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Homework Help: Wavefunction integral

  1. Sep 1, 2008 #1
    1. The problem statement, all variables and given/known data

    A wavefunction for a hydrogen electron is given by [tex]\Psi = -

    \sqrt{\frac{3}{8 \pi}} sin\theta e^{i \phi} (\frac{1}{2a^3})^{3/2}

    \frac{re^{-r/2a}}{a \sqrt{3}}[/tex]

    Prove that the electron exists in space, ie, [tex]\int {\Psi}^2= 1 [/tex]

    2. Relevant equations & attempt at solution

    Apologise in advance for the shortcuts, these equations are terrible to type

    out.

    Expressed in spherical polar coordinates, [tex] dV = r^2 sin \theta dr

    d\theta d\phi [/tex]

    The squared wavefunction,
    [tex]
    {\Phi}^2 = \frac{1}{64\pi a^5} r^2 {sin}^2 \theta e^{2i\phi}
    [/tex]

    With respect to r, [tex] \int^{\infty}{\0} r^4 e^{r/a} = 24 a^5 [/tex]

    This is a pain to do due to iterated application of integration by parts, but

    by inspection,

    [tex]

    \int^{\infty}{\0} r^4 e^{r/a} = 4a \int^{\infty}{\0} r^3 e^{r/a} = 4.3a^2

    \int^{\infty}{\0} r^2 e^{r/a}... = 24a^5

    [/tex]


    With respect to [tex]\theta[/tex],

    [tex] \int^{\pi}{\0} {sin}^3 \theta d\theta = \frac{4}{3} [/tex]

    This gives us,

    [tex]
    \int {\Phi}^2 dV = \frac{1}{2\pi} \int^{2\pi}{\0} e^{2i\phi} d\phi
    [/tex]

    I'm stuck at this point. How do I proceed? Was my earlier working correct?

    If the earlier integration was right, then the last integral must be equal to 2pi.


    Exploration
    From using traditional methods the answer I actually get is 0. How does the pi term come about.
     
  2. jcsd
  3. Sep 2, 2008 #2

    HallsofIvy

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    Science Advisor

    Shouldn't that be
    [tex]\int |\Psi|^2 dV[/tex]?

    In other words, you want the absolute value squared, not the function. And
    [tex]|e^{i\phi}|= 1[/itex]
     
  4. Sep 2, 2008 #3
    Thanks, HallsofIvy, I think that explains that one. (Sorry, I mixed up psi/phi.) The rest of my solution is ok, right?

    My text explicitly gave the condition required as

    [tex] \int {\Psi}^2 dV = 1[/tex]

    Which leads me to another question. What is the physical significance of the modulus? (I'm not studying a text on quantum physics at the moment, I'm working on a mathematical physics text.)
     
  5. Sep 2, 2008 #4

    Ygggdrasil

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    2017 Award

    [tex]|\Psi |^2[/tex] is interpreted to be a probability density function describing the probability of finding your particle in a specific state.
     
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