Let $V:\mathbb R\to[0,\infty)$ be (for simplicity) a smooth function. Consider the Schrödinger operator $$H:=-\Delta/2+V$$ acting on $L^2(\mathbb R)$. By the Feynman-Kac formula, the semigroup of this operator $T_t:=e^{-t H},~(t\geq0)$, is an integral operator with integral kernel $$u_t(x,y):=\frac{e^{-(x-y)^2/2t}}{\sqrt{2\pi t}}\mathbb E\left[\exp\left(-\int_0^tV(x+W_s)~ds\right)\bigg|x+W_t=y\right],$$ where $W$ is a standard Brownian motion.
Question. Are there known sufficient and necessary conditions on the potential $V$ for $T_t$ to be a Hilbert-Schmidt operator, that is $$\int_{\mathbb R\times\mathbb R}u_t(x,y)^2~dxdy=\int_{\mathbb R}u_{2t}(x,x)~dx<\infty?$$ (Equivalently, we may ask for conditions for $T_{2t}$ to be trace-class.)
If we use elementary computations on the maxima of Brownian bridges, it is not too difficult to find sufficient conditions in terms of $V(x)$'s growth as $x\to\infty$. However, I was not able to find any results which attempted to identify optimal conditions.