Section 7-4
Section 7-4 Linear Variation: The Polarizability of the Hydrogen Atom
that the hamiltonian operator is a linear operator [i.e., ˆ
H (c
ˆ
1φ1 + c2φ2) = (c1H φ1 + c ˆ 2H φ2)] makes a linear variation procedure more convenient for most purposes.
7-4 Linear Variation: The Polarizabilityof the Hydrogen Atom Suppose we wish to express a wavefunction ψ (which may be approximate) as a linear combination of two known functions φ1 and φ2: ψ (c1, c2) = c1φ1 + c2φ2
(7-36)
The question is, what values of c1 and c2 give a ψ that best approximates the exact wavefunction for a particular system? The usual approach is to determine which values of c1 and c2 give the ψ associated with the minimum average energy attainable. The technique for achieving this, called the linear variation method, is by far the most common type of quantum chemical calculation performed.
An example of a problem that can be treated by this method is the polarizability of the hydrogen atom. The wavefunction for the unperturbed hydrogen atom in its ground state is spherically symmetrical. But, when a uniform z-directed external electric field of strength F is imposed, the positive nucleus and the negative electron are attracted in opposite directions, which leads to an electronic distribution that is skewed with respect to the nucleus. The wavefunction describing this skewed distribution may be approximated by mixing with the unperturbed 1s function some 2pz function: ψ = c11s + c22pz. As indicated in Fig. 7-2, this produces a skewed wavefunction because the 2pz function is of the same sign as the 1s function on one side of the nucleus and of the opposite sign on the other. We will work out the details of this example after developing the method for the general case.
Let the generalized trial function ψ be a linear combination of known functions φ1, φ2, . . . , φn. (This set of functions is called the basis set for the calculation.)
ψ = c1φ1 + c2φ2 + · · · + cnφn
(7-37)
where the coefficients c are to be determined so that ψ∗ ˆHψ dτ
= ¯E
(7-38)
ψ∗ψ dτ is minimized. Substituting Eq. (7-37) into Eq. (7-38) gives
ˆ
¯
c∗φ∗ + c∗φ∗ + · · · + c∗ H (c1φ1 + c2φ2 + · · · + cnφn) dτ
E = 1 1
2 2
nφ∗
n
c∗φ∗ + c∗φ∗ + · · · + c∗ (c 1 1
2 2
nφ∗
n
1φ1 + c2φ2 + · · · + cnφn) dτ = num
(7-39)
denom
Since we will be dealing with cases in which the c’s and φ’s are real, we will temporarily omit the complex conjugate notation to simplify the derivation. At the minimum value of ¯
E,
∂ ¯ E = ∂ ¯E =···= ∂ ¯E =0
(7-40)
∂c1
∂c2
∂cn
Chapter 7 The Variation MethodFigure 7-2 Values of ψ versus z for 1s state of H atom (—-) and for approximate wavefunction given by 0.982 1s −0.188 2pz (- - -). The nucleus is at z = 0 for each case.
The partial derivative of Eq. (7-39) with respect to c1 is
∂ ¯
E
φ ˆ
H
=
1
(c1φ1 + · · · + cnφn) dτ + (c1φ1 +···+cnφn) ˆHφ1 dτ
∂c1
denom
denom
− (num) (denom)−2 φ1 (c1φ1 + · · · + cnφn) dτ
+
(c1φ1 + · · · + cnφn) φ1dτ = 0
(7-41)
Section 7-4 Linear Variation: The Polarizability of the Hydrogen Atom
Multiplying through by denom, recalling that num/denom equals ¯ E, and rearranging, gives
c
ˆ
1
φ1H φ1 dτ − ¯
E
φ1φ1 dτ
+c
ˆ
2
φ1H φ2 dτ − ¯
E
φ1φ2 dτ
+ · · · + c
ˆ
n
φ1H φn dτ − ¯
E
φ1φn dτ = 0
(7-42)
At this point, it is convenient to switch to an abbreviated notation:
φ ˆ i H φj dτ ≡ Hij
(7-43)
φiφj dτ ≡ Sij
(7-44)
The integral Sij is normally called an overlap integral since its value is, in certain cases, an indication of the extent to which the two functions φi and φj occupy the same space. Use of this abbreviated notation produces, for Eq. (7-42),
c1 H11 − ¯ ES11 + c2 H12 − ¯ ES12 + · · · + cn H1n − ¯ ES1n = 0
(7-45)
A similar treatment for ∂ ¯ E/∂ci gives a similar equation:
c1 Hi1 − ¯ ESi1 + c2 Hi2 − ¯ ESi2 + · · · + cn Hin − ¯ ESin = 0
(7-46)
Thus, requiring that ∂ ¯ E/∂ci vanish for all coefficients produces n homogeneous linear equations (homogeneous, all equal zero; linear, all ci’s to first power). If one chooses a value for ¯ E, there remain n unknowns—the coefficients ci. (The integrals Hij and Sij are presumably knowable since ˆ H and the functions φi are known.) Of course, one trivial solution for Eqs. (7-46) is always possible, namely, c1 = c2 = · · · = cn = 0. But this corresponds to ψ = 0, a case of no physical interest. Are there nontrivial solutions as well? In quantum chemical calculations, nontrivial solutions usually exist only for
certain discrete values of ¯ E. This provides the approach for solving the problem.
