Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

Helium atom ground state energy estimation notes.

Posted by peeterjoot on October 1, 2011

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

Motivation.

In section 24.2.1 of the text [1] is an expectation value calculation associated with the Helium atom. One of the equations (24.76) seems wrong (according to hand calculation as shown below and according to mathematica). Is there another compensating error somewhere? Here I work the entire calculation in detail to attempt to find this.

Guts

We start with

\begin{aligned}\left\langle{{\frac{e^2}{{\left\lvert{\mathbf{r}_1 - \mathbf{r}_2}\right\rvert}}}}\right\rangle&=\left( \frac{Z^3}{\pi a_0^3}\right)^2 e^2\int d^3 k d^3 r_1 d^3 r_2 \frac{1}{{2 \pi^2 k^2}} e^{i \mathbf{k} \cdot (\mathbf{r}_1 - \mathbf{r}_2) } e^{ -2 Z (r_1 + r_2)/a_0} \\ &= \left( \frac{Z^3}{\pi a_0^3}\right)^2 e^2\frac{1}{{2 \pi^2}} \int d^3 k \frac{1}{{k^2}}d^3 r_1 e^{i \mathbf{k} \cdot \mathbf{r}_1 } e^{ -2 Z r_1 /a_0} \int d^3 r_2 e^{-i \mathbf{k} \cdot \mathbf{r}_2 } e^{ -2 Z r_2/a_0} \\ \end{aligned}

To evaluate the two last integrals, I figure the author has aligned the axis for the d^3 r_1 volume elements to make the integrals easier. Specifically, for the first so that \mathbf{k} \cdot \mathbf{r}_1 = k r_1 \cos\theta, so the integral takes the form

\begin{aligned}\int d^3 r_1 e^{i \mathbf{k} \cdot \mathbf{r}_1 } e^{ -2 Z r_1 /a_0} &=-\int r_1^2 d r_1 d\phi d(\cos\theta)e^{i k r_1 \cos\theta } e^{ -2 Z r_1 /a_0} \\ &=- 2 \pi \int_{r=0}^\infty \int_{u=1}^{-1}r^2 dr due^{i k r u } e^{ -2 Z r /a_0} \\ &=- 2 \pi \int_{r=0}^\infty r^2 dr \frac{1}{{i k r}} \left( e^{-i k r } - e^{i k r} \right) e^{ -2 Z r /a_0} \\ &=\frac{4 \pi}{k } \int_{r=0}^\infty r dr \frac{1}{{2i}} \left( e^{i k r } - e^{-i k r} \right) e^{ -2 Z r /a_0} \\ &=\frac{4 \pi}{k } \int_{r=0}^\infty r dr \sin(k r) e^{ -2 Z r /a_0} \\ \end{aligned}

For this last, mathematica gives me (24.75) from the text

\begin{aligned}\int d^3 r_1 e^{i \mathbf{k} \cdot \mathbf{r}_1 } e^{ -2 Z r_1 /a_0} =\frac{ 16 \pi Z a_0^3 }{(k^2 a_0^2 + 4 Z^2)^2}\end{aligned} \hspace{\stretch{1}}(2.1)

For the second integral, if we align the axis so that -\mathbf{k} \cdot \mathbf{r}_2 = k r \cos\theta and repeat, then we have

\begin{aligned}\left\langle{{\frac{e^2}{{\left\lvert{\mathbf{r}_1 - \mathbf{r}_2}\right\rvert}}}}\right\rangle&=\left( \frac{Z^3}{\pi a_0^3}\right)^2 e^2\frac{1}{{2 \pi^2}} 16^2 \pi^2 Z^2 a_0^6 \int d^3 k \frac{1}{{k^2}}\frac{ 1 }{(k^2 a_0^2 + 4 Z^2)^4} \\ &=\frac{128 Z^8}{\pi^2 } e^2\int dk d\Omega \frac{ 1 }{(k^2 a_0^2 + 4 Z^2)^4} \\ &=\frac{512 Z^8}{\pi} e^2\int dk \frac{ 1 }{(k^2 a_0^2 + 4 Z^2)^4} \\ &=\frac{512 Z^8}{\pi} e^2\int dk \frac{ 1 }{(k^2 a_0^2 + 4 Z^2)^4} \\ \end{aligned}

