Section 12-6
Section 12-6 Perturbation Theory for a Degenerate State
SOLUTION The HOMO-LUMO contribution to the charge shift favors atom 4, since its coefficients in these MOs are larger than those for atoms 2 or 3. However, the actual data (not given in Appendix 6) are: π1,2 = 0.21342, π1,3 = −0.01771, π1,4 = 0.13937. So atom 2 experiences the greatest population change, atom 3 the least. Why is this? It results from the fact that we are summing products that can be positive or negative. Two atoms near each other are less likely to have nodal planes separating them, so their coefficients are more likely to be of the same sign giving fewer sign changes in their products. So the sum has less tendency to cancel itself. This is the way in which MO theory predicts a tendency for polarization effects to die off as distance increases—the well-known “inductive effect.” This dying-off is not monotonic, as can be seen in the remaining atom-atom polarizabilities: π1,5 = 0.02325, π1,6 = −0.00643, π1,7 = 0.03228, π1,8 = −0.02674, π1,9 = 0.08889, π1,10 = −0.00356.
In general, πlk can have either sign. This means that making atom k more attractive for electrons will, to first order, cause the π electron density to decrease at some carbons and increase at others. (This is necessary since the total π-charge density must be conserved.)
12-6 Perturbation Theory for a Degenerate State Equations (12-19), (12-21), and (12-24) have denominators containing Ei − Ej . If ψi and ψj are degenerate, this leads to difficulty, alerting us to the fact that our earlier derivations do not take proper account of states of interest that are degenerate. The problem results from our initial expansion of φ˙ as a power series in λ. The leading
l
term in the expansion, φ(0), is the wavefunction that φ
i
i becomes in the limit when λ = 0. In nondegenerate systems there is no is ambiguity; φ(0) has to be ψ i
i . But if ψi is degenerate with ψj , any linear combination of them is also an eigenfunction. We need a method to determine how ψi and ψj should be mixed together to form the correct zeroth-order functions φ(0) and φ(0).
i
j
To find the conditions that φ(0) and φ(0) must satisfy, we return to the first-order i
j perturbation equation (12-11) but with φ(0) in place of ψ
i i : (H − W (1))φ(0) + (H = 0
(12-73)
i
i 0 − Ei )φ(1)
i
Multiplying from the left by φ(0)∗ and integrating gives (taking φ(0) and φ(0) to be
j
i
j
orthogonal and H0 hermitian) φ(0)|H|φ(0) + (E |φ(1) = 0
(12-74)
j
i
j − Ei )φ(0)
j
i
If Ei = Ej , this gives φ(0)|H |φ(0) = 0
(12-75)
j
i
This means that the first-order perturbation equation is satisfied for degenerate states only when the wavefunctions “diagonalize” the perturbation operator H . Therefore, our problem reduces to finding those linear combinations of ψi and ψj that will diagonalize H . We have already seen two ways to accomplish this in earlier chapters. One way is to construct the matrix H, where (H)ij = ψi|H |ψj , and then find the unitary matrix C that diagonalizes H in the similarity transformation C†HC. The elements in C are then the coefficients for the linear combinations of ψi and ψj , and the diagonal they are the first-order corrections to the energy of φ(0) and φ(0). The other method,
i
j
equivalent to the above, is to set up the determinantal equation
H − E
H
11
12
H H − E = 0
(12-76)
21
22
and solve for the roots E. Substituting these back into the simultaneous equations corresponding to the determinant gives the coefficients with which ψi and ψj should be combined. The roots E are the first-order corrections to the energy and are identical to the diagonal matrix elements mentioned above.
In the event that the roots are equal, the perturbation has failed to split the degeneracy to first order and it is then necessary to proceed to higher order. We will not pursue this problem to higher orders, however.
If there are n degenerate states, the above procedure is simply carried out with an n × n matrix or determinant.
We will now give an example of perturbation theory for a degenerate state.
12-7 Polarizability of the HydrogenAtom in the n = 2 States In Chapter 7, the polarizability of the hydrogen atom in the 1s state was calculated by the variation method (see Section 7-4 and Problem 7-13). We now use perturbation theory for degenerate states to calculate to first order the polarizabilities of the n = 2 states.
The perturbation due to a z-directed uniform electric field is, in atomic units, H = −F r cos θ
(12-77)
There are four degenerate states at the n = 2 level, giving us a 4 × 4 secular determinant.