First, find those values of ¯ E for which nontrivial coefficients exist. Second, substitute into Eqs. (7-46) whichever of these values of ¯ E one is interested in and solve for the coefficients. (Each value of ¯ E has its own associated set of coefficients.) But how do we find these particular values of ¯ E that yield nontrivial solutions to Eqs. (7-46)? The answer is given in Appendix 2, where it is shown that the condition which must be met
by the coefficients of a set of linear homogeneous equations in order that nontrivial
solutions exist is that their determinant vanish. Notice that, in the standard treatment given in Appendix 2, the coefficients are known and x, y, and z are unknown. Here, however, the coefficients ci are unknown, and Hij and Sij are known. Therefore, it is the determinant of the H ’s, S’s and ¯ E in Eqs. (7-46) that must equal zero:
H
11 − ¯ES11 H12 − ¯ES12 · · · H1n − ¯ES1n
H21 − ¯ES21 H22 − ¯ES22 · · · H2n − ¯ES2n
.
.
.
= 0
(7-47)
.
.
.
.
.
.
H
n1 − ¯ ESn1 Hn2 − ¯ ESn2 · · · Hnn − ¯
ESnn
Chapter 7 The Variation Method
Expansion of this determinant gives a single equation containing the unknown ¯ E. Any value of ¯
E satisfying this equation is associated with a nontrivial set of coefficients.
The lowest of these values of ¯ E is the minimum average energy achievable by variation of the coefficients. Substitution of this value of ¯ E back into Eqs. (7-46) produces n simultaneous equations for the n coefficients. Equation (7-47) is referred to as the secular equation, and the determinant on the left-hand side is called the secular
determinant.
This method is best illustrated by example, and we will now proceed with the prob lem of a hydrogen atom in a z-directed uniform electric field of strength F a.u. As mentioned earlier, a suitable choice of functions to mix together to approximate the accurate wavefunction is the 1s and 2p, hydrogenlike functions. The choice of two basis functions leads to a 2 × 2 secular determinantal equation:
H 11 − ¯ES11 H12 − ¯ES12
= 0
(7-48)
H21 − ¯ ES21 H22 − ¯
ES22
If we arbitrarily associate the 1s function with the index 1 and the 2pz function with index 2 (consistent with ψ = c11s + c22pz), then the terms in the determinant are (returning to general complex conjugate notation):
H11 =
1s∗ ˆ
H 1s dτ
H12 =
1s∗ ˆ
H 2pz dτ
H
ˆ
ˆ 21 = 2p∗zH 1s dτ H22 = 2p∗zH 2pz dτ
(7-49)
(The electron label has been omitted since there is only one electron.) The corresponding S integrals are obtained from these if ˆ H is omitted in each case. The hamiltonian operator is just that for a hydrogen atom with an additional term to account for the z-directed field:
ˆ
H = − 1 ∇2 − (1/r) − F r cos θ
(7-50)
2
[The energy of a charge −e in a uniform electric field of strength F and direction z is −eF z. In atomic units, one unit of field strength is e/a2 = 5.142 × 1011 V /m. The unit
0
of charge in atomic units is e, so this symbol does not appear explicitly in Eq. (7-50).
Also, the identity z = r cos θ has been used.] This may also be written
ˆ
H = ˆ Hhyd − F r cos θ
(7-51)
where ˆ Hhyd is the hamiltonian for the unperturbed hydrogen atom.
The secular determinant contains four H -type and four S-type terms. However, evaluating these eight integrals turns out to be much easier than one might expect.
In the first place, S12 = S21 since these integrals differ only in the order of the two functions in the integrand, and the functions commute. Also, because ˆ H is hermitian, it follows immediately that H21 = H ∗ . This leaves us with three S terms and three H
12
terms to evaluate. The three S terms are simple. Because the hydrogenlike functions are orthonormal, S11 and S22 equal unity, and S12 vanishes. These points have already reduced the secular determinantal equation to
H
11 − ¯E
H12
=
0
(7-52)
H ∗
H
12
22 − ¯
E
Section 7-4 Linear Variation: The Polarizability of the Hydrogen Atom Consider next the term H11. This may be written as
H11 =
1s∗ ˆ
Hhyd1s dτ − 1s∗ (F r cos θ ) 1s dτ
(7-53)
But the 1s function is an eigenfunction of ˆ Hhyd with eigenvalue − 1 a.u. Therefore, the
2
first integral on the right-hand side of Eq. (7-53) is
1s∗ ˆ
Hhyd1s dτ = − 1 1s∗1s dτ = − 1 a.u.
