Subsections


4.6 The Hydrogen Molecular Ion

The hydrogen atom studied earlier is where full theoretical analysis stops. Larger systems are just too difficult to solve analytically. Yet, it is often quite possible to understand the solution of such systems using approximate arguments. As an example, this section considers the H$_2^+$-​ion. This ion consists of two protons and a single electron circling them. It will be shown that a chemical bond forms that holds the ion together. The bond is a “covalent” one, in which the protons share the electron.

The general approach will be to compute the energy of the ion, and to show that the energy is less when the protons are sharing the electron as a molecule than when they are far apart. This must mean that the molecule is stable: energy must be expended to take the protons apart.

The approximate technique to be used to find the state of lowest energy is a basic example of what is called a “variational method.”


4.6.1 The Hamiltonian

First the Hamiltonian is needed. Since the protons are so much heavier than the electron, to good approximation they can be considered fixed points in the energy computation. That is called the “Born-Oppenheimer approximation”. In this approximation, only the Hamiltonian of the electron is needed. It makes things a lot simpler, which is why the Born-Oppenheimer approximation is a common assumption in applications of quantum mechanics.

Compared to the Hamiltonian of the hydrogen atom of section 4.3.1, there are now two terms to the potential energy, the electron experiencing attraction to both protons:

\begin{displaymath}
H = -\frac{\hbar^2}{2m_{\rm e}}\nabla^2
- \frac{e^2}{4\p...
...\rm {l}}}
- \frac{e^2}{4\pi\epsilon_0}\frac{1}{r_{\rm {r}}}
\end{displaymath} (4.74)

where $r_{\rm {l}}$ and $r_{\rm {r}}$ are the distances from the electron to the left and right protons,
\begin{displaymath}
r_{\rm {l}}\equiv \vert{\skew0\vec r}- {\skew0\vec r}_{\rm...
...r}}\equiv \vert{\skew0\vec r}- {\skew0\vec r}_{\rm {rp}}\vert
\end{displaymath} (4.75)

with ${\skew0\vec r}_{\rm {lp}}$ the position of the left proton and ${\skew0\vec r}_{\rm {rp}}$ that of the right one.

The hydrogen ion in the Born-Oppenheimer approximation can be solved analytically using prolate spheroidal coordinates. However, approximations will be used here. For one thing, you learn more about the physics that way.


Key Points
$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
In the Born-Oppenheimer approximation, the electronic structure is computed assuming that the nuclei are at fixed positions.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
The Hamiltonian in the Born-Oppenheimer approximation has been found. It is above.


4.6.2 Energy when fully dissociated

The fully dissociated state is when the protons are very far apart and there is no coherent molecule, as in figure 4.14. The best the electron can do under those circumstances is to combine with either proton, say the left one, and form a hydrogen atom in the ground state of lowest energy. In that case the right proton will be alone.

Figure 4.14: Hydrogen atom plus free proton far apart.
\begin{figure}
\centering
{}%
\epsffile{h2-leftf.eps}
\end{figure}

According to the solution for the hydrogen atom, the electron loses 13.6 eV of energy by going in the ground state around the left proton. Of course, it would lose the same energy going into the ground state around the right proton, but for now, assume that it is around the left proton.

The wave function describing this state is just the ground state $\psi_{100}$ derived for the hydrogen atom, equation (4.40), but the distance should be measured from the position ${\skew0\vec r}_{\rm {lp}}$ of the left proton instead of from the origin:

\begin{displaymath}
\psi=\psi_{100}(\vert{\skew0\vec r}- {\skew0\vec r}_{\rm {lp}}\vert)
\end{displaymath}

To shorten the notations, this wave function will be denoted by $\psi_{\rm {l}}$:
\begin{displaymath}
\psi_{\rm {l}}({\skew0\vec r}) \equiv \psi_{100}(\vert{\skew0\vec r}- {\skew0\vec r}_{\rm {lp}}\vert)
\end{displaymath} (4.76)

Similarly the wave function that would describe the electron as being in the ground state around the right proton will be denoted as $\psi_{\rm {r}}$, with

\begin{displaymath}
\psi_{\rm {r}}({\skew0\vec r}) \equiv \psi_{100}(\vert{\skew0\vec r}- {\skew0\vec r}_{\rm {rp}}\vert)
\end{displaymath} (4.77)


Key Points
$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
When the protons are far apart, there are two lowest energy states, $\psi_{\rm {l}}$ and $\psi_{\rm {r}}$, in which the electron is in the ground state around the left, respectively right, proton. In either case there is an hydrogen atom plus a free proton.


