I've to compute this expression

$$ \hat{H} ~=~\frac{1}{4}g_2\int d^3R\int d^3r\ \bar{\Psi}(\vec{R}+\frac{\vec{r}}{2})\bar{\Psi}(\vec{R}-\frac{\vec{r}}{2}) $$$$ \times \left[ \delta(\vec{r})\nabla_{\vec{r}}^2 +\nabla_{\vec{r}}^2\delta(\vec{r}) \right]\Psi(\vec{R}+\frac{\vec{r}}{2}) \Psi(\vec{R}-\frac{\vec{r}}{2}), \tag{15} $$

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 $$ \hat{H} = \frac{1}{4}g_2\int d^3R\ \bar{\Psi}(\vec{R})\left[ \nabla^2(\bar{\Psi}(\vec{R})\ \Psi(\vec{R}))\right]\Psi(\vec{R}). \tag{16} $$ 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!

This integrals (15) comes from the paper Phys. Rev. A 67 053612 and authors say they do integration for part and then over $\textbf{r}$.

Does anyone have ideas how to calculate this integral?

/// Update ///

I tried to calculate the integral using yours suggestions. I'm near the solution! At the last there is an extra term and an extra $1/2$. In the followed images, the conjugate is $\phi^*$ and I indicate $\phi_+ = \Psi(\vec{R}+\frac{\vec{r}}{2})$ and $\phi_- = \Psi(\vec{R}-\frac{\vec{r}}{2})$

The first term is firstep

and the second is secondstep


$$ \int d^3\vec{R}\int d^3\vec{r}\nabla^2_{\vec{r}}\left( \phi_+^* \phi^*_-\phi_-\phi_+ \right)\delta(\vec{r}) $$

and then, at the last $$ \int d^3\vec{R}\ \frac{1}{2}(\phi^*\phi(\phi\nabla^2\phi^*+\phi^*\nabla^2\phi) - \phi^2|\nabla\phi^*|^2-\phi^{*2}|\nabla\phi|^2) $$ and completing it by adding and subtracting $2\phi\phi^*\nabla\phi^*\cdot\nabla\phi$

$$ \int d^3\vec{R}\ \frac{1}{2}\phi^*\left( \nabla^2(\phi^*\phi)\right)\phi - \int d^3\vec{R}\ \frac{1}{2} (\nabla(\phi\phi^*))^2 $$ I did all calculations by hand and then i checked them with mathematica.

is the term $\int d^3\vec{R}\ (\nabla(\phi\phi^*))^2 = 0$ for any reasons? I hope yes.

Why is there the constant $1/2$?

  • 1
    $\begingroup$ related: physics.stackexchange.com/q/38424 $\endgroup$
    – ACuriousMind
    Commented Jul 5, 2014 at 17:51
  • $\begingroup$ @ACuriousMind thanks. The curiosity is that the result of the integral depends on the laplacian of product between the conjugate of $\Psi$ and itself. And, integrating by parts, i don't understand why the conjugate is inside the laplacian $\endgroup$
    – apt45
    Commented Jul 5, 2014 at 18:01

1 Answer 1


Hints to the sought-for formula (16) for $\hat{H}$:

  1. Use integration by parts in ${\bf r}$-space to remove derivatives from the Dirac delta distributions, cf. comment by user ACuriousMind.

  2. Work on the problem from both ends (15) and (16). Use Leibniz rule $$\tag{*}\nabla^2 (fg)~=~ g\nabla^2 f + f \nabla^2 g+ 2 \nabla f\cdot\nabla g,$$ so that $\nabla$ only acts on single objects everywhere. Let's call the last term in eq. (*) for a 'cross-term'.

  3. Change the derivative $\nabla_{\bf r}~$ to $~\pm\frac{1}{2}\nabla_{\bf R}$.

  4. Perform the $\bf r$-integration.

  5. For cross-terms that act on $\Psi\Psi$ or $\bar{\Psi}\bar{\Psi}$, integrate by parts in ${\bf R}$-space, so that there are only cross-terms that act on $\bar{\Psi}\Psi$.

  6. Compare!

  • $\begingroup$ Thanks Qmechanic. I've just edited my question with few calculations. I'm near the solution... can you help me? $\endgroup$
    – apt45
    Commented Jul 5, 2014 at 23:01
  • $\begingroup$ I see that you edited your answer thank you! Are my calculation corrects? When you say "For cross-terms that act on $\Psi\Psi$ or $\bar{\Psi}\bar{\Psi}$, integrate by parts in R-space, so that there are only cross-terms that act on $\bar{\Psi}\Psi$ does it mean that $\int \nabla\Psi \cdot \nabla\Psi = 0$ under assumptions that $\Psi(\infty) \rightarrow 0$ ? $\endgroup$
    – apt45
    Commented Jul 6, 2014 at 9:05
  • $\begingroup$ Concerning your last sentence: No, only total divergence terms are omitted. $\endgroup$
    – Qmechanic
    Commented Jul 6, 2014 at 11:09
  • $\begingroup$ I solved using the fact that $\int d^3\vec{R} \phi\nabla^2\phi = -\int d^3\vec{R} \nabla\phi\cdot\nabla\phi$ thank you!! $\endgroup$
    – apt45
    Commented Jul 6, 2014 at 11:22

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