7-5 Linear Combination of Atomic Orbitals:
Chapter 7 The Variation Method plane, they do interact, H23 is not zero, and so all three basis functions show up in the lowest-energy solution. 1s and 3d0 are indirectly linked because they are each directly linked to 2p0. Physically, one can argue that a 2p0 AO that has been polarized by admixture with 3d0 can better polarize the 1s AO than can the pure 2p0 AO. (One could say that 3d0 is brought in “on the coat-tails” of 2p0.)
Note also that 1s has zero angular momentum along the z axis before the electric field is turned on. The z-directed field distorts the ground state, but has no effect on the z component of angular momentum. Therefore, only eigenfunctions having ml = 0 can contribute to the polarized ground state.
7-5 Linear Combination of Atomic Orbitals:The H+ Molecule–Ion
2
We are now ready to consider using the linear variation method on molecular systems.
We begin with the simplest case, H+. This molecule–ion has enough symmetry so that
2
we could guess many important features of the solution without calculation. However, to demonstrate the method, we shall first simply plunge ahead mathematically, and discuss symmetry later.
The H+ system consists of two protons separated by a variable distance R, and a
2
single electron (see Fig. 7-5). The hamiltonian for this molecule is, in atomic units
ˆ
∇2
∇2
1
1
1
H (r
A
B
A, rB , r1) = − 1 ∇2 +
+
−
−
+
(7-74)
2
1
1836
1836
rA1
rB1
R
Since all the particles in the system are capable of motion, the exact eigenfunction of ˆ H will be a function of the coordinates of the electron and the protons. However, the protons are each 1836 times as heavy as the electron, and in states of chemical interest their velocity is much smaller than that of the electron. This means that, to a very good degree of approximation, the electron can respond instantly to changes in internuclear separation. In other words, whenever the nuclei are separated by a given distance R, no matter how they got there, the motion of the electron will always be described in the same way by ψ. This means that we can separate the electronic and nuclear motions with little loss of accuracy.3 Given an internuclear separation, we can solve for the electronic wavefunction by ignoring nuclear motion [i.e., omitting ∇2 and
A Figure 7-5 The H+ molecule–ion. A and B are protons. The electron is numbered “1.”
2
3However, there are times when coupling of electronic and nuclear motions becomes important.
Section 7-5 Linear Combination of Atomic Orbitals: The H+ Molecule–Ion2
∇2 in Eq. (7-74)]. This gives us a hamiltonian for the electronic energy and nuclear
B repulsion energy of the system:
ˆ
1
1
1
H (rA, rB , r1) = − 1 ∇2 −
−
+
(7-75)
2 1
rA1
rB1
R
For a given internuclear separation R, the internuclear repulsion 1/R is a constant, and we can omit it and merely add it on again after we have found the electronic energy. If we let ˆ Helec stand for the first three terms on the right-hand side of Eq. (7-75), we can write
ˆ
Helecψelec (r1) = Eelecψelec (r1)
(7-76)
and Eelec + Enuc rep = Eelec + 1/R
(7-77)
Solving Eq. (7-76) for Eelec for every value of R allows us to plot the electronic and also the total energy of the system as a function of R. But Eelec(R) + 1/R is just the potential energy for nuclear motion. Therefore, this quantity can be inserted as the potential in the hamiltonian operator for nuclear motion:
ˆ
∇2
∇2
H
A
B
nuc(rA, rB ) = − 1
+
+ Eelec(R) + 1/R
(7-78)
2
1836
1836
ˆ
Hnuc(rA, rB )ψnuc(rA, rB ) = Enucψnuc (rA, rB)
(7-79)
The eigenfunctions of Eq. (7-79) describe the translational, vibrational, and rotational states of the molecule. Note that the eigenvalues of the hamiltonian for nuclear motion are total energies for the system because they contain the electronic energy in their potential parts. Hence, Etot = Enuc
(7-80)
But ψtot(rA, rB , r1) = ψelec(r1)ψnuc (rA, rB)
(7-81)
This approximation—that the electronic wavefunction depends only on the positions of nuclei and not on their momenta—is called the Born–Oppenheimer approximation.4 Only to the extent that this approximation holds true is it valid, for example, to separate electronic and vibrational wavefunctions and treat various vibrational states as a subset existing in conjunction with a given electronic state. We will assume the Born–Oppenheimer approximation to be valid in all cases treated in this book.
