{\Lambda^2}+M^2(\gamma-1)\Big)+O\left(1\right)$$ And the $$O(t)$$ term translates into $$O(1)$$ in terms of $$M$$. To get the next orders of expansion one needs to use higher orders terms in the heat kernel expansion and also corrections to factorization (2).