For a many-electron atom, the self-consistent-field (SCF) method is used to construct an approximate wave function as a Slater determinant of (one-electron) spin-orbitals. The one-electron spatial part of a spin-orbital is an atomic orbital (AO). We took each AO as a product of a spherical harmonic and a radial factor. As an initial approximation to the radial factors, we can use hydrogenlike radial functions with effective nuclear charges.
For many-electron molecules, which (unlike $\mathrm{H}_{2}^{+}$) cannot be solved exactly, we want to use many of the ideas of the SCF treatment of atoms. We shall write an approximate molecular electronic wave function as a Slater determinant of (one-electron) spin-orbitals. The one-electron spatial part of a molecular spin-orbital is a molecular orbital (MO).
Because of the Pauli principle, each MO can hold no more than two electrons, just as for AOs. What kind of functions do we use for the MOs? Ideally, the analytic form of each MO is found by an SCF calculation (Section 14.3). In this section, we seek simple approximations for the MOs that will enable us to gain some qualitative understanding of chemical bonding. Just as we took the angular part of each AO to be the same kind of function (a spherical harmonic) as in the one-electron hydrogenlike atom, we shall take the angular part of each diatomic MO to be $(2 \pi)^{-1 / 2} e^{i m \phi}$, as in $\mathrm{H}{2}^{+}$. However, the $\xi$ and $\eta$ factors in the $\mathrm{H}{2}^{+}$wave functions are complicated functions not readily usable in MO calculations. We therefore seek simpler functions that will provide reasonably accurate approximations to the $\mathrm{H}{2}^{+}$wave functions and that can be used to construct molecular orbitals for manyelectron diatomic molecules. With this discussion as motivation for looking at approximate solutions in a case where the Schrödinger equation is exactly solvable, we consider approximate treatments of $\mathrm{H}{2}^{+}$.
We shall use the variation method, writing down some function containing several parameters and varying them to minimize the variational integral. This will give an approximation to the ground-state wave function and an upper bound to the ground-state energy. By use of the factor $e^{i m \phi}$ in the trial function, we can get an upper bound to the energy of the lowest $\mathrm{H}_{2}^{+}$level for any given value of $m$ (see Section 8.2). By using linear variation functions, we can get approximations for excited states.
The $\mathrm{H}_{2}^{+}$ground state has $m=0$, and the wave function depends only on $\xi$ and $\eta$. We could try any well-behaved function of these coordinates as a trial variation function. We shall, however, use a more systematic approach based on the idea of a molecule as being formed from the interaction of atoms.
Consider what the $\mathrm{H}{2}^{+}$wave function would look like for large values of the internuclear separation $R$. When the electron is near nucleus $a$, nucleus $b$ is so far away that we essentially have a hydrogen atom with origin at $a$. Thus, when $r{a}$ is small, the groundstate $\mathrm{H}{2}^{+}$electronic wave function should resemble the ground-state hydrogen-atom wave function of Eq. (6.104). We have $Z=1$, and the Bohr radius $a{0}$ has the numerical value 1 in atomic units; hence (6.104) becomes
\(
\begin{equation}
\pi^{-1 / 2} e^{-r_{a}} \tag{13.39}
\end{equation}
\)
Similarly, we conclude that when the electron is near nucleus $b$, the $\mathrm{H}_{2}^{+}$ground-state wave function will be approximated by
\(
\begin{equation}
\pi^{-1 / 2} e^{-r_{b}} \tag{13.40}
\end{equation}
\)
This suggests that we try as a variation function
\(
\begin{equation}
c{1} \pi^{-1 / 2} e^{-r{a}}+c{2} \pi^{-1 / 2} e^{-r{b}} \tag{13.41}
\end{equation}
\)
where $c{1}$ and $c{2}$ are variational parameters. When the electron is near nucleus $a$, the variable $r{a}$ is small and $r{b}$ is large, and the first term in (13.41) predominates, giving a function resembling (13.39). The function (13.41) is a linear variation function, and we are led to solve a secular equation, which has the form (8.56), where the subscripts 1 and 2 refer to the functions (13.39) and (13.40).