4.6.3 Energy when closer together

Figure 4.15: Hydrogen atom plus free proton closer together.
\begin{figure}
\centering
{}%
\epsffile{h2-left.eps}
\end{figure}

When the protons get a bit closer to each other, but still well apart, the distance $r_{\rm {r}}$ between the electron orbiting the left proton and the right proton decreases, as sketched in figure 4.15. The potential that the electron sees is now not just that of the left proton; the distance $r_{\rm {r}}$ is no longer so large that the $\vphantom0\raisebox{1.5pt}{$-$}$$e^2$$\raisebox{.5pt}{$/$}$$4\pi\epsilon_0r_{\rm {r}}$ potential can be completely neglected.

However, assuming that the right proton stays sufficiently clear of the electron wave function, the distance $r_{\rm {r}}$ between electron and right proton can still be averaged out as being the same as the distance $d$ between the two protons. Within that approximation, it simply adds the constant $\vphantom0\raisebox{1.5pt}{$-$}$$e^2$$\raisebox{.5pt}{$/$}$$4\pi\epsilon_0d$ to the Hamiltonian of the electron. And adding a constant to a Hamiltonian does not change the eigenfunction; it only changes the eigenvalue, the energy, by that constant. So the ground state $\psi_{\rm {l}}$ of the left proton remains a good approximation to the lowest energy wave function.

Moreover, the decrease in energy due to the electron/right proton attraction is balanced by an increase in energy of the protons by their mutual repulsion, so the total energy of the ion remains the same. In other words, the right proton is to first approximation neither attracted nor repelled by the neutral hydrogen atom on the left. To second approximation the right proton does change the wave function of the electron a bit, resulting in some attraction, but this effect will be ignored.

So far, it has been assumed that the electron is circling the left proton. But the case that the electron is circling the right proton is of course physically equivalent. In particular the energy must be exactly the same by symmetry.


Key Points
$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
To first approximation, there is no attraction between the free proton and the neutral hydrogen atom, even somewhat closer together.


4.6.4 States that share the electron

The approximate energy eigenfunction $\psi_{\rm {l}}$ that describes the electron as being around the left proton has the same energy as the eigenfunction $\psi_{\rm {r}}$ that describes the electron as being around the right one. Therefore any linear combination of the two,

\begin{displaymath}
\psi = a \psi_{\rm {l}} + b \psi_{\rm {r}}
\end{displaymath} (4.78)

is also an eigenfunction with the same energy. In such combinations, the electron is shared by the protons, in ways that depend on the chosen values of $a$ and $b$.

Note that the constants $a$ and $b$ are not independent: the wave function should be normalized, $\langle\psi\vert\psi\rangle$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. Since $\psi_{\rm {l}}$ and $\psi_{\rm {r}}$ are already normalized, and assuming that $a$ and $b$ are real, this works out to

\begin{displaymath}
\langle a \psi_{\rm {l}} + b \psi_{\rm {r}}\vert
a \psi_...
...2 ab \langle\psi_{\rm {l}}\vert
\psi_{\rm {r}}\rangle = 1 %
\end{displaymath} (4.79)

As a consequence, only the ratio the coefficients $a$$\raisebox{.5pt}{$/$}$$b$ can be chosen freely.

A particularly interesting case is the antisymmetric one, $b$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$$a$. As figure 4.16 shows, in this state there is zero probability of finding the electron at the symmetry plane midway in between the protons.

Figure 4.16: The electron being anti-symmetrically shared.
\begin{figure}
\centering
{}%
\epsffile{h2-asym.eps}
\end{figure}

The reason is that $\psi_{\rm {l}}$ and $\psi_{\rm {r}}$ are equal at the symmetry plane, making their difference zero.

