Let us begin with recalling Bohr’s treatment of hydrogen-like atoms. He assumed that the electron is moving on a circular orbit around a proton or a point-like ion of charge +Ze. It follows then from classical mechanics and the quantization rule that the orbital radius r and velocity v of the electron are related to one another so that

$\frac{Z{e}^2}{r^2}=\frac{m{v}^2}{r}$ and $2\pi rmv=nh$ or $rp=n\hslash$

The Coulomb force is expressed in natural units ( $4\pi {\epsilon }_o=1$) and the energy of this system is

$E=V+T=\frac{-Z{e}^2}{r}+\frac{m{v}^2}{2}$ where $\frac{m{v}^2}{2}=\frac{n^2}{2r^2}\frac{{\hslash }^2}{m}=\frac{Z{e}^2}{2r}$

Thus, $r=\frac{n^2}{Z}{a}_o$ where $a_o=\frac{{\hslash }^2}{m{e}^2}=52.918\text{pm}$

$E=\frac{-Z}{n^2}{E}_o$ where $E_o=\frac{e^2}{2a_o}=13.606\text{eV}$

In honor of Niels Bohr and to facilitate comparisons with H atoms, we adopt the Bohr radius ( $a_o$) and the ionization energy of hydrogen atoms ( $E_o=1\text{Ry}$) as natural units. So-called “atomic units” are defined by setting $\hslash =m=e=\text{1}$, which yields $a_o=1$, but $2E_o=1$ Hartree. We note that

$V=-2T$ and $E=-T$(4)

What happens for H₂ molecules? Bohr assumed that the protons are bound to one another when both electrons are moving on the same circular orbit, situated in the median plane of the protons. The electrons are there always opposite, to minimize their mutual repulsion. This model is represented in Figure 1(a) . Dynamical equilibrium and the quantization rule for $n=1$ require that every electron is attracted towards the symmetry axis by the Newtonian force

$F=\frac{2r}{d^3}-\frac{1}{4r^2}=\frac{a_o}{r^3}$ where $d^2=r^2+x^2$ and $d^3=\frac{8r^4}{r+4a_o}$

Thus $d=d\left(r\right)$ and $x=x\left(r\right)$, but $r\ge \left(4/7\right)a_o$. The energy of the molecule is

$E\left(r\right)=\frac{1}{x}+\frac{1}{r}-\frac{8}{d}+\frac{2a_o}{r^2}$(5)

This function displays a minimum when

$r=r_o=0.9532a_o$ $x=x_o=0.550a_o$ $E_{\min }=-2.201E_o$

In parametric representation, we get the red curve of Figure 1(b) where $R=2x$. The measured separation of the protons is $R=1.40a_o$ and the measured dissociation energy for ${\text{H}}_2\to 2\text{H}$ is $0.329E_o$. This energy is determined with respect to the energy level $-2E_o$ of two separated hydrogen atoms [14] . Today, the dissociation energy of hydrogen molecules is known with very high precision [15] . Figure 1 implies that both electrons are always moving in the medium plane of the protons. The orbital radius r and the distance d would thus be constantly increasing until the molecule is completely decomposed ( ${\text{H}}_2\to 2{\text{H}}^{+}+2{\text{e}}^{-}$). When this state defines the energy level $E=0$, the ground state energy $E\to -2.329E_o$.

The first wave-mechanical treatment of H₂ molecules was performed by Heitler and London and published in 1927. They belonged to Schrödinger’s group and demonstrated the power of his equation [16] . It was sufficient, indeed, to consider the sum and difference of atomic wave functions, to get constructive or destructive interference in the median plane of the protons. Linear Combination of Atomic Orbitals (LCAO) could thus account for the bound state and the first excited state. According to this approximation, the ground state would be situated at $E=-2.23E_o$ and $R=1.51a_o$. These results were reported by Pauling and Wilson [17] , who did extensively use this method for other molecular structures.

