# Expectation Values of Radii in the Hydrogen Atom

1. Dec 31, 2011

### Lunar_Lander

1. The problem statement, all variables and given/known data

Determine for the hydrogen atom states 1s and 2p the expectation value of the radius r and the associated mean square error Δr.

2. Relevant equations

Wave Functions for 1s and 2p from Demtroeder's Experimental Physics Volume 3 (it says "The normalized complete eigenfunctions of an electron in the Coulomb potential $V(r)=-Z\cdot e^2/(4\pi\epsilon_0r)$", is this what we need?) :

$\psi_{1s}(r,\theta,\phi)=\frac{1}{\sqrt{\pi}}(Z/a_0)^{3/2}\cdot e^{-Zr/a_0}$
$\psi_{2p_0}(r,\theta,\phi)=\frac{1}{4\sqrt{2\pi}}(Z/a_0)^{3/2}\cdot\frac{Zr}{a_0}\cdot e^{-Zr/2a_0}\cos(\theta)$

$\Delta r=\sqrt{<(r-\overline{r})^2>}$

3. The attempt at a solution
Well, as in the neighbour thread (https://www.physicsforums.com/showthread.php?t=562573), the expectation value for r seems to be $<\psi|r|\psi>=\int_0^r\int_0^{\pi}\int_0^{2\pi} r\cdot\psi^2\cdot r^2\sin(\theta)~dr~d\theta~d\phi$, which I would, being naive, simplify to $<r>=\int_0^r\int_0^{\pi}\int_0^{2\pi} \cdot\psi^2\cdot r^3\sin(\theta)~dr~d\theta~d\phi$. As we are in Hydrogen, Z should be 1.

Then I tried 1s. I get
$\psi^2=\frac{Z^3\exp(-\frac{2rZ}{a_0})}{\pi\cdot a_0^3}=\frac{1}{a_0^3\pi}\exp(-\frac{2r}{a_0})$
Phi Integration:
$\int_0^r\int_0^{\pi} 2\pi \frac{1}{a_0^3\pi}\exp(-\frac{2r}{a_0}) \cdot r^3\sin(\theta)~dr~d\theta$
Theta Integration:
$\int_0^r 4\pi \frac{1}{a_0^3\pi}\exp(-\frac{2r}{a_0}) \cdot r^3~dr$
$\int_0^r \frac{4}{a_0^3}\exp(-\frac{2r}{a_0}) \cdot r^3~dr$
r Integration:
$\frac{4}{a_0^3} \int_0^r \exp(-\frac{2r}{a_0}) \cdot r^3~dr$
$<r>_{1s}=\frac{4}{a_0^3} 6a_0^4-a_0\cdot\exp(-r/a_0)(6a_0^3+6a_0^2 r+3a_0r^2+r^3)$

Is this the right track?

PS: There is a third wave function for 2p, for m=+/-1: $\psi_{2p_{\pm 1}}(r,\theta,\phi)=\frac{1}{8\sqrt{2\pi}}(Z/a_0)^{3/2}\cdot\frac{Zr}{a_0}\cdot e^{-Zr/2a_0}\sin(\theta)\cdot e^{\pm i\varphi}$
So two more calculations to run?

Last edited: Dec 31, 2011
2. Jan 2, 2012

### Lunar_Lander

I also attempted a run at the 2p (for m=0) and it looks like this:

$\psi_{2p_0}^2=\frac{1}{32\pi\cdot a_0^5}\cdot r^2Z^5\exp(-\frac{2Z\cdot r}{a_0})\cdot\cos^2(\theta)$

$<r>=\frac{1}{32\pi\cdot a_0^5}\int_0^r \int_0^{\pi} \int_0^{2\pi} r^2Z^5\exp(-\frac{2Z\cdot r}{a_0})\cdot\cos^2(\theta)\cdot r^2\sin(\theta)~dr~d\theta~d\phi$

$<r>=\frac{1}{32\pi\cdot a_0^5}\int_0^r \int_0^{\pi} \int_0^{2\pi} r^2Z^5\exp(-\frac{2Z\cdot r}{a_0})\cdot\cos^2(\theta)\cdot r^2\sin(\theta)~dr~d\theta~d\phi~|Z=1$

