Not a physicist, and I'm having trouble understanding how to apply the Laplacian-like operator described in this paper and the original. We let:
$$ \hat{f}(x) = f(x) + \frac{\int H(x,y)\psi(y) dy}{\sqrt{\pi(x)}} $$
Where $H$ is a Hermitian operator (real symmetric, actually), $\pi(x)$ is an un-normalized density function (measurable positive), $\psi(x)$ is arbitrary integrable function, and f(x) is an arbitrary measurable function. $H$ is chosen to ensure that the following condition holds:
$$ \int H(x,y)\sqrt{\pi(x)} dx = 0 \tag{1} $$
This is done so that $\hat{f}(x)$ and $f(x)$ have the same expectation under $\pi(x)$: $\int f(x)\pi(x) dx = \int \hat{f}(x)\pi(x) dx $.
The paper goes on to define a Schrödinger-like $H$ for $\{x \in \mathbb{R}^d\}$:
$$H = -\frac{1}{2}\sum_{i=1}^d \frac{\partial^2}{\partial x_i^2} + V(x)$$
Where $V(x)$ is constructed to meet (1):
$$ V(X) = \frac{1}{2\sqrt{\pi(x)}} \sum_{i=1}^d \frac{\partial^2 \sqrt{\pi(x)}}{\partial x_i^2} $$
Which allows us to verify $(1)$ as follows:
$$ \begin{align} H \sqrt{\pi} =& \int H(x,y)\sqrt{\pi(y)} dy \\ =& \int -\frac{1}{2}\sum_{i=1}^d \frac{\partial^2 \sqrt{\pi(y)}}{\partial y_i^2} + \frac{\sqrt{\pi(y)}}{2\sqrt{\pi(y)}} \sum_{i=1}^d \frac{\partial^2 \sqrt{\pi(y)}}{\partial y_i^2} dy = 0\\ \end{align} $$
I got the right result above, but I'm not sure I applied the H operator correctly...
The paper considers the following form for $\psi(x)$:
$$ \psi(x) = P(x)\sqrt{\pi(x)}$$
The authors then derive the following:
$$ \hat{f}(x) = f(x) -\frac{1}{2} \Delta P(x) + \nabla P(x) \cdot \left(-\frac{1}{2}\nabla \ln \pi(x)\right) $$
Where $\nabla$ denotes the gradient $(\frac{\partial}{\partial x_1},...,\frac{\partial}{\partial x_d})$ and $\Delta$ denotes the Laplacian operator $\sum_{i=1}^d \frac{\partial^2}{\partial x_i^2}$. I'm not able to re-derive this equation. Again, I'm not sure I'm applying the H operator correctly, because it looks like I should end up with something in terms of $x$. My obviously incorrect attempt follows:
$$ \begin{align} \int H(x,y) \psi(y) dy =& \int -\frac{1}{2}\sum_{i=1}^d \frac{\partial^2 \psi(y)}{\partial y_i^2} + \frac{\psi(y)}{2\sqrt{\pi(y)}} \sum_{i=1}^d \frac{\partial^2 \sqrt{\pi(y)}}{\partial y_i^2} dy\\ =& \int -\frac{1}{2}\sum_{i=1}^d \frac{\partial^2 P(y)\sqrt{\pi(y)}}{\partial y_i^2} + \frac{P(y)}{2} \sum_{i=1}^d \frac{\partial^2 \sqrt{\pi(y)}}{\partial y_i^2} dy\\ =& \int -\frac{\sqrt{\pi(y)}}{2} \Delta P(y) - \frac{1}{2\sqrt{\pi(y)}}\nabla P(y) \cdot \nabla \pi(y) dy \end{align} $$
Edit:
I was able to reproduce the result by applying the operator by just swapping $x$ and $y$ and not explicitly solving the integral. here is what I got:
$$ \begin{align} H\psi =& -\frac{1}{2}\sum_{i=1}^d \frac{\partial^2 \psi(x)}{\partial x_i^2} + \frac{\psi(x)}{2\sqrt{\pi(x)}} \sum_{i=1}^d \frac{\partial^2 \sqrt{\pi(x)}}{\partial y_i^2}\\ =& -\frac{\sqrt{\pi(x)}}{2} \Delta P(x) - \frac{1}{2\sqrt{\pi(x)}} \nabla P(x) \cdot \nabla \pi(y) \\ \end{align} $$ Dividing by $\sqrt{\pi(x)}$ yields: $$ = -\frac{1}{2} \Delta P(x) - \frac{1}{2} \nabla P(x) \cdot \nabla \ln \pi(x) $$ Which is the same. Does this mean that the definition of $H$ is describing the solution to the integral? What is $H(x,y)$?