1
$\begingroup$

The approximate ground state for the two electrons of the He atom is given by $$\psi_0(r_1,r_2) = \frac{8}{\pi a^3}e^{-2(r_1+r_2)/a}$$, where $a$ is the Bohr radius.

I want to calculate the expectation value $\big\langle \frac{1}{|\mathbf r_1-\mathbf r_2|} \big\rangle$, but I really don't see how to proceed. The integrand seems to be correct.

We have

\begin{align} \require{cancel} \bigg\langle \frac{1}{|r_1-r_2|} \bigg\rangle &= \bigg( \frac{8}{\pi a^3} \bigg)^2 \cancel{\int_{r_1=0}^{\infty}\int_{r_2=0}^{\infty} \frac{1}{\sqrt{(r_1-r_2)^2}} e^{-4(r_1+r_2)/a} dr_1 dr_2} \\ &= \ldots \end{align}

Any hints would be appreciated.


The integrals $$\int xe^{-\alpha x}dx = e^{-\alpha x}\bigg( -\frac{x}{\alpha} - \frac{1}{\alpha} \bigg)$$ and $$ \int x^2 e^{-\alpha x} dx = e^{-\alpha x} \bigg( -\frac{x^2}{\alpha}-\frac{2x}{\alpha^2} - \frac{2}{\alpha^3} \bigg) $$ are given as additional information.


As @pawel_winzig pointed out, we need to integrate over all space:

\begin{align} \bigg\langle \frac{1}{|\mathbf r_1-\mathbf r_2|} \bigg\rangle &= \bigg( \frac{8}{\pi a^3} \bigg)^2 \int_{\mathbf{r}_1 \in \mathbb R^3} \int_{\mathbf{r}_2 \in \mathbb R^3} \frac{1}{\sqrt{(\mathbf r_1-\mathbf r_2)^2}} e^{-4(r_1+ r_2)/a} d\mathbf r_1 d \mathbf r_2 \end{align}

$\endgroup$
  • 3
    $\begingroup$ You have to integrate over the whole space, i.e., $r^2 \sin(\theta)drd\theta d\phi$. See e.g. this presentation: web.stanford.edu/group/fayer/lectures/lecturespdfs/… $\endgroup$ – pawel_winzig Feb 9 '17 at 13:01
  • $\begingroup$ Just to double-check, are you sure you want the mean of $\dfrac{1}{|r_1-r_2|}$ (which depends on the difference between two electrons' orbital radii) instead of $\dfrac{1}{|\mathbf{r}_1-\mathbf{r}_2|}$ (which depends on the distance between the electrons)? The latter seems more important, because it's proportional to the electron-electron interaction's mean electrostatic potential. $\endgroup$ – J.G. Feb 9 '17 at 14:03
  • $\begingroup$ @J.G. Yes, you're right; I mean $\frac{1}{\mathbf r_1 - \mathbf r_2|}$ instead of $\frac{1}{|r_1-r_2}$. I'll make an edit. $\endgroup$ – Mussé Redi Feb 9 '17 at 14:05
  • $\begingroup$ The argument of the exponential probably depends only on the orbital radii though. $\endgroup$ – Tony Feb 9 '17 at 14:36
  • $\begingroup$ @Tony That's correct. I'll make an edit. $\endgroup$ – Mussé Redi Feb 9 '17 at 14:37
1
$\begingroup$

Let us first take a look at the integrand $$ \frac{e^{-4(r_1+r_2)/a}}{|\mathbf r_1 - \mathbf r_2|}d\mathbf r_1 d\mathbf r_2 $$ and its Jacobian $$ d\mathbf r_1 d\mathbf r_2 = \big( r_1^2 \sin(\theta_1) \operatorname{dr_1 d\theta_1 d\phi_1}\big) \cdot \big(\operatorname{r_2^2 \sin(\theta_2) dr_2 d\theta_2 d\phi_2} \big). $$ The reciprocal can be rewritten as $$ \frac{1}{|\mathbf r_1 - \mathbf r_2|}=\frac{1}{\sqrt{r_1^2+r_2^2-2r_1r_2\cos(\gamma)}} $$ where $\gamma =\angle(\mathbf r_1, \mathbf r_2)$ is the angle between the vectors.

Since the order of integration doesn't matter, we can assume that $\theta_1=0$, while we integrate over $\theta_2$; effectively placing the first particle on the $z$-axis, at first. Hence, we can set $\color{blue}{\gamma := \theta_2 }$.

We can now write our integrand with a substitution $ \color{green}{u = r_1^2+r_2^2-2r_1 r_2 \cos(\theta_2) \ , \ du = 2r_1r_2\sin(\theta_2)}: $ $$ \frac{e^{-4(r_1+r_2)/a}}{|\mathbf r_1 - \mathbf r_2|}d\mathbf r_1 d\mathbf r_2 = \frac{1}{2} \frac{e^{-4(r_1+r_2)/a}}{\sqrt{u}}r_1r_2 \sin(\theta_1) \operatorname{dr_1 dr_2 d \theta_1 du d\phi_1 d\phi_2} . $$ After integrating over the azimuthal angles $\phi_1, \phi_2$, which yields $4\pi^2$; our substitution $u$, which yields $\frac{1}{2} \sqrt{u}$; our polar angle $\theta_1$, which yields $2$; and we are left with:

$$ \bigg( \frac{8}{\pi^3}\bigg) ^2 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot 4\pi^2 \cdot 2 \int_{u=(r_1-r_2)^2}^{(r_1+r_2)^2}\sqrt{u} e^{-4(r_1+r_2)/a} \operatorname{dr_1 dr_2 du} = \frac{32}{\pi^4} \int_{r_1=0}^\infty \int_{r_2=0}^\infty r_2^2 r_1 e^{-4(r_1+r_2)/a}\operatorname{dr_1 dr_2} $$ for which we have our standard integrals mentioned in the post.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.