We can also approach the problem using perturbation theory. We take the unperturbed system as the $\mathrm{H}{2}^{+}$molecule with $R=\infty$. For $R=\infty$, the electron can be bound to nucleus $a$ with wave function (13.39), or it can be bound to nucleus $b$ with wave function (13.40). In either case the energy is $-\frac{1}{2}$ hartree, and we have a doubly degenerate unperturbed energy level. Bringing the nuclei in from infinity gives rise to a perturbation that splits the doubly degenerate unperturbed level into two levels. This is illustrated by the $U(R)$ curves for the two lowest $\mathrm{H}{2}^{+}$electronic states, which both dissociate to a ground-state hydrogen atom (see Fig. 13.5). The correct zeroth-order wave functions for the perturbed levels are linear
combinations of the form (13.41), and we are led to a secular equation of the form (8.56), with $W$ replaced by $E^{(0)}+E^{(1)}$ (see Prob. 9.20).
Before solving (8.56), let us improve the trial function (13.41). Consider the limiting behavior of the $\mathrm{H}_{2}^{+}$ground-state electronic wave function as $R$ goes to zero. In this limit we get the $\mathrm{He}^{+}$ion, which has the ground-state wave function [put $Z=2$ in (6.104)]
\(
\begin{equation}
2^{3 / 2} \pi^{-1 / 2} e^{-2 r} \tag{13.42}
\end{equation}
\)
From Fig. 13.3 we see that as $R$ goes to zero, both $r{a}$ and $r{b}$ go to $r$. Hence as $R$ goes to zero, the trial function (13.41) goes to $\left(c{1}+c{2}\right) \pi^{-1 / 2} e^{-r}$. Comparing with (13.42), we see that our trial function has the wrong limiting behavior at $R=0$; it should go to $e^{-2 r}$, not $e^{-r}$. We can fix things by multiplying $r{a}$ and $r{b}$ in the exponentials by a variational parameter $k$, which will be some function of $R ; k=k(R)$. For the correct limiting behavior at $R=0$ and at $R=\infty$, we have $k(0)=2$ and $k(\infty)=1$ for the $\mathrm{H}_{2}^{+}$ground electronic state. Physically, $k$ is some sort of effective nuclear charge, which increases as the nuclei come together. We thus take the trial function as
\(
\begin{equation}
\phi=c{a} 1 s{a}+c{b} 1 s{b} \tag{13.43}
\end{equation}
\)
where the $c$ 's are variational parameters and
\(
\begin{equation}
1 s{a}=k^{3 / 2} \pi^{-1 / 2} e^{-k r{a}}, \quad 1 s{b}=k^{3 / 2} \pi^{-1 / 2} e^{-k r{b}} \tag{13.44}
\end{equation}
\)
The factor $k^{3 / 2}$ normalizes $1 s{a}$ and $1 s{b}$ [see Eq. (6.104)]. The molecular-orbital function (13.43) is a linear combination of atomic orbitals, an LCAO-MO. The trial function (13.43) was first used by Finkelstein and Horowitz in 1928.
