Is there a Feynman-Kac formula applicable to Dirichlet problems for Schrödinger operators?

On pg. 133 of Heat Kernels and Spectral Theory, Davies is studying the heat kernel $K(x,y,t)$ of the operator $H = -\Delta + |x|^{\alpha}$ for $\alpha > 0$. He wishes to prove a lower bound, and writes, "If $H_{B}$ is the operator obtained from $H$ by imposing Dirichlet boundary conditions on the surface of the ball with center $x$ and radius $1$, then $$K(x,x,t) \geq K_{B}(x,x,t)$$ for all $t > 0$." (This I see follows from the parabolic maximum principle and nonnegativity of $K$.) He then continues, "Moreover, $$|y|^{\alpha} \leq (|x| + 1)^{\alpha}$$ for all $y \in B$ so $$K_{B}(x,x,t) \geq K_{0}(x,x,t)\exp\left\{-(|x| + 1)^{\alpha}t\right\}$$ for all $t > 0$ where $K_{0}$ is the heat kernel of $-\Delta$ on $B$ with Dirichlet boundary conditions."

This is where he loses me. I'm aware that for, say, $V_{1}, V_{2} \in L^{\infty}(\mathbb{R}^{n})$ with $V_{1} \leq V_{2}$, the Feynman-Kac formula implies that the operator $H_{1} = -\Delta + V_{1}$ has heat kernel greater than or equal to that of $H_{2} = -\Delta + V_{2}$.

But I'm not sure if something analogous is the justification in Davies' argument, because I can't find a version of the Feynman-Kac formula that would apply in this setting.

Does anyone happen to know the proper explanation? Thank you.

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Perhaps one could reverse the order of the two reductions? Instead of going from $-\Delta+|y|^\alpha$ on $R^n$ to $-\Delta+|y|^\alpha$ on $B$ with Dirichlet conditions to $-\Delta+(|x|+1)^\alpha$ on $B$ with Dirichlet conditions, one could use $\Delta + \max(|x|+1,|y|)^\alpha$ on $R^n$ as the intermediate operator and then the Feynman-Kac formula on $R^n$ could be used. –  Terry Tao Aug 26 '13 at 11:06
Regarding the question of whether there is a Feynman-Kac formula assuming Dirichlet conditions, it seems that the formula would be signed rather than unsigned (basically one picks up a minus sign whenever one bounces off the boundary) which looks not particularly useful for proving comparison theorems. On the other hand, the heat kernel is still non-negative, so there may still be some probabilistic interpretation, perhaps after performing some additional cancellation... –  Terry Tao Aug 26 '13 at 11:10
@TerryTao: thanks for the reply! What would you consider the best reference for the Feynman-Kac formula on $\mathbb{R}^{n}$? (I have seen Broderix, K. & Hundertmark, D. & Leschke, H.: Continuity properties of Schrödinger semigroups with magnetic fields mentioned.) –  Michael Tinker Aug 26 '13 at 13:26
I guess Davies may have just used a second application of the maximum principle; namely, $u = K_{B} - K_{0}\exp\left\{(|x| + 1)^{\alpha}\right\}$ is a supersolution of $-\Delta + (|x|+1)^{\alpha}$ which vanishes on $\partial B$, so $u \geq 0$ inside $B$. –  Michael Tinker Aug 26 '13 at 13:44
Yes, if $H = -\Delta + V$ on a smooth domain $D$ with Dirichlet boundary conditions, then the solution to $\partial_t u = - H u$ with initial condition $u_0$ is given by $$u(t,x) = E_x \Bigl[ \exp\Bigl(-\int_0^t V(X_s)\,ds\Bigr) u_0(X_s) 1_{t < \tau}\Bigr]\;,$$ where $X$ is a Brownian motion started at $x$ and $\tau$ is the first time when $X$ hits the boundary of $D$. I don't have a reference off the top of my head, but you can see this quite easily by arguing that $H$ is the limit (in resolvent sense) of $H_n$ where $H_n$ is defined on all of $\mathbf{R}^d$ and has potential $V_n$ which is equal to $V$ inside $D$ and $n$ outside $D$. (The intuition is that the Dirichlet boundary condition really corresponds to having a potential that is infinite outside of $D$.) For Neumann boundary conditions, you get a similar formula by using a reflected Brownian motion instead of a killed Brownian motion.
Thanks Martin! Would it suffice for $V$ to be in $L^{q}(D)$ for some $q > 1$, or is something stronger required? Also, is $1_{\tau < t}$ supposed to read $1_{t < \tau}$? –  Michael Tinker Aug 26 '13 at 22:50
So I see that we can take $u_{0}(x) = \delta(y - x)$ for fixed $y \in D$, and the appropriate inequality on kernels would follow...do you think this is what Davies had in mind? –  Michael Tinker Aug 26 '13 at 22:52
You are right, it should be $t < \tau$, I fixed it. I suppose that Davies had something like this in mind, but of course I cannot speak for him ;-) Not sure what the minimal conditions on $V$ are, probably something like $L^q$ for $q > d/2$, see the paper "Schrödinger semigroups" by Barry Simon. –  Martin Hairer Aug 27 '13 at 8:36