Recent quantum-mechanical calculations for H₂ provide excellent agreement with observations [18] . The probability density is then maximal near the proton, but increased in the median plane. Bohr’s model indicates that it is sufficient to assume that the electrons are only moving in the median plane to get a bound state. This approximation applies to average positions, of course, though they are assumed to be well-defined. Schrödinger attributed Bohr’s concept of “stationary states” to the existence of stationary waves. Since they yield a static charge distribution, they restored agreement with Maxwell’s theory, but led also to the idea that covalent bonds do merely result from electrostatic effects. Ruedenberg insisted therefore in l962 on the fact that motions of both electrons are essential for these bonds [19] . He did that by computing the quantum-mechanical average values of the potential energy V and the kinetic energy T. It appeared that $\langle V\rangle =-2\langle T\rangle$ and thus that $\langle E\rangle =-\langle T\rangle$. This is a generalization of (4) and the virial theorem, which was established already by Clausius in 1870 for any isolated system that is composed of charged particles and has reached its equilibrium state. Bacskay and Nordholm [20] did also perform tedious computations of quantum-mechanical average values for other divalent molecules. However, it is easy to verify that SQM accounts already for the virial theorem. It is merely due to Coulomb forces and independent of more or less detailed quantum effects. The word “virial” stems even from the Latin word for force, which survived in the adjective “virile”.

SQM has the advantage to describe what is essential during the dissociation of hydrogen molecules by means of a modified Bohr model (MBM). The two electrons are again assumed to move on circular orbits, but the orbital planes are separated, as shown in Figure 2(a) . The electrons are always moving as far as possible from one another. Figure 2(b) represents the limit of applicability of the BM. The value of $r_1$ is unknown, but $x_1=x\left(r_1\right)$ as for the BM of Figure 1 . The energy curve $E\left(R\right)$ is then bifurcating, since $E\to 0$ for the BM, while $E\to -2E_o$ for the MBM. Figure 2(c) shows that the MBM should allow for a metastable state when $x=x_2$, since two hydrogen atoms could polarize one another. The orbital plane is then separated from the median plane by the distance $z_2<x_2$ and the orbital radius $\rho ={\rho }_2$.

Even the MBM of Figure 2(a) led to difficulties and it is useful to see the reasons. Svidzinsky and his collaborators treated this problem in 2005, by means of dimensional scaling [21] . The result was amazing, since the energy curve of Figure 1(b) was modified so that the minimum did coincide with the red dot. Such a precise agreement with experimental results without wave mechanics, could have been suspicious, but was published in several scientific journals. A didactical version appeared in Physics Today [22] . However, D-scaling was merely a hypothetical procedure, suggested by Witten to solve the riddle of elementary particles [23] .

We mentioned already in the introduction that they can be distinguished from one another by means of empirically defined quantum numbers. Since this fact cannot be explained by conventional theories, it has been proposed that elementary particles could be strings, instead of points. These strings should even have peculiar structures, to allow for unobservable degrees of freedom. Another hypothesis was that space could have more than 3 dimensions. Momentum vectors do then have more components and the expression of kinetic energies is modified. However, when the number D of dimension is increasing, the limit $D\to \infty$ yields finite results. Witten recognized that “so far there has not been much tangible progress”, but since it was possible to apply this formalism to hydrogen atoms [24] , it seemed credible that it could also account for dissociating hydrogen molecules. However, the viral theorem cannot be satisfied when the kinetic energy T is modified, while the potential energy V is still defined by Coulomb forces. Moreover, the riddle of elementary particle physics can be solved by means of space-time quantization [7] .

Yang and Hou did also examine the problem of dissociating hydrogen molecules [25] . Though their article was published in 2019, they ignored Svidzinsky’s theory. They tried also to modify the BM, to get perfect agreement with experimental results, but they modified the potential energy V instead of the kinetic energy T. This would also contradict the virial theorem. Actually, they proposed that Coulomb interactions would be affected by screening effects. However, there are only 4 interacting particles for SQM.