For the function (13.43), the secular equation (8.56) is
\(
\left|\begin{array}{ll}
H{a a}-W S{a a} & H{a b}-W S{a b} \tag{13.45}\
H{b a}-W S{b a} & H{b b}-W S{b b}
\end{array}\right|=0
\)
The integrals $H{a a}$ and $H{b b}$ are
\(
\begin{equation}
H{a a}=\int 1 s{a}^{} \hat{H} 1 s{a} d v, \quad H{b b}=\int 1 s{b}^{*} \hat{H} 1 s{b} d v \tag{13.46}
\end{}
\)
where the $\mathrm{H}{2}^{+}$electronic Hamiltonian operator $\hat{H}$ is given by (13.32). We can relabel the variables in a definite integral without affecting its value. Changing $a$ to $b$ and $b$ to $a$ changes $1 s{a}$ to $1 s{b}$ but leaves $\hat{H}$ unaffected (this would not be true for a heteronuclear diatomic molecule). Hence $H{a a}=H_{b b}$. We have
\(
\begin{equation}
H{a b}=\int 1 s{a}^{} \hat{H} 1 s{b} d v, \quad H{b a}=\int 1 s{b}^{*} \hat{H} 1 s{a} d v \tag{13.47}
\end{}
\)
Since $\hat{H}$ is Hermitian and the functions in these integrals are real, we conclude that $H{a b}=H{b a}$. The integral $H{a b}$ is called a resonance (or bond) integral. Since $1 s{a}$ and $1 s_{b}$ are normalized and real, we have
\(
\begin{gather}
S{a a}=\int 1 s{a}^{} 1 s{a} d v=1=S{b b} \
S{a b}=\int 1 s{a}^{} 1 s{b} d v=S{b a} \tag{13.48}
\end{gather}
\)
The overlap integral $S_{a b}$ lies between 1 and 0 , and decreases as the distance between the two nuclei increases.
The secular equation (13.45) becomes
\(
\begin{gather}
\left|\begin{array}{cc}
H{a a}-W & H{a b}-S{a b} W \
H{a b}-S{a b} W & H{a a}-W
\end{array}\right|=0 \tag{13.49}\
H{a a}-W= \pm\left(H{a b}-S{a b} W\right) \tag{13.50}\
W{1}=\frac{H{a a}+H{a b}}{1+S{a b}}, \quad W{2}=\frac{H{a a}-H{a b}}{1-S_{a b}} \tag{13.51}
\end{gather}
\)
These two roots are upper bounds to the energies of the ground and first excited electronic states of $\mathrm{H}{2}^{+}$. We shall see that $H{a b}$ is negative, so $W_{1}$ is the lower-energy root.
We now find the coefficients in (13.43) for each of the roots of the secular equation. From Eq. (8.54), we have
\(
\begin{equation}
\left(H{a a}-W\right) c{a}+\left(H{a b}-S{a b} W\right) c_{b}=0 \tag{13.52}
\end{equation}
\)
Substituting in $W_{1}$ from (13.51) [or using (13.50)], we get
\(
\begin{gather}
c{a} / c{b}=1 \tag{13.53}\
\phi{1}=c{a}\left(1 s{a}+1 s{b}\right) \tag{13.54}
\end{gather}
\)
We fix $c_{a}$ by normalization:
\(
\begin{gather}
\left|c{a}\right|^{2} \int\left(1 s{a}^{2}+1 s{b}^{2}+2 \cdot 1 s{a} 1 s{b}\right) d v=1 \tag{13.55}\
\left|c{a}\right|=\frac{1}{\left(2+2 S_{a b}\right)^{1 / 2}} \tag{13.56}
\end{gather}
\)
The normalized trial function corresponding to the energy $W_{1}$ is therefore
\(
\begin{equation}
\phi{1}=\frac{1 s{a}+1 s{b}}{\sqrt{2}\left(1+S{a b}\right)^{1 / 2}} \tag{13.57}
\end{equation}
\)
For the root $W{2}$, we find $c{b}=-c_{a}$ and
\(
\begin{equation}
\phi{2}=\frac{1 s{a}-1 s{b}}{\sqrt{2}\left(1-S{a b}\right)^{1 / 2}} \tag{13.58}
\end{equation}
\)
Equations (13.57) and (13.58) come as no surprise. Since the nuclei are identical, we expect $|\phi|^{2}$ to remain unchanged on interchanging $a$ and $b$; in other words, we expect no polarity in the bond.
