Comprehensive Study Notes
Most of this chapter deals with the electronic Schrödinger equation for diatomic molecules, but this section examines nuclear motion in a bound electronic state of a diatomic molecule. From (13.10) and (13.11), the Schrödinger equation for nuclear motion in a diatomicmolecule bound electronic state is
\(
\begin{equation}
\left[-\frac{\hbar^{2}}{2 m{\alpha}} \nabla{\alpha}^{2}-\frac{\hbar^{2}}{2 m{\beta}} \nabla{\beta}^{2}+U(R)\right] \psi{N}=E \psi{N} \tag{13.13}
\end{equation}
\)
where $\alpha$ and $\beta$ are the nuclei, and the nuclear-motion wave function $\psi{N}$ is a function of the nuclear coordinates $x{\alpha}, y{\alpha}, z{\alpha}, x{\beta}, y{\beta}, z_{\beta}$.
The potential energy $U(R)$ is a function of only the relative coordinates of the two nuclei, and the work of Section 6.3 shows that the two-particle Schrödinger equation (13.13) can be reduced to two separate one-particle Schrödinger equations-one for translational energy of the entire molecule and one for internal motion of the nuclei relative to each other. We have
\(
\begin{equation}
\psi{N}=\psi{N, \mathrm{tr}} \psi{N, \text { int }} \text { and } E=E{\mathrm{tr}}+E_{\mathrm{int}} \tag{13.14}
\end{equation}
\)
The translational energy levels can be taken as the energy levels (3.72) of a particle in a three-dimensional box whose dimensions are those of the container holding the gas of diatomic molecules.
The Schrödinger equation for $\psi_{N, \text { int }}$ is [Eq. (6.43)]
\(
\begin{equation}
\left[-\frac{\hbar^{2}}{2 \mu} \nabla^{2}+U(R)\right] \psi{N, \mathrm{int}}=E{\mathrm{int}} \psi{N, \mathrm{int}}, \quad \mu \equiv m{\alpha} m{\beta} /\left(m{\alpha}+m_{\beta}\right) \tag{13.15}
\end{equation}
\)
where $\psi{N, \text { int }}$ is a function of the coordinates of one nucleus relative to the other. The best coordinates to use are the spherical coordinates of one nucleus relative to the other (Fig. 6.5 with $m{N}$ and $m{e}$ replaced by $m{\alpha}$ and $m{\beta}$ ). The radius $r$ in relative spherical coordinates is the internuclear distance $R$, and we shall denote the relative angular coordinates by $\theta{N}$ and $\phi_{N}$. Since the potential energy in (13.15) depends on $R$ only, this is a central-force problem, and the work of Section 6.1 shows that
\(
\begin{equation}
\psi{N, \text { int }}=P(R) Y{J}^{M}\left(\theta{N}, \phi{N}\right), \quad J=0,1,2, \ldots, \quad M=-J, \ldots, J \tag{13.16}
\end{equation}
\)
where the $Y_{J}^{M}$ functions are the spherical harmonic functions with quantum numbers $J$ and $M$.
From (6.17), the radial function $P(R)$ is found by solving
\(
\begin{equation}
-\frac{\hbar^{2}}{2 \mu}\left[P^{\prime \prime}(R)+\frac{2}{R} P^{\prime}(R)\right]+\frac{J(J+1) \hbar^{2}}{2 \mu R^{2}} P(R)+U(R) P(R)=E_{\mathrm{int}} P(R) \tag{13.17}
\end{equation}
\)
This differential equation is simplified by defining $F(R)$ as
\(
\begin{equation}
F(R) \equiv R P(R) \tag{13.18}
\end{equation}
\)
Substitution of $P=F / R$ into (13.17) gives [Eq. (6.137)]
\(
\begin{equation}
-\frac{\hbar^{2}}{2 \mu} F^{\prime \prime}(R)+\left[U(R)+\frac{J(J+1) \hbar^{2}}{2 \mu R^{2}}\right] F(R)=E_{\mathrm{int}} F(R) \tag{13.19}
\end{equation}
\)
which is a one-dimensional Schrödinger equation with the effective potential energy $U(R)+J(J+1) \hbar^{2} / 2 \mu R^{2}$.
The most fundamental way to solve (13.19) is as follows: (a) Solve the electronic Schrödinger equation (13.7) at several values of $R$ to obtain $E{\mathrm{el}}$ of the particular molecular electronic state you are interested in; (b) add $Z{\alpha} Z{\beta} e^{2} / 4 \pi \varepsilon{0} R$ to each $E_{\text {el }}$ value to obtain $U$ at these $R$ values; (c) devise a mathematical function $U(R)$ whose parameters are adjusted to give a good fit to the calculated $U$ values; (d) insert the function $U(R)$ found in (c) into the nuclear-motion radial Schrödinger equation (13.19) and solve (13.19) by numerical methods.