Allmendinger considered the MBM in 2018, without modifying the expressions of V or T. This approach was realistic [26] and required that the energy is defined by

$E_1\left(x,z,\rho \right)=\left(\frac{1}{x}+\frac{1}{\delta }-\frac{4}{d_{+}}-\frac{4}{d_{-}}+\frac{2a_o}{{\rho }^2}\right)E_o$(6)

The last term accounts for the kinetic energy of both electrons and the quantization rule for circular motions. It results also from Figure 2(a) that

${\delta }^2=z^2+{\rho }^2$ and $d_{\pm }^2={\left(x\pm z\right)}^2+{\rho }^2$

Since the electron orbits have to be radially and laterally stable,

$\frac{\rho }{d_{+}^3}+\frac{\rho }{d_{-}^3}-\frac{2\rho }{{\left(2\delta \right)}^3}=\frac{a_o}{{\rho }^3}$ and $\frac{x+z}{d_{+}^3}=\frac{2z}{{\left(2\delta \right)}^3}+\frac{x-z}{d_{-}^3}$(7)

Unfortunately, Allmendinger replaced the first condition (7) by the requirement that the forces applied on the protons should equilibrate one another. This is not necessary, since the energy (6) is calculated for various separations of positions of motionless protons. Computing the energy for several structures by Exel, Allmendiger found that the energy is minimal when

$E_1\left(x,z,\rho \right)=E_1\left(0.705,0.488,0.892\right)=-2.127E_o$

All distances are expressed in units $a_o$. Allmendiger did not consider the BM and was thus not aware that it yields a lower energy. To determine the metastable state, we have to account for both conditions (7) and require that the energy (6) yields a minimum for this state. Thus, $\partial E_1\left(x,z,\rho \right)/\partial x=0$. Solving these equations by Mathcad, we get

$E_1\left(x_2,z_2,{\rho }_2\right)=E_1\left(0.817,0.706,0.9755\right)=-2.130E_o$(8)

These values define the metastable state of Figure 2(c) , but we would also like to know the value of x₁ in Figure 2(b) . SQM does even allow to describe the dissociation process, by introducing a new variable y to define the position of the proton for the MBM. The separation z between the orbital planes and the median plane of the protons is increasing so that $z\to x$ in an asymptotic way. The initial value of the orbital radius ρ is r₁ and its final value is $a_o=1$, but a metastable state would require that the total energy E is decreased by reducing the mutual repulsion of the electrons. These ideas are expressed by setting

$x\left(y\right)=x_1+y$ where $x_1=x\left(r_1\right)$(9)

$z\left(y\right)=x\left(y\right)\left(1-{\text{e}}^{-\alpha y}\right)$(10)

$\rho \left(y\right)=1+\left(r_1-1+Ay\right)\left(1-{\text{e}}^{-\alpha y}\right)$(11)

We get thus 4 adaptable parameters ( $\alpha ,r_1,A,y_2$), where $y_2=x_2-x_1$. These parameters can be determined by solving the Equations (9)-(11) and (6) for the metastable state (8). These conditions yield

$\alpha =42.468$ $r_1=1.134$ $A=12.545$ $y_2=0.047$

It follows that the branching point of the energy curves is situated at $x_1=x\left(r_1\right)=0.770$, well beyond $x_o=0.550$ for the ground state in Figure 1(b) . Svidzinsky’s assumption led on the contrary to $x_1<x_o$ and did not account for a metastable state. The present results lead to Figure 3(a) for the functions $z\left(x\right)$ and $\rho \left(x\right)$. We see that dissociation occurs in an explosive way when $x=x_1$, since the orbital planes do then rapidly join the protons. The orbital radius ρ ceases to increase as for the BM and becomes rapidly equal to $a_o=1$ as in isolated hydrogen atoms, though the dissociation is not yet finished. The orbital radius ρ does not immediately decrease. There appears even a small initial increase. Since the mutual repulsion of the electrons is then decreased, it causes the existence of a metastable state, which is clearly indicated by the small potential well for the red curve in Figure 3(b) . The black curve shows that the BM remains valid in a large domain, since bifurcation of the energy curves occurs only when the separation $R=2x_1=1.54a_o$. Figure 3(b) accounts also for the fact that $E\to -2E_o$. This energy level is now replaced by $E=0$. It is approached in progressive way, since Coulomb forces have a quasi-infinite range. Their effects become only negligible when $R\approx 15$.