Making the Born–Oppenheimer approximation for H+, we seek to solve for the
2
electronic eigenfunctions and eigenvalues with the nuclei fixed at various separation distances R. We already know these solutions for the two extremes of R. When the two 4It is analogous to the concept of reversibility in thermodynamics: The piston moves so slowly in the cylinder that the gas can always maintain equilibrium, so pressure depends only on piston position; the nuclei move so slowly in a molecule that the electrons can always maintain their optimum motion at each R, so electronic energy depends only on nuclear position.
Chapter 7 The Variation Method nuclei are widely separated, the lowest-energy state is a 1s hydrogen atom and a distant proton. (Since there is a choice about which nucleus is “the distant proton,” there are really two degenerate lowest-energy states.) When R = 0, the system becomes He+
These two extremes are commonly referred to as the separated-atom and united-atom
limits, respectively.
In carrying out a linear variation calculation on H+, our first problem is choice of
2
basis. In the separated-atom limit, the appropriate basis for the ground state would be a 1s atomic orbital (AO) on each proton. Then, regardless of which nucleus the electron resided at, the basis could describe the wavefunction correctly. The appropriate basis at the united-atom limit is a hydrogenlike 1s wavefunction with Z = 2. At intermediate values of R, choice of an appropriate basis is less obvious. One possible choice is a large number of hydrogenlike orbitals or, alternatively, Stater-type orbitals (STOs), all centered at the molecular midpoint. Such a basis is capable of approximating the exact wavefunction to a high degree of accuracy, provided a sufficiently large number of basis functions is used.5 Calculations using such a basis are called single-center expansions. A different basis, and one that is much more popular among chemists, is the separated-atom basis—a 1s hydrogen AO centered on each nucleus. At finite values of R, this basis can produce only an approximation to the true wavefunction.
One way to improve this approximation is to allow additional AOs on each nucleus, 2s, 2p, 3s, etc., thereby increasing the mathematical flexibility of the basis. If we restrict our basis to AOs that are occupied in the separated-atom limit ground state (the 1s AOs in this case), then we are performing what is called a minimal basis set calculation. For now, we will use a minimal basis set.
The wavefunction that we produce by linear variation will extend over the whole H+ molecule, and its square will tell us how the electron density is distributed in the
2
molecule. Hence, the one-electron molecular wavefunction is referred to as a molecular
orbital (MO) just as the one-electron atomic wavefunction is referred to as an atomic
orbital (AO). With a basis set of the type we have selected, the MOs are expressed as linear combinations of AOs. For this reason, this kind of calculation is referred to as a minimal basis set linear combination of atomic orbitals–molecular orbital (LCAO– MO) calculation.
Our second problem, now that we have selected a basis, is construction of the secular determinant. Since we have only two basis functions (1sA, 1sB), we expect a 2 × 2 determinant:
H
AA − ¯ESAA HAB − ¯ESAB
HBA − ¯ ESBA HBB − ¯ ESBB = 0
(7-82)
Here we have used the notation developed earlier, where HAB = 1s∗A (1) ˆ Helec (1) 1sB (1) dτ (1)
(7-83) etc. ¯
E of Eq. (7-82) is the average value of the energy. Henceforth, the bar is omitted.
If we take our basis functions to be normalized, SAA = SBB = 1. Since our basis functions and hamiltonian are all real, their integrals will be real. Therefore, SAB = SBA 5The hydrogenlike orbitals are a complete set if the continuum functions are included. Hence, this set can allow one to approach arbitrarily close to the exact eigenfunction and eigenvalue. The Slater-type orbitals do not constitute a complete set.
