# Integral with Dirac Delta

I've to compute this expression

$$\hat{H} = \frac{1}{4}g_2\int d^3R\int d^3r\ \bar{\Psi}(\vec{R}+\frac{\vec{r}}{2})\bar{\Psi}(\vec{R}-\frac{\vec{r}}{2})\left[ \delta(\vec{r})\nabla_{\vec{r}}^2 +\nabla_{\vec{r}}^2\delta(\vec{r}) \right]\Psi(\vec{R}+\frac{\vec{r}}{2})\Psi(\vec{R}-\frac{\vec{r}}{2})$$

where $\bar{\Psi}$ is the conjugate of $\Psi$.

Using dirac delta properties, can i say that $\left[ \delta(\vec{r})\nabla_{\vec{r}}^2 +\nabla_{\vec{r}}^2\delta(\vec{r}) \right] = 2 \delta(\vec{r})\nabla_{\vec{r}}^2$ ?

If not, how can i calculate this integral?

I should obtain $$\hat{H} = \frac{1}{4}g_2\int d^3R\ \bar{\Psi}(\vec{R})\left[ \nabla^2(\bar{\Psi}(\vec{R})\ \Psi(\vec{R}))\right]\Psi(\vec{R})$$

I calculate the laplacian in this way, but i'm not sure about my hypothesis ($\textbf{r}$ module dependency) (here, $\phi$ it's the same of $\Psi$) (Note: $\textbf{R}$ or $\textbf{r}$ is a 3D-Vector) $$\frac{\partial}{\partial x}\left\{\frac{\partial}{\partial x}\left[\phi(\textbf{R}+\frac{\textbf{r}}{2})\phi(\textbf{R}-\frac{\textbf{r}}{2})\right] \right\} = \frac{\partial}{\partial x}\left\{ \phi_+'\phi_-\frac{x}{2r} - \frac{x}{2r}\phi_-'\phi_+ \right\} = -\frac{1}{2}\left( \frac{r-x^2/r}{r^2}\right) \phi_-'\phi_+ + \left( \frac{x}{2r}\right)^2 \phi_-''\phi_+ - \left( \frac{x}{2r}\right)^2 \phi_-'\phi_+' + + \frac{1}{2}\left( \frac{r-x^2/r}{r^2}\right) \phi_-\phi_+' - \left( \frac{x}{2r}\right)^2 \phi_-' \phi_+' + \left( \frac{x}{2r}\right)^2 \phi_-\phi_+''$$ and summing all over the coordinates \begin{align} \nabla^2_{\textbf{r}}\phi_+\phi_- = -\frac{1}{r} \phi_-'\phi_+ + \frac{1}{4} \phi_-''\phi_+ - \frac{1}{4} \phi_-'\phi_+' + \frac{1}{r} \phi_-\phi_+' - \frac{1}{4} \phi_-' \phi_+' + \frac{1}{4} \phi_-\phi_+'' . \end{align}

I think it is incorrect. A method should be expanding $\Phi = V^{-1/2} \sum_\alpha a_\alpha e^{i\textbf{k}_\alpha\cdot\textbf{r}}$, but i don't have any idea what doing!

Does anyone have ideas how to calculate this integral?

• Its not entirely clear to me what the exact nature of the rest of your question is. Are you trying to show that the integral simplifies to "$\hat{H} = \frac{1}{4}g_2\int d^3R\ \bar{\Psi}(\vec{R})\left[ \nabla^2(\bar{\Psi}(\vec{R})\ \Psi(\vec{R}))\right]\Psi(\vec{R})$"? – Spencer Jul 5 '14 at 0:34
• @Spencer yes, exactly – apt45 Jul 5 '14 at 7:33
• It would be helpful to know what can we assume about $\Psi$ and the range of integration. For instance in these situations integration by parts usually gets used quite liberally, but that only makes sense if $\Psi$ is zero on the boundary of our region. – Spencer Jul 5 '14 at 14:50
• We Know just the wave function is zero on the boundary of the region. Anything else... It'a a calculus of a paper that i'm reading – apt45 Jul 5 '14 at 15:03
• @Spencer the article is Phys. Rev. A 67 053612 and the range of integration is all the space, $R^3$ – apt45 Jul 5 '14 at 17:38

Note that, $$\frac{\partial}{\partial x} \delta(x) = - \delta(x) \frac{\partial}{\partial x},$$
$$\frac{\partial^2}{\partial x^2} \delta(x) = \frac{\partial}{\partial x} \frac{\partial}{\partial x} \delta(x) = - \frac{\partial}{\partial x} \delta(x) \frac{\partial}{\partial x} = + \delta(x) \frac{\partial}{\partial x} \frac{\partial}{\partial x} = \delta(x) \frac{\partial^2}{\partial x^2} ,$$
$$\nabla^2 \delta(\vec{r}) = \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}\right) \delta(x) \delta(y) \delta(z) = \delta(x)\delta(y)\delta(z)\left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}\right) = \delta(\vec{r}) \nabla^2.$$
• This was meant to answer to the first question you listed: "Using dirac delta properties, can i say that $\left[ \delta(\vec{r} )\nabla^2_{\vec{r}} +\nabla^2_{\vec{r}}\delta(\vec{r}) \right]=2\delta(\vec{r})\nabla^2_{\vec{r}}$?". – Spencer Jul 5 '14 at 0:28