At present, it is possible to solve Schrödinger’s equation for hydrogen molecules, by accounting even for correlated motions, as well as relativistic and quantum-electrodynamic effects [27] . This yields an energy curve that agrees with precise measurements of the dissociation energy and even with the infrared spectrum for vibrating hydrogen molecules [28] . Schrödinger’s equation could also be solved for excited states of H₂ molecules [29] . However, it is astonishing that wave mechanics yields an energy curve that becomes suddenly flat, as if the dissociation ${\text{H}}_2\to 2\text{H}$ were immediately complete. There is not even the slightest sign of a metastable sate. It might thus be useful to launch a more specific search in this regard.

Niels Bohr considered already the apparently simple case where only one electron is attracted by two protons. He adopted of course the model of Figure 4(a) , but found that ${\text{H}}_2^{+}$ would then be unstable [1, part III, p. 867]. It was not yet known that hydrogen molecular ions are slightly stable. The search of an explanation led to enormous difficulties. We begin therefore with Bohr’s model. Dynamical stability requires that the force

$F=\frac{2r{e}^2}{d^3}=\frac{m{v}^2}{r}$ where $d^2=x^2+r^2$

The values of r and x are also determined by the quantization rule $rmv=n\hslash$. Thus,

$d^3=\frac{2r^4}{n^2{a}_o}$

For the ground state $n=1$, the energy of a hydrogen molecular ion is

$E_i=\left(\frac{1}{x}-\frac{4}{d}+\frac{a_o}{r^2}\right)E_o=E_i\left(r\right)$(12)

Since $x=x\left(r\right)$, parametric representation yields the red curve of Figure 4(b) , where $R=2x$ is the separations of the protons. The ground state would thus be situated at the white dot, where $R=1.731a_o$ and $E_i=-0.878E_o>-E_o$. According to the BM, hydrogen molecular ions would therefore be unstable. It should be noted that x has to be real, but that concerns only small values of R. The LCAO approximation did yield the blue dot, while experimental results are represented by the red dot.

Since ${\text{H}}_2^{+}$ molecular ions are at least slightly stable, the initial BM has to modified. This can be done by adopting the model of Figure 5(a) . The modified energy is then

$E_m\left(x,z,r\right)=\left(\frac{1}{x}-\frac{2}{d_{+}}-\frac{2}{d_{-}}+\frac{a_o}{r^2}\right)E_o$(13)

Radial and lateral stability of the electron orbit requires that

$\frac{1}{d_{+}^3}+\frac{1}{d_{-}^3}=\frac{a_o}{r^4}$ and $\frac{x+z}{d_{+}^3}=\frac{x-z}{d_{-}^3}$

Bifurcation of the energy curves occurs when $E_m\left(x_1,0,r_1\right)=E_i\left(r_1\right)$, while the lowest possible energy for the MBM is reached when $\partial E_m\left(x,z,\rho \right)/\partial x=0$. The solution of these equations is

$x_1=0.583$ $r_1=0.871$ $x_2=46.668$ $z_2=x_2$ and ${\rho }_2=1$

Since $x_1<0.866$, bifurcation of the energy curves occurs now before reaching the white dot in Figure 4(b) . However, the dissociation process is again described by of (9)-(11). Thus,

$\alpha =99.413$ $A=-1.624$ and $y_2=x_2-x_1\approx 46$

The resulting effects are represented in Figure 5(b) . Dissociation occurs when $R=R_1=2x_1=1.166a_o$. This process is explosive, since the displacement z of the orbital plane with respect to the median plane of the protons is suddenly increased. The electron does even immediately join the chosen proton ( $z\to x$). The orbital radius ρ becomes immediately equal to $a_o$ as in an isolated hydrogen atoms. Nevertheless, this electron is still attracted by the other proton. Their separation $d_{+}$ is increasing and the energy $E_m\to -E_o$ in a progressive way. Since the value of x₂ is enormous, polarization effects are negligible. Nevertheless, there does exist a metastable state, since the (black) energy curve $E_i$ for the BM is replaced by the red curve when $R\ge R_1$.

The sudden increase of the orbital radius ρ does only reduce the attraction of the electron by the departing proton. Hydrogen molecular ion would thus be slightly stable, but for a peculiar reason. We ignore if the resulting metastable state can be observed or not.