(7-54)
2
2
The second term on the right-hand side of Eq. (7-53) is zero by symmetry since 1s∗1s is symmetric for reflection in the xy plane while r cos θ (= z) is antisymmetric. Thus, H11 = − 1 a.u. Similarly, H a.u. [Recall that the 8 eigenvalues of hydrogen
2
22 = − 1
8
are equal to −1/(2n2), and here n = 2.] All that remains is H12:
H12 =
1s∗ ˆ Hhyd2pz dτ − F 1s∗ (r cos θ ) 2pz dτ
(7-55)
The first term on the right-hand side is easily shown to be zero:
1s∗ ˆ
Hhyd2pz dτ = 1s∗ − 1 2pz dτ = 0
(7-56)
8
Here we employ the fact that 2pz is an eigenfunction of ˆ Hhyd and then the fact that 1s and 2pz AOs are orthogonal. The second term on the right-hand side must be written out in full and integrated by “brute force.” It is remarkable that, of the eight terms originally considered, only one needs to be done by detailed integration. Proceeding, we substitute formulas for 1s and 2pz in this last integral to obtain
−F (π )−1/2 exp(−r) r cos θ (32π)−1/2r exp(−r/2) cos θ(r2 sin θ) dr dθ dφ
(7-57)
We consider the integration over spin to have been carried out already, giving a factor of unity. Integrating over φ to obtain 2π , and regrouping terms gives
√
∞
π
−2πF/(4 2π) r4 exp(−3r/2) dr cos2 θ sin θ dθ
(7-58)
0
0
√
= −
4!
2
F /2 2
= −215/2F a.u.
(7-59)
3 5
3
35
2
This completes the task of evaluating the terms in the secular determinant.2 The final result is (in atomic units)
−1 − ¯
E
−215/2(F/35)
2
= 0
(7-60)
−215/2(F/35) −1 − ¯E
8
2Since H12 is real, it is clear that H21 = H12.
Chapter 7 The Variation Method which expands to 1 +(5 ¯E/8)+ ¯E2 −215F2/310 =0
(7-61)
16
This is quadratic in ¯ E having roots
9
1/2
¯
+ 217F 2/310
E = − 5 ± 64
(7-62)
16
2
When no external field is present, F = 0 and the roots are just − 1 and − 1 a.u., the
2
8
1s and 2pz eigenvalues for the unperturbed hydrogen atom. As F increases from zero, the roots change, as indicated in Fig. 7-3. We see that, for a given field strength, there are only two values of ¯ E that will cause the determinant to vanish. If either of these two values of ¯ E is substituted into the simultaneous equations related to Eq. (7-52), then nontrivial values for c1 and c2 can be found. Thus, at F = 0.1 a.u., ¯
E = −0.51425 a.u., and ¯ E = −0.1107 a.u. are values of ¯ E for which ∂ ¯ E/∂c1, and ∂ ¯ E/∂c2 both vanish. The former is the minimum, the latter the maximum in the curve of ¯
E versus c1 (Fig. 7-4). (The normality requirement results in the two c’s being dependent, and so ¯
E may be plotted against either of them.) Since the variation principle tells us that ¯E ≥ Elowest exact, we can say immediately that the energy of the hydrogen atom in a uniform electric field of 0.1 a.u. is −0.51425 a.u. or lower. That is, −0.51425 a.u. is an upper bound to the true energy.
Now that we have the value of the lowest ¯ E1 we can solve for c1 and c2 and obtain the approximate ground state wavefunction. The homogeneous equations related to the determinant in Eq. (7-52) are c1(H11 − ¯ E) + c2H12 = 0
(7-63)
c1H12 + c2 H22 − ¯ E = 0
(7-64)
Substituting −0.51425 for ¯ E, and inserting the values for H11, H22, and H12 found earlier gives (when F = 0.1 a.u.)
0.01425c1 − 0.074493c2 = 0
(7-65) −0.074493c1 + 0.38925c2 = 0
(7-66)
Equation (7-65) gives c1 = 5.2275c2
(7-67)
If we substitute this expression for c1 into Eq. (7-66) we get −0.3892c2 + 0.3892c2 = 0
(7-68)
This is useless for evaluating c2. It is one of the properties of such a set of homogeneous equations that the last unused equation is useless for determining coefficients. This arises because an equation like (7-63) still equals zero when c1 and c2 are both multiplied by the same arbitrary constant. Therefore, these equations are inherently capable of telling us the ratio of c1 and c2 only, and not their absolute values. We shall determine
Section 7-4 Linear Variation: The Polarizability of the Hydrogen AtomFigure 7-3 Average energies for a hydrogen atom in a uniform electric field of strength F as given by a linear variation calculation using a 1s, 2pz basis. (- - -) Results from accurate calculations.
absolute values by invoking the requirement that ψ be normalized. In this case, this means that c2 +
=
1
c22
1
(7-69)
or (5.2275c2)2 + c2 =
2
1
(7-70)
which gives c2 = ± 0.18789
(7-71)
If we arbitrarily choose the positive root for c2, it follows from Eq. (7-67) that c1 = 0.98219.
