A (relatively) simple QM Problem, but seeking my mistake

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1. Dec 7, 2014

Emspak

1. The problem statement, all variables and given/known data
Find <r> and <r2> for an electron in the ground state of hydrogen. Express in terms of Bohr radius.

2. Relevant equations
We know the relevant wave functions are:

$R_{10} = \frac{c_0}{a}e^{r/a}Y^0_0$
and $Y^0_0 = \frac{1}{\sqrt{4\pi}}$
3. The attempt at a solution

As I understand it the way to get <r> is to do the following integral

$\langle r \rangle = \int^{2\pi}_0 \int^{\pi}_0 \int^{\infty}_0 r |R_{10}|^2 r^2|Y^0_0|^2 dr d\theta d\phi$

when I pug everything in, knowing that $c_0 = \frac{2}{\sqrt{a}}$ I get

$\langle r \rangle = \int^{2\pi}_0 \int^{\pi}_0 \int^{\infty}_0 r^3 \frac{2}{a^3}e^{-2r/a} \frac{1}{4\pi} dr d\theta d\phi$

and i can ignore (for the moment) the outer two integrals because the quantity <r> doesn't depend on either phi or theta.

So that leaves me integrating this:

$\langle r \rangle = \int^{\infty}_0 r^3 \frac{2}{a^3}e^{-2r/a} \frac{1}{4\pi} dr = \frac{1}{2 \pi a^3} \int^{\infty}_0 r^3 \frac{2}{a^3}e^{-2r/a}dr$

Integrating by parts with $u = r^3$, $du = 3r^2 dr$, $dv = e^{2r/a}$ and $v = \frac{-ae^{2r/a}}{2}$ that gets me to

$r^3 \frac{-ae^{-2r/a}}{2} - \int^{\infty}_0 \frac{-ae^{2r/a}}{2} 3r^2 dr = -r^3 \frac{ae^{-2r/a}}{2} + \frac{3}{2} \int^{\infty}_0 ae^{-2r/a} r^2 dr = r^3 \frac{ae^{-2r/a}}{2} + \frac{3a}{2} \int^{\infty}_0 e^{-2r/a} r^2 dr$

the first term goes to zero. integrating by parts again, with $u = r^2$ and $du = 2rdr$ the same thing happens:
$\frac{1}{2 \pi a^3} \frac{3}{2} \left[ -ar^2 \frac{e^{-2r/a}}{2} + \int^{\infty}_0 \frac{ae^{-2r/a}}{2} 2r dr \right]$

that first term goes to zero again. OK so next up is another integration by parts. $u = 2r$ and $du = 2dr$ so
$\frac{1}{2 \pi a^3} \frac{3}{4} \left[ -2ar \frac{e^{-2r/a}}{2} + \int^{\infty}_0 \ ae^{-2r/a} dr \right]$

and we do it all once more, to get

$\langle r \rangle = \frac{3}{8 \pi a^2} \frac{-ae^{-2r/a}}{2} |^{\infty}_0 = \frac{3}{8 \pi a^2} \left[0 + \frac{a}{2} \right] = \frac{3}{16\pi a}$

Problem is this isn't what the various solutions I have seen to the problem gets, which is 3a/2.

So, my question is where the mistake I am making is. I posted it here because I suspect the problem is some elementary operation (rather than a deep physics problem) I am not doing right, some algebraic thing I screwed up but I went through this twice so far and can't seem to pin it down. That factor of pi has to go away somehow, for example. (In the two outer integrals maybe?)

thanks.

2. Dec 8, 2014

Staff: Mentor

Instead of ignoring them, you could calculate them. They just give constant prefactors and make the solution much closer to the right answer.

At some point in your substitution, your units start to get wrong.

3. Dec 16, 2014

Emspak

thanks -sorry i didn't get back to you sooner - i figured out what the problem was eventually (part of it was realizing that when you integrate in a sphere there are "extra" sine terms. Whoops.
btw there doesn't seem to be a thanks button anymore. :-(

4. Dec 16, 2014

Nathanael

It was replaced by the "Like" button :s