A general introduction that insists on basic concepts, also for hydrogen molecules and molecular ions, was provided by Beiser [30] . Schrödinger’s equation was solved in 2006 for ${\text{H}}_2^{+}$ with remarkable precision [31] . In 2021, the energy curves were computed for the ground state and several excited states with an accuracy of 40 digits [32] . The BM of Figure 4(a) was already abandoned very soon, to search for another orbit. This adventure was initiated by Sommerfeld, since he considered elliptical orbits for the single electron in hydrogen atoms. His excellent results suggested that the single electron, attracted by two protons in ${\text{H}}_2^{+}$ could be moving around both protons. It would then be sufficient to consider a particular sagittal plane. The ground state seemed to require a lemniscate. It is an 8-like orbit, but is not defined by some nice mathematical formula.

It results from Coulomb forces, when the electron is turning around both protons, but the electron can also keep them together by passing through the center. Since all sagittal planes are equally probable, this would generate two closed conical shells that touch one another. They would generalize Bohr’s concept of spherical shells for hydrogen atoms. Sommerfeld advised thus his brilliant student Wolfgang Pauli to examine this problem for his PHD thesis. He accepted and used elliptical coordinates [33] . When the electron is separated from the protons by the distances r₁ and r₂ and when R is the separation of these protons, they are defined by

$\xi =\frac{r_1+r_2}{R}$ and $\eta =\frac{r_1-r_2}{R}$

Any possible position of the electron is thus determined by the intersections of confocal ellipses and confocal hyperbola. However, Pauli used the first quantization rule (2) as Sommerfeld had done for elliptical orbits in hydrogen atoms. This approach led only to a classification of possible motions in 1922. The actual shape and size of the electron orbit for the ground state could not be determined. Karl Niessen came to the same conclusion [34] . These results did contribute to abandoning “old quantum mechanics”, but the essential reason was the development of wave mechanics. Heitler and London did solve Schrodinger’s equation by means of the LCAO approximation [10] . That is even simpler for one electron than for two electrons and presented in most textbooks on quantum mechanics. Since this procedure requires to calculate some integrals, it is useful to know how that can be done [35] .

Burrau [36] did even solve Schrödinger’s equation for stationary states of ${\text{H}}_2^{+}$ without any approximation. He did also use elliptical coordinates, since Schrödinger’s equation yields then two separable equations. Computations were not easy with available tools, but Burrau concluded that the ground state would be situated at $E=-1.20E_o$ and $R=2.0a_o$. Grivet presented in 2002 a method for simplifying the numerical calculations [37] . It appeared then that the differential equation for ξ led everywhere to well-behaved solutions, but the other differential equation did only yield precise results for low and high values of η. The computed curves did even meet with an unphysical discontinuity for an intermediate value of η. Grivet proposed an interpolation procedure to overcome this difficulty. Actually, he discovered a much more profound problem.

Strand and Reinhardt [38] presented in 1979 a very thorough and detailed study of this issue. They thought that the semiclassical approximation of wave mechanics is more realistic than the assumption of sharply defined positions in old quantum mechanics. The WKB approximation of wave mechanics leads, indeed, to Newtonian laws of motion for any trajectory of an electron. Since the average position is then defined by a wave packet, it accounts for Heisenberg’s uncertainty relations. However, a wave packet can penetrate into the walls of a potential well. The separation of the turning points for the most probable position of the electron is then greater than when the electron is merely reflected by these walls. Strand and Reinhardt adopted therefore the quantization rule

${\displaystyle \oint p_i\text{d}{q}_i}=\left(n_i+{\alpha }_i\right)h$ (14)

It is a generalization of the first relation (2). For parabolic potential wells, ${\alpha }_i=1/2$. Rotation around the symmetry axis requires an angular variable φ, but ${\alpha }_i=0$ for this degree of freedom. Even these precautions did merely yield a classification of possible orbital motions. The expected 8-like orbit could be partially determined, but there appeared a “gap” in the region where the separation of the protons $R\approx 2a_o$. This is precisely the region where the ground state is situated, according to experimental results. That was a disappointing result and the cause of this failure was not explained.