Section 7-5 Linear Combination of Atomic Orbitals: The H+ Molecule–Ion2
and HAB = H ∗ = H
BA
BA. Since the hamiltonian is invariant to an interchange of the labels A and B, it follows that HAA = HBB. (HAA is the energy of an electron when it is in a 1s AO on one side of the molecule, HBB when it is on the other side.) This leaves us with
H AA − E HAB − ESAB =
0
(7-84)
HAB − ESAB HAA − E For Eq. (7-84) to be satisfied, the product of the diagonal terms must equal that of the off-diagonal terms, which means that HAA − E = ±(HAB − ESAB)
(7-85)
This gives two values for E: E± = HAA ± HAB
(7-86)
1 ± SAB
To arrive at numerical values for E+ and E− requires that we choose a value for R and explicitly evaluate HAA, HAB, and SAB. We know in advance that SAB increases monotonically from zero at R = ∞ to unity at R = 0 because 1sA and 1sB are each normalized and everywhere positive. HAA is the average energy of an electron in a 1s AO on nucleus A, subject also to an attraction by nucleus B. Hence, HAA should be lower than the energy of the isolated H atom (− 1 a.u.) whenever R is finite. H
2
AB is easily expanded to
HAB = 1sA − 1 ∇2 − 1/rB 1sBdv + 1sA(−1/rA)1sB dv
(7-87)
2
The operator in the first integrand is simply the hamiltonian operator for a hydrogen atom centered at nucleus B. This operator operates on 1sB to give − 1 1s
2
B . Hence, the first integral becomes simply − 1 S 2 AB . The second integral in Eq. (7-87) gives the attraction between a nucleus and the “overlap charge.” Thus, HAB is zero at R = ∞ and negative for finite R. The formulas for these terms are (after nontrivial mathematical evaluation)
SAB = 1sA1sB dv = exp(−R) 1 + R + R2/3
(7-88)
H
ˆ
AA = 1sAHelec1sA dv = − 1 − (1/R) 1 − e−2R (1 + R)
(7-89)
2
H
ˆ
AB = 1sAHelec1sB dv = −SAB/2 − e−R (1 + R)
(7-90)
When R = 2 a.u., SAB = 0.586, HAA = −0.972 a.u. and HAB = −0.699 a.u. Inserting these values into Eq. (7-86) gives E+ = −1.054 a.u. and E− = −0.661 a.u. These are electronic energies. Internuclear repulsion energy (+ 1 a.u.) can be added to these
2
values to give −0.554 a.u. and −0.161 a.u., respectively.
The ways in which HAA, HAB, SAB, and 1/R contribute to the energy are illustrated for R = 2 a.u. in Fig. 7-6. HAA is lower than the energy of an isolated H atom because
Chapter 7 The Variation MethodFigure 7-6 Contributions to the energy of H+ at R = 2 a.u. in a minimal basis LCAO–MO cal 2
culation.
the electron experiences additional nuclear attraction at R = 2. The effect of the HAB interaction element is to split the energy into two levels equally spaced above and below HAA. The SAB term has the effect of partially negating this splitting. The internuclear repulsion energy 1/R merely raises each level by 1 a.u. The lower energy, E
2
+ + 1/R, has a final value that is lower than the separated-atom energy of − 1 a.u. Since the exact
2
energy at R = 2 must be as low or lower than our value of −0.554 a.u., we can conclude that the H+ molecule–ion has a state that is stable, with respect to dissociation into
2
H + H +, by at least 0.054 a.u., or 1.47 eV, or 33.9 kcal/mole, neglecting vibrational energy effects.
The data depicted in Fig. 7-6 are sometimes presented in the abbreviated form of Fig. 7-7. The energy levels for the pertinent AOs of the separated atoms are indicated on the left and right, and the final energies (either electronic or electronic plus internuclear) are shown in the center.
Section 7-5 Linear Combination of Atomic Orbitals: The H+ Molecule–Ion2
Figure 7-7 Separated atom energies and energies at an intermediate R for H+.
2
Figure 7-8 E± + 1/R versus R for H+. (—) Calculation described in text. (- - -) Exact
2
calculation.
The behavior of these energies as a function of R is plotted in Fig. 7-8. Included for comparison are the exact energies for the two lowest-energy states of H+. Only
2
the lower of these shows stability with respect to molecular dissociation. Both energy levels approach infinity asymptotically as R approaches zero because of internuclear repulsion. (The zero of energy corresponds to complete separation of the protons and electron.)
Having found the roots E± for the secular determinant, we can now solve for the coefficients which describe the approximate wavefunctions in terms of our basis set.
