schroedinger equation - Correct application of Laplacian Operator

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:

$$\int H(x,y)\sqrt{\pi(x)} dx = 0 \ \ \ \ \ \ \ (1)$$

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:

$$\begin{eqnarray*} H \sqrt{\pi} = \int H(x,y)\sqrt{\pi(y)} dy \\ = \int -\frac{1}{2}\sum_{i=1}^d \frac{\partial^2 \sqrt{\pi(y)}}{\partial y_i^2} + \frac{\sqrt{\pi(y)}}{2\sqrt{\pi(y)}} \sum_{i=1}^d \frac{\partial^2 \sqrt{\pi(y)}}{\partial y_i^2} dy = 0\\ \end{eqnarray*}$$

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:

$$\hat{f}(x) = f(x) -\frac{1}{2} \Delta P(x) + \nabla P(x) \cdot (-\frac{1}{2}\nabla \ln \pi(x))$$

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:

$$\begin{eqnarray*} \int H(x,y) \psi(y) dy \\ = \int -\frac{1}{2}\sum_{i=1}^d \frac{\partial^2 \psi(y)}{\partial y_i^2} + \frac{\psi(y)}{2\sqrt{\pi(y)}} \sum_{i=1}^d \frac{\partial^2 \sqrt{\pi(y)}}{\partial y_i^2} dy\\ = \int -\frac{1}{2}\sum_{i=1}^d \frac{\partial^2 P(y)\sqrt{\pi(y)}}{\partial y_i^2} + \frac{P(y)}{2} \sum_{i=1}^d \frac{\partial^2 \sqrt{\pi(y)}}{\partial y_i^2} dy\\ = \int -\frac{\sqrt{\pi(y)}}{2} \Delta P(y) - \frac{1}{2\sqrt{\pi(y)}}\nabla P(y) \cdot \nabla \pi(y) dy \end{eqnarray*}$$

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:

$$\begin{eqnarray*} H\psi = -\frac{1}{2}\sum_{i=1}^d \frac{\partial^2 \psi(x)}{\partial x_i^2} + \frac{\psi(x)}{2\sqrt{\pi(x)}} \sum_{i=1}^d \frac{\partial^2 \sqrt{\pi(x)}}{\partial y_i^2}\\ = -\frac{\sqrt{\pi(x)}}{2} \Delta P(x) - \frac{1}{2\sqrt{\pi(x)}} \nabla P(x) \cdot \nabla \pi(y) \\ \end{eqnarray*}$$ 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 $H(x,y)$?

Answer

In the original paper it is said that equation (10): $$H = -\frac{1}{2}\sum_{i=1}^d \frac{\partial^2}{\partial x_i^2} + V(x) \qquad (10)$$ "is written using the standard quantum-mechanical notation for a local Hamiltonian in the x-space realization".

This means that actually $H$ from (10) is the operator and $H(x,y)$ from your eq. (1) is the matrix of this operator in the linear space of integrable functions with the following basis: $$f_y(\xi) = \delta(\xi-y).$$

So one can write: $$H\psi(x) = \int H(x,y)\psi(y) \; dy.$$

The elements of the matrix can be calculated as follows: $$H(x,y) = H_{x,y} = (f_x, Hf_y) = \int f_x^*(\xi) H(\xi) f_y(\xi) \; d\xi =$$ $$\int f_x^*(\xi) \left(-\frac{1}{2}\sum_{i=1}^d \frac{\partial^2}{\partial \xi_i^2} + V(\xi)\right) f_y(\xi) \; d\xi =$$ $$\int \delta(\xi-x) \left(-\frac{1}{2}\sum_{i=1}^d \delta''_i(\xi-y) + V(\xi) \delta(\xi-y) \right) \; d\xi.$$ So $$H(x,y) = -\frac{1}{2}\sum_{i=1}^d \delta''_i(x-y) + V(x) \delta(x-y),$$ where $$\delta''_i(r) = \frac{\partial^2}{\partial r_i^2} \delta(r).$$