Chapter 7 The Variation MethodFigure 7-4 ¯E versus c1 for a hydrogen atom in a uniform electric field of strength 0.1 a.u.
Thus, when F = 0.1 a.u., the linear variation method using a 1s, 2pz basis set gives an upper bound to the energy of −0.51425 a.u. and a corresponding approximate wavefunction of ψ = 0.98219 1s + 0.18789 2pz
(7-72)
As mentioned earlier, the admixture of 2pz with 1s produces the skewed charge distri bution shown in Fig. 7-2.
It is important to note that the extent of mixing between 1s and 2pz depends partly on the size of the off-diagonal determinantal element H12. When H12 is zero (no external field), no mixing occurs. As H12 increases, mixing increases. Generally speaking, the larger the size of the off-diagonal element connecting two basis functions in the secular determinant, the greater the degree of mixing of these basis functions in the final solution, other factors being equal. Hij is sometimes referred to as the interaction
element between basis functions i and j .
If we carried through the same procedure using the maximum ¯ E of −0.1107 a.u., we would obtain the approximate wavefunction ψ = 0.98219 2pz − 0.18789 1s
(7-73)
This wavefunction is orthogonal to ψ.
It may be proved that the second root, ¯E = −0.1107 is an upper bound for the energy of the second-lowest state of the hydrogen atom in the field. However, ψ is probably not too good an approximation to the exact wavefunction for that state. This is partly because the wavefunction ψ is one that maximizes ¯ E. Hence, there is no particular tendency for the procedure to isolate the second-lowest state from the infinite manifold of states. Also, our basis set was chosen with an eye toward its appropriateness for approximating the lowest state.
The true second-lowest state might be expected to contain significant amounts of the 2s AO, which is not included in this basis.
The values of ¯ E versus c1 are plotted in Fig. 7-4. The values of ¯ E that we obtained by the variation procedure correspond to the extrema in this figure. The low-energy extreme corresponds to a wavefunction that shifts negative charge in the direction it is attracted by the field. The high-energy extreme corresponds to a wavefunction that shifts charge in the opposite direction. (When a calculation is performed over a more extensive basis to produce more than two roots, the highest and lowest roots correspond, respectively, to the maximum and minimum, the other roots to saddle points, on the energy hypersurface.)
The detailed treatment just completed is rather involved, and so we now summarize the main points. Step 1 involved selection of a basis set of functions which is capable of approximating the exact solution. Step 2 was the construction of the secular determinant, including evaluation of all the Hij - and Sij -type integrals. Step 3 was the conversion of the determinantal equation into its equivalent equation in powers of ¯E and solution for the roots ¯E. Step 4 was the substitution of an ¯E of interest into the simultaneous equations that are related to the secular determinant and solution for
c1/c2. Finally, we used the normality requirement to arrive at convenient values for c1 and c2.
There are many ways one could increase the flexibility of the trial function in an effort to increase the accuracy of the calculation. By adding additional basis functions, one would stay within the linear variation framework, merely increasing the size of the secular determinant. If these additional basis functions are of appropriate symmetries, they will cause the minimum energy root to be lowered further and will mix into the corresponding wavefunction to make it a better approximation to the lowest-energy eigenfunction for the system. Also, the additional functions will increase the number of roots ¯ E, thereby providing upper bounds for the energies of the third-, fourth-, etc.
lowest states of the system. Another possibility is to allow nonlinear variation of the 1s and 2pz orbital exponents, in combination with linear variation. This would be more involved than the calculation we have shown here, but could easily be accomplished with the aid of a computer.
EXAMPLE 7-1 Suppose the variational process just described were performed using ψ(c1, c2, c3) = c11s + c22p0 + c33d0. Would all three of theseAOs be present in the lowest-energy solution?
SOLUTION We already know that 1s and 2p0 will be mixed. They differ in reflection symmetry in just the right manner to provide a wavefunction skewed in the z direction, and this is manifested as a nonzero interaction element H12. 3d0, however, is, like 1s, symmetric for reflection through the x, y plane. Therefore, it cannot skew 1s in the z direction, and its interaction element with 1s, H13, equals zero (by symmetry). Therefore, it is tempting to think that that 3d0 will not contribute. However, because 3d0 and 2p0 have opposite reflection symmetries through the x, y
that the hamiltonian operator is a linear operator [i.e., ˆ
H (c
ˆ
1φ1 + c2φ2) = (c1H φ1 + c ˆ 2H φ2)] makes a linear variation procedure more convenient for most purposes.
