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Homework Help: Hydrogenic Atom emitting electron - probability of remaining in ground state

  1. Sep 21, 2009 #1
    1. The problem statement, all variables and given/known data

    Hi all - I've done all but the last part of this problem, for which I have the solution, but I was hoping someone could explain why the solution is what it is :)

    q:The Hamiltonian of a quantum system suddenly changes by a finite amount. Show that the wavefunction must change continuously if the time-dependent Schrodinger equation is to be valid throughout the change.

    Show that the ground-state wavefunction for a hydrogenic atom (a bound state of one electron and a nucleus of charge Ze, with Z a positive integer) takes the form

    [tex]\psi_{0}(r) = \frac{1}{\sqrt{\pi a^3}}exp(\frac{-r}{a})[/tex]

    and determine the dependence of the constant a on Z. Such a hydrogenic atom is in its ground state. The nucleus emits an electron, suddenly changing to one with charge (Z + 1)e.
    Compute the probability that if the energy is now measured, the atom will still be found in its ground state.

    Now it's just the red bit I'm stuck on - I've done the rest, but to save rewriting the formula every time you use it, you can just call it [tex]\psi_{0}(r,a)[/tex] if it's easier.

    I have from the previous parts [tex]a=\frac{4 \pi \epsilon_{0} h^2}{z e^2 m}[/tex], where h is actually the 'h-bar' version of Planck's constant, but I'm not sure how to LaTeX that!

    Now in the solution I have, the probability given is

    [tex]\frac{1}{\sqrt{\pi a^3}}}\frac{1}{\sqrt{\pi \tilde{a}^3}}}\int_0^{\infty} exp(-\frac{r}{a})exp(-\frac{r}{\tilde{a}})r^2\, dr \,sin(\theta)\,d\theta\,d\phi[/tex]

    (possibly with some mod bars somewhere), the calculation of which is fairly simple - where '[tex]\tilde{a}[/tex]' is [tex]\frac{a z}{z+1}[/tex], i.e. the new value of 'a' you'd have if z was z+1 - which makes the integral

    [tex]\int_0^{\infty} \psi_{0}(r,a)\psi_{0}(r,\tilde{a})r^2\, dr \,sin(\theta)\,d\theta\,d\phi[/tex] I believe.

    Why is it this to begin with? It doesn't seem to be the standard 'probability is mod wavefunction squared' because there's one [tex]a[/tex] and one [tex]\tilde{a}[/tex] - can anyone explain?

    One other thing, I'm not good with bra-ket notation(Dirac notation) so unless absolutely necessary words/other notation would be much more appreciated!

    Thanks very much all :)
    Last edited: Sep 21, 2009
  2. jcsd
  3. Sep 23, 2009 #2


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    It's very simple; if a particle is known to be in the (normalized) state [itex]\psi_1(\textbf{r})[/itex] (In the position basis), Then the probability of measuring it in a different (normalized) state [itex]\psi_2(\textbf{r})[/itex] is defined as

    [tex]P(\psi_2)=\int_{\text{all space}}\left|\psi_1(\textbf{r})\psi_2(\textbf{r})\right|dV[/tex]

    In Bra-Ket notation, this is the same as [itex]P(\psi_2)=|\langle \psi_2|\psi_1\rangle|[/itex]. You should have a similar definition in your text.

    Your wavefunctions are real-valued, so taking the norm of their product is trivial, and the volume integral is done in spherical coordinates.
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