2.7. Linear Rigid Rotor

Overview

This section covers the quantum mechanical treatment of a linear rigid rotor (e.g., a diatomic molecule) and shows how to derive thermodynamic properties from its rotational partition function.

Review of the Linear Rigid Rotor

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import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d.art3d import Poly3DCollection
from scipy.constants import k, eV
from labellines import labelLines
from myst_nb import glue

fig, axs = plt.subplot_mosaic([[0]], figsize=(4, 4))

# Plot the energy levels (blue lines)
for J in range(0, 3):
    g_J = 2 * J + 1
    x_min = -0.04 - 0.1 * (g_J - 1) / 2
    for i in range(g_J):
        # Plot each degenerate sub-level horizontally
        if i == (g_J - 1) / 2:
            energy_line = axs[0].plot(
                [x_min, x_min + 0.08], 
                [J * (J + 1), J * (J + 1)], 
                color='blue',
                label=r'$E_{%d}$' % J
            )
        else:
            axs[0].plot(
                [x_min, x_min + 0.08], 
                [J * (J + 1), J * (J + 1)], 
                color='blue'
            )
        x_min += 0.1
    # Label only one line at each J for clarity
    labelLines(energy_line, zorder=2.5)

axs[0].set_ylabel('Energy (arb. units)')
axs[0].set_xticks([])
axs[0].set_yticks([])
axs[0].spines['top'].set_visible(False)
axs[0].spines['bottom'].set_visible(False)
axs[0].spines['right'].set_visible(False)
axs[0].spines['left'].set_visible(False)

plt.tight_layout()
glue('linear_rigid_rotor_energy_levels', fig, display=False)
plt.close(fig)

Fig. 21 Energy levels for a linear rigid rotor. The level labeled \(E_J\) has degeneracy \(g_J = 2J + 1\). For example, the \(J=2\) level is split into five degenerate microstates corresponding to \(m = -2, -1, 0, 1, 2\).

For a rigid, linear rotor of moment of inertia \(I\), the energy levels are:

\[E_J \;=\; \frac{\hbar^2}{2I}\,J\bigl(J+1\bigr) \quad\text{for}\quad J = 0, 1, 2, \dots\]

Here, \(\hbar\) is the reduced Planck constant, and \(g_J = 2J + 1\) is the degeneracy of level \(E_J\).

Physical Significance of the Moment of Inertia

The moment of inertia \(I\) measures how mass is distributed around the rotation axis. For a diatomic molecule of atoms A and B,

\[ I \;=\; \mu\,r^2, \quad \mu = \frac{m_A\,m_B}{m_A + m_B},\]

where \(r\) is the bond length and \(\mu\) is the reduced mass of the two atoms.

Partition Function for a Linear Rigid Rotor

In the canonical ensemble, the rotational partition function is:

\[q_{\mathrm{rot}} \;=\; \sum_{i=0}^\infty e^{-\beta E_i} \;=\; \sum_{J=0}^\infty g_J \, e^{-\beta\,E_J} \;=\; \sum_{J=0}^\infty (2J+1)\,\exp\!\Bigl[-\beta \,\frac{\hbar^2}{2I}\,J(J+1)\Bigr].\]

Summation vs. Levels

• Sum over \(i\): sums over individual microstates (each degenerate sub-level).

• Sum over \(J\): reorganizes the sum by energy level \(E_J\), factoring in the degeneracy \(g_J = 2J+1\).

High-Temperature Approximation

When \(k_\mathrm{B} T \gg \frac{\hbar^2}{2I}\), we can approximate the discrete sum by converting it into an integral. Let us set

\[x \;=\; J(J+1), \quad dx \;=\; (2J+1)\,dJ.\]

Hence,

\[q_{\mathrm{rot}} \;\approx\; \int_{0}^{\infty} (2J+1)\,\exp\Bigl[-\beta \,\frac{\hbar^2}{2I}\,J(J+1)\Bigr]\;dJ \;=\; \int_{x=0}^{\infty} \exp\Bigl[-\beta \,\tfrac{\hbar^2}{2I}\,x\Bigr]\;dx.\]

Evaluating this integral,

\[\int_{0}^{\infty} \exp\Bigl[-\beta \,\tfrac{\hbar^2}{2I}\,x\Bigr]\;dx \;=\; \frac{1}{\beta \,\frac{\hbar^2}{2I}} \;=\; \frac{2I\,k_\mathrm{B} T}{\hbar^2}.\]

We define the rotational temperature \(\Theta_{\mathrm{rot}}\) by

\[\Theta_{\mathrm{rot}} \;=\; \frac{\hbar^2}{2k_{\mathrm{B}}\,I}.\]

Thus, for a heteronuclear diatomic rotor (symmetry factor \(\sigma = 1\)),

\[q_{\mathrm{rot}} \;\approx\; \frac{2I\,k_\mathrm{B} T}{\hbar^2} \;=\; \frac{T}{\Theta_{\mathrm{rot}}}.\]

Symmetry Factor \(\sigma\)

For homonuclear diatomics, or other symmetric linear rotors, identical orientations in space might be indistinguishable, leading to \(\sigma=2\). This modifies the partition function to

\[q_{\mathrm{rot}} \;\approx\; \frac{T}{\sigma\,\Theta_{\mathrm{rot}}}.\]

Whether or not you include \(\sigma\) depends on the level of detail needed (e.g., for absolute entropy calculations).

Ensemble Averages

Natural Logarithm of the Partition Function

From the high-\(T\) approximation (with \(\sigma=1\) for simplicity),

\[\ln q_{\mathrm{rot}} \;=\; \ln \Bigl(\tfrac{T}{\Theta_{\mathrm{rot}}}\Bigr) \;=\; \ln T \;-\;\ln \Theta_{\mathrm{rot}}.\]

Internal Energy

The (rotational) internal energy \(U_{\mathrm{rot}}\) is given by

\[U_{\mathrm{rot}} \;=\; - \left(\frac{\partial \ln q_{\mathrm{rot}}}{\partial \beta}\right)_{N,V} \;=\; k_{\mathrm{B}}\,T^2 \left(\frac{\partial \ln q_{\mathrm{rot}}}{\partial T}\right)_{N,V}.\]

Since \(\ln q_{\mathrm{rot}} = \ln T - \ln \Theta_{\mathrm{rot}},\) we get

\[\frac{\partial \ln q_{\mathrm{rot}}}{\partial T} \;=\; \frac{\partial}{\partial T}\bigl(\ln T\bigr) \;=\; \frac{1}{T}.\]

Hence,

\[U_{\mathrm{rot}} \;=\; k_{\mathrm{B}} \, T.\]

Heat Capacity at Constant Volume

The rotational contribution to the heat capacity is

\[C_V^{(\mathrm{rot})} \;=\; \left(\frac{\partial U_{\mathrm{rot}}}{\partial T}\right)_{N,V} \;=\; k_{\mathrm{B}}.\]

Physically, this means one linear rotor contributes \(k_{\mathrm{B}}\) to the heat capacity in the classical (high-\(T\)) limit, corresponding to its two rotational degrees of freedom (each contributes \(\tfrac{1}{2}k_{\mathrm{B}}\)).

Note

In a more rigorous quantum treatment (and for lower temperatures), the partition function and the resulting averages must use the full sum over \(J\). At sufficiently low \(T\), only the \(J=0\) and \(J=1\) levels are significantly populated, which reduces the effective heat capacity below \(k_\mathrm{B}\).