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Suppose $V:\mathbb{R}^{n} \to \mathbb{R}$ is just a positive polynomial and $K_{t}(x,y)$ is the heat kernel of $H = -\Delta + V$. Then does it follow

$$\int_{\mathbb{R}^{n}} K_{t}(x, \cdot)\,dy = 1?$$

Or would this at least follow for the kernel $\tilde{K}_{t}$ after transference of $H$ to an appropriate $L^{2}$ space?

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up vote 2 down vote accepted

If $V$ is a positive polynomial then $\int_{\mathbb{R}^n} K_t(x,y)dy$ is strictly less than $1$. This follows from the Feynman-Kac formula, $$ \int_{\mathbb{R}^n} K_t(x,y)dy = \mathbb{E}_x \left ( \exp\left ( -\int_0^t V(b(s))ds\right ) \right )$$ where the expectation on the right hand side is over Brownian motion $b$ starting at $x$. Stochastically you may interpret a positive potential as a local "killing rate" for the process.

On the other hand, since you are concerned with $V$ polynomial and positive, provided $V$ is non-constant, it is standard that $H$ has discrete spectrum and that the ground state eigenvalue $E_0$ is non-degenerate and that the ground state eigenfunction $\psi_0(x)$ vanishes nowhere and may be taken to be positive. Clearly $$\int_{\mathbb{R}^n} e^{tE_0} K_t(x,y)\psi_0(y)dy = \psi_0(x),$$ which is to say that the heat kernel for $H-E_0$ is stochastically complete on a suitable space.

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