About a dozen other authors tried thus to circumvent the enigmatic obstacle, but we mention only the enduring effort of Knudson. In 2006, he published a review for chemists [39] , where he mentioned that “the motivation for such studies is the insight into the quantum behavior offered by the classical viewpoint”. However, new computations could only confirm the existence of a “gap” in the critical region. It was even quite large ( $0.83a_o<R<2.30a_o$) for ${\text{H}}_2^{+}$. Knudson suggested that this defect might be due to the “inability to account for electron tunneling”. He reanalyzed this problem [40] by modifying (14) to get two quantum numbers ( $n_i$ and $m_j$). This increased the flexibility of the semiclassical approximation, but the “gap” was still there, even for HeH²⁺ and for LiH³⁺. Knudson returned to this nagging problem in 2022, as emeritus [41] . More computations were presented, but the gap did stubbornly resist. Knudson recognized then that this “disagreement… is not just quantitative, but qualitative in nature”. He concluded even that “old quantum mechanics does not work… for the ground state near its known equilibrium position”. However, he added that the concept of electron orbits is not totally useless. Orbital motions are more intuitive than probability distributions.

It is also noteworthy that the problem of dynamically stable closed orbits was already raised by Euler in 1760 for gravitational forces in classical mechanics. This apparently simple problem was treated by eminent scientists, to no avail. Wild [42] has shown that the differential equations of motion can be solved by means of personal computers. It is sufficient to proceed step by step for given initial conditions, but this applies to open orbits. It is even well-known that continued motions will become chaotic. Stability requires closed orbits. They were computed by Arensdorf for satellites turning around the Earth and the Moon [43] . They were not quantized and could have different sizes and display several lobes of variable shape.

The LCAO approximation of wave mechanics did solve Schrödinger’s equation for stationary states, but implied that the electron can be close to both protons with equal probabilities. The resulting probability distribution was static. However, Feynman proved that it can oscillate [44] . This requires to solve the time-dependent Schrödinger equation, which is possible and indicates that the electron is not simultaneously, but alternatively close to these protons. Feynmann called this a “flip-flop model”. It did innovate quantum-chemistry, but provides also a powerful justification of orbital motions. However, Pauli and Niessen, Strand and Reinhardt, as well as Knudson and other authors had not succeeded to determine the orbit when the electron does successively move around the protons.

There are several reasons. The use of elliptical coordinates suggested that separated differential equations do allow for different quantization rules. This led to imposing the first relation (2), even when the electron is attracted by two protons. The second relation (2) requires only one quantum number and is justified by de Broglie’s interpretation of Bohr’s quantization rule. It applied to circular orbits, but can be generalized for any closed orbit. The momentum vector p is everywhere tangential to this orbit and applies also to a progressively varying wavelength $\lambda =h/p$. It is thus sufficient that

${\displaystyle \oint p\text{d}s}=nh$ where ${\left(p\text{d}s\right)}^2={\left(\stackrel{\dot{}}{x}\text{d}x\right)}^2+{\left(\stackrel{\dot{}}{y}\text{d}y\right)}^2$

This applies to Cartesian coordinates and the chosen unit of the electron mass ( $m=1$). Since $\text{d}x=\stackrel{\dot{}}{x}\text{d}t$ and $\text{d}y=\stackrel{\dot{}}{y}\text{d}t$, the quantization condition is

${\displaystyle {\int }_0^{T_0}{V}_o^2\text{d}t}=nh$ where $V_o^2=V_x^2+V_y^2$(15)

$V_o$ is the speed of the electron and $T_o$ is the period for motions on the closed orbit. We might object that Newton’s laws of motion are not compatible with Heisenberg’s uncertainty relations. That is not true, since the electron can move on a well-defined orbit, though we ignore where it is situated at some particular instant. In other words, we have to abandon the idea that Newton’s laws of motion can only be solved by providing initial conditions. That is not necessary, even for a vibrating string, since it is possible to consider the general solution with adaptable parameters. Their values are then determined by boundary conditions and not by initial conditions.

