Before solving the Schrödinger equation for the hydrogen atom, we will first deal with the two-particle rigid rotor. This is a two-particle system with the particles held at a fixed distance from each other by a rigid massless rod of length $d$. For this problem, the vector $\mathbf{r}$ in Fig. 6.1 has the constant magnitude $|\mathbf{r}|=d$. Therefore (see Section 6.3), the kinetic energy of internal motion is wholly rotational energy. The energy of the rotor is entirely kinetic, and
\(
\begin{equation}
V=0 \tag{6.44}
\end{equation}
\)
Equation (6.44) is a special case of Eq. (6.41), and we may therefore use the results of the last section to separate off the translational motion of the system as a whole. We will concern ourselves only with the rotational energy. The Hamiltonian operator for the rotation is given by the terms in brackets in (6.43) as
\(
\begin{equation}
\hat{H}=\frac{\hat{p}{\mu}^{2}}{2 \mu}=-\frac{\hbar^{2}}{2 \mu} \nabla^{2}, \quad \mu=\frac{m{1} m{2}}{m{1}+m_{2}} \tag{6.45}
\end{equation}
\)
where $m{1}$ and $m{2}$ are the masses of the two particles. The coordinates of the fictitious particle of mass $\mu$ are the relative coordinates of $m{1}$ and $m{2}$ [Eq. (6.28)].
Instead of the relative Cartesian coordinates $x, y, z$, it will prove more fruitful to use the relative spherical coordinates $r, \theta, \phi$. The $r$ coordinate is equal to the magnitude of
the $\mathbf{r}$ vector in Fig. 6.1, and since $m{1}$ and $m{2}$ are constrained to remain a fixed distance apart, we have $r=d$. Thus the problem is equivalent to a particle of mass $\mu$ constrained to move on the surface of a sphere of radius $d$. Because the radial coordinate is constant, the wave function will be a function of $\theta$ and $\phi$ only. Therefore the first two terms of the Laplacian operator in (6.8) will give zero when operating on the wave function and may be omitted. Looking at things in a slightly different way, we note that the operators in (6.8) that involve $r$ derivatives correspond to the kinetic energy of radial motion, and since there is no radial motion, the $r$ derivatives are omitted from $\hat{H}$.
Since $V=0$ is a special case of $V=V(r)$, the results of Section 6.1 tell us that the eigenfunctions are given by (6.16) with the $r$ factor omitted:
\(
\begin{equation}
\psi=Y_{J}^{m}(\theta, \phi) \tag{6.46}
\end{equation}
\)
where $J$ rather than $l$ is used for the rotational angular-momentum quantum number.
The Hamiltonian operator is given by Eq. (6.8) with the $r$ derivatives omitted and $V(r)=0$. Thus
\(
\hat{H}=\left(2 \mu d^{2}\right)^{-1} \hat{L}^{2}
\)
Use of (6.13) gives
\(
\begin{align}
\hat{H} \psi & =E \psi \
\left(2 \mu d^{2}\right)^{-1} \hat{L}^{2} Y{J}^{m}(\theta, \phi) & =E Y{J}^{m}(\theta, \phi) \
\left(2 \mu d^{2}\right)^{-1} J(J+1) \hbar^{2} Y{J}^{m}(\theta, \phi) & =E Y{J}^{m}(\theta, \phi) \
E=\frac{J(J+1) \hbar^{2}}{2 \mu d^{2}}, \quad J & =0,1,2 \cdots \tag{6.47}
\end{align}
\)
The moment of inertia $I$ of a system of $n$ particles about some particular axis in space as defined as
\(
\begin{equation}
I \equiv \sum{i=1}^{n} m{i} \rho_{i}^{2} \tag{6.48}
\end{equation}
\)
where $m{i}$ is the mass of the $i$ th particle and $\rho{i}$ is the perpendicular distance from this particle to the axis. The value of $I$ depends on the choice of axis. For the two-particle rigid rotor, we choose our axis to be a line that passes through the center of mass and is perpendicular to the line joining $m{1}$ and $m{2}$ (Fig. 6.2). If we place the rotor so that the center of mass, point $C$, lies at the origin of a Cartesian coordinate system and the line joining $m{1}$ and $m{2}$ lies on the $x$ axis, then $C$ will have the coordinates $(0,0,0), m{1}$ will have the coordinates $\left(-\rho{1}, 0,0\right)$, and $m{2}$ will have the coordinates $\left(\rho{2}, 0,0\right)$. Using these coordinates in (6.30), we find
\(
\begin{equation}
m{1} \rho{1}=m{2} \rho{2} \tag{6.49}
\end{equation}
\)
FIGURE 6.2 Axis (dashed line) for calculating the moment of inertia of a two-particle rigid rotor. $C$ is the center of mass.