7-4 Linear Variation: The Polarizabilityof the Hydrogen Atom Suppose we wish to express a wavefunction ψ (which may be approximate) as a linear combination of two known functions φ1 and φ2: ψ (c1, c2) = c1φ1 + c2φ2
(7-36)
The question is, what values of c1 and c2 give a ψ that best approximates the exact wavefunction for a particular system? The usual approach is to determine which values of c1 and c2 give the ψ associated with the minimum average energy attainable. The technique for achieving this, called the linear variation method, is by far the most common type of quantum chemical calculation performed.
An example of a problem that can be treated by this method is the polarizability of the hydrogen atom. The wavefunction for the unperturbed hydrogen atom in its ground state is spherically symmetrical. But, when a uniform z-directed external electric field of strength F is imposed, the positive nucleus and the negative electron are attracted in opposite directions, which leads to an electronic distribution that is skewed with respect to the nucleus. The wavefunction describing this skewed distribution may be approximated by mixing with the unperturbed 1s function some 2pz function: ψ = c11s + c22pz. As indicated in Fig. 7-2, this produces a skewed wavefunction because the 2pz function is of the same sign as the 1s function on one side of the nucleus and of the opposite sign on the other. We will work out the details of this example after developing the method for the general case.
Let the generalized trial function ψ be a linear combination of known functions φ1, φ2, . . . , φn. (This set of functions is called the basis set for the calculation.)
ψ = c1φ1 + c2φ2 + · · · + cnφn
(7-37)
where the coefficients c are to be determined so that ψ∗ ˆHψ dτ
= ¯E
(7-38)
ψ∗ψ dτ is minimized. Substituting Eq. (7-37) into Eq. (7-38) gives
ˆ
¯
c∗φ∗ + c∗φ∗ + · · · + c∗ H (c1φ1 + c2φ2 + · · · + cnφn) dτ
E = 1 1
2 2
nφ∗
n
c∗φ∗ + c∗φ∗ + · · · + c∗ (c 1 1
2 2
nφ∗
n
1φ1 + c2φ2 + · · · + cnφn) dτ = num
(7-39)
denom
Since we will be dealing with cases in which the c’s and φ’s are real, we will temporarily omit the complex conjugate notation to simplify the derivation. At the minimum value of ¯
E,
∂ ¯ E = ∂ ¯E =···= ∂ ¯E =0
(7-40)
∂c1
∂c2
∂cn
Chapter 7 The Variation MethodFigure 7-2 Values of ψ versus z for 1s state of H atom (—-) and for approximate wavefunction given by 0.982 1s −0.188 2pz (- - -). The nucleus is at z = 0 for each case.
The partial derivative of Eq. (7-39) with respect to c1 is
∂ ¯
E
φ ˆ
H
=
1
(c1φ1 + · · · + cnφn) dτ + (c1φ1 +···+cnφn) ˆHφ1 dτ
∂c1
denom
denom
− (num) (denom)−2 φ1 (c1φ1 + · · · + cnφn) dτ
+
(c1φ1 + · · · + cnφn) φ1dτ = 0
(7-41)
Section 7-4 Linear Variation: The Polarizability of the Hydrogen Atom
Multiplying through by denom, recalling that num/denom equals ¯ E, and rearranging, gives
c
ˆ
1
φ1H φ1 dτ − ¯
E
φ1φ1 dτ
+c
ˆ
2
φ1H φ2 dτ − ¯
E
φ1φ2 dτ
+ · · · + c
ˆ
n
φ1H φn dτ − ¯
E
φ1φn dτ = 0
(7-42)
At this point, it is convenient to switch to an abbreviated notation:
φ ˆ i H φj dτ ≡ Hij
(7-43)
φiφj dτ ≡ Sij
(7-44)
The integral Sij is normally called an overlap integral since its value is, in certain cases, an indication of the extent to which the two functions φi and φj occupy the same space. Use of this abbreviated notation produces, for Eq. (7-42),
c1 H11 − ¯ ES11 + c2 H12 − ¯ ES12 + · · · + cn H1n − ¯ ES1n = 0
(7-45)
A similar treatment for ∂ ¯ E/∂ci gives a similar equation:
c1 Hi1 − ¯ ESi1 + c2 Hi2 − ¯ ESi2 + · · · + cn Hin − ¯ ESin = 0
(7-46)
Thus, requiring that ∂ ¯ E/∂ci vanish for all coefficients produces n homogeneous linear equations (homogeneous, all equal zero; linear, all ci’s to first power). If one chooses a value for ¯ E, there remain n unknowns—the coefficients ci. (The integrals Hij and Sij are presumably knowable since ˆ H and the functions φi are known.) Of course, one trivial solution for Eqs. (7-46) is always possible, namely, c1 = c2 = · · · = cn = 0. But this corresponds to ψ = 0, a case of no physical interest. Are there nontrivial solutions as well? In quantum chemical calculations, nontrivial solutions usually exist only for
certain discrete values of ¯ E. This provides the approach for solving the problem.