We now consider evaluation of the integrals $H{a a}, H{a b}$, and $S{a b}$. From (13.44) and (13.33), the integrand of $S{a b}$ is $1 s{a} 1 s{b}=k^{3} \pi^{-1} e^{-k\left(r{a}+r{b}\right)}=k^{3} \pi^{-1} e^{-k R \xi}$. The volume element in confocal elliptic coordinates is (Eyring, Walter, and Kimball, Appendix III)
\(
\begin{equation}
d v=\frac{1}{8} R^{3}\left(\xi^{2}-\eta^{2}\right) d \xi d \eta d \phi \tag{13.59}
\end{equation}
\)
Substitution of these expressions for $1 s{a} 1 s{b}$ and $d v$ into $S_{a b}$ and use of the Appendix integral (A.11) to do the $\xi$ integral gives (Prob. 13.16)
\(
\begin{equation}
S_{a b}=e^{-k R}\left(1+k R+\frac{1}{3} k^{2} R^{2}\right) \tag{13.60}
\end{equation}
\)
The evaluation of $H{a a}$ and $H{a b}$ is considered in Prob. 13.17. The results are
\(
\begin{equation}
H_{a a}=\frac{1}{2} k^{2}-k-R^{-1}+e^{-2 k R}\left(k+R^{-1}\right) \tag{13.61}
\end{equation}
\)
\(
\begin{equation}
H{a b}=-\frac{1}{2} k^{2} S{a b}-k(2-k)(1+k R) e^{-k R} \tag{13.62}
\end{equation}
\)
where $\hat{H}$ is given by (13.32) and so omits the internuclear repulsion.
Substituting the values for the integrals into (13.51), we get
$W{1,2}=-\frac{1}{2} k^{2}+\frac{k^{2}-k-R^{-1}+R^{-1}(1+k R) e^{-2 k R} \pm k(k-2)(1+k R) e^{-k R}}{1 \pm e^{-k R}\left(1+k R+k^{2} R^{2} / 3\right)}$
where the upper signs are for $W{1}$. Since $\hat{H}$ in (13.32) omits the internuclear repulsion $1 / R$, $W{1}$ and $W{2}$ are approximations to the purely electronic energy $E{\mathrm{el}}$, and $1 / R$ must be added to $W{1,2}$ to get $U_{1,2}(R)$ [Eq. (13.8)].
The final task is to vary the parameter $k$ at many fixed values of $R$ so as to minimize first $U{1}(R)$ and then $U{2}(R)$. This can be done numerically using a computer (Prob. 13.19) or analytically. The results are that, for the $1 s{a}+1 s{b}$ function (13.57), $k$ increases almost monotonically from 1 to 2 as $R$ decreases from $\infty$ to 0 ; for the $1 s{a}-1 s{b}$ function (13.58), $k$ decreases almost monotonically from 1 to 0.4 as $R$ decreases from $\infty$ to 0 . Since $0
We might ask why the variational parameter $k$ for the $\sigma{u}^{*} 1 s$ state goes to 0.4 , rather than to 2 , as $R$ goes to zero. The answer is that this state of $\mathrm{H}{2}^{+}$does not go to the ground state ( $1 s$ ) of $\mathrm{He}^{+}$as $R$ goes to zero. The $\sigma{u}^{*} 1 s$ state has odd parity and must correlate with an odd state of $\mathrm{He}^{+}$. The lowest odd states of $\mathrm{He}^{+}$are the $2 p$ states (Section 11.5); since the $\sigma{u}^{*} 1 s$ state has zero electronic orbital angular momentum along the internuclear $(z)$ axis, this state must go to an atomic $2 p$ state with $m=0$, that is, to the $2 p{0}=2 p{z}$ state.
