Section 4-2
Section 4-2 Separation of Variables
wavefunctions are not used in most quantum chemical applications, so we will not discuss them further in this book.
EXAMPLE 4-5 A nodal plane is one through which a wavefunction is antisymmetric for reflection. Consider the 3dxy and 3dyz orbitals of Fig. 4-8. Through which planes do these two orbitals have different reflection symmetries?
SOLUTION 3dxy is antisymmetric for reflection only through the x, z and y, z planes. 3dyz
is antisymmetric for reflection only through the x, z and x, y planes. Hence, these orbitals differ in their symmetries for reflection through the y, z and x, y planes.
4-2 Separation of Variables
We shall indicate in some detail the way in which the Schr¨odinger equation (4-6) is solved. Recall the strategy of separating variables which we used in Section 2-7: 1. Express ψ as a product of functions, each depending on only one variable.
2. Substitute this product into the Schr¨odinger equation and try to manipulate it so that the equation becomes a sum of terms, each depending on a single variable. These terms must sum to a constant.
3. Since terms for different variables are independent of each other, the terms for each variable must equal a constant. This enables one to set up an equation in each variable. If this can be done, the initial assumption (1) is justified.
In this case we begin by assuming that ψ(r, θ, φ) = R(r) (θ)(φ)
(4-31)
Substituting into Eq. (4-6) gives
−h2
d
1
d
d
1
d2
r2 dR
+ R sin θ + R 8π 2µr2
dr
dr
sin θ dθ
dθ
sin2 θ dφ2 − Ze2 R = ER (4-32) 4π ε0r
Since each derivative operator now acts on a function of a single coordinate, we use total, rather than partial, derivative notation.
Let us first see if we can isolate the φ dependence. Multiplying Eq. (4-32) by (−8πµr2 sin2 θ/ h2R ) and rearranging gives
sin2 θ d
r2 dR + 8π2µr2 sin2 θ E + Ze2 R
dr
dr
h2
4π ε0r
+ sin θ d
d
d2
sin θ + 1 = 0 (4-33)
dθ
dθ
dφ2
The r and θ dependence is still mixed in the first two terms, but we now have a rather simple term in the coordinate φ. Now we can argue, as in Section 2-7, that, as φ alone Chapter 4 The Hydrogenlike Ion, Angular Momentum, and the Rigid Rotor changes, the first three terms in Eq. (4-33) do not change. That is, if only φ changes, Eq. (4-33) may be written constant + constant + constant + (1/)(d2/dφ2) = 0
(4-34)
so that (1/)(d2/dφ2) = −m2 (a constant)
(4-35)
We call the constant −m2 for future mathematical convenience. We can rearrange Eq. (4-35) into the more familiar form for an eigenvalue equation: d2/dφ2 = −m2
(4-36)
We arrived at Eq. (4-36) by assuming that only φ changes while r and θ are constant.
However, it should be obvious that the behavior of the term in φ is uninfluenced by changes in r and θ since it has no dependence on these coordinates. Thus, by establishing that this term is constant under certain circumstances, we have actually shown that it must be constant under all circumstances, and we have produced an eigenvalue equation for .
We can now proceed with further separation of variables. Since we know that the last term in Eq. (4-33) is a constant, we can write
1 d
d
d
r2 dR + 8π2µr2 E + Ze2 +
1
sin θ − m2 = 0
R dr
dr
h2
4π ε0r sin θ dθ
dθ
sin2 θ
(4-37)
Note that we have separated the θ and r dependences by dividing through by sin2 θ .
We now have two terms wholly dependent on r and two wholly dependent on θ , their sum being zero. Hence, as before, the sum of the two r-dependent terms must equal a constant, β, and the sum of the θ -dependent terms must equal −β. Thus
d
r2 dR + 8π2µr2 E + Ze2 R = βR
(4-38)
dr
dr
h2
4π ε0r
1
d
d
sin θ − m2 = −β
(4-39)
sin θ dθ
dθ
sin2 θ where we have multiplied through by R in the first equation and by in the second.
The assumption that ψ = R has led to separate equations for R, , and . This indicates that the assumption of separability was valid. However, there is some linkage between R and via β, and between and via m.
4-3 Solution of the R, , and Equations4-3.A The Equation The solution of Eq. (4-36) is similar to that of the particle in a ring problem of Section 2-6. The normalized solutions are
√
= (1/ 2π) exp(imφ), m = 0, ±1, ±2, . . .
(4-40)
As shown in Section 2-6, the constant m must be an integer if is to be a single-valued function.
Section 4-3 Solution of the R, , and Equations4-3.B The Equation There is great similarity between the mathematical techniques used in solving the R and equations and those used to solve the one-dimensional harmonic oscillator problem of Chapter 3. Hence, we will only summarize the steps involved in these solutions and make a few remarks about the results. More detailed treatments are presented in many
texts.5
The equation can be solved as follows:
1. Change the variable to obtain a more convenient form for the differential equation.
2. Express the solution as a power series and obtain a recursion relation.
3. Observe that the series diverges for certain values of the variables, producing nonsquare-integrable wavefunctions. Correct this by requiring that the series terminate. This requires that the truncated series be either symmetric or antisymmetric in the variable and also that β of Eq. (4-38) and (4-39) be equal to l(l + 1) with l an integer.
4. Recognize these truncated series as being the associated Legendre functions.
5. Return to the original variable to obtain an expression for in terms of the starting coordinate.
Reference to the end of Section 3-4 will illustrate the similarity between this and the harmonic oscillator case.
The final result for m 0 is
(2l + 1) (l − |m|)! 1/2 |
m|
l,m(θ ) = (−1)m
P (cos θ )
(4-41)
2
(l + |m|)!
l
For m
wavefunctions are not used in most quantum chemical applications, so we will not discuss them further in this book.