This is actually a quite weird result. You combine two states, in both of which the electron has some probability of being at the symmetry plane, and in the combination the electron has zero probability of being there. The probability of finding the electron at any position, including the symmetry plane, in the first state is given by $\vert\psi_{\rm {l}}\vert^2$. Similarly, the probability of finding the electron in the second state is given by $\vert\psi_{\rm {r}}\vert^2$. But for the combined state nature does not do the logical thing of adding the two probabilities together to come up with $\frac12\vert\psi_{\rm {l}}\vert^2+\frac12\vert\psi_{\rm {r}}\vert^2$.

Instead of adding physically observable probabilities, nature squares the unobservable wave function $a\psi_{\rm {l}}-a\psi_{\rm {r}}$ to find the new probability distribution. The squaring adds a cross term, $-2a^2\psi_{\rm {l}}\psi_{\rm {r}}$, that simply adding probabilities does not have. This term has the physical effect of preventing the electron to be at the symmetry plane, but it does not have a normal physical explanation. There is no force repelling the electrons from the symmetry plane or anything like that. Yet it looks as if there is one in this state.

The most important combination of $\psi_{\rm {l}}$ and $\psi_{\rm {r}}$ is the symmetric one, $b$ $\vphantom0\raisebox{1.5pt}{$=$}$ $a$. The approximate wave function then takes the form $a(\psi_{\rm {l}}+\psi_{\rm {r}})$. That can be written out fully in terms of the hydrogen ground state wave function as:

\begin{displaymath}
\fbox{$\displaystyle
\Psi \approx a
\left[
\psi_{100...
...{100}(r) \equiv \frac{1}{\sqrt{\pi a_0^3}} e^{-r/a_0}
$} %
\end{displaymath} (4.80)

where $a_0$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0.53 Å is the Bohr radius and ${\skew0\vec r}$, ${\skew0\vec r}_{\rm {lp}}$, and ${\skew0\vec r}_{\rm {rp}}$ are again the position vectors of electron and protons. In this case, there is increased probability for the electron to be at the symmetry plane, as shown in figure 4.17.

Figure 4.17: The electron being symmetrically shared.
\begin{figure}
\centering
{}%
\epsffile{h2-sym.eps}
\end{figure}

A state in which the electron is shared is truly a case of the electron being in two different places at the same time. For if instead of sharing the electron, each proton would be given its own half electron, the expression for the Bohr radius, $a_0$ $\vphantom0\raisebox{1.5pt}{$=$}$ $4\pi\epsilon_0\hbar^2$$\raisebox{.5pt}{$/$}$${m_{\rm e}}e^2$, shows that the eigenfunctions $\psi_{\rm {l}}$ and $\psi_{\rm {r}}$ would have to blow up in radius by a factor four. (Because of $m_{\rm e}$ and $e$; the second factor $e$ is the proton charge.) The energy would then reduce by the same factor four. That is simply not what happens. You get the physics of a complete electron being present around each proton with 50% probability, not the physics of half an electron being present for sure.


Key Points
$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
This subsection brought home the physical weirdness arising from the mathematics of the unobservable wave function.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
In particular, within the approximations made, there exist states that all have the same ground state energy, but whose physical properties are dramatically different.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
The protons may share the electron. In such states there is a probability of finding the electron around either proton.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
Even if the protons share the electron equally as far as the probability distribution is concerned, different physical states are still possible. In the symmetric case that the wave functions around the protons have the same sign, there is increased probability of the electron being found in between the protons. In the antisymmetric case of opposite sign, there is decreased probability of the electron being found in between the protons.


4.6.5 Comparative energies of the states

The previous two subsections described states of the hydrogen molecular ion in which the electron is around a single proton, as well as states in which it is shared between protons. To the approximations made, all these states have the same energy. Yet, if the expectation energy of the states is more accurately examined, it turns out that increasingly large differences show up when the protons get closer together. The symmetric state has the least energy, the antisymmetric state the highest, and the states where the electron is around a single proton have something in between.

