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Homework Help: Is the wave function normalized?

  1. Jan 27, 2009 #1
    1. The problem statement, all variables and given/known data

    The ground state wave function for the electron in a hydrogen atom is:

    [tex]\psi(r) = \frac{1}{\sqrt (\pi a_o^3)} e^\frac{-r}{a_o}[/tex]

    where r is the radial coordinate of the electron and a_o is the Bohr radius.

    Show that the wave function as given is normalized.


    2. Relevant equations

    Any wave function satisfying the following equation is said to be normalized:

    [tex]\int_{-\infty}^{+\infty} |\psi|^2\dx = 1[/tex]


    3. The attempt at a solution

    Because the sum of all probabilities over all values of r must be 1,

    [tex]\int_{-\infty}^{+\infty} (\frac{1}{\sqrt (\pi a_o^3)} e^\frac{-r}{a_o})^2 dr = \frac{1}{\pi a_o^3} \int_{-\infty}^{+\infty} e^\frac{-2r}{a_o^2} dr = 1[/tex]

    Since the integral can be expressed as the sum of two integrals, we have,

    [tex]\frac{1}{\pi a_o^3} \int_{-\infty}^{+\infty} e^\frac{-2r}{a_o^2} dr = \frac{2}{\pi a_o^3} \int_0^{+\infty} e^\frac{-2r}{a_o^2} dr = 1 [/tex]

    After integrating, I obtain,

    [tex]\frac{1}{\pi a_o^3} = 1[/tex]

    which is definitely incorrect. However, I do not see any other way to proceed. could someone give some assitance.

    Thanks for your kind assistance

    jg370
     
  2. jcsd
  3. Jan 27, 2009 #2
    you integrated wrong -
    [itex]\int_0^{\infty} e^{-\frac{2r}{a_0^2}} dr = [-\frac{a_0^2}{2} e^{-\frac{2r}{a_0^2}]_0^{\infty}=\frac{a_0^2}{2} [/itex]

    which gives [itex]\frac{1}{2 \pi a_0}=1[/itex]
     
  4. Jan 27, 2009 #3
    the a_0 in your exponent shouldn't be squared, also the normalization condition says you should integrate over the bounds of your function. if r is the radial coordinate then how can it have a value of negative infinity? similarly your no longer normalizing in 1-d but 3D I recommend using spherical coordinates in which case your integral inherits a r^2 sin theta
     
  5. Jan 28, 2009 #4
    Tks. I wil try again using spherical coordinates
     
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