EXAMPLE 4-5 A nodal plane is one through which a wavefunction is antisymmetric for reflection. Consider the 3dxy and 3dyz orbitals of Fig. 4-8. Through which planes do these two orbitals have different reflection symmetries?
SOLUTION 3dxy is antisymmetric for reflection only through the x, z and y, z planes. 3dyz
is antisymmetric for reflection only through the x, z and x, y planes. Hence, these orbitals differ in their symmetries for reflection through the y, z and x, y planes.
4-2 Separation of Variables
We shall indicate in some detail the way in which the Schr¨odinger equation (4-6) is solved. Recall the strategy of separating variables which we used in Section 2-7: 1. Express ψ as a product of functions, each depending on only one variable.
2. Substitute this product into the Schr¨odinger equation and try to manipulate it so that the equation becomes a sum of terms, each depending on a single variable. These terms must sum to a constant.
3. Since terms for different variables are independent of each other, the terms for each variable must equal a constant. This enables one to set up an equation in each variable. If this can be done, the initial assumption (1) is justified.
In this case we begin by assuming that ψ(r, θ, φ) = R(r) (θ)(φ)
(4-31)
Substituting into Eq. (4-6) gives
−h2
d
1
d
d
1
d2
r2 dR
+ R sin θ + R 8π 2µr2
dr
dr
sin θ dθ
dθ
sin2 θ dφ2 − Ze2 R = ER (4-32) 4π ε0r
Since each derivative operator now acts on a function of a single coordinate, we use total, rather than partial, derivative notation.
Let us first see if we can isolate the φ dependence. Multiplying Eq. (4-32) by (−8πµr2 sin2 θ/ h2R ) and rearranging gives
sin2 θ d
r2 dR + 8π2µr2 sin2 θ E + Ze2 R
dr
dr
h2
4π ε0r
+ sin θ d
d
d2
sin θ + 1 = 0 (4-33)
dθ
dθ
dφ2
The r and θ dependence is still mixed in the first two terms, but we now have a rather simple term in the coordinate φ. Now we can argue, as in Section 2-7, that, as φ alone Chapter 4 The Hydrogenlike Ion, Angular Momentum, and the Rigid Rotor changes, the first three terms in Eq. (4-33) do not change. That is, if only φ changes, Eq. (4-33) may be written constant + constant + constant + (1/)(d2/dφ2) = 0
(4-34)
so that (1/)(d2/dφ2) = −m2 (a constant)
(4-35)
We call the constant −m2 for future mathematical convenience. We can rearrange Eq. (4-35) into the more familiar form for an eigenvalue equation: d2/dφ2 = −m2
(4-36)
We arrived at Eq. (4-36) by assuming that only φ changes while r and θ are constant.
However, it should be obvious that the behavior of the term in φ is uninfluenced by changes in r and θ since it has no dependence on these coordinates. Thus, by establishing that this term is constant under certain circumstances, we have actually shown that it must be constant under all circumstances, and we have produced an eigenvalue equation for .
We can now proceed with further separation of variables. Since we know that the last term in Eq. (4-33) is a constant, we can write
1 d
d
d
r2 dR + 8π2µr2 E + Ze2 +
1
sin θ − m2 = 0
R dr
dr
h2
4π ε0r sin θ dθ
dθ
sin2 θ
(4-37)
Note that we have separated the θ and r dependences by dividing through by sin2 θ .
We now have two terms wholly dependent on r and two wholly dependent on θ , their sum being zero. Hence, as before, the sum of the two r-dependent terms must equal a constant, β, and the sum of the θ -dependent terms must equal −β. Thus
d
r2 dR + 8π2µr2 E + Ze2 R = βR
(4-38)
dr
dr
h2
4π ε0r
1
d
d
sin θ − m2 = −β
(4-39)
sin θ dθ
dθ
sin2 θ where we have multiplied through by R in the first equation and by in the second.
The assumption that ψ = R has led to separate equations for R, , and . This indicates that the assumption of separability was valid. However, there is some linkage between R and via β, and between and via m.
4-3 Solution of the R, , and Equations4-3.A The Equation The solution of Eq. (4-36) is similar to that of the particle in a ring problem of Section 2-6. The normalized solutions are
√
= (1/ 2π) exp(imφ), m = 0, ±1, ±2, . . .
(4-40)
As shown in Section 2-6, the constant m must be an integer if is to be a single-valued function.
Section 4-3 Solution of the R, , and Equations4-3.B The Equation There is great similarity between the mathematical techniques used in solving the R and equations and those used to solve the one-dimensional harmonic oscillator problem of Chapter 3. Hence, we will only summarize the steps involved in these solutions and make a few remarks about the results. More detailed treatments are presented in many
texts.5
The equation can be solved as follows:
1. Change the variable to obtain a more convenient form for the differential equation.
2. Express the solution as a power series and obtain a recursion relation.
3. Observe that the series diverges for certain values of the variables, producing nonsquare-integrable wavefunctions. Correct this by requiring that the series terminate. This requires that the truncated series be either symmetric or antisymmetric in the variable and also that β of Eq. (4-38) and (4-39) be equal to l(l + 1) with l an integer.
4. Recognize these truncated series as being the associated Legendre functions.
5. Return to the original variable to obtain an expression for in terms of the starting coordinate.
Reference to the end of Section 3-4 will illustrate the similarity between this and the harmonic oscillator case.
The final result for m 0 is
(2l + 1) (l − |m|)! 1/2 |
m|
l,m(θ ) = (−1)m
P (cos θ )
(4-41)
2
(l + |m|)!
l
For m