A commonly used fitting procedure for step (c) is the method of cubic splines, for which computer programs exist (see Press et al., Chapter 3; Shoup, Chapter 6).
As for step (d), numerical solution of the one-dimensional Schrödinger equation (13.19) is done using either the Cooley-Numerov method [see J. Tellinghuisen, J. Chem. Educ., 66,
51 (1989)], which is a modification of the Numerov method (Sections 4.4 and 6.9), or the finite-element method [see D. J. Searles and E. I. von Nagy-Felsobuki, Am. J. Phys., 56, 444 (1988)].
The solutions $F(R)$ of the radial equation (13.19) for a given $J$ are characterized by a quantum number $v$, where $v$ is the number of nodes in $F(R) ; v=0,1,2, \ldots$. The energy levels $E{\text {int }}$ [which are found from the condition that $P(R)=F(R) / R$ be quadratically integrable] depend on the quantum number $J$, which occurs in (13.19), and depend on $v$, which characterizes $F(R) ; E{\text {int }}=E{v, J}$. The angular factor $Y{J}^{M}\left(\theta{N}, \phi{N}\right)$ in (13.16) is a function of the angular coordinates. Changes in $\theta{N}$ and $\phi{N}$ with $R$ held fixed correspond to changes in the spatial orientation of the diatomic molecule, which is rotational motion. The quantum numbers $J$ and $M$ are rotational quantum numbers. Note that $Y_{J}^{M}$ is the wave function of a rigid two-particle rotor [Eq. (6.46)]. A change in the $R$ coordinate is a change in the internuclear distance, which is a vibrational motion, and the quantum number $v$, which characterizes $F(R)$, is a vibrational quantum number.
Since accurate solution of the electronic Schrödinger equation [step (a)] is hard, one often uses simpler, less accurate procedures than that of steps (a) to (d). The simplest approach is to expand $U(R)$ in a Taylor series about $R_{e}$ (Prob. 4.1):
\(
\begin{align}
U(R)= & U\left(R{e}\right)+U^{\prime}\left(R{e}\right)\left(R-R{e}\right)+\frac{1}{2} U^{\prime \prime}\left(R{e}\right)\left(R-R{e}\right)^{2} \
& +\frac{1}{6} U^{\prime \prime \prime}\left(R{e}\right)\left(R-R_{e}\right)^{3}+\cdots \tag{13.20}
\end{align}
\)
At the equilibrium internuclear distance $R{e}$, the slope of the $U(R)$ curve is zero (Fig. 13.1), so $U^{\prime}\left(R{e}\right)=0$. We can anticipate that the molecule will vibrate about the equilibrium distance $R{e}$. For $R$ close to $R{e},\left(R-R{e}\right)^{3}$ and higher powers will be small, and we shall neglect these terms. Defining the equilibrium force constant $k{e}$ as $k{e} \equiv U^{\prime \prime}\left(R{e}\right)$, we have
\(
\begin{gather}
U(R) \approx U\left(R{e}\right)+\frac{1}{2} k{e}\left(R-R{e}\right)^{2}=U\left(R{e}\right)+\frac{1}{2} k{e} x^{2} \tag{13.21}\
k{e} \equiv U^{\prime \prime}\left(R{e}\right) \quad \text { and } \quad x \equiv R-R{e}
\end{gather}
\)
We have approximated $U(R)$ by a parabola [Fig. 4.6 with $V \equiv U(R)-U\left(R{e}\right)$ and $\left.x \equiv R-R{e}\right]$.