It is not that easy to see physically why the symmetric state has the lowest energy. An argument is often made that in the symmetric case, the electron has increased probability of being in between the protons, where it is most effective in pulling them together. However, actually the potential energy of the symmetric state is higher than for the other states: putting the electron midway in between the two protons means having to pull it away from one of them.

The Feynman lectures on physics, [22], argue instead that in the symmetric case, the electron is somewhat less constrained in position. According to the Heisenberg uncertainty relationship, that allows it to have less variation in momentum, hence less kinetic energy. Indeed the symmetric state does have less kinetic energy, but this is almost totally achieved at the cost of a corresponding increase in potential energy, rather than due to a larger area to move in at the same potential energy. And the kinetic energy is not really directly related to available area in any case. The argument is not incorrect, but in what sense it explains, rather than just summarizes, the answer is debatable.


Key Points
$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
The energies of the discussed states are not the same when examined more closely.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
The symmetric state has the lowest energy, the antisymmetric one the highest.


4.6.6 Variational approximation of the ground state

The objective of this subsection is to use the rough approximations of the previous subsections to get some very concrete data on the hydrogen molecular ion.

The idea is simple but powerful: since the true ground state is the state of lowest energy among all wave functions, the best among approximate wave functions is the one with the lowest energy. In the previous subsections, approximations to the ground state were discussed that took the form $a\psi_{\rm {l}}+b\psi_{\rm {r}}$, where $\psi_{\rm {l}}$ described the state where the electron was in the ground state around the left proton, and $\psi_{\rm {r}}$ where it was around the right proton. The wave function of this type with the lowest energy will produce the best possible data on the true ground state, {N.6}.

Note that all that can be changed in the approximation $a\psi_{\rm {l}}+b\psi_{\rm {r}}$ to the wave function is the ratio of the coefficients $a$$\raisebox{.5pt}{$/$}$$b$, and the distance between the protons $d$. If the ratio $a$$\raisebox{.5pt}{$/$}$$b$ is fixed, $a$ and $b$ can be computed from it using the normalization condition (4.79), so there is no freedom to chose them individually. The basic idea is now to search through all possible values of $a$$\raisebox{.5pt}{$/$}$$b$ and $d$ until you find the values that give the lowest energy.

This sort of method is called a “variational method” because at the minimum of energy, the derivatives of the energy must be zero. That in turn means that the energy does not vary with infinitesimally small changes in the parameters $a$$\raisebox{.5pt}{$/$}$$b$ and $d$.

To find the minimum energy is nothing that an engineering graduate student could not do, but it does take some effort. You cannot find the best values of $a$$\raisebox{.5pt}{$/$}$$b$ and $d$ analytically; you have to have a computer find the energy at a lot of values of $d$ and $a$$\raisebox{.5pt}{$/$}$$b$ and search through them to find the lowest energy. Or actually, simply having a computer print out a table of values of energy versus $d$ for a few typical values of $a$$\raisebox{.5pt}{$/$}$$b$, including $a$$\raisebox{.5pt}{$/$}$$b$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 and $a$$\raisebox{.5pt}{$/$}$$b$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$1, and looking at the print-out to see where the energy is most negative works fine too. That is what the numbers below came from.

You do want to evaluate the energy of the approximate states accurately as the expectation value. If you do not find the energy as the expectation value, the results may be less dependable. Fortunately, finding the expectation energy for the given approximate wave functions can be done exactly; the details are in derivation {D.21}.

If you actually go through the steps, your print-out should show that the minimum energy occurs when $a$ $\vphantom0\raisebox{1.5pt}{$=$}$ $b$, the symmetric state, and at a separation distance between the protons equal to about 1.3 Å. This separation distance is called the “bond length”. The minimum energy is found to be about 1.8 eV below the energy of -13.6 eV when the protons are far apart. So it will take at least 1.8 eV to take the ground state with the protons at a distance of 1.3 Å completely apart into well separated protons. For that reason, the 1.8 eV is called the “binding energy”.