It was also believed that the semiclassical approximation of wave mechanics has to be used with the quantization rule (14). That leads to complications, which are not necessary to account for Heisenberg’s uncertainty relations. According to classical mechanics, a harmonic oscillator can be at rest at its equilibrium position. That is excluded in wave mechanics, by means of a zero-point energy and ${\alpha }_i$ in (14). However, for orbital motions it is automatically excluded that an electron could be somewhere at rest. Condition (15) is sufficient.

In 2008, it was claimed [45] that old quantum mechanics was abandoned for “wrong reasons”. That is true, but not because of the proposed linear oscillations of the electron along the interproton axis. They are mathematically possible, but the turning points would be situated beyond the protons and thus require that the electron can pass through these protons. That is impossible. The electron has to turn around them, but that requires specific developments.

To determine the orbit of a single electron, moving around two protons in an arbitrarily chosen sagittal plane, we can use Cartesian coordinates ( $x,y$). The motionless protons are situated at ( $\pm c,0$) and attract the electron by Coulomb forces. The resulting potential energy of the electron is

$V\left(x,y\right)=-\frac{1}{d}-\frac{1}{D}$ where $d^2={\left(c-x\right)}^2+y^2$ and $D^2={\left(c+x\right)}^2+y^2$

Near the center, we get thus

$V\left(x,0\right)=-\frac{1}{c-x}-\frac{1}{c+x}\approx -\frac{2}{c^2}-\frac{2x^2}{c^4}$ and $V\left(0,y\right)\approx -\frac{2}{c^2}+\frac{2y^2}{c^4}$

At the center (0, 0), the potential is $-2/c^2$, but it is parabolically decreasing along the x-axis and parabolically increasing along the y-axis. The center is thus a saddle point. Oscillations on the y-axis are excluded, since they would be unstable. The coordinates x(t) and y(t) have thus to describe orbital motions. Newton’s laws of motion yield two coupled equations:

$F_x=\frac{c-x}{d^3}-\frac{c+x}{D^3}=\ddot{x}$ and $F_y=\frac{-y}{d^3}+\frac{-y}{D^3}=\ddot{y}$(16)

Since the functions x(t) and y(t) are periodic, they can be expressed by Fourier series. Closed orbits are thus Lissajous curves. Such an orbit can pass through the center, but near the center

$F_x=\frac{4x}{c^3}=\ddot{x}$ and $F_y=\frac{-2y}{c^3}=\ddot{y}$(17)

The first equation does not allow for a single sine function and it would be arbitrary to assume that this is possible for y(t). The simplest general solution of (16) is thus

$x\left(t\right)=a\left[\sin \left(\theta \right)-k\sin \left(3\theta \right)\right]$ and $y\left(t\right)=b\left[\sin \left(2\theta \right)-K\sin \left(4\theta \right)\right]$(18)

It is not necessary to assume that this motion results from initial condition, as for a projectile that would be thrown or shot from the center or some other point with a precisely computed velocity to get a closed orbit. According to (18), the electron is passing thought the origin when $\theta =0$, but we can set $\theta =\omega t+\varphi$, where $\varphi$ has any constant value. The period would always be $T_o=1/2\omega$. We consider thus motions that are characterized by adaptable parameters. Since (18) has to be compatible with (17), it is necessary that

$k=\frac{4+\mu }{3\left(4+3\mu \right)}$ and $K=\frac{1-2\mu }{2\left(1-8\mu \right)}$ where $\mu ={\omega }^2{c}^3$(19)

The parameter μ is a constant of motion, which indicates that smaller separations of the protons require higher angular frequencies. The value of k is always quite small and positive, while the value of K can diverge and change sign. The general solution (18) implies that x(t) will reach its maximal value $x_o=a\left(1+k\right)$ when $\theta =\pi /2$. Since (16) has to be satisfied at this turning point,

$\frac{1}{{\left(x_o-c\right)}^2}+\frac{1}{{\left(x_o+c\right)}^2}={\omega }^{2}a\left[1+9k\right]$(20)

The turning points for y(t) are reached when $V_y=0$ and thus already when $\theta =\tau$ and

$\cos \left(2\tau \right)-2K\cos \left(4\tau \right)=0$(21)

