CSU East Bay logo

Chemistry 352

The Eigenvalues

Having obtained the angular wavefunctions for the rigid rotor, we now turn to the energy eigenvalues. These eigenvalues arise directly from the angular part of the Schrödinger equation and describe the allowed rotational energy levels of a free molecule.


The rigid rotor Schrödinger equation

For a rigid rotor with fixed bond length, the Hamiltonian contains only angular kinetic energy:

\[ \hat{H} = -\frac{\hbar^2}{2I} \left[ \frac{1}{\sin\theta}\frac{\partial}{\partial\theta} \left( \sin\theta\,\frac{\partial}{\partial\theta} \right) + \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\phi^2} \right], \]

where \(I\) is the moment of inertia of the molecule. The time-independent Schrödinger equation is therefore

\[ \hat{H}\psi(\theta,\phi)=E\psi(\theta,\phi). \]

The solutions of this equation are the spherical harmonics \(Y_\ell^{m_\ell}(\theta,\phi)\).


Rotational energy eigenvalues

Substitution of the spherical harmonics into the Schrödinger equation yields the rotational energy levels

\[ E_\ell = \frac{\hbar^2}{2I}\,