First, find those values of ¯ E for which nontrivial coefficients exist. Second, substitute into Eqs. (7-46) whichever of these values of ¯ E one is interested in and solve for the coefficients. (Each value of ¯ E has its own associated set of coefficients.) But how do we find these particular values of ¯ E that yield nontrivial solutions to Eqs. (7-46)? The answer is given in Appendix 2, where it is shown that the condition which must be met
by the coefficients of a set of linear homogeneous equations in order that nontrivial
solutions exist is that their determinant vanish. Notice that, in the standard treatment given in Appendix 2, the coefficients are known and x, y, and z are unknown. Here, however, the coefficients ci are unknown, and Hij and Sij are known. Therefore, it is the determinant of the H ’s, S’s and ¯ E in Eqs. (7-46) that must equal zero:
H
11 − ¯ES11 H12 − ¯ES12 · · · H1n − ¯ES1n
H21 − ¯ES21 H22 − ¯ES22 · · · H2n − ¯ES2n
.
.
.
= 0
(7-47)
.
.
.
.
.
.
H
n1 − ¯ ESn1 Hn2 − ¯ ESn2 · · · Hnn − ¯
ESnn
Chapter 7 The Variation Method
Expansion of this determinant gives a single equation containing the unknown ¯ E. Any value of ¯
E satisfying this equation is associated with a nontrivial set of coefficients.
The lowest of these values of ¯ E is the minimum average energy achievable by variation of the coefficients. Substitution of this value of ¯ E back into Eqs. (7-46) produces n simultaneous equations for the n coefficients. Equation (7-47) is referred to as the secular equation, and the determinant on the left-hand side is called the secular
determinant.
This method is best illustrated by example, and we will now proceed with the prob lem of a hydrogen atom in a z-directed uniform electric field of strength F a.u. As mentioned earlier, a suitable choice of functions to mix together to approximate the accurate wavefunction is the 1s and 2p, hydrogenlike functions. The choice of two basis functions leads to a 2 × 2 secular determinantal equation:
H 11 − ¯ES11 H12 − ¯ES12
= 0
(7-48)
H21 − ¯ ES21 H22 − ¯
ES22
If we arbitrarily associate the 1s function with the index 1 and the 2pz function with index 2 (consistent with ψ = c11s + c22pz), then the terms in the determinant are (returning to general complex conjugate notation):
H11 =
1s∗ ˆ
H 1s dτ
H12 =
1s∗ ˆ
H 2pz dτ
H
ˆ
ˆ 21 = 2p∗zH 1s dτ H22 = 2p∗zH 2pz dτ
(7-49)
(The electron label has been omitted since there is only one electron.) The corresponding S integrals are obtained from these if ˆ H is omitted in each case. The hamiltonian operator is just that for a hydrogen atom with an additional term to account for the z-directed field:
ˆ
H = − 1 ∇2 − (1/r) − F r cos θ
(7-50)
2
[The energy of a charge −e in a uniform electric field of strength F and direction z is −eF z. In atomic units, one unit of field strength is e/a2 = 5.142 × 1011 V /m. The unit
0
of charge in atomic units is e, so this symbol does not appear explicitly in Eq. (7-50).
Also, the identity z = r cos θ has been used.] This may also be written
ˆ
H = ˆ Hhyd − F r cos θ
(7-51)
where ˆ Hhyd is the hamiltonian for the unperturbed hydrogen atom.
The secular determinant contains four H -type and four S-type terms. However, evaluating these eight integrals turns out to be much easier than one might expect.
In the first place, S12 = S21 since these integrals differ only in the order of the two functions in the integrand, and the functions commute. Also, because ˆ H is hermitian, it follows immediately that H21 = H ∗ . This leaves us with three S terms and three H
12
terms to evaluate. The three S terms are simple. Because the hydrogenlike functions are orthonormal, S11 and S22 equal unity, and S12 vanishes. These points have already reduced the secular determinantal equation to
H
11 − ¯E
H12
=
0
(7-52)
H ∗
H
12
22 − ¯
E
Section 7-4 Linear Variation: The Polarizability of the Hydrogen Atom Consider next the term H11. This may be written as
H11 =
1s∗ ˆ
Hhyd1s dτ − 1s∗ (F r cos θ ) 1s dτ
(7-53)
But the 1s function is an eigenfunction of ˆ Hhyd with eigenvalue − 1 a.u. Therefore, the
2
first integral on the right-hand side of Eq. (7-53) is
1s∗ ˆ
Hhyd1s dτ = − 1 1s∗1s dτ = − 1 a.u.