If we choose the real AOs as our basis set and let s, x, y, z represent these functions, the determinantal equation is
H − E
H
ss
sz
H sy
H sx
H − E
H
zs
H zz
zy
H zx
=
0
(12-78)
H
−
ys
H yz
H yy
E
H yx
H
−
xs
H xz
H xy
H xx
E
where
H =
yz
y|H |z, etc.
(12-79)
The operator H is antisymmetric for reflection in the xy plane, but symmetric for reflection in the xz or yz planes. It follows from this that all integrals in Eq. (12-78)
vanish except for H sz and H zs , which are equal. Assigning a value of x (not to be confused with the x coordinate) to these integrals, our equation becomes
−E
x
0
0
x
−E
0
0
=
0
(12-80)
0
0 −E
0
0
0
0 −E
Section 12-7 Polarizability of the Hydrogen Atom in the n = 2 States
The block diagonal form indicates that this is a product of one 2 × 2 determinant and two 1 × 1’s. Evidently, the proper zeroth-order wavefunctions must be a mixture of 2s and 2pz AOs, with 2px and 2py being acceptable as they stand. The roots of the 1 × 1 determinants are zero, the roots of the 2 × 2 are E+ = +x, E− = −x. These lead to coefficients that produce the zeroth-order wave-functions
√
φ(0) + = 1/ 2 (2s + 2pz)
(12-81)
√
φ(0) − = 1/ 2 (2s − 2pz)
(12-82)
The roots +x and −x are the first-order corrections to the energy. We now proceed to calculate these quantities: x = 2s|H |2pz = · · · = 3F
(12-83)
Our perturbation calculation indicates that the n = 2 level splits into three levels under the influence of an electric field. Since this splitting occurs to first order, it is sometimes called a first-order Stark effect. Since x is proportional to F , the splitting is linear in F , as indicated in Fig. 12-6.
It is instructive to compare these results with the behavior of the 1s state. In the first place, the effect of a uniform electric field on the 1s level is zero, to first order, because 1s|H |1s vanishes for reasons of symmetry. Only when first-order corrections are made to the 1s wavefunction are energy effects seen, and these occur in the second-order energy terms. Therefore, the 1s state gives a second-order Stark effect, but no firstorder effect. The 1s state gives no first-order effect because the spherically symmetric zeroth-order wavefunction has no electric dipole to interact with the field. But the proper zeroth-order wave functions for some of the n = 2 states, given by Eqs. (12-81) and (12-82), do provide electric dipoles in opposite directions that interact with the field to produce first-order energies of opposite signs.
Notice that the energy change for the second-order effect goes as F 2 [(Eq. (7-62)], whereas that for the first-order effect goes as F . These dependences are indicative of an induced dipole and a permanent dipole, respectively. The induced dipole for the 1s state depends on the field strength F , since, as F increases, the dipole moment increases (due to mixing in higher states). This induced dipole, which depends on F ,
Figure 12-6 Energy to first order of n = 2 level of hydrogen as a function of uniform electric field strength (z-directed field).
Chapter 12 Time-Independent Rayleigh–Schr ¨odinger Perturbation Theory then interacts with the field of strength F . Since both the size of the induced dipole and the energy of interaction with the field depend on F to the first power, the energy goes as F 2. For the n = 2 wavefunctions φ(0) + and φ(0) − , however, the dipole is permanent.7 (The mixing to produce these states occurs even in the limit as F goes to zero.) This permanent dipole interacts with the field to give an energy depending on F instead of F 2.
The 2px and 2py AOs are not affected to first-order in the perturbation. At higher orders, these AOs (as well as φ(0) + and φ(0) − ) mix in higher-energy AOs of symmetries appropriate to produce induced dipoles. However, due to the equivalence of 2px and 2py with respect to H , the induced dipole is the same in both cases so that the degeneracy is still not lifted.
The qualitatively different behaviors of the n = 1 and n = 2 levels of hydrogen result from the fact that degenerate eigenfunctions can be mixed together at no energy expense, whereas the n = 1 state can produce a dipole only by mixing in higher-energy states. This suggests that the polarizability of an atom or molecule should be strongly dependent on the availability of fairly low-energy unoccupied orbitals or wavefunctions, and this is indeed the case. As we proceed down the series H−, He, Li+, Be+2, . . . , etc.,
we find that the distance in energy between the occupied state and the higher-energy states increases. Also, we find that the systems become less and less polarizable. Again, if we compare helium and beryllium, both of which are ordinarily considered to have s2 valence-state configurations, we find beryllium to be much more polarizable. This is due to the presence of empty 2p orbitals lying fairly close in energy to the 2s AO in beryllium. In fact, these two atoms are strongly reminiscent of the n = 1, n = 2 polarizabilities in hydrogen, the chief difference being that the 2s and 2p levels of beryllium are only close in energy—not actually degenerate.
