# What's the lowest nuclear charge $Z < 1$ that will support a bound two-electron ion $(Z,2e^-)$?

In my programming project I calculate the minimal energy of an atom with 2 electrons in the $L=0, S=0$ state, using a Hylleraas wave function.

The values I find for $Z=2$ (He) and $Z=1$ (H$^-$) are in good correspondence with what can be found in literature (Pekeris, 1962):

• $E_0({\rm He}) \approx -2.903~{\rm a.u.}$
• $E_0({\rm H^-}) \approx -0.528~{\rm a.u.}$

Here a.u. are the atomic units in which the energy of the hydrogen atom is $-0.5$. However, the next part of the project is to find a lower limit for $Z$, e.i. imagine a theoretical ion with $0<Z<1$ $(Z\in \mathbb{R})$, what is the minimal value for $Z$ in order for the two-electron system to remain bound? This is equivalent with asking: for which value of $Z$ is $E_0=0$?

I cannot find any references about this theoretical lower limit online, but I don't think my results are correct: the $E(Z)$ relation, also shown in the graphs below, looks like a power law. This means that $E_0\to0$ for $Z\to0$, but in this case you just have two electrons and you would expect a very positive energy as the two will strongly repel each other.

I do not expect you to help me with the programming project, but maybe someone could provide some useful thoughts or a reference about this theoretical lower limit? I keep finding it strange that the results for $Z=1,2$ are correct with errors of only 0.01% and everything seems to go wrong for small $Z$.

Note: how are these energies exactly calculated? We start with a Hylleraas wavefunction subjected to a coordinate rescaling factor $\alpha$: $$\langle\vec{r}_1\vec{r}_2|\Psi_\alpha\rangle = \sum_{STU}C_{STU}N_{STU}{\rm e}^{-\alpha s/2}(\alpha s)^S(\alpha t)^T(\alpha u)^U$$ with $s = r_1 + r_2, t = r_1-r_2, u=r_{12}=|\vec{r}_1-\vec{r}_2|$, $S,U\in\mathbb{N}$, $T\in2\mathbb{N}$ and $\alpha\in\mathbb{R}^+$. Three matrices can be calculated: the overlap matrix $[M]$, the kinetic energy $[T]$ and potential energy $[V].$ They scale like: $$\langle\Psi_\alpha|\Psi_\alpha\rangle = \langle\Psi|\Psi\rangle/\alpha^6$$ $$\langle\Psi_\alpha|T|\Psi_\alpha\rangle = \langle\Psi|T|\Psi\rangle/\alpha^4$$ $$\langle\Psi_\alpha|V|\Psi_\alpha\rangle = \langle\Psi|V|\Psi\rangle/\alpha^5$$ Variation to the expansion coefficients in $\langle\Psi_\alpha|(T+V)|\Psi_\alpha\rangle/\langle\Psi_\alpha|\Psi_\alpha\rangle$ leads to a generalized eigenvalue problem: $$\left(\alpha^2[T]+\alpha[V]\right) C_\alpha = E_\alpha[M]C_\alpha$$ The lowest energy eigenvalue $E_\alpha^0$ gives a function of $\alpha$ of which the minimum needs to be determined. This minimum is the best variational approximation of the ground state energy.

UPDATE: It is interesting to plot the evolution of the coordinate rescaling factor $\alpha$ as a function of $Z$. Apparently, there is a discontinuity around $Z\approx 0.89841$, which is close to the critical value of $Z\approx 0.91$ found in literature. I would suppose this point has a significant meaning, but I have failed so far to give it a physical interpretation (I hope to find an interpretation which has to do with the ionization of at least 1 electron).

The relationship $\alpha(Z)$ is plotted below. The red graphs are added to show how this discontinuity arises (a second local minimum arises and takes over at the critical $Z$) and to explain why no discontinuity arises in the energy relation $E(Z)$.

• arXiv:1102.4493 appears to at least be relevant literature, with Fig. 7 putting the critical charge at maybe $Z_c\approx0.9$. – Emilio Pisanty Apr 28 '16 at 21:39
• @EmilioPisanty, this seems to be useful, thank you. I do not completely understand how they determine $Z_c$, however. It is not the point where the sign of the total energy changes from negative (a bound system) to positive (an unbound system)? – Zdenko Heyvaert May 2 '16 at 9:40
• I'm not sure why that curve stops at negative energy - presumably it's in one of their references. This one seems to address it directly, though - I would look at both its references and the papers that cite it. – Emilio Pisanty May 2 '16 at 10:40
• This is a good question; have a bounty to help attract attention :-). – Emilio Pisanty May 2 '16 at 10:41
• "you would expect a very positive energy as the two will strongly repel each other": except that the orbital becomes bigger as $Z$ decreases, so the repulsion decreases. Can you plot the electron-electron contribution to total energy, vs. $Z$? – L. Levrel May 2 '16 at 20:21

Capturing more of the (always negative) correlation energy due to more variational degrees of freedom in the solution, the MCHF code seems to find bound solutions down to basically Z=0.8. Due to the error bars, an exact critical value for $$Z$$ cannot be determined from these results. With the given resolution or total energy accuracy for smaller Z, there doesn't seem to be any discontinuity in the energy range considered.
All to be taken with a grain of salt due to the self-consistency convergence problems for $$Z$$<1.
Since one electron will always bind to the nucleus with an energy of $-Z/2$ Atomic units, looking for the zero of energy is not correct. You want to locate the charge where only one electron binds. At this point, the other electron will be far away, and the correct ground state wave function will look like $\psi_0(r_1)+\psi_0(r_2)$, with $\psi_0(r)$ the single electron hydrogen like ground state with charge Z. This form, is of course quite different from your form, but you can see that your code is trying to change $\alpha$ to get closer.
In any case, for an estimate of the critical $Z$, look for the point where the curve E_0 = -Z/2 crosses your curve. Looking at your plot, it's roughly $Z=0.9$. That's the point where the separated wave wave function above gives the same energy as your wave function. It will give a lower energy for all smaller $Z$ values.