Having found $k(R)$ for each root, one calculates $W{1}$ and $W{2}$ from (13.63) and adds $1 / R$ to get the $U(R)$ curves. The calculated ground-state $U(R)$ curve has a minimum at 2.00 bohrs (Prob. 13.20), in agreement with the true $R{e}$ value 2.00 bohrs, and has $U\left(R{e}\right)=-15.96 \mathrm{eV}$, giving a predicted $D{\mathrm{e}}$ of 2.36 eV , as compared with the true value 2.79 eV . (If we omit varying $k$ but simply set it equal to 1 , we get $R{e}=2.49$ bohrs and $D_{e}=1.76 \mathrm{eV}$.)
Now consider the appearance of the trial functions for the $\sigma{g} 1 s$ and $\sigma{u}^{*} 1 s$ states at intermediate values of $R$. Figure 13.6 shows the values of the functions $\left(1 s{a}\right)^{2}$ and $\left(1 s{b}\right)^{2}$ at points on the internuclear axis (see also Fig. 6.7). For the $\sigma{g} 1 s$ function $1 s{a}+1 s_{b}$, we get a buildup of electronic probability density between the nuclei, as shown in Fig. 13.7. It is especially significant that the buildup of charge between the nuclei is greater than that obtained by simply taking the sum of the separate atomic charge densities. The probability
FIGURE 13.6 Atomic probability densities for $\mathrm{H}_{2}^{+}$. Note the cusps at the nuclei.
FIGURE 13.7 Probability density along the internuclear axis for the LCAOMO function $N\left(1 s{a}+1 s{b}\right)$.
density for an electron in a $1 s{a}$ atomic orbital is $\left(1 s{a}\right)^{2}$. If we add the probability density for half an electron in a $1 s{a} \mathrm{AO}$ and half an electron in a $1 s{b} \mathrm{AO}$, we get
\(
\begin{equation}
\frac{1}{2}\left(1 s{a}^{2}+1 s{b}^{2}\right) \tag{13.64}
\end{equation}
\)
However, in quantum mechanics, we do not add the separate atomic probability densities. Instead, we add the wave functions, as in (13.57). The $\mathrm{H}_{2}^{+}$ground-state probability density is then
\(
\begin{equation}
\phi{1}^{2}=\frac{1}{2\left(1+S{a b}\right)}\left[1 s{a}^{2}+1 s{b}^{2}+2\left(1 s{a} 1 s{b}\right)\right] \tag{13.65}
\end{equation}
\)
The difference between (13.65) and (13.64) is
\(
\begin{equation}
\phi{1}^{2}-\frac{1}{2}\left(1 s{a}^{2}+1 s{b}^{2}\right)=\frac{1}{2\left(1+S{a b}\right)}\left[2\left(1 s{a} 1 s{b}\right)-S{a b}\left(1 s{a}^{2}+1 s_{b}^{2}\right)\right] \tag{13.66}
\end{equation}
\)
Putting $R=2.00$ and $k=1.24$ in Eq. (13.60), we find that $S{a b}=0.46$ at $R{e}$. (It might be thought that because of the orthogonality of different AOs, the overlap integral $S{a b}$ should be zero. However, the AOs $1 s{a}$ and $1 s_{b}$ are eigenfunctions of different Hamiltonian operators-one for a hydrogen atom at $a$ and one for a hydrogen atom at $b$. Hence the orthogonality theorem does not apply.)
Consider now the relative magnitudes of the two terms in brackets in (13.66) for points on the molecular axis. To the left of nucleus $a$, the function $1 s{b}$ is very small; to the right of nucleus $b$, the function $1 s{a}$ is very small. Hence outside the region between the nuclei, the product $1 s{a} 1 s{b}$ is small, and the second term in brackets in (13.66) is dominant. This gives a subtraction of electronic charge density outside the internuclear region, as compared with the sum of the densities of the individual atoms. Now consider the region between the nuclei. At the midpoint of the internuclear axis (and anywhere on the plane perpendicular to the axis and bisecting it), we have $1 s{a}=1 s{b}$, and the bracketed terms in (13.66) become $2\left(1 s{a}\right)^{2}-0.92\left(1 s{a}\right)^{2} \approx 1 s_{a}^{2}$, which is positive. We thus get a buildup of charge probability density between the nuclei in the molecule, as compared with the sum of the densities of the individual atoms. This buildup of electronic charge between the nuclei allows the electron to feel the attractions of both nuclei at the same time, which lowers its potential energy. The greater the overlap in the internuclear region between the atomic orbitals forming the bond, the greater the charge buildup in this region.