12-8 Degenerate-Level Perturbation Theory by Inspection Degenerate and nondegenerate perturbation theories share the feature that the first-order correction to the energy is the expectation value of the perturbation operator using the unperturbed wavefunction. Often we can evaluate or estimate this expectation value by inspection. The special problem for degenerate wavefunctions or orbitals is that there is an infinite number of ways to express the unperturbed wavefunctions, and only one of these ways is proper for finding W (1). The mathematical procedure for finding the proper set of functions tends to complicate and obscure the situation. However, it is
often easy to deduce the proper unperturbed wavefunctions by inspection.
The clue for doing this comes from Eq. (12-76). It is apparent that this equation, which leads to the first-order corrections to the energy and to the proper unperturbed (or zeroth-order) wavefunctions, is exactly like the secular determinantal equation we would use for a variational calculation. Since the secular equation comes from a calculus-based determination of energy extrema, we know that the energies that come out of Eq. (12-76) must be the minimum and maximum possible values and that the proper zeroth-order functions must be those that give the maximum and minimum expectation values for H . This leads immediately to a powerful insight: For a given
7
(0)
(0)
In the language of hybridization (Chapter 13), φ+ and φ− are sp hybrids.
Section 12-8 Degenerate-Level Perturbation Theory by Inspectionperturbation, the proper zeroth-order degenerate wavefunctions to use for predicting
the effect on energies are that set giving the greatest difference in response.
This rule can be seen operating in the hydrogen atom polarization example of the preceding section. The proper combination of 2s and 2pz AOs is the pair of sp hybrid combinations giving the greatest energy increase and decrease. (The 2px and 2py AOs do not mix any 2s or 2pz into themselves because to do so will have no energy-raising or -lowering effect.)
Another example is seen when changing the coulomb integral at an atom in the HMO level treatment of a cyclic molecule. Such molecules always have some degenerate MOs, and so one must decide which linear combination is “proper” for evaluating the effect of this perturbation. The situation is analyzed most easily by recognizing that mixing such degenerate MOs together corresponds to rotating the positions of the nodes and antinodes in the molecule. So the question of finding the proper MOs becomes one of asking where the nodes and antinodes should be located. The greatest difference in response to changing the coulomb integral at an atom comes when one of the degenerate pair of MOs has a node (i.e., is zero) at that atom and the other MO has an antinode (i.e., has maximum absolute value) there. Suppose, for example, you are asked to evaluate the effect on a particular degenerate pair of MOs of such a perturbation at carbon number 3 in a cyclic molecule. When you look at those MOs in a tabulation, you may find that atom 3 is not at the node or antinode position for either MO. Instead, you might observe that the node of one MO and antinode of the other occur at atom number 4. You could set about mixing these MOs to move the node-antinode positions to atom 3, but it is much easier and just as correct to pretend that the perturbation is occurring at atom 4, since, in a cycle, all atoms are equivalent.
This allows you to use the tabulated MOs to evaluate the perturbation (now at atom 4) without any modification. Once this is recognized, the problem can be completed by inspection, just as in nondegenerate cases: The MO energy for the MO that has its node at atom 4 is unaffected to first order by the perturbation, and the other is affected by an amount proportional to the square of the coefficient at atom 4. Thus, the MOs become nondegenerate and are described to zeroth order by these “proper” functions.
It is important to realize that the identity of the proper unperturbed wavefunctions depends on the nature of the perturbation. Thus, for the cyclic system discussed above (which we now assume has an even number of atoms), stretching the molecule so that two bonds on opposite sides of the cycle get longer is treated perturbatively with a different pair of proper MOs. We can quickly guess which MOs those are by seeking the pair that respond most differently to the bond stretching. Since any degenerate MO in this even–alternant cycle has at least one nodal plane perpendicular to the molecular plane, there is always a rotational position that places a nodal plane across the center of the two bonds being stretched. The mate to this MO then has an antinode in these bonds. That means that one MO is antibonding in these bonds and the other MO is bonding. This in turn means that one MO’s energy will drop as the bond lengthens and the other MO’s energy will rise. This, then, is the proper MO set to use for evaluating this perturbation to first order. Furthermore, these MOs are nondegenerate after the perturbation and are the zeroth-order approximation to the MOs of the perturbed system.
