Section 12-4
Section 12-4 The Ground-State Energy to First-Order of Heliumlike Systems
Therefore, ψ3 contributes to φ(1) with a coefficient about 3 the magnitude of the con 2
5
tribution from ψ1. The sign of c(1) is positive and a sketch of ψ
32
2 plus a small amount of ψ3 will demonstrate that this causes charge shifting to the left in accord with our general statement above.
Since ψ4 contributes nothing (by symmetry) and ψ5 is fairly distant in energy, we will neglect all contributions to φ(2) above ψ
2
3.
We have, then, two sizable contributions to φ(1), each favoring charge shifts in
2
opposite directions. Let us see how W (2) reflects this:
2
W (2) ∼ = c(1)2(E
(E
2
12
2 − E1) + c(1)2
32
2 − E3)
(12-42a)
∼
3
= c(1)2(3) + ( c(1))2(−5) π2/2L2
(12-42b)
12
5 12
∼
= 1.2c(1)2π2/2L2
(12-42c)
12
W (2) is the difference between energy contributions of opposite sign. The net result
2
(energy increases) comes about in this case because ψ1 contributes more heavily than ψ3.
The fact that contributing wavefunctions from lower and higher energies affect W (2)
i
oppositely is made evident by Eq. (12-22). Hence, we can extend our general statement above by adding that the second-order contribution to the energy from states below ψi in energy cause the energy of the j th state to go up, contributions from above cause it to go down.
Because the unperturbed energies of the particle in the box increase as n2, any state (except the lowest) is closer in energy to states below than to states above. Hence, for at least some kinds of perturbation, we might expect these states to “feel” the effects of states at lower energies more strongly and to rise in energy (as far as W (2) is concerned).
This is what happens in this example. There are no states below the lowest, and so W (2) cannot be positive, but W (2) is positive and it turns out that W (2) is positive for
1
2
i all higher i as well.
It is interesting to compare these results with those from classical physics. Classi cally, the particle moves most slowly at the top of the potential gradient and therefore spends most of its time there. The lowest-energy state has responded in the opposite manner, in a way we might call anticlassical. The second and all higher states have responded classically.
12-4 The Ground-State Energy to First-Orderof Heliumlike Systems
The hamiltonian for a two-electron atom or ion with nuclear charge Z a.u. is (neglecting relativistic effects and assuming infinite nuclear mass)
H (1, 2) = − 1 ∇2 + ∇2 − Z/r1 − Z/r2 + 1/r12
(12-43)
2
1
2
This may be written as a sum of one-electron operators and a two-electron operator: H (1, 2) = H (1) + H (2) + 1/r12
(12-44)
Chapter 12 Time-Independent Rayleigh–Schr ¨odinger Perturbation Theory where H (i) is simply the hamiltonian for the hydrogenlike system with nuclear charge Z: H (i) = − 1 ∇2 − Z/ri
(12-45)
2 i
Since we know the eigenvalues and eigenfunctions for H (i), we can let H (1) + H (2)
be the unperturbed hamiltonian with 1/r12 the perturbation. Such a perturbation is not very small, but it is of interest to see how well the method works in such a case. We have, therefore, for the lowest-energy state of the system: H0 = H (1) + H (2)
(12-46)
H = 1/r12
(12-47)
ψ1 = (Z3/π) exp [−Z(r1 + r2)]
(12-48)
E1 = −Z2/2 − Z2/2 = −Z2
(12-49)
W (1) = ψ
1
1|1/r12|ψ1
(12-50)
All quantities are in atomic units. The unperturbed wavefunction [Eq. (12-48)] is simply the product of two one-electron 1s AOs. Because this is an eigenfunction of the system in the absence of interelectronic repulsion, it is too contracted about the nucleus.
The first-order correction to the energy is the repulsion between the two electrons in this overly contracted eigenfunction [Eq. (12-50)]. We have encountered this same repulsion integral in our earlier variational calculation on helium-like systems. The evaluation of this integral is described in Appendix 3. Its value is 5Z/8. Hence, to first order, W1 = E1 + W (1) = −Z2 + 5Z/8
(12-51)
1
In Table 12-1, this result is compared with exact energies for the first ten members of this series. Several points should be noted:
1. The effect of the perturbation to first order is to increase the ground-state energy.
This is expected since 1/r12 is always positive (i.e., repulsive).
2. The energy to first order is never below the exact ground-state energy. This is a general property of perturbation calculations as is easily proved (Problem 12-1).
3. The energy to first order is in error by a fairly constant amount throughout the series.
For H −, this gives a substantial percentage of error and fails to show H − stable compared with an H atom and an unbound electron. For higher Z, this error becomes relatively smaller since 1/r12 becomes relatively less important compared with Z/r˙ι.
The assumption that H is a small perturbation is thus better fulfilled at large Z and results in W (1) being a much smaller correction relative to the total energy
1
W1. Because “large” and “small” are relative terms, they can be misleading. Since energies of chemical interest are often small differences between large numbers (see the discussion at the end of Chapter 7), errors that were originally relatively small can become relatively large after the subtraction. Therefore, even though perturbation terminology would suggest that the results at Z = 10 are better than those at Z = 1, this may not be the case for some practical applications.