The preceding discussion seems to attribute the bonding in $\mathrm{H}{2}^{+}$mainly to the lowering in the average electronic potential energy that results from having the shared electron interact with two nuclei instead of one. This, however, is an incomplete picture. Calculations on $\mathrm{H}{2}^{+}$by Feinberg and Ruedenberg show that the decrease in electronic potential energy due to the sharing is of the same order of magnitude as the nuclear repulsion energy $1 / R$ and hence is insufficient by itself to give binding. Two other effects also contribute to the bonding. The increase in atomic orbital exponent $\left(k=1.24\right.$ at $R_{e}$ versus 1.0 at $\left.\infty\right)$ causes charge to accumulate near the nuclei (as well as in the internuclear region), and this further lowers the electronic potential energy. Moreover, the buildup of charge in the internuclear region makes $\partial \psi / \partial z$ zero at the midpoint of the molecular axis and small in the
\(
\frac{\left[1 s{a}(0,0, z)-1 s{b}(0,0, z)\right]^{2}}{2\left(1-S_{a b}\right)}
\)
region close to this point. Hence the $z$ component of the average electronic kinetic energy [which can be expressed as $\frac{1}{2} \int|\partial \psi / \partial z|^{2} d \tau$; Prob. 7.7b] is lowered as compared with the atomic $\left\langle T_{z}\right\rangle$. (However, the total average electronic kinetic energy is raised; see Section 14.5.) For details, see M. J. Feinberg and K. Ruedenberg, J. Chem. Phys., 54, 1495 (1971); M. P. Melrose et al., Theor. Chim. Acta, 88, 311 (1994); see also K. Ruedenberg and M. W. Schmidt, J. Phys. Chem. A, 113, 1954 (2009); J. Comput. Chem., 28, 391 (2007).
Bader, however, has criticized the views of Feinberg and Ruedenberg. Bader states (among other points) that $\mathrm{H}{2}^{+}$and $\mathrm{H}{2}$ are atypical and that, in contrast to the increase of charge density in the immediate vicinity of the nuclei in $\mathrm{H}{2}$ and $\mathrm{H}{2}^{+}$, molecule formation that involves atoms other than H is usually accompanied by a substantial reduction in charge density in the immediate vicinity of the nuclei. See R. F. W. Bader in The Force Concept in Chemistry, B. M. Deb, ed., Van Nostrand Reinhold, 1981, pp. 65-67, 71, 95-100, 113-115. Further study is needed before the origin of the covalent bond can be considered a settled question.
The $\sigma{u}^{*} 1 s$ trial function $1 s{a}-1 s{b}$ is proportional to $e^{-k r{a}}-e^{-k r{b}}$. On the plane perpendicular to the internuclear axis and midway between the nuclei, we have $r{a}=r{b}$, so this plane is a nodal plane for the $\sigma{u}^{} 1 s$ function. We do not get a buildup of charge between the nuclei for this state, and the $U(R)$ curve has no minimum. We say that the $\sigma{g} 1 s$ orbital is bonding and the $\sigma{u}^{} 1 s$ orbital is antibonding. (See Fig. 13.8.)