(7-54)
2
2
The second term on the right-hand side of Eq. (7-53) is zero by symmetry since 1s∗1s is symmetric for reflection in the xy plane while r cos θ (= z) is antisymmetric. Thus, H11 = − 1 a.u. Similarly, H a.u. [Recall that the 8 eigenvalues of hydrogen
2
22 = − 1
8
are equal to −1/(2n2), and here n = 2.] All that remains is H12:
H12 =
1s∗ ˆ Hhyd2pz dτ − F 1s∗ (r cos θ ) 2pz dτ
(7-55)
The first term on the right-hand side is easily shown to be zero:
1s∗ ˆ
Hhyd2pz dτ = 1s∗ − 1 2pz dτ = 0
(7-56)
8
Here we employ the fact that 2pz is an eigenfunction of ˆ Hhyd and then the fact that 1s and 2pz AOs are orthogonal. The second term on the right-hand side must be written out in full and integrated by “brute force.” It is remarkable that, of the eight terms originally considered, only one needs to be done by detailed integration. Proceeding, we substitute formulas for 1s and 2pz in this last integral to obtain
−F (π )−1/2 exp(−r) r cos θ (32π)−1/2r exp(−r/2) cos θ(r2 sin θ) dr dθ dφ
(7-57)
We consider the integration over spin to have been carried out already, giving a factor of unity. Integrating over φ to obtain 2π , and regrouping terms gives
√
∞
π
−2πF/(4 2π) r4 exp(−3r/2) dr cos2 θ sin θ dθ
(7-58)
0
0
√
= −
4!
2
F /2 2
= −215/2F a.u.
(7-59)
3 5
3
35
2
This completes the task of evaluating the terms in the secular determinant.2 The final result is (in atomic units)
−1 − ¯
E
−215/2(F/35)
2
= 0
(7-60)
−215/2(F/35) −1 − ¯E
8
2Since H12 is real, it is clear that H21 = H12.
Chapter 7 The Variation Method which expands to 1 +(5 ¯E/8)+ ¯E2 −215F2/310 =0
(7-61)
16
This is quadratic in ¯ E having roots
9
1/2
¯
+ 217F 2/310
E = − 5 ± 64
(7-62)
16
2
When no external field is present, F = 0 and the roots are just − 1 and − 1 a.u., the
2
8
1s and 2pz eigenvalues for the unperturbed hydrogen atom. As F increases from zero, the roots change, as indicated in Fig. 7-3. We see that, for a given field strength, there are only two values of ¯ E that will cause the determinant to vanish. If either of these two values of ¯ E is substituted into the simultaneous equations related to Eq. (7-52), then nontrivial values for c1 and c2 can be found. Thus, at F = 0.1 a.u., ¯
E = −0.51425 a.u., and ¯ E = −0.1107 a.u. are values of ¯ E for which ∂ ¯ E/∂c1, and ∂ ¯ E/∂c2 both vanish. The former is the minimum, the latter the maximum in the curve of ¯
E versus c1 (Fig. 7-4). (The normality requirement results in the two c’s being dependent, and so ¯
E may be plotted against either of them.) Since the variation principle tells us that ¯E ≥ Elowest exact, we can say immediately that the energy of the hydrogen atom in a uniform electric field of 0.1 a.u. is −0.51425 a.u. or lower. That is, −0.51425 a.u. is an upper bound to the true energy.
Now that we have the value of the lowest ¯ E1 we can solve for c1 and c2 and obtain the approximate ground state wavefunction. The homogeneous equations related to the determinant in Eq. (7-52) are c1(H11 − ¯ E) + c2H12 = 0
(7-63)
c1H12 + c2 H22 − ¯ E = 0
(7-64)
Substituting −0.51425 for ¯ E, and inserting the values for H11, H22, and H12 found earlier gives (when F = 0.1 a.u.)
0.01425c1 − 0.074493c2 = 0
(7-65) −0.074493c1 + 0.38925c2 = 0
(7-66)
Equation (7-65) gives c1 = 5.2275c2
(7-67)
If we substitute this expression for c1 into Eq. (7-66) we get −0.3892c2 + 0.3892c2 = 0
(7-68)
This is useless for evaluating c2. It is one of the properties of such a set of homogeneous equations that the last unused equation is useless for determining coefficients. This arises because an equation like (7-63) still equals zero when c1 and c2 are both multiplied by the same arbitrary constant. Therefore, these equations are inherently capable of telling us the ratio of c1 and c2 only, and not their absolute values. We shall determine
Section 7-4 Linear Variation: The Polarizability of the Hydrogen AtomFigure 7-3 Average energies for a hydrogen atom in a uniform electric field of strength F as given by a linear variation calculation using a 1s, 2pz basis. (- - -) Results from accurate calculations.
absolute values by invoking the requirement that ψ be normalized. In this case, this means that c2 +
=
1
c22
1
(7-69)
or (5.2275c2)2 + c2 =
2
1
(7-70)
which gives c2 = ± 0.18789
(7-71)
If we arbitrarily choose the positive root for c2, it follows from Eq. (7-67) that c1 = 0.98219.