The two perturbations described above bring up a final important point. Suppose we have two different perturbations occurring simultaneously in a system involving degenerate MOs or state functions. The question then arises as to whether the proper Chapter 12 Time-Independent Rayleigh–Schr ¨odinger Perturbation Theory zeroth-order functions are the same or different for the two perturbations. We have just seen that, if the perturbations involve substituting a new atom for a carbon and stretching a pair of bonds in a cyclic molecule, the proper zeroth-order MOs are different. This means that one of these perturbations favors a postperturbative pair of nondegenerate MOs that is different from the pair being favored by the other perturbation. In other words, these perturbations compete to produce different final MOs. The other possibility is that two perturbations have the same set of proper zeroth-order wavefunctions, which means that they cooperate to produce the same final MOs. The cooperative situation is the one that gives the largest energy effects, since a single set of proper zeroth-order functions is selected that produces the greatest difference in response to both perturbations simultaneously. Competing perturbations require a compromise among zeroth-order functions which involves diminution in the differences in energy response.
The reader may have realized that there is another way to select the proper zeroth order wavefunctions, and that is to use symmetry.
Recall that the proper wave functions ψ(0) and ψ(0) are those that “diagonalize the determinant,” i.e., that cause
i
j
ψ(0)|H|ψ(0) to equal zero. In each of the examples cited above, this can be seen to i
j be true. The H operator corresponding to changing the coulomb integral at an atom, and nowhere else, is unaffected by reflection through a plane containing that atom. The MO placing a maximum at that atom is symmetric for such a reflection, and the MO placing a node there is antisymmetric. Hence the integrand is overall antisymmetric and the integral vanishes, in accord with the requirement that must be satisfied by proper zeroth-order wavefunctions. A similar argument applies for stretching or shrinking bonds, with the reflection plane now occurring at the bond midpoints. As a useful guide, then, proper zeroth-order wavefunctions may be recognized as those having opposite
symmetry for an operation by which the perturbation operator is unaffected.
EXAMPLE 12-7 Consider the cyclopentadienyl radical in the HMO approximation.
Assume it to be planar and pentagonal. (See Appendix 6 for data.) If α3 is perturbed to become α + 0.1000β, what are the MO energy changes, to first order?
SOLUTION The lowest-energy MO is nondegenerate, and all atoms have the same coefficient, so E1 = (0.4472)2(0.1000β) = 0.0200β. The second two MOs are degenerate, so we need to identify the proper zeroth-order combinations. One should be symmetric with respect to reflection through atom 3, the other antisymmetric. None of the tabulated MOs have these properties about atom 3, but they do have these properties for atom 1. Therefore, we proceed by pretending that atom 1 is the one being perturbed (it makes no difference). Then, E2 = (0.6325)2(0.1000β) = 0.0400β, E3 = 0.0000β, E4 = E2, E5 = E3.
12-9 Interaction Between Two Orbitals: An ImportantChemical Model
Qualitative quantum-chemical discussion often relies on a simplified model wherein all interactions are neglected except for the primary one. For example, in considering how a Lewis base and a Lewis acid interact, one might consider only the highest occupied Section 12-9 Interaction Between Two Orbitals: An Important Chemical Model MO (HOMO) of the base (electron donor) and the lowest unfilled MO (LUMO) of the acid (electron acceptor). Consideration of the ways two levels interact will provide some useful rules of thumb and also consolidate some of our earlier findings.8 We label the unperturbed orbitals and energies ψa, ψb, and Ea, Eb. We will work within the H¨uckel type of framework so that state energy differences may be written as orbital energy differences. The orbitals are allowed to interact with each other by virtue of close approach. We will assume that the perturbation felt by the orbitals is proportional to the overlap between them. (If the systems involved are ions, strong electrostatic interactions will exist as well. Therefore, we assume all systems to be neutral.)
Since the overlap of ψa on ψb is the same as that of ψb on ψa, the energy change to first order is the same for both levels and hence we ignore it (unless the levels are degenerate—an eventuality we discuss shortly).
