# Hydrogenic Atom emitting electron - probability of remaining in ground state

1. Sep 21, 2009

### Mathmos6

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

$$\psi_{0}(r) = \frac{1}{\sqrt{\pi a^3}}exp(\frac{-r}{a})$$

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 $$\psi_{0}(r,a)$$ if it's easier.

I have from the previous parts $$a=\frac{4 \pi \epsilon_{0} h^2}{z e^2 m}$$, 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

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

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

$$\int_0^{\infty} \psi_{0}(r,a)\psi_{0}(r,\tilde{a})r^2\, dr \,sin(\theta)\,d\theta\,d\phi$$ 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 $$a$$ and one $$\tilde{a}$$ - 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. Sep 23, 2009

### gabbagabbahey

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

$$P(\psi_2)=\int_{\text{all space}}\left|\psi_1(\textbf{r})\psi_2(\textbf{r})\right|dV$$

In Bra-Ket notation, this is the same as $P(\psi_2)=|\langle \psi_2|\psi_1\rangle|$. 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.