Key Points
$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
The best approximation to the ground state using approximate wave functions is the one with the lowest energy.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
Making such an approximation is called a variational method.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
The energy should be evaluated as the expectation value of the Hamiltonian.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
Using combinations of $\psi_{\rm {l}}$ and $\psi_{\rm {r}}$ as approximate wave functions, the approximate ground state turns out to be the one in which the electron is symmetrically shared between the protons.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
The binding energy is the energy required to take the molecule apart.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
The bond length is the distance between the nuclei.

4.6.6 Review Questions
1.

The solution for the hydrogen molecular ion requires elaborate evaluations of inner product integrals and a computer evaluation of the state of lowest energy. As a much simpler example, you can try out the variational method on the one-di­men­sion­al case of a particle stuck inside a pipe, as discussed in chapter 3.5. Take the approximate wave function to be:

\begin{displaymath}
\psi = a x(\ell -x)
\end{displaymath}

Find $a$ from the normalization requirement that the total probability of finding the particle integrated over all possible $x$ positions is one. Then evaluate the energy $\langle{E}\rangle$ as $\langle\psi\vert H\vert\psi\rangle$, where according to chapter 3.5.3, the Hamiltonian is

\begin{displaymath}
H = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}
\end{displaymath}

Compare the ground state energy with the exact value,

\begin{displaymath}
E_1=\hbar^2\pi^2/2m\ell^2
\end{displaymath}

(Hints: $\int_0^{\ell}x(\ell -x){\,\rm d}{x}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ell^3$$\raisebox{.5pt}{$/$}$​6 and $\int_0^{\ell}x^2(\ell -x)^2{\,\rm d}{x}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ell^5$$\raisebox{.5pt}{$/$}$​30)

Solution hione-a


4.6.7 Comparison with the exact ground state

The variational solution derived in the previous subsection is only a crude approximation of the true ground state of the hydrogen molecular ion. In particular, the assumption that the molecular wave function can be approximated using the individual atom ground states is only valid when the protons are far apart, and is inaccurate if they are 1.3 Å apart, as the solution says they are.

Yet, for such a poor wave function, the main results are surprisingly good. For one thing, it leaves no doubt that a bound state really exists. The reason is that the true ground state must always have a lower energy than any approximate one. So, the binding energy must be at least the 1.8 eV predicted by the approximation.

In fact, the experimental binding energy is 2.8 eV. The found approximate value is only a third less, pretty good for such a simplistic assumption for the wave function. It is really even better than that, since a fair comparison requires the absolute energies to be compared, rather than just the binding energy; the approximate solution has $\vphantom0\raisebox{1.5pt}{$-$}$15.4 eV, rather than $\vphantom0\raisebox{1.5pt}{$-$}$16.4. This high accuracy for the energy using only marginal wave functions is one of the advantages of variational methods {A.7}.

The estimated bond length is not too bad either; experimentally the protons are 1.06 Å apart instead of 1.3 Å. (The analytical solution using spheroidal coordinates mentioned earlier gives 2.79 eV and 1.06 Å, in good agreement with the experimental values. But even that solution is not really exact: the electron does not bind the nuclei together rigidly, but more like a spring force. As a result, the nuclei behave like a harmonic oscillator around their common center of gravity. Even in the ground state, they will retain some uncertainty around the 1.06 Å position of minimal energy, and a corresponding small amount of additional molecular kinetic and potential energy. The improved Born-Oppenheimer approximation of chapter 9.2.3 can be used to compute such effects.)

The qualitative properties of the approximate wave function are correct. For example, it can be seen that the exact ground state wave function must be real and positive {A.8}; the approximate wave function is real and positive too.

It can also be seen that the exact ground state must be symmetric around the symmetry plane midway between the protons, and rotationally symmetric around the line connecting the protons, {A.9}. The approximate wave function has both those properties too.

Incidentally, the fact that the ground state wave function must be real and positive is a much more solid reason that the protons must share the electron symmetrically than the physical arguments given in subsection 4.6.5, even though it is more mathematical.


Key Points
$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
The obtained approximate ground state is pretty good.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
The protons really share the electron symmetrically in the ground state.