Reflection of the electron's coordinates in the $\sigma{h}$ symmetry plane perpendicular to the molecular axis and midway between the nuclei converts $r{a}$ to $r{b}$ and $r{b}$ to $r{a}$ and leaves $\phi$ unchanged [Eq. (13.79)]. The operator $\hat{O}{\sigma{h}}$ (Section 12.1) commutes with the electronic Hamiltonian (13.32) and with the parity (inversion) operator. Hence we can choose the $\mathrm{H}{2}^{+}$wave functions to be eigenfunctions of this reflection operator as well as of the parity operator. Since the square of this reflection operator is the unit operator, its eigenvalues must be +1 and -1 (Section 7.5). States of $\mathrm{H}_{2}^{+}$for which the wave function changes sign upon reflection in this plane (eigenvalue -1 ) are indicated by a star as a superscript to the letter that specifies $\lambda$. States whose wave functions are unchanged on reflection in this plane are left unstarred. Since orbitals with eigenvalue -1 for this reflection have a nodal plane between the nuclei, starred orbitals are antibonding.
Instead of using graphs, we can make contour diagrams of the orbitals (Section 6.7); see Fig. 13.9.
FIGURE 13.8 Probability density along the internuclear axis for the LCAO-MO function $N^{\prime}\left(1 s{a}-1 s{b}\right)$.
FIGURE 13.9 Contours of constant $|\psi|$ for the $\sigma{g} 1$ s and $\sigma{u}^{*} 1 s$ MOs. The threedimensional contour surfaces are generated by rotating these figures about the $z$ axis. Note the resemblance of the antibonding-MO contours to those of a $2 p_{z} \mathrm{AO}$.
Sometimes the binding in $\mathrm{H}{2}^{+}$is attributed to the resonance integral $H{a b}$, since in the approximate treatment we have given, it provides most of the binding energy. This viewpoint is misleading. In the exact treatment of Section 13.4, there arose no such resonance integral. The resonance integral simply arises out of the nature of the LCAO approximation we used.
In summary, we have formed the two $\mathrm{H}{2}^{+}$MOs (13.57) and (13.58), one bonding and one antibonding, from the $\mathrm{AOs} 1 s{a}$ and $1 s_{b}$. The MO energies are given by Eq. (13.51) as
\(
\begin{equation}
W{1,2}=H{a a} \pm \frac{H{a b}-H{a a} S{a b}}{1 \pm S{a b}} \tag{13.67}
\end{equation}
\)
where $H{a a}=\left\langle 1 s{a}\right| \hat{H}\left|1 s{a}\right\rangle$, with $\hat{H}$ being the purely electronic Hamiltonian of $\mathrm{H}{2}^{+}$. The integral $H{a a}$ would be the molecule's purely electronic energy if the electron's wave function in the molecule were $1 s{a}$. In a sense, $H{a a}$ is the energy of the $1 s{a}$ orbital in the molecule. In the limit $R=\infty, H{a a}$ becomes the $1 s \mathrm{AO}$ energy in the H atom. In the molecule, $H{a a}$ is substantially lower than the electronic energy of an H atom because the electron is attracted to both nuclei. A diagram of MO formation from AOs is given in Fig. 13.22. To get $U(R)$, the electronic energy including nuclear repulsion, we must add $1 / R$ to (13.67).
Problem 13.21 outlines the use of Mathcad to create an animation showing how contour plots of the $\mathrm{H}{2}^{+}$LCAO MOs $\phi{1}$ and $\phi_{2}$ change as $R$ changes.
We have described the lowest two $\mathrm{H}{2}^{+}$electronic states according to the state of the hydrogen atom obtained on dissociation. This is a separated-atoms description. Alternatively, we can use the state of the atom formed as the internuclear distance goes to zero. This is a united-atom description. We saw that for the two lowest electronic states of $\mathrm{H}{2}^{+}$ the united-atom states are the $1 s$ and $2 p{0}$ states of $\mathrm{He}^{+}$. The united-atom designation is put on the left of the symbol for $\lambda$. The $\sigma{g} 1 s$ state thus has the united-atom designation $1 s \sigma{g}$. The $\sigma{u}^{} 1 s$ state has the united-atom designation $2 p \sigma_{u}^{}$. It is not necessary to write this state as $2 p{0} \sigma{u}^{*}$, because the fact that it is a $\sigma$ state tells us that it correlates with the united-atom $2 p{0}$ state. For the united-atom description, the subscripts $g$ and $u$ are not needed, since molecular states correlating with $s, d, g, \ldots$ atomic states must be $g$, while states correlating with $p, f, h, \ldots$ atomic states must be $u$. From the separated-atoms states, we cannot tell whether the molecular wave function is $g$ or $u$. Thus from the $1 s$ separated-atoms state we formed both a $g$ and a $u$ function for $\mathrm{H}{2}^{+}$.
