Hydrogen-like wavefunctions have the form:

$$R_{10} = \left(\frac{Z}{a_0}\right)^{\frac{3}{2}} 2\space e^{-\frac{Zr}{a_0}}$$ $$Y_{00} = \frac{1}{\sqrt {4\pi}} $$

where $a_0 = \frac{4\pi \epsilon_0 \hbar^2}{\mu e^2}$

Suppose Tritium $^3$H that's in its ground state beta decays to $^3$He suddenly. I want to find the probability the helium atom is in the 1s state.

Firstly, is the reduced mass $\mu$ the same for both helium and $^3$He? I assumed it is $m_e$.

Wavefunction of helium is $R_{10}' = \left(\frac{2}{a_0}\right)^{\frac{3}{2}}2\space e^{-\frac{2r}{a_0}} $. Wavefunction of tritium is $R_{10} = \left(\frac{1}{a_0}\right)^{\frac{3}{2}}2\space e^{-\frac{r}{a_0}} $.

To find the probability, do I simply overlap the two and square them?

$$\langle (1,0,0)_H| (1,0,0)_T \rangle = \left(\frac{1}{a_0}\right)^3 (2)^{\frac{3}{2}} (2)^2\left(\frac{1}{4\pi}\right) \int e^{-\frac{3r}{a_0}} r^2 sin\theta \space d\theta d\phi $$

$$ = \frac{2^{\frac{9}{2}}}{27}$$

Thus probabiliy is $\frac{2^9}{27^2} = \frac{512}{729} $?

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    $\begingroup$ This is called the "sudden approximation", you assume that the transition happens so fast that the wavefunction does not change. Note that there is a factor of 2 missing in the overlap you calculated (both the R10 functions yield a factor 2). This will make the probability 512/729. $\endgroup$ – Count Iblis Jun 5 '14 at 18:47
  • $\begingroup$ But the wavefunction changed, since Z changed. What's the difference between this and Adiabatic approximation? An adiabatic approximation is that when the system is originally in state k, if the hamiltonian changes slowly enough, it remains in state k $\endgroup$ – user44840 Jun 5 '14 at 18:52
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    $\begingroup$ Adiabatic approximation is the opposite limit of infinitely slow changes. In that case if the Hamiltonian is H(t), to a first approxamation the system at time t will be in the eigenstate of H(t). Had this approximation been appropriate here, then the system would have evolved to the 1s state of Helium with probability 1. But, of course, the adiabatic approximation does not apply here. $\endgroup$ – Count Iblis Jun 5 '14 at 19:19
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    $\begingroup$ @user44840: Yes, it does, but the electron wave function under such sudden perturbation does not have time to change so the $r$-dependence remains the same at $t=0$. Then this form "evolves" as a superposition of different Helium wave functions with different energy exponentials. $\endgroup$ – Vladimir Kalitvianski Jun 5 '14 at 20:07
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    $\begingroup$ thanks alot, I got it now. I read somewhere that for an adiabatic change, $\tau >> \frac{\hbar}{\delta E}$ and for a sudden change, $\tau << \frac{\hbar}{\delta E}$ $\endgroup$ – user44840 Jun 5 '14 at 20:09

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