The second-order contributions to the energies are:
|ψa|H|ψb|2
W (2) =
a
(12-84)
Ea − Eb
|ψa|H|ψb|2
W (2) =
(12-85)
b
Eb − Ea Thus, the second-order energy W (2)
a is positive if Ea > Eb, negative if Ea
SOLUTION The HOMO-LUMO contribution to the charge shift favors atom 4, since its coefficients in these MOs are larger than those for atoms 2 or 3. However, the actual data (not given in Appendix 6) are: π1,2 = 0.21342, π1,3 = −0.01771, π1,4 = 0.13937. So atom 2 experiences the greatest population change, atom 3 the least. Why is this? It results from the fact that we are summing products that can be positive or negative. Two atoms near each other are less likely to have nodal planes separating them, so their coefficients are more likely to be of the same sign giving fewer sign changes in their products. So the sum has less tendency to cancel itself. This is the way in which MO theory predicts a tendency for polarization effects to die off as distance increases—the well-known “inductive effect.” This dying-off is not monotonic, as can be seen in the remaining atom-atom polarizabilities: π1,5 = 0.02325, π1,6 = −0.00643, π1,7 = 0.03228, π1,8 = −0.02674, π1,9 = 0.08889, π1,10 = −0.00356.
In general, πlk can have either sign. This means that making atom k more attractive for electrons will, to first order, cause the π electron density to decrease at some carbons and increase at others. (This is necessary since the total π-charge density must be conserved.)
12-6 Perturbation Theory for a Degenerate State Equations (12-19), (12-21), and (12-24) have denominators containing Ei − Ej . If ψi and ψj are degenerate, this leads to difficulty, alerting us to the fact that our earlier derivations do not take proper account of states of interest that are degenerate. The problem results from our initial expansion of φ˙ as a power series in λ. The leading
l
term in the expansion, φ(0), is the wavefunction that φ
i
i becomes in the limit when λ = 0. In nondegenerate systems there is no is ambiguity; φ(0) has to be ψ i
i . But if ψi is degenerate with ψj , any linear combination of them is also an eigenfunction. We need a method to determine how ψi and ψj should be mixed together to form the correct zeroth-order functions φ(0) and φ(0).
i
j
To find the conditions that φ(0) and φ(0) must satisfy, we return to the first-order i
j perturbation equation (12-11) but with φ(0) in place of ψ
i i : (H − W (1))φ(0) + (H = 0
(12-73)
i
i 0 − Ei )φ(1)
i
Multiplying from the left by φ(0)∗ and integrating gives (taking φ(0) and φ(0) to be
j
i
j
orthogonal and H0 hermitian) φ(0)|H|φ(0) + (E |φ(1) = 0
(12-74)
j
i
j − Ei )φ(0)
j
i
If Ei = Ej , this gives φ(0)|H |φ(0) = 0
(12-75)
j
i
This means that the first-order perturbation equation is satisfied for degenerate states only when the wavefunctions “diagonalize” the perturbation operator H . Therefore, our problem reduces to finding those linear combinations of ψi and ψj that will diagonalize H . We have already seen two ways to accomplish this in earlier chapters. One way is to construct the matrix H, where (H)ij = ψi|H |ψj , and then find the unitary matrix C that diagonalizes H in the similarity transformation C†HC. The elements in C are then the coefficients for the linear combinations of ψi and ψj , and the diagonal they are the first-order corrections to the energy of φ(0) and φ(0). The other method,
i
j
equivalent to the above, is to set up the determinantal equation
H − E
H
11
12
H H − E = 0
(12-76)
21
22
and solve for the roots E. Substituting these back into the simultaneous equations corresponding to the determinant gives the coefficients with which ψi and ψj should be combined. The roots E are the first-order corrections to the energy and are identical to the diagonal matrix elements mentioned above.
In the event that the roots are equal, the perturbation has failed to split the degeneracy to first order and it is then necessary to proceed to higher order. We will not pursue this problem to higher orders, however.
If there are n degenerate states, the above procedure is simply carried out with an n × n matrix or determinant.
We will now give an example of perturbation theory for a degenerate state.
12-7 Polarizability of the HydrogenAtom in the n = 2 States In Chapter 7, the polarizability of the hydrogen atom in the 1s state was calculated by the variation method (see Section 7-4 and Problem 7-13). We now use perturbation theory for degenerate states to calculate to first order the polarizabilities of the n = 2 states.
The perturbation due to a z-directed uniform electric field is, in atomic units, H = −F r cos θ
(12-77)
There are four degenerate states at the n = 2 level, giving us a 4 × 4 secular determinant.
