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? 



Is $R^n$ stochastically complete for the heat kernel of a Schrödinger operator?
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. 