With the change of independent variable $x \equiv R-R_{e}$, (13.19) becomes
\(
\begin{gather}
-\frac{\hbar^{2}}{2 \mu} S^{\prime \prime}(x)+\left[U\left(R{e}\right)+\frac{1}{2} k{e} x^{2}+\frac{J(J+1) \hbar^{2}}{2 \mu\left(x+R{e}\right)^{2}}\right] S(x) \approx E{\text {int }} S(x) \tag{13.22}\
\text { where } \quad S(x) \equiv F(R) \tag{13.23}
\end{gather}
\)
Expanding $1 /\left(x+R_{e}\right)^{2}$ in a Taylor series, we have (Prob. 13.7)
\(
\begin{equation}
\frac{1}{\left(x+R{e}\right)^{2}}=\frac{1}{R{e}^{2}\left(1+x / R{e}\right)^{2}}=\frac{1}{R{e}^{2}}\left(1-2 \frac{x}{R{e}}+3 \frac{x^{2}}{R{e}^{2}}-\cdots\right) \approx \frac{1}{R_{e}^{2}} \tag{13.24}
\end{equation}
\)
We are assuming that $R-R{e}=x$ is small compared with $R{e}$, so all terms after the 1 have been neglected in (13.24). Substitution of (13.24) into (13.22) and rearrangement gives
\(
\begin{equation}
-\frac{\hbar^{2}}{2 \mu} S^{\prime \prime}(x)+\frac{1}{2} k{e} x^{2} S(x) \approx\left[E{\mathrm{int}}-U\left(R{e}\right)-\frac{J(J+1) \hbar^{2}}{2 \mu R{e}^{2}}\right] S(x) \tag{13.25}
\end{equation}
\)
Equation (13.25) is the same as the Schrödinger equation for a one-dimensional harmonic oscillator with coordinate $x$, mass $\mu$, potential energy $\frac{1}{2} k{e} x^{2}$, and energy eigenvalues $E{\text {int }}-U\left(R{e}\right)-J(J+1) \hbar^{2} / 2 \mu R{e}^{2}$. [The boundary conditions for (13.25) and (4.32) are not the same, but this difference is unimportant and can be ignored (Levine, Molecular
Spectroscopy, p. 147).] We can therefore set the terms in brackets in (13.25) equal to the harmonic-oscillator eigenvalues, and we have
\(
\begin{gather}
E{\mathrm{int}}-U\left(R{e}\right)-J(J+1) \hbar^{2} / 2 \mu R{e}^{2} \approx\left(v+\frac{1}{2}\right) h \nu{e} \
E{\mathrm{int}} \approx U\left(R{e}\right)+\left(v+\frac{1}{2}\right) h \nu{e}+J(J+1) \hbar^{2} / 2 \mu R{e}^{2} \tag{13.26}\
\nu{e}=\left(k{e} / \mu\right)^{1 / 2} / 2 \pi, \quad v=0,1,2, \ldots \tag{13.27}
\end{gather}
\)
where (4.23) was used for $\nu{e}$, the equilibrium (or harmonic) vibrational frequency. The molecular internal energy $E{\text {int }}$ is approximately the sum of the electronic energy $U\left(R{e}\right) \equiv E{\text {elec }}$ (which differs for different electronic states of the same molecule), the vibrational energy $\left(v+\frac{1}{2}\right) h \nu{e}$, and the rotational energy $J(J+1) \hbar^{2} / 2 \mu R{e}^{2}$. The approximations (13.21) and (13.24) correspond to a harmonic-oscillator, rigid-rotor approximation. From (13.26) and (13.14), the molecular energy $E=E{\text {tr }}+E{\text {int }}$ is approximately the sum of translational, rotational, vibrational, and electronic energies:
\(
E \approx E{\mathrm{tr}}+E{\mathrm{rot}}+E{\mathrm{vib}}+E{\mathrm{elec}}
\)
From (13.14), (13.16), (13.18), and (13.23), the nuclear-motion wave function is
\(
\begin{equation}
\psi{N} \approx \psi{N, \mathrm{tr}} S{v}\left(R-R{e}\right) R^{-1} Y{J}^{M}\left(\theta{N}, \phi_{N}\right) \tag{13.28}
\end{equation}
\)
where $S{v}\left(R-R{e}\right)$ is a harmonic-oscillator eigenfunction with quantum number $v$.
The approximation (13.26) gives rather poor agreement with experimentally observed vibration-rotation energy levels of diatomic molecules. The accuracy can be improved by the addition of the first- and second-order perturbation-theory energy corrections due to the terms neglected in (13.21) and (13.24). When this is done (see Levine, Molecular Spectroscopy, Section 4.2), the energy contains additional terms corresponding to vibrational anharmonicity [Eq. (4.60)], vibration-rotation interaction, and rotational centrifugal distortion of the molecule (Section 6.4), where vibrational anharmonicity is the largest of these corrections and centrifugal distortion is the smallest.