If we choose the real AOs as our basis set and let s, x, y, z represent these functions, the determinantal equation is
H − E
H
ss
sz
H sy
H sx
H − E
H
zs
H zz
zy
H zx
=
0
(12-78)
H
−
ys
H yz
H yy
E
H yx
H
−
xs
H xz
H xy
H xx
E
where
H =
yz
y|H |z, etc.
(12-79)
The operator H is antisymmetric for reflection in the xy plane, but symmetric for reflection in the xz or yz planes. It follows from this that all integrals in Eq. (12-78)
vanish except for H sz and H zs , which are equal. Assigning a value of x (not to be confused with the x coordinate) to these integrals, our equation becomes
−E
x
0
0
x
−E
0
0
=
0
(12-80)
0
0 −E
0
0
0
0 −E
Section 12-7 Polarizability of the Hydrogen Atom in the n = 2 States
The block diagonal form indicates that this is a product of one 2 × 2 determinant and two 1 × 1’s. Evidently, the proper zeroth-order wavefunctions must be a mixture of 2s and 2pz AOs, with 2px and 2py being acceptable as they stand. The roots of the 1 × 1 determinants are zero, the roots of the 2 × 2 are E+ = +x, E− = −x. These lead to coefficients that produce the zeroth-order wave-functions
√
φ(0) + = 1/ 2 (2s + 2pz)
(12-81)
√
φ(0) − = 1/ 2 (2s − 2pz)
(12-82)
The roots +x and −x are the first-order corrections to the energy. We now proceed to calculate these quantities: x = 2s|H |2pz = · · · = 3F
(12-83)
Our perturbation calculation indicates that the n = 2 level splits into three levels under the influence of an electric field. Since this splitting occurs to first order, it is sometimes called a first-order Stark effect. Since x is proportional to F , the splitting is linear in F , as indicated in Fig. 12-6.
It is instructive to compare these results with the behavior of the 1s state. In the first place, the effect of a uniform electric field on the 1s level is zero, to first order, because 1s|H |1s vanishes for reasons of symmetry. Only when first-order corrections are made to the 1s wavefunction are energy effects seen, and these occur in the second-order energy terms. Therefore, the 1s state gives a second-order Stark effect, but no firstorder effect. The 1s state gives no first-order effect because the spherically symmetric zeroth-order wavefunction has no electric dipole to interact with the field. But the proper zeroth-order wave functions for some of the n = 2 states, given by Eqs. (12-81) and (12-82), do provide electric dipoles in opposite directions that interact with the field to produce first-order energies of opposite signs.
Notice that the energy change for the second-order effect goes as F 2 [(Eq. (7-62)], whereas that for the first-order effect goes as F . These dependences are indicative of an induced dipole and a permanent dipole, respectively. The induced dipole for the 1s state depends on the field strength F , since, as F increases, the dipole moment increases (due to mixing in higher states). This induced dipole, which depends on F ,
Figure 12-6 Energy to first order of n = 2 level of hydrogen as a function of uniform electric field strength (z-directed field).
Chapter 12 Time-Independent Rayleigh–Schr ¨odinger Perturbation Theory then interacts with the field of strength F . Since both the size of the induced dipole and the energy of interaction with the field depend on F to the first power, the energy goes as F 2. For the n = 2 wavefunctions φ(0) + and φ(0) − , however, the dipole is permanent.7 (The mixing to produce these states occurs even in the limit as F goes to zero.) This permanent dipole interacts with the field to give an energy depending on F instead of F 2.
The 2px and 2py AOs are not affected to first-order in the perturbation. At higher orders, these AOs (as well as φ(0) + and φ(0) − ) mix in higher-energy AOs of symmetries appropriate to produce induced dipoles. However, due to the equivalence of 2px and 2py with respect to H , the induced dipole is the same in both cases so that the degeneracy is still not lifted.
The qualitatively different behaviors of the n = 1 and n = 2 levels of hydrogen result from the fact that degenerate eigenfunctions can be mixed together at no energy expense, whereas the n = 1 state can produce a dipole only by mixing in higher-energy states. This suggests that the polarizability of an atom or molecule should be strongly dependent on the availability of fairly low-energy unoccupied orbitals or wavefunctions, and this is indeed the case. As we proceed down the series H−, He, Li+, Be+2, . . . , etc.,
we find that the distance in energy between the occupied state and the higher-energy states increases. Also, we find that the systems become less and less polarizable. Again, if we compare helium and beryllium, both of which are ordinarily considered to have s2 valence-state configurations, we find beryllium to be much more polarizable. This is due to the presence of empty 2p orbitals lying fairly close in energy to the 2s AO in beryllium. In fact, these two atoms are strongly reminiscent of the n = 1, n = 2 polarizabilities in hydrogen, the chief difference being that the 2s and 2p levels of beryllium are only close in energy—not actually degenerate.
