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Expectation Values of Radii in the Hydrogen Atom

  1. Dec 31, 2011 #1
    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 [itex]V(r)=-Z\cdot e^2/(4\pi\epsilon_0r)[/itex]", is this what we need?) :

    [itex]\psi_{1s}(r,\theta,\phi)=\frac{1}{\sqrt{\pi}}(Z/a_0)^{3/2}\cdot e^{-Zr/a_0}[/itex]
    [itex]\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)[/itex]

    [itex]\Delta r=\sqrt{<(r-\overline{r})^2>}[/itex]

    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 [itex]<\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[/itex], which I would, being naive, simplify to [itex]<r>=\int_0^r\int_0^{\pi}\int_0^{2\pi} \cdot\psi^2\cdot r^3\sin(\theta)~dr~d\theta~d\phi[/itex]. As we are in Hydrogen, Z should be 1.

    Then I tried 1s. I get
    [itex]\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})[/itex]
    Phi Integration:
    [itex]\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[/itex]
    Theta Integration:
    [itex]\int_0^r 4\pi \frac{1}{a_0^3\pi}\exp(-\frac{2r}{a_0}) \cdot r^3~dr[/itex]
    [itex]\int_0^r \frac{4}{a_0^3}\exp(-\frac{2r}{a_0}) \cdot r^3~dr[/itex]
    r Integration:
    [itex]\frac{4}{a_0^3} \int_0^r \exp(-\frac{2r}{a_0}) \cdot r^3~dr[/itex]
    [itex]<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)[/itex]

    Is this the right track?

    PS: There is a third wave function for 2p, for m=+/-1: [itex]\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}[/itex]
    So two more calculations to run?
     
    Last edited: Dec 31, 2011
  2. jcsd
  3. Jan 2, 2012 #2
    I also attempted a run at the 2p (for m=0) and it looks like this:

    [itex]\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)[/itex]

    [itex]<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[/itex]

    [itex]<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[/itex]

    [itex]=\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[/itex]

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

    [itex]=\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})[/itex]
     
    Last edited: Jan 2, 2012
  4. Jan 3, 2012 #3
    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:

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

    As we are discussing hydrogen, [itex]Z=1[/itex].

    [itex]<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[/itex]

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

    [itex]=\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[/itex]

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

    [itex]=\int_0^{\infty} \frac{4}{a_0^3}\exp(-\frac{2r}{a_0}) r^3 ~dr[/itex]

    [itex]<r>=\frac{3}{2}a_0[/itex]

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

    2p:
    [itex]\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)[/itex]

    [itex]<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[/itex]

    [itex]\int_0^{\infty}\int_0^{\pi}\int_0^{2\pi} \psi^2 r^3 \sin(\theta)~dr~d\theta~d\varphi[/itex]

    [itex]\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[/itex]

    [itex]\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[/itex]

    [itex]\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[/itex]

    [itex]\int_0^{\infty} \frac{1}{24 a_0^5} r^5\exp(-\frac{r}{a_0})~dr[/itex]

    [itex]<r>=5a_0[/itex]

    [itex]<r^2>=30a_0[/itex], [itex]\Delta r_{2p}=30a_0^2-5a_0[/itex]

    Please let me know if there are any comments.
     
    Last edited: Jan 3, 2012
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