EXAMPLE
An approximate representation of the potential-energy function of a diatomic molecule is the Morse function
\(
U(R)=U\left(R{e}\right)+D{e}\left[1-e^{-a\left(R-R_{e}\right)}\right]^{2}
\)
Use of $U^{\prime \prime}\left(R{e}\right)=k{e}$ [Eq. (4.59)] and (13.27) gives (see Prob. 4.29; the Morse functions in Prob. 4.29 and in this example differ because of different choices for the zero of energy)
\(
a=\left(k{e} / 2 D{e}\right)^{1 / 2}=2 \pi \nu{e}\left(\mu / 2 D{e}\right)^{1 / 2}
\)
Use the Morse function and the Numerov method (Section 4.4) to (a) find the lowest six vibrational energy levels of the ${ }^{1} \mathrm{H}{2}$ molecule in its ground electronic state, which has $D{e} / h c=38297 \mathrm{~cm}^{-1}, \nu{e} / c=4403.2 \mathrm{~cm}^{-1}$, and $R{e}=0.741 \AA$, where $h$ and $c$ are Planck's constant and the speed of light; (b) find $\langle R\rangle$ for each of these vibrational states.
(a) The vibrational energy levels correspond to states with the rotational quantum number $J=0$. Making the change of variables $x \equiv R-R_{e}$ and $S(x) \equiv F(R)$ [Eq. (13.23)] and substituting the Morse function into the nuclear-motion Schrödinger equation (13.19), we get for $J=0$
\(
-\left(\hbar^{2} / 2 \mu\right) S^{\prime \prime}(x)+D{e}\left(1-e^{-a x}\right)^{2} S(x)=\left[E{\text {int }}-U\left(R{e}\right)\right] S(x)=E{\text {vib }} S(x)
\)
since for $J=0, E{\text {int }}=E{\text {elec }}+E{\text {vib }}=U\left(R{e}\right)+E{\text {vib }}$ [Eq. (13.26)]. As usual in the Numerov method, we switch to the dimensionless reduced variables $E{\text {vib }, r} \equiv E{\text {vib }} / A$ and $x{r} \equiv x / B$, where $A$ and $B$ are products of powers of the constants $\hbar, \mu$, and $a$. The procedure of Section 4.4 gives (Prob. 13.8a) $A=\hbar^{2} a^{2} / \mu$ and $B=a^{-1}$, so
\(
x{r} \equiv x / B=a x, \quad E{\mathrm{vib}, r} \equiv E{\mathrm{vib}} / A=\mu E{\mathrm{vib}} / \hbar^{2} a^{2}=\left(2 D{e} / h^{2} \nu{e}^{2}\right) E_{\mathrm{vib}}
\)
where we used $a=2 \pi \nu{e}\left(\mu / 2 D{e}\right)^{1 / 2}$. Substitution of
\(
\begin{gathered}
x=x{r} / a, \quad E{\mathrm{vib}}=\hbar^{2} a^{2} E{\mathrm{vib}, r} / \mu, \quad D{e, r}=D{e} /\left(\hbar^{2} a^{2} / \mu\right), \quad S(x)=S{r}\left(x{r}\right) B^{-1 / 2} \
S^{\prime \prime}=B^{-5 / 2} S{r}^{\prime \prime}=B^{-1 / 2} B^{-2} S{r}^{\prime \prime}=B^{-1 / 2} a^{2} S{r}^{\prime \prime}
\end{gathered}
\)
\(
S{r}^{\prime \prime}\left(x{r}\right)=\left[2 D{e, r}\left(1-e^{-x{r}}\right)^{2}-2 E{\mathrm{vib}, r}\right] S{r}\left(x{r}\right) \equiv G{r} S{r}\left(x{r}\right)
\)
This last equation has the form of (4.82) with $G{r} \equiv 2 D{e, r}\left(1-e^{-x{r}}\right)^{2}-2 E{\mathrm{vib}, r}$, so we are now ready to apply the Numerov procedure of Section 4.4. For the $\mathrm{H}_{2}$ ground electronic state, we find (Prob. 13.8b)
\(
\begin{gathered}
A=h^{2} \nu{e}^{2} / 2 D{e}=h^{2} c^{2}\left(4403.2 \mathrm{~cm}^{-1}\right)^{2} / 2 h c\left(38297 \mathrm{~cm}^{-1}\right)=\left(253.12{9} \mathrm{~cm}^{-1}\right) h c \
B=0.51412 \AA, \quad D{e, r}=D{e} / A=151.29{4}
\end{gathered}
\)
We want to start and end the Numerov procedure in the classically forbidden regions. If we used the harmonic-oscillator approximation for the vibrational levels, the energies of the first six vibrational levels would be $\left(v+\frac{1}{2}\right) h \nu{e}, v=0,1, \ldots, 5$. The reduced energy of the sixth harmonic-oscillator vibrational level would be $5.5 h \nu{e} / A=5.5 h \nu{e} /\left(253 \mathrm{~cm}^{-1}\right) h c=5.5\left(4403 \mathrm{~cm}^{-1}\right) /\left(253 \mathrm{~cm}^{-1}\right)=95.7$. Because of anharmonicity (Section 4.3), the sixth vibrational level will actually occur below 95.7, so we are safe in using 95.7 to find the limits of the classically allowed region. We have $D{e, r}\left(1-e^{-x{r}}\right)^{2}=95.7$, and