12-8 Degenerate-Level Perturbation Theory by Inspection Degenerate and nondegenerate perturbation theories share the feature that the first-order correction to the energy is the expectation value of the perturbation operator using the unperturbed wavefunction. Often we can evaluate or estimate this expectation value by inspection. The special problem for degenerate wavefunctions or orbitals is that there is an infinite number of ways to express the unperturbed wavefunctions, and only one of these ways is proper for finding W (1). The mathematical procedure for finding the proper set of functions tends to complicate and obscure the situation. However, it is
often easy to deduce the proper unperturbed wavefunctions by inspection.
The clue for doing this comes from Eq. (12-76). It is apparent that this equation, which leads to the first-order corrections to the energy and to the proper unperturbed (or zeroth-order) wavefunctions, is exactly like the secular determinantal equation we would use for a variational calculation. Since the secular equation comes from a calculus-based determination of energy extrema, we know that the energies that come out of Eq. (12-76) must be the minimum and maximum possible values and that the proper zeroth-order functions must be those that give the maximum and minimum expectation values for H . This leads immediately to a powerful insight: For a given
7
(0)
(0)
In the language of hybridization (Chapter 13), φ+ and φ− are sp hybrids.
Section 12-8 Degenerate-Level Perturbation Theory by Inspectionperturbation, the proper zeroth-order degenerate wavefunctions to use for predicting
the effect on energies are that set giving the greatest difference in response.
This rule can be seen operating in the hydrogen atom polarization example of the preceding section. The proper combination of 2s and 2pz AOs is the pair of sp hybrid combinations giving the greatest energy increase and decrease. (The 2px and 2py AOs do not mix any 2s or 2pz into themselves because to do so will have no energy-raising or -lowering effect.)
Another example is seen when changing the coulomb integral at an atom in the HMO level treatment of a cyclic molecule. Such molecules always have some degenerate MOs, and so one must decide which linear combination is “proper” for evaluating the effect of this perturbation. The situation is analyzed most easily by recognizing that mixing such degenerate MOs together corresponds to rotating the positions of the nodes and antinodes in the molecule. So the question of finding the proper MOs becomes one of asking where the nodes and antinodes should be located. The greatest difference in response to changing the coulomb integral at an atom comes when one of the degenerate pair of MOs has a node (i.e., is zero) at that atom and the other MO has an antinode (i.e., has maximum absolute value) there. Suppose, for example, you are asked to evaluate the effect on a particular degenerate pair of MOs of such a perturbation at carbon number 3 in a cyclic molecule. When you look at those MOs in a tabulation, you may find that atom 3 is not at the node or antinode position for either MO. Instead, you might observe that the node of one MO and antinode of the other occur at atom number 4. You could set about mixing these MOs to move the node-antinode positions to atom 3, but it is much easier and just as correct to pretend that the perturbation is occurring at atom 4, since, in a cycle, all atoms are equivalent.
This allows you to use the tabulated MOs to evaluate the perturbation (now at atom 4) without any modification. Once this is recognized, the problem can be completed by inspection, just as in nondegenerate cases: The MO energy for the MO that has its node at atom 4 is unaffected to first order by the perturbation, and the other is affected by an amount proportional to the square of the coefficient at atom 4. Thus, the MOs become nondegenerate and are described to zeroth order by these “proper” functions.
It is important to realize that the identity of the proper unperturbed wavefunctions depends on the nature of the perturbation. Thus, for the cyclic system discussed above (which we now assume has an even number of atoms), stretching the molecule so that two bonds on opposite sides of the cycle get longer is treated perturbatively with a different pair of proper MOs. We can quickly guess which MOs those are by seeking the pair that respond most differently to the bond stretching. Since any degenerate MO in this even–alternant cycle has at least one nodal plane perpendicular to the molecular plane, there is always a rotational position that places a nodal plane across the center of the two bonds being stretched. The mate to this MO then has an antinode in these bonds. That means that one MO is antibonding in these bonds and the other MO is bonding. This in turn means that one MO’s energy will drop as the bond lengthens and the other MO’s energy will rise. This, then, is the proper MO set to use for evaluating this perturbation to first order. Furthermore, these MOs are nondegenerate after the perturbation and are the zeroth-order approximation to the MOs of the perturbed system.