The values of the coordinates x and y are then

$X=a\left[\sin \left(\tau \right)-k\sin \left(3\tau \right)\right]$ and $Y=b\left[\sin \left(2\tau \right)-K\sin \left(4\tau \right)\right]$

The distances d and D are there defined by

$d^2={\left(c-X\right)}^2+Y^2$ and $D^2={\left(c+X\right)}^2+Y^2$

The equations of motion (16) require therefore that

$\frac{c-X}{d^3}-\frac{c+X}{D^3}=-a{\omega }^2\left[\sin \left(\tau \right)-9k\sin \left(3\tau \right)\right]$(22)

$\frac{Y}{d^3}+\frac{Y}{D^3}=4b{\omega }^2\left[\sin \left(2\tau \right)-4K\sin \left(4\tau \right)\right]$(23)

The actual orbit has also to be compatible with the quantization condition (15). It requires merely integration of squares and products of sine or cosine functions over one period. It is thus sufficient to consider average values. They yield the relation

${\omega }^2\left[a^2\left(1+9k^2\right)+4b^2\left(1+4K^2\right)\right]=n\hslash$(24)

Using units where $E_o=a_o=m=1$, it follows that ${\hslash }^2=e^2=2$. Since k and K are function of μ, there remain only 5 parameters ( $a,b,c,\mu ,\tau$). Their values are determined by Equations (20)-(24). It follows that

$a=1.026$ $b=0.031$ $c=0.622$ $\mu =0.133$ $\tau =0.407$ for $n=1$

$a=1.728$ $b=0.88$ $c=1.413$ $\mu =1.044$ $\tau =0.857$ for $n=2$

The resulting orbits are represented in Figure 6 . It is thus possible to determine electron orbits that are dynamically stable and quantized, but we get only a simple 8-like orbit for the exited state.

Figure 6(a) SQM determines the orbit of the electron for the ground state of hydrogen molecular ions. This orbit (red line) defines all possible positions of the moving electron and not where it is situated at a particular instant. (b) The first excited state allows for an 8-like orbit and the separation of the protons (represented by black dots) is greater. The scales are indicated by light gray circles of radius a_o. The speed of the electron (blue curves) varies more for the ground state than for the excited state.

The ground state requires thus a double 8-like orbit and that the electron does pass very close to the protons. It is then strongly accelerated but slows down when it approaches the potential barrier between the adjacent wells. The predicted separation $R=2c$ of the protons is then only $1.24a_o$ for the ground state and thus notably smaller than the measured value. The BM or SQM is much less useful for a single electron than for electron pairs. Nevertheless, it is possible to determine the orbit. The total energy E is then

$E=\frac{1}{2c}-\frac{1}{d}-\frac{1}{D}+\frac{1}{2}{V}_o^2=E\left(a,b,c,\mu ,\theta \right)$

It is not a constant of motion, but when all positions are equally probable on the computed orbits, the average value $\langle E\rangle =-2.53E_o$ for the ground state. This value is determined with respect to the energy level $E=0$ for complete dissociation, but is much lower than the experimentally measured value in Figure 4(b) . Because of Figure 1(b) , it is obvious that SQM yields a better approximation for an electron pair in the median plane than for a single electron that is alternatively turning around both protons. For the excited state, the separation of the protons would be $2.8a_o$ and the average energy $\langle E\rangle =0.06E_o$. The angular frequencies are respectively $\omega =0.743$ and $\omega =0.220$ for the ground state and the excited state. The essential result is, of course, that the electron orbit in ${\text{H}}_2^{+}$ could not be determined during more than 100 years, but is possible by means of conceptual clarifications.

We considered also motions that are symmetric with respect to the median plane, instead of being antisymmetric with respect to the center. They would be described by the functions

$x\left(t\right)=a\sin \left(\theta \right)$ and $y\left(t\right)=b\left[\cos \left(\theta \right)-\sigma \cos \left(3\theta \right)\right]$

The electron does then pass through the median plane when $y=b\left(1-\sigma \right)$. This could happen at the center when $\sigma =1$, but also without touching the interproton axis or by crossing it. We established the required equations, but found no acceptable solution for this kind of quantized motions. That is not required anymore. It is more useful to spare some time for basic considerations.