Let us first find the approximate wavefunction corresponding to the lower energy. To do this, we substitute the expression for E+ [Eq. (7-86)] into the simultaneous equations associated with the secular determinant (7-84): cA(HAA − E+) + cB(HAB − E+SAB) = 0
(7-91)
cA(HAB − E+SAB) + cB(HAA − E+) = 0
(7-92)
Chapter 7 The Variation Method Equation (7-91) leads to
cA HAA − (HAA + HAB)/ 1 + S = −c H
1 + S AB
B
AB − (HAA + HAB )SAB /
AB
(7-93)
which ultimately gives cA = cB
(7-94)
The same procedure for E− produces the result cA = −cB
(7-95)
The normality requirement is
(7-96)
so that c2 +
+
A
c2B
2cAcBSAB = 1
(7-97)
For cA = cB, this gives cA = 1/ [2 (1 + SAB)]1/2 = cB
(7-98)
For cA = −cB, cA = 1/ [2 (1 − SAB)]1/2 = −cB
(7-99)
Therefore, the LCAO–MO wavefunction corresponding to the lower energy E+ is
ψ+ =
1
√
(1sA + 1sB)
(7-100) 2(1 + SAB)
The higher-energy solution is
ψ− =
1
√
(1sA − 1sB)
(7-101) 2(1 − SAB) Just as was true for AOs, there are a number of ways to display these MOs pictorially.
One possibility is to plot the value of ψ± or ψ2± along a ray passing through both nuclei.
(Other rays could also be chosen if we were especially interested in other regions.)
Another approach is to plot contours of ψ or ψ2 on a plane containing the internuclear axis, Still another way is to sketch a three dimensional view of a surface of constant value of ψ or ψ2 containing about 90–95% of the wavefunction or the electron charge.
Finally, one can plot the value of ψ as distance above or below the plane containing the internuclear axis. All of these schemes are shown in Fig. 7-9 for ψ+ and ψ−.
Section 7-5 Linear Combination of Atomic Orbitals: The H+ Molecule–Ion2
Figure 7-9 (a) Plot of ψ+, along the z axis. [Eq. (7-100)]. (b) Plot of ψ− along the z axis [Eq. (7-101)].
The wavefunctions ψ+ and ψ− may be seen from Fig. 7-9 to be, respectively, sym metric and antisymmetric for inversion through the molecular midpoint. [They are commonly called gerade (German for even) and ungerade, respectively.] This would be expected for nondegenerate eigenfunctions of the hamiltonian, since it is invariant to inversion. But ψ+ and ψ− are not eigenfunctions. They are only approximations to eigenfunctions. How is it that they show the proper symmetry characteristics of eigenfunctions? The reason is that the symmetry of the H+ molecule is manifested as
2
a symmetry in our secular determinant of Eq. (7-84). Note that the determinant is symmetric for reflection across either diagonal. The symmetry for reflection through the principal diagonal (which runs from upper left to lower right) is due to the hermiticity of ˆ H , and is always present in the secular determinant for any molecule regardless of symmetry.6 Symmetry for reflection through the other diagonal is due to the fact that the hamiltonian is invariant to inversion and also to the fact that the basis functions at 6Hermiticity requires that H
∗
ij = Hj i . If Hij is imaginary, the determinant will be antisymmetric for reflection through the principal diagonal.
Chapter 7 The Variation MethodFigure 7-9 (Continued)
(c) Contour diagram of ψ+. (d) Contour diagram of ψ−.
the two ends of the molecule are identical. When the AOs in a basis set are interchanged (times ±1) by a symmetry operation of the molecule, the basis set is said to be balanced
for that operation. Thus, a 1sA and 1sB basis is balanced for inversion in H+, but a
2
1sA and 2sB basis is not. Whenever a symmetry-balanced basis is used for a molecule, the symmetry of the molecule is manifested in the secular determinant and ultimately leads to approximate MOs that show the proper symmetry characteristics.
Since the eigenfunctions for H+ must be gerade or ungerade, and since we started
2
with the simple balanced basis 1sA and 1sB, it should be evident that it is unnecessary to go through the linear variation procedure for this case. With such a simple basis set, there is only one possible gerade linear combination of AOs, namely 1sA + 1sB.
Similarly, 1sA − 1sB is the only possible ungerade combination. Therefore, we could have used symmetry to guess our solutions at the outset. Usually, however, we are not so limited in our basis. We shall see that, while symmetry is useful in such circumstances, it does not suffice to produce the variationally best solution.
Section 7-5 Linear Combination of Atomic Orbitals: The H+ Molecule–Ion2
Figure 7-9 (Continued)
(e) Three-dimensional sketch of contour envelope for ψ+ and (f) for ψ−.