The two perturbations described above bring up a final important point. Suppose we have two different perturbations occurring simultaneously in a system involving degenerate MOs or state functions. The question then arises as to whether the proper Chapter 12 Time-Independent Rayleigh–Schr ¨odinger Perturbation Theory zeroth-order functions are the same or different for the two perturbations. We have just seen that, if the perturbations involve substituting a new atom for a carbon and stretching a pair of bonds in a cyclic molecule, the proper zeroth-order MOs are different. This means that one of these perturbations favors a postperturbative pair of nondegenerate MOs that is different from the pair being favored by the other perturbation. In other words, these perturbations compete to produce different final MOs. The other possibility is that two perturbations have the same set of proper zeroth-order wavefunctions, which means that they cooperate to produce the same final MOs. The cooperative situation is the one that gives the largest energy effects, since a single set of proper zeroth-order functions is selected that produces the greatest difference in response to both perturbations simultaneously. Competing perturbations require a compromise among zeroth-order functions which involves diminution in the differences in energy response.
The reader may have realized that there is another way to select the proper zeroth order wavefunctions, and that is to use symmetry.
Recall that the proper wave functions ψ(0) and ψ(0) are those that “diagonalize the determinant,” i.e., that cause
i
j
ψ(0)|H|ψ(0) to equal zero. In each of the examples cited above, this can be seen to i
j be true. The H operator corresponding to changing the coulomb integral at an atom, and nowhere else, is unaffected by reflection through a plane containing that atom. The MO placing a maximum at that atom is symmetric for such a reflection, and the MO placing a node there is antisymmetric. Hence the integrand is overall antisymmetric and the integral vanishes, in accord with the requirement that must be satisfied by proper zeroth-order wavefunctions. A similar argument applies for stretching or shrinking bonds, with the reflection plane now occurring at the bond midpoints. As a useful guide, then, proper zeroth-order wavefunctions may be recognized as those having opposite
symmetry for an operation by which the perturbation operator is unaffected.
EXAMPLE 12-7 Consider the cyclopentadienyl radical in the HMO approximation.
Assume it to be planar and pentagonal. (See Appendix 6 for data.) If α3 is perturbed to become α + 0.1000β, what are the MO energy changes, to first order?
SOLUTION The lowest-energy MO is nondegenerate, and all atoms have the same coefficient, so E1 = (0.4472)2(0.1000β) = 0.0200β. The second two MOs are degenerate, so we need to identify the proper zeroth-order combinations. One should be symmetric with respect to reflection through atom 3, the other antisymmetric. None of the tabulated MOs have these properties about atom 3, but they do have these properties for atom 1. Therefore, we proceed by pretending that atom 1 is the one being perturbed (it makes no difference). Then, E2 = (0.6325)2(0.1000β) = 0.0400β, E3 = 0.0000β, E4 = E2, E5 = E3.
12-9 Interaction Between Two Orbitals: An ImportantChemical Model
Qualitative quantum-chemical discussion often relies on a simplified model wherein all interactions are neglected except for the primary one. For example, in considering how a Lewis base and a Lewis acid interact, one might consider only the highest occupied Section 12-9 Interaction Between Two Orbitals: An Important Chemical Model MO (HOMO) of the base (electron donor) and the lowest unfilled MO (LUMO) of the acid (electron acceptor). Consideration of the ways two levels interact will provide some useful rules of thumb and also consolidate some of our earlier findings.8 We label the unperturbed orbitals and energies ψa, ψb, and Ea, Eb. We will work within the H¨uckel type of framework so that state energy differences may be written as orbital energy differences. The orbitals are allowed to interact with each other by virtue of close approach. We will assume that the perturbation felt by the orbitals is proportional to the overlap between them. (If the systems involved are ions, strong electrostatic interactions will exist as well. Therefore, we assume all systems to be neutral.)
Since the overlap of ψa on ψb is the same as that of ψb on ψa, the energy change to first order is the same for both levels and hence we ignore it (unless the levels are degenerate—an eventuality we discuss shortly).
The second-order contributions to the energies are:
|ψa|H|ψb|2
W (2) =
a
(12-84)
Ea − Eb
|ψa|H|ψb|2
W (2) =
(12-85)
b
Eb − Ea Thus, the second-order energy W (2)
a is positive if Ea > Eb, negative if Ea