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Therefore, ψ3 contributes to φ(1) with a coefficient about 3 the magnitude of the con 2
5
tribution from ψ1. The sign of c(1) is positive and a sketch of ψ
32
2 plus a small amount of ψ3 will demonstrate that this causes charge shifting to the left in accord with our general statement above.
Since ψ4 contributes nothing (by symmetry) and ψ5 is fairly distant in energy, we will neglect all contributions to φ(2) above ψ
2
3.
We have, then, two sizable contributions to φ(1), each favoring charge shifts in
2
opposite directions. Let us see how W (2) reflects this:
2
W (2) ∼ = c(1)2(E
(E
2
12
2 − E1) + c(1)2
32
2 − E3)
(12-42a)
∼
3
= c(1)2(3) + ( c(1))2(−5) π2/2L2
(12-42b)
12
5 12
∼
= 1.2c(1)2π2/2L2
(12-42c)
12
W (2) is the difference between energy contributions of opposite sign. The net result
2
(energy increases) comes about in this case because ψ1 contributes more heavily than ψ3.
The fact that contributing wavefunctions from lower and higher energies affect W (2)
i
oppositely is made evident by Eq. (12-22). Hence, we can extend our general statement above by adding that the second-order contribution to the energy from states below ψi in energy cause the energy of the j th state to go up, contributions from above cause it to go down.
Because the unperturbed energies of the particle in the box increase as n2, any state (except the lowest) is closer in energy to states below than to states above. Hence, for at least some kinds of perturbation, we might expect these states to “feel” the effects of states at lower energies more strongly and to rise in energy (as far as W (2) is concerned).
This is what happens in this example. There are no states below the lowest, and so W (2) cannot be positive, but W (2) is positive and it turns out that W (2) is positive for
1
2
i all higher i as well.
It is interesting to compare these results with those from classical physics. Classi cally, the particle moves most slowly at the top of the potential gradient and therefore spends most of its time there. The lowest-energy state has responded in the opposite manner, in a way we might call anticlassical. The second and all higher states have responded classically.
12-4 The Ground-State Energy to First-Orderof Heliumlike Systems
The hamiltonian for a two-electron atom or ion with nuclear charge Z a.u. is (neglecting relativistic effects and assuming infinite nuclear mass)
H (1, 2) = − 1 ∇2 + ∇2 − Z/r1 − Z/r2 + 1/r12
(12-43)
2
1
2
This may be written as a sum of one-electron operators and a two-electron operator: H (1, 2) = H (1) + H (2) + 1/r12
(12-44)
Chapter 12 Time-Independent Rayleigh–Schr ¨odinger Perturbation Theory where H (i) is simply the hamiltonian for the hydrogenlike system with nuclear charge Z: H (i) = − 1 ∇2 − Z/ri
(12-45)
2 i
Since we know the eigenvalues and eigenfunctions for H (i), we can let H (1) + H (2)
be the unperturbed hamiltonian with 1/r12 the perturbation. Such a perturbation is not very small, but it is of interest to see how well the method works in such a case. We have, therefore, for the lowest-energy state of the system: H0 = H (1) + H (2)
(12-46)
H = 1/r12
(12-47)
ψ1 = (Z3/π) exp [−Z(r1 + r2)]
(12-48)
E1 = −Z2/2 − Z2/2 = −Z2
(12-49)
W (1) = ψ
1
1|1/r12|ψ1
(12-50)
All quantities are in atomic units. The unperturbed wavefunction [Eq. (12-48)] is simply the product of two one-electron 1s AOs. Because this is an eigenfunction of the system in the absence of interelectronic repulsion, it is too contracted about the nucleus.
The first-order correction to the energy is the repulsion between the two electrons in this overly contracted eigenfunction [Eq. (12-50)]. We have encountered this same repulsion integral in our earlier variational calculation on helium-like systems. The evaluation of this integral is described in Appendix 3. Its value is 5Z/8. Hence, to first order, W1 = E1 + W (1) = −Z2 + 5Z/8
(12-51)
1
In Table 12-1, this result is compared with exact energies for the first ten members of this series. Several points should be noted:
1. The effect of the perturbation to first order is to increase the ground-state energy.
This is expected since 1/r12 is always positive (i.e., repulsive).
2. The energy to first order is never below the exact ground-state energy. This is a general property of perturbation calculations as is easily proved (Problem 12-1).
3. The energy to first order is in error by a fairly constant amount throughout the series.
For H −, this gives a substantial percentage of error and fails to show H − stable compared with an H atom and an unbound electron. For higher Z, this error becomes relatively smaller since 1/r12 becomes relatively less important compared with Z/r˙ι.
The assumption that H is a small perturbation is thus better fulfilled at large Z and results in W (1) being a much smaller correction relative to the total energy
1
W1. Because “large” and “small” are relative terms, they can be misleading. Since energies of chemical interest are often small differences between large numbers (see the discussion at the end of Chapter 7), errors that were originally relatively small can become relatively large after the subtraction. Therefore, even though perturbation terminology would suggest that the results at Z = 10 are better than those at Z = 1, this may not be the case for some practical applications.
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28
)
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exactE
[4].
1
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Knight
=
and
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when
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= = = = = = = = =
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+
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e +
+
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3
4
5
6
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10
405