Chapter 7 The Variation MethodFigure 7-4 ¯E versus c1 for a hydrogen atom in a uniform electric field of strength 0.1 a.u.
Thus, when F = 0.1 a.u., the linear variation method using a 1s, 2pz basis set gives an upper bound to the energy of −0.51425 a.u. and a corresponding approximate wavefunction of ψ = 0.98219 1s + 0.18789 2pz
(7-72)
As mentioned earlier, the admixture of 2pz with 1s produces the skewed charge distri bution shown in Fig. 7-2.
It is important to note that the extent of mixing between 1s and 2pz depends partly on the size of the off-diagonal determinantal element H12. When H12 is zero (no external field), no mixing occurs. As H12 increases, mixing increases. Generally speaking, the larger the size of the off-diagonal element connecting two basis functions in the secular determinant, the greater the degree of mixing of these basis functions in the final solution, other factors being equal. Hij is sometimes referred to as the interaction
element between basis functions i and j .
If we carried through the same procedure using the maximum ¯ E of −0.1107 a.u., we would obtain the approximate wavefunction ψ = 0.98219 2pz − 0.18789 1s
(7-73)
This wavefunction is orthogonal to ψ.
It may be proved that the second root, ¯E = −0.1107 is an upper bound for the energy of the second-lowest state of the hydrogen atom in the field. However, ψ is probably not too good an approximation to the exact wavefunction for that state. This is partly because the wavefunction ψ is one that maximizes ¯ E. Hence, there is no particular tendency for the procedure to isolate the second-lowest state from the infinite manifold of states. Also, our basis set was chosen with an eye toward its appropriateness for approximating the lowest state.
The true second-lowest state might be expected to contain significant amounts of the 2s AO, which is not included in this basis.
The values of ¯ E versus c1 are plotted in Fig. 7-4. The values of ¯ E that we obtained by the variation procedure correspond to the extrema in this figure. The low-energy extreme corresponds to a wavefunction that shifts negative charge in the direction it is attracted by the field. The high-energy extreme corresponds to a wavefunction that shifts charge in the opposite direction. (When a calculation is performed over a more extensive basis to produce more than two roots, the highest and lowest roots correspond, respectively, to the maximum and minimum, the other roots to saddle points, on the energy hypersurface.)
The detailed treatment just completed is rather involved, and so we now summarize the main points. Step 1 involved selection of a basis set of functions which is capable of approximating the exact solution. Step 2 was the construction of the secular determinant, including evaluation of all the Hij - and Sij -type integrals. Step 3 was the conversion of the determinantal equation into its equivalent equation in powers of ¯E and solution for the roots ¯E. Step 4 was the substitution of an ¯E of interest into the simultaneous equations that are related to the secular determinant and solution for
c1/c2. Finally, we used the normality requirement to arrive at convenient values for c1 and c2.
There are many ways one could increase the flexibility of the trial function in an effort to increase the accuracy of the calculation. By adding additional basis functions, one would stay within the linear variation framework, merely increasing the size of the secular determinant. If these additional basis functions are of appropriate symmetries, they will cause the minimum energy root to be lowered further and will mix into the corresponding wavefunction to make it a better approximation to the lowest-energy eigenfunction for the system. Also, the additional functions will increase the number of roots ¯ E, thereby providing upper bounds for the energies of the third-, fourth-, etc.
lowest states of the system. Another possibility is to allow nonlinear variation of the 1s and 2pz orbital exponents, in combination with linear variation. This would be more involved than the calculation we have shown here, but could easily be accomplished with the aid of a computer.
EXAMPLE 7-1 Suppose the variational process just described were performed using ψ(c1, c2, c3) = c11s + c22p0 + c33d0. Would all three of theseAOs be present in the lowest-energy solution?
SOLUTION We already know that 1s and 2p0 will be mixed. They differ in reflection symmetry in just the right manner to provide a wavefunction skewed in the z direction, and this is manifested as a nonzero interaction element H12. 3d0, however, is, like 1s, symmetric for reflection through the x, y plane. Therefore, it cannot skew 1s in the z direction, and its interaction element with 1s, H13, equals zero (by symmetry). Therefore, it is tempting to think that that 3d0 will not contribute. However, because 3d0 and 2p0 have opposite reflection symmetries through the x, y