The moment of inertia of the rotor about the axis we have chosen is
\(
\begin{equation}
I=m{1} \rho{1}^{2}+m{2} \rho{2}^{2} \tag{6.50}
\end{equation}
\)
Using (6.49), we transform Eq. (6.50) to (see Prob. 6.14)
\(
\begin{equation}
I=\mu d^{2} \tag{6.51}
\end{equation}
\)
where $\mu \equiv m{1} m{2} /\left(m{1}+m{2}\right)$ is the reduced mass of the system and $d \equiv \rho{1}+\rho{2}$ is the distance between $m{1}$ and $m{2}$. The allowed energy levels (6.47) of the two-particle rigid rotor are
\(
\begin{equation}
E=\frac{J(J+1) \hbar^{2}}{2 I}, \quad J=0,1,2, \ldots \tag{6.52}
\end{equation}
\)
The lowest level is $E=0$, so there is no zero-point rotational energy. Having zero rotational energy and therefore zero angular momentum for the rotor does not violate the uncertainty principle; recall the discussion following Eq. (5.105). Note that $E$ increases as $J^{2}+J$, so the spacing between adjacent rotational levels increases as $J$ increases.
Are the rotor energy levels (6.52) degenerate? The energy depends on $J$ only, but the wave function (6.46) depends on $J$ and $m$, where $m \hbar$ is the $z$ component of the rotor's angular momentum. For each value of $J$, there are $2 J+1$ values of $m$, ranging from $-J$ to $J$. Hence the levels are $(2 J+1)$-fold degenerate. The states of a degenerate level have different orientations of the angular-momentum vector of the rotor about a space-fixed axis.
The angles $\theta$ and $\phi$ in the wave function (6.46) are relative coordinates of the two point masses. If we set up a Cartesian coordinate system with the origin at the rotor's center of mass, $\theta$ and $\phi$ will be as shown in Fig. 6.3. This coordinate system undergoes the same translational motion as the rotor's center of mass but does not rotate in space.
The rotational angular momentum $\left[J(J+1) \hbar^{2}\right]^{1 / 2}$ is the angular momentum of the two particles with respect to an origin at the system's center of mass $C$.
The rotational levels of a diatomic molecule can be well approximated by the two-particle rigid-rotor energies (6.52). It is found (Levine, Molecular Spectroscopy, Section 4.4) that when a diatomic molecule absorbs or emits radiation, the allowed pure-rotational transitions are given by the selection rule
\(
\begin{equation}
\Delta J= \pm 1 \tag{6.53}
\end{equation}
\)
In addition, a molecule must have a nonzero dipole moment in order to show a pure-rotational spectrum. A pure-rotational transition is one where only the rotational
FIGURE 6.3 Coordinate system for the two-particle rigid rotor.
quantum number changes. [Vibration-rotation transitions (Section 4.3) involve changes in both vibrational and rotational quantum numbers.] The spacing between adjacent low-lying rotational levels is significantly less than that between adjacent vibrational levels, and the pure-rotational spectrum falls in the microwave (or the far-infrared) region. The frequencies of the pure-rotational spectral lines of a diatomic molecule are then (approximately)
\(
\begin{gather}
\nu=\frac{E{J+1}-E{J}}{h}=\frac{[(J+1)(J+2)-J(J+1)] h}{8 \pi^{2} I}=2(J+1) B \tag{6.54}\
B \equiv h / 8 \pi^{2} I, \quad J=0,1,2, \ldots \tag{6.55}
\end{gather}
\)
$B$ is called the rotational constant of the molecule.
The spacings between the diatomic rotational levels (6.52) for low and moderate values of $J$ are generally less than or of the same order of magnitude as $k T$ at room temperature, so the Boltzmann distribution law (4.63) shows that many rotational levels are significantly populated at room temperature. Absorption of radiation by diatomic molecules having $J=0$ (the $J=0 \rightarrow 1$ transition) gives a line at the frequency $2 B$; absorption by molecules having $J=1$ (the $J=1 \rightarrow 2$ transition) gives a line at $4 B$; absorption by $J=2$ molecules gives a line at $6 B$; and so on. See Fig. 6.4.
Measurement of the rotational absorption frequencies allows $B$ to be found. From $B$, we get the molecule's moment of inertia $I$, and from $I$ we get the bond distance $d$. The value of $d$ found is an average over the $v=0$ vibrational motion. Because of the asymmetry of the potential-energy curve in Figs. 4.6 and 13.1, $d$ is very slightly longer than the equilibrium bond length $R_{e}$ in Fig. 13.1.