According to the theorem mentioned earlier (but not proved), the nth lowest root of a linear variation calculation for a state function must lie above the nth lowest exact eigenvalue for the system. However, the two states we are dealing with have different symmetries. In such a case, a more powerful boundedness theorem holds— one that holds even if we are not using a linear variation procedure. To prove this, we first recognize that every H+ eigenfunction is either gerade (i.e., symmetric for
2
inversion) or ungerade. Since the lowest-energy approximate wavefunction ψ+ is gerade, it must be expressible as a linear combination of these gerade eigenfunctions.
Hence, its average energy E+ cannot be lower than the lowest eigenvalue for the gerade eigenfunctions. Similarly, E− cannot be lower than the lowest eigenvalue for the ungerade eigenfunctions, and so we have a separate lower bound for the average energy of trial functions of each symmetry type. For this reason, our approximate energies in Fig. 7-8 must lie above the exact energies for both states. If we were to make further efforts to lower the average energy of the ungerade function, even by going outside the linear variation procedure, we could never fall below the exact energy for the lowest-energy ungerade state unless we somehow allowed our trial function to change symmetry. This means that a lowest average energy criterion can be used in attempting to find the lowest-energy state of each symmetry type for a system, by either linear or nonlinear variation methods.
Chapter 7 The Variation MethodFigure 7-9 (Continued)
(g) Value of ψ+ and (h) of ψ− versus position in plane containing the nuclei at points a and b.
EXAMPLE 7-2 Suppose one were to do a variational calculation on the hydrogen atom using the unnormalized trial wavefunction φ = exp(−αr2) cos θ, with α being varied. What lower bound could we expect for the average energy?
SOLUTION This function has a node in the x, y plane (where θ = π/2), and this nodal plane exists regardless of the value of α. Therefore, our trial function is like the 2p0 AO in symmetry. It will be represented by a linear combination of p0 AOs–a combination that changes as α is varied.
The average value of energy cannot be lower than the lowest eigenvalue in the set, which is −1/8 a.u.
(for the 2p0 AO).
Inspection of Fig. 7-9 shows that the charge distribution for the state described by ψ+ is augmented at the molecular midpoint compared to the charge due to unperturbed atoms. This state is also the one that gives H+ stability at finite R. Because this MO
2
puts electronic charge into the bond region and stabilizes the molecule, it is commonly called a bonding MO. In the state described by ψ−, charge is shifted out of the bond region and the molecule is unstable at finite R, and so ψ− is called an antibonding MO.
Because the potential in H+ (or in any linear molecule) is independent of φ, the
2
angle about the internuclear axis, the φ dependence of the wavefunctions is always of the form
√
(φ) = (1/ 2π) exp(imφ), m = 0, ±1, ±2, . . .
(7-102)
This fact may be arrived at in two ways. One way is to write down the Schr¨odinger equation for H+ using spherical polar coordinates or elliptical coordinates. (φ is a
2
coordinate in each of these coordinate systems.) Then one attempts to separate coordinates and finds that the φ coordinate is indeed separable from the others and yields the equation d2 (φ) =−m2(φ)
(7-103)
dφ2
The acceptable solutions of this are the functions (7-102). The other approach is to note that, since the potential in ˆ H has no φ dependence, ˆ H commutes with ˆ Lz, the angular momentum operator, so that the eigenfunctions of ˆ H are simultaneously eigenfunctions of ˆ Lz. We know the eigenfunctions of ˆ Lz are the functions (7-102), and thus, we know that these functions must also give the φ dependence of the eigenfunctions of ˆ H .
Two important conclusions emerge. First, each nondegenerate H+ wavefunction
2
must have a φ dependence given by one of the functions (7-102). This tells us something about the shapes of the wavefunctions. Second, the nondegenerate H+ wavefunctions
2
are eigenfunctions of ˆ Lz, which means that an electron in any one of these states has a definite, unvarying (i.e., sharp) component of angular momentum of value m ¯h mks units (m a.u.) along the internuclear axis. It is thus useful to know the m value associated with a given one-electron wavefunction. A standard notation is used, which is analogous to the atomic orbital notation wherein s, p, d, f, correspond to l values of 0, 1, 2, 3, respectively. The corresponding Greek lower-case letters σ , π , δ, φ indicate values of |m| of 0, 1, 2, 3, respectively, in one-electron orbitals of linear molecules.