$=\frac{1}{16\cdot a_0^5}\int_0^r \int_0^{\pi} r^4\exp(-\frac{2r}{a_0})\cdot\cos^2(\theta)\cdot \sin(\theta)~dr~d\theta$

$=\frac{1}{24\cdot a_0^5}\int_0^r r^4\exp(-\frac{2r}{a_0})~dr$

$=\frac{1}{96}(3-\frac{\exp(-\frac{2r}{a_0})(3a_0^4+6a_0^3r+6a_0^2r^2+4a_0r^3+2r^4)}{a_0})$

Last edited: Jan 2, 2012
3. Jan 3, 2012

### Lunar_Lander

I think I got it now. First of all I learned that the 2p energy level is degenerate, thus m should have no influence on it and there is only one calculation to do for 2p.

Then I tried to calculate 1s:

$\psi_{1s}=\frac{1}{\sqrt{\pi}}(\frac{Z}{a_0})^{3/2}\exp(-\frac{Zr}{a_0})$

As we are discussing hydrogen, $Z=1$.

$<r>=<\psi|r|\psi>=\int_0^{\infty}\int_0^{\pi}\int_0^{2\pi} r\psi^2 r^2 \sin(\theta)~dr~d\theta~d\varphi$

$=\int_0^{\infty}\int_0^{\pi}\int_0^{2\pi} \psi^2 r^3 \sin(\theta)~dr~d\theta~d\varphi$

$=\int_0^{\infty}\int_0^{\pi}\int_0^{2\pi} \frac{1}{\pi a_0^3}\exp(-\frac{2r}{a_0}) r^3 \sin(\theta)~dr~d\theta~d\varphi$

$=\int_0^{\infty}\int_0^{\pi} \frac{2}{a_0^3}\exp(-\frac{2r}{a_0}) r^3 \sin(\theta)~dr~d\theta$

$=\int_0^{\infty} \frac{4}{a_0^3}\exp(-\frac{2r}{a_0}) r^3 ~dr$

$<r>=\frac{3}{2}a_0$

To get the mean squared error which is $\Delta r=<r^2>-<r>$, I did the calculation above with $<\psi|r^2|\psi>$ and got as an result $<r^2>=3a_0^2$, so $\Delta r_{1s}=3a_0^2-\frac{3}{2}a_0$.

2p:
$\psi_{2p}=\frac{1}{4\sqrt{2\pi}}(\frac{Z}{a_0})^{3/2}\frac{Zr}{a_0}\exp(-\frac{Zr}{2a_0})\cos(\theta)$

$<r>=<\psi|r|\psi>=\int_0^{\infty}\int_0^{\pi}\int_0^{2\pi} r\psi^2 r^2 \sin(\theta)~dr~d\theta~d\varphi$

$\int_0^{\infty}\int_0^{\pi}\int_0^{2\pi} \psi^2 r^3 \sin(\theta)~dr~d\theta~d\varphi$

$\int_0^{\infty}\int_0^{\pi}\int_0^{2\pi} \frac{1}{32\pi a_0^5} r^2\exp(-\frac{r}{a_0})\cos^2(\theta) r^3 \sin(\theta)~dr~d\theta~d\varphi$

$\int_0^{\infty}\int_0^{\pi}\int_0^{2\pi} \frac{1}{32\pi a_0^5} r^5\exp(-\frac{r}{a_0})\cos^2(\theta) \sin(\theta)~dr~d\theta~d\varphi$

$\int_0^{\infty}\int_0^{\pi} \frac{1}{16 a_0^5} r^5\exp(-\frac{r}{a_0})\cos^2(\theta) \sin(\theta)~dr~d\theta$

$\int_0^{\infty} \frac{1}{24 a_0^5} r^5\exp(-\frac{r}{a_0})~dr$

$<r>=5a_0$

$<r^2>=30a_0$, $\Delta r_{2p}=30a_0^2-5a_0$