As noted in Section 4.3, isotopic species such as ${ }^{1} \mathrm{H}^{35} \mathrm{Cl}$ and ${ }^{1} \mathrm{H}^{37} \mathrm{Cl}$ have virtually the same electronic energy curve $U(R)$ and so have virtually the same equilibrium bond distance. However, the different isotopic masses produce different moments of inertia and hence different rotational absorption frequencies.
Because molecules are not rigid, the rotational energy levels for diatomic molecules differ slightly from rigid-rotor levels. From (6.52) and (6.55), the two-particle rigid-rotor levels are $E{\mathrm{rot}}=B h J(J+1)$. Because of the anharmonicity of molecular vibration (Fig. 4.6), the average internuclear distance increases with increasing vibrational quantum number $v$, so as $v$ increases, the moment of inertia $I$ increases and the rotational constant $B$ decreases. To allow for the dependence of $B$ on $v$, one replaces $B$ in $E{\text {rot }}$ by $B{v}$. The mean rotational constant $B{v}$ for vibrational level $v$ is $B{v}=B{e}-\alpha{e}(v+1 / 2)$, where $B{e}$ is calculated using the equilibrium internuclear separation $R_{e}$ at the bottom
FIGURE 6.4 Two-particle rigid-rotor absorption transitions.
of the potential-energy curve in Fig. 4.6, and the vibration-rotation coupling constant $\alpha{e}$ is a positive constant (different for different molecules) that is much smaller than $B{e}$. Also, as the rotational energy increases, there is a very slight increase in average internuclear distance (a phenomenon called centrifugal distortion). This adds the term $-h D J^{2}(J+1)^{2}$ to $E{\text {rot }}$, where the centrifugal-distortion constant $D$ is an extremely small positive constant, different for different molecules. For example, for ${ }^{12} \mathrm{C}^{16} \mathrm{O}$, $B{0}=57636 \mathrm{MHz}, \alpha{e}=540 \mathrm{MHz}$, and $D=0.18 \mathrm{MHz}$. As noted in Section 4.3 , for lighter diatomic molecules, nearly all the molecules are in the ground $v=0$ vibrational level at room temperature, and the observed rotational constant is $B{0}$.
For more discussion of nuclear motion in diatomic molecules, see Section 13.2. For the rotational energies of polyatomic molecules, see Townes and Schawlow, chaps. 2-4.
EXAMPLE
The lowest-frequency pure-rotational absorption line of ${ }^{12} \mathrm{C}^{32} \mathrm{~S}$ occurs at 48991.0 MHz . Find the bond distance in ${ }^{12} \mathrm{C}^{32} \mathrm{~S}$.
The lowest-frequency rotational absorption is the $J=0 \rightarrow 1$ line. Equations (1.4), (6.52), and (6.51) give
\(
h \nu=E{\text {upper }}-E{\text {lower }}=\frac{1(2) \hbar^{2}}{2 \mu d^{2}}-\frac{0(1) \hbar^{2}}{2 \mu d^{2}}
\)
which gives $d=\left(h / 4 \pi^{2} \nu \mu\right)^{1 / 2}$. Table A. 3 in the Appendix gives
\(
\mu=\frac{m{1} m{2}}{m{1}+m{2}}=\frac{12(31.97207)}{(12+31.97207)} \frac{1}{6.02214 \times 10^{23}} \mathrm{~g}=1.44885 \times 10^{-23} \mathrm{~g}
\)
The SI unit of mass is the kilogram, and
\(
\begin{aligned}
d=\frac{1}{2 \pi}\left(\frac{h}{\nu_{0 \rightarrow 1} \mu}\right)^{1 / 2} & =\frac{1}{2 \pi}\left[\frac{6.62607 \times 10^{-34} \mathrm{~J} \mathrm{~s}}{\left(48991.0 \times 10^{6} \mathrm{~s}^{-1}\right)\left(1.44885 \times 10^{-26} \mathrm{~kg}\right)}\right]^{1 / 2} \
& =1.5377 \times 10^{-10} \mathrm{~m}=1.5377 \AA
\end{aligned}
\)
EXERCISE The $J=1$ to $J=2$ pure-rotational transition of ${ }^{12} \mathrm{C}^{16} \mathrm{O}$ occurs at $230.538 \mathrm{GHz} .\left(1 \mathrm{GHz}=10^{9} \mathrm{~Hz}\right.$. ) Find the bond distance in this molecule. (Answer: $1.1309 \times 10^{-10} \mathrm{~m}$.)