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Let us forget about quantum mechanics and confine ourselves to classical mechanics.

The Hamiltonian for a classical helium atom would be

$$ H = \frac{p_1^2 + p_2^2}{2m } - \frac{Z}{r_1} - \frac{Z}{r_2} +\frac{1}{r_{12}}. $$

Here the central charge $Z $ is a variable. The problem is, if $Z $ is too small, the two electrons might cannot be bound around the nuclear forever, i.e., whatever the initial condition is, in the limit of $t \rightarrow \infty $, at least one electron will go to infinity.

So, what is the critical value of $Z $?

Similar problem also exists in the quantum case, it concerns the existence of bound states of the two electrons.

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    $\begingroup$ The three body system will always be potentially unstable no matter what the value of $Z$ is because as $r_1 \rightarrow 0$ the energy can go to $-\infty$ so the energy of the other electron can go to $+\infty$. $\endgroup$ – John Rennie Jun 2 '16 at 15:43
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    $\begingroup$ But the question is whether there are stable solutions $\endgroup$ – Jiang-min Zhang Jun 2 '16 at 23:52
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There are surely solutions for any $Z$ for which the electrons will remain bound forever. For example, the electrons may orbit the nucleus exactly on the opposite sides and the same distance.

On the other hand, for any $Z$, almost all initial states will lead to the escape of one of the electrons because the motion is chaotic and will have a nonzero probability for one electron to fly close enough to the nucleus, reach a high velocity, and be "kicked" by another kick by the other electron from the rear side.

Once the first electron leaves, the other electron will probably revolve around an elliptic orbit forever.

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