Before constructing approximate molecular orbitals for other $\mathrm{H}{2}^{+}$states, we consider how the trial function (13.57) can be improved. From the viewpoint of perturbation theory, (13.57) is the correct zeroth-order wave function. We know that the perturbation of molecule formation will mix in other hydrogen-atom states besides 1s. Dickinson in 1933 used a trial function with some $2 p{0}$ character mixed in (since the ground state of $\mathrm{H}{2}^{+}$is a $\sigma$ state, it would be wrong to mix in $2 p{ \pm 1}$ functions); he took
\(
\begin{equation}
\phi=\left[1 s{a}+c\left(2 p{0}\right){a}\right]+\left[1 s{b}+c\left(2 p{0}\right){b}\right] \tag{13.68}
\end{equation}
\)
where $c$ is a variational parameter and where (Table 6.2)
\(
1 s{a}=k^{3 / 2} \pi^{-1 / 2} e^{-k r{a}}, \quad\left(2 p{0}\right){a}=\left(2 p{z}\right){a}=\frac{\beta^{5 / 2}}{4(2 \pi)^{1 / 2}} r{a} e^{-\beta r{a} / 2} \cos \theta_{a}
\)
with $k$ and $\beta$ being two other variational parameters. We have similar expressions for $1 s{b}$ and $\left(2 p{0}\right){b}$. The angles $\theta{a}$ and $\theta{b}$ refer to two sets of spherical coordinates, one set at each nucleus; see Fig. 13.10. The definitions of $\theta{a}$ and $\theta_{b}$ correspond to using a right-handed coordinate system on atom $a$ and a left-handed system on atom $b$. The coefficient $c$ goes to zero as $R$ goes to either zero or infinity.
The mixing together of two or more AOs on the same atom is called hybridization. The function $1 s+c 2 p{0}$ is a hybridized atomic orbital. Since the $2 p{0}$ function is positive in one lobe and negative in the other, the inclusion of $2 p{0}$ produces additional charge buildup between the nuclei, giving a greater binding energy. The hybridization allows for the polarization of the $1 s{a}$ and $1 s{b}$ atomic orbitals that occurs on molecule formation. The function (13.68) gives a $U(R)$ curve with a minimum at 2.01 bohrs. At this distance, the parameters have the values $k=1.246, \beta=2.965$, and $c=0.138$ [F. Weinhold, J. Chem. Phys., 54, 530 (1971)]. The calculated $D{e}$ is 2.73 eV , close to the true value 2.79 eV .
The quantum mechanics and spectroscopy of $\mathrm{H}_{2}^{+}$are reviewed in C. A. Leach and R. E. Moss, Annu. Rev. Phys. Chem., 46, 55 (1995).
One final point. The approximate wave functions in this chapter are written in atomic units. When rewriting these functions in ordinary units, we must remember that wave functions are not dimensionless. A one-particle wave function $\psi$ has units of length ${ }^{-3 / 2}$ (Section 3.5). The AOs $1 s{a}$ and $1 s{b}$ that occur in the functions (13.57) and (13.58) are given by (13.44) in atomic units. In ordinary units, $1 s{a}=\left(k / a{0}\right)^{3 / 2} \pi^{-1 / 2} e^{-k r{a} / a{0}}$.