with $D{e, r}=151.29$, we find $x{r}=-0.58$ and $x{r}=1.58$ as the limits of the classically allowed region for a reduced energy of 95.7. Extending the range by 1.2 at each end, we would start the Numerov procedure at $x{r}=-1.8$ and end at $x{r}=2.8$. However, $x{r}=\left(R-R{e}\right) / B=$ $(R-0.741 \AA) /(0.514 \AA)$ and the minimum possible internuclear distance $R$ is 0 , so the minimum possible value of $x{r}$ is -1.44 . We therefore start at $x{r}=-1.44$ and end at 2.8. If we take an interval of $s{r}=0.04$, we will have about 106 points, which is adequate, but we will try for higher accuracy by taking $s{r}=0.02$ to give about 212 points. With these choices, we set up the Numerov spreadsheet in the usual manner (or use Mathcad or the computer program of Table 4.1). We find (Prob. 13.9) the following lowest six $E{\text {vib }, r}$ values: 8.572525, 24.967566, 40.362582, 54.757570, 68.152531, 80.547472. Using $E{\mathrm{vib}, r} \equiv E{\text {vib }} / A$, we find the lowest levels to be $E{\text {vib }} / h c=2169.95,6320.01,10216.94,13860.73,17251.38,20388.90 \mathrm{~cm}^{-1}$. Note the reduced spacings between levels as the vibrational quantum number increases. (For comparison, the harmonic-oscillator approximation gives the following values: $2201.6,6604.8,11008.0,15411.2,19814.4,24217.6 \mathrm{~cm}^{-1}$.)
It happens that the Schrödinger equation for the Morse function can be analytically solved virtually exactly, and the analytic solution (Prob. 13.11) gives the following lowest eigenvalues: 2169.96, 6320.03, 10216.97, 13860.78, 17251.47, $20389.02 \mathrm{~cm}^{-1}$. Agreement between the Numerov Morse-function values and the analytic Morse-function values is very good. The experimentally observed lowest vibrational levels of $\mathrm{H}{2}$ are 2170.08, 6331.22, 10257.19, 13952.43, 17420.44, $20662.00 \mathrm{~cm}^{-1}$. The deviations of the Morse-function values from the experimental values indicate that the Morse function is not a very accurate representation of the ground-state $\mathrm{H}{2} U(R)$ function.
FIGURE 13.2 The $v=5$ Morse vibrational wave function for $\mathrm{H}_{2}$ as found by the Numerov method.
(b) We have $\left\langle x{r}\right\rangle \approx \int{-1.44}^{2.8} x{r}\left|S{r}\right|^{2} d x{r} \approx \sum{x{r}=1.44}^{2.8} x{r}\left|S{r}\right|^{2} s{r}$, where $s{r}=0.02$ is the interval spacing (not to be confused with the vibrational wave function $S{r}$ ), and where the vibrational wave function $S{r}$ must be normalized. (See also Prob. 13.12.) We normalize $S{r}$ as described in Section 4.4, and then create a column of $x{r}\left|S{r}\right|^{2} s{r}$ values. Next we sum these values to find the following results for the six lowest vibrational states (Prob. 13.9b) $\left\langle x{r}\right\rangle=0.0440,0.1365,0.2360,0.3435,0.4605,0.5884$. Using $x{r}=\left(R-R{e}\right) / B$, we find the following values: $\langle R\rangle=0.763,0.811,0.862,0.918$, $0.978,1.044 \AA$ A. (To get accurate $\left\langle x{r}\right\rangle$ values, $E{\text {vib, } r}$ must be found to many more decimal places than given in (a)-enough places to make the wave function close to zero at 2.8 . If the spreadsheet does not allow you to enter enough decimal places to do this for $v=0$, you can take the right-hand limit as 2.5 instead of 2.8.) Because of vibrational anharmonicity, the molecule gets larger as the vibrational quantum number increases. This effect is rather large for light atoms such as hydrogen. The $v=5$ Numerov-Morse vibrational wave function (Fig. 13.2) shows marked asymmetry about the origin ( $x{r}=0$, which corresponds to $R=R{e}$ ). For a spectacular example of the effect of anharmonicity on bond length, see the discussion of $\mathrm{He}_{2}$ (the world's largest diatomic molecule) near the end of Section 13.7.