Thus, for the case at hand, ψ+ and ψ− are both σ MOs because they are cylindrically symmetrical (i.e., no φ dependence), which requires that m be zero. Because ψ+ is a gerade function, it is symbolized σg. ψ−, then, is a σu MO. (A simple way to determine whether an MO is σ , π , or δ is to imagine viewing it (in its real form) end-on (i.e., along a projection of the internuclear axis.) If the MO from this view looks like an s AO, it is a σ MO. If it looks like a p AO, it is a π MO. A δ MO has the four-leaf-clover appearance of a d AO.)
Let us now examine the dependence of our LCAO–MO results on our original choice of basis. The 1s AOs we have used are capable of giving the exact energy when R = ∞, but become increasingly inadequate as R decreases. As a result, Fig. 7-8 shows that, for both states, the approximate energy deviates more and more from the exact energy as R decreases. At R = 0, our σg function becomes a single 1s H atom (Z = 1) AO centered on a doubly positive nucleus. Yet we know that the lowestenergy eigenfunction for that situation is a single 1s He+(Z = 2) AO. An obvious way to improve our wavefunction, then, is to allow the 1s basis functions to change
Chapter 7 The Variation Method their orbital exponents as R changes. This adds a nonlinear variation, and the calculation is more complicated. It is very easily performed with the aid of a computer, however, and we summarize the results in Figs. 7-10 and 7-11. The internuclear repulsion has been omitted from the energies in Fig. 7-10. The lowest approximate energy curve is now in perfect agreement with the exact electronic energy both at R = 0 and R = ∞, and shows improved, though still not perfect, agreement at intermediate R.
The R dependence of the orbital exponent for this wavefunction (Fig. 7-11) varies smoothly from 1 at R = ∞ to 2 at R = 0, as expected. In contrast, the σu function fails to reach a well-defined energy at R = 0 because the function becomes indeterminate at that point. [At R = 0, 1sA − 1sB becomes (1sA − 1sA) = 0.] However, Figs. 7-10 and 7-11 indicate that the exact energy E− is − 1 a.u. at R = 0, corre 2
sponding to the n = 2 level of He+, and that the orbital exponent in our 1s basis functions approaches 0.4 in the effort to approximate this state function at small R.
To understand this behavior, we must once again consider the symmetries of these functions.
We have already seen that the nondegenerate eigenfunctions of H+ must be either
2
gerade or ungerade, and we note in Fig. 7-10 that the energy curve for the gerade state is continuous as is the one for the ungerade state. In other words, as we move along a given curve, we are always referring to a wavefunction of the same symmetry. This continuity of symmetry along an energy curve is central to many applications of quantum chemistry. The reason for continuity of symmetry can be seen by considering a molecule having some element of symmetry, and having a nondegenerate wavefunction or molecular orbital (which must be symmetric or antisymmetric with respect to the symmetry operations of the molecule). If we change the molecule infinitesimally (without destroying its symmetry), we expect the wavefunction to change infinitesimally also.
Figure 7-10 E+ and E− for H+ from (- - -) exact, (· · · ) variable ζ , and (—) fixed ζ (ζ = 1)
2
treatments.
Section 7-5 Linear Combination of Atomic Orbitals: The H+ Molecule–Ion2
Figure 7-11 Values of ζ minimizing E+ and E− as a function of R.
In particular it should not change symmetry because this is generally not an infinitesimal change. (To change symmetry requires adding or removing nodes, changing signs in parts of the function. Such changes have more than infinitesimal effects on ψ and on kinetic and potential parts of the energy.) The entire curve can be traversed by an infinite number of such infinitesimal but symmetry conserving steps.
The continuity of symmetry enables the σg state of H+ to correlate with an s-type
2
AO of He+ as R goes to zero. This correlation is symmetry allowed because the s-type AOs of He+ have the proper symmetry characteristics—they are gerade and have no dependence on angle about the axis that is the internuclear axis when R > 0 (see Fig. 7-12). In contrast, the σu MO cannot correlate with an s-type AO. It must correlate with an AO that is cylindrically symmetrical about the old internuclear axis but is antisymmetric for inversion. A p-type AO pointing along the old internuclear axis (called a pσ AO) satisfies these requirements.
Figure 7-12
Sketches demonstrating how separated atom functions can be related to united-atom functions through symmetry invariance. These functions are not drawn to a common scale.