With k a_0 = 2 Z \kappa this is

\begin{aligned}\left\langle{{\frac{e^2}{{\left\lvert{\mathbf{r}_1 - \mathbf{r}_2}\right\rvert}}}}\right\rangle&=\frac{512 Z^8}{\pi} e^2\int d\kappa \frac{ 2 Z }{a_0}\frac{ 1 }{(2 Z)^8 (\kappa^2 + 1)^4} \\ &=\frac{4 Z}{\pi a_0} e^2\int d\kappa \frac{ 1 }{(\kappa^2 + 1)^4} \\ \end{aligned}

Here I note that

\begin{aligned}\frac{d}{d\kappa}\frac{-1}{ 3 (1+x)^3 }=\frac{1}{{ (1+x)^4 }}\end{aligned}

so the definite integral has the value

\begin{aligned}\int_0^\infty d\kappa \frac{ 1 }{ (\kappa^2 + 1)^4} =\frac{-1}{ 3 (1+\infty)^3 }-\frac{-1}{ 3 (1+0)^3 }= \frac{1}{{3}}\end{aligned}

(not 5 \pi/3 as claimed in the text). This gives us

\begin{aligned}\left\langle{{\frac{e^2}{{\left\lvert{\mathbf{r}_1 - \mathbf{r}_2}\right\rvert}}}}\right\rangle&=\frac{4 Z e^2} { \pi a_0 } \int d\kappa \frac{ 1 }{(\kappa^2 + 1)^4} \\ &=\frac{4 Z e^2} { 3 \pi a_0 }  \\ &\approx 0.424 \frac{Z e^2 }{a_0} \ne \frac{5}{8} \frac{Z e^2 }{a_0}\end{aligned}

So, no compensating error is found, yet the end result of the calculation, which requires the 5/8 result matches with the results obtained other ways (as in problem set III). How would this be accounted for? Is there an error above?

References

[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.

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3 Responses to “Helium atom ground state energy estimation notes.”

  1. Hi Peeter,

    Loving your blog so far. I ran through the calculations you did above and I also came to the same result as you. I also tried using Mathematica to integrate all the equations and it came to the same end result. It’s quite possible that your book has an error in it.

    Unlikely as it may be, it could be the reason why you’re arriving at a different answer.

    • peeterjoot said

      This book has many many errors, so finding another is not really a suprise. What’s surprising is that when the result is calculated a different way, the end result \frac{5 Z e^2 }{8 a_0} is still obtained. Did you have any success getting Mathematica to evaluate the following directly:

      \begin{aligned}\int d^3 r_1 e^{i \mathbf{k} \cdot \mathbf{r}_1 } e^{ -2 Z r_1 /a_0}  \end{aligned}

      ? If I try this one, Mathematica just spits out a pretty printed version. The other source of possible error is the inverse Fourier transform that is used as a starting point, where the author writes

      \frac{1}{{{\left\lvert{\mathbf{r}_1 - \mathbf{r}_2}\right\rvert}}} = \frac{1}{2 \pi^2} \int d^3 k \frac{1}{k^2} e^{i \mathbf{k} \cdot (\mathbf{r}_1 - \mathbf{r}_2) }

      We were asked to verify this equation in exersizes in class and it also works out okay (direct integration when the two vectors are not equal is possible, and it can be used nicely to verify the Poisson potential solution to the Coulomb potential problem). Somewhere in the mix it appears there must be another error, but I am yet to find it.

      • Peeter, unfortunately there’s no easy way to perform the integration in Mathematica. Even with the vector calculus package, it has trouble integrating the resulting equations. It can perform the radial integration, but it generates a rather complicated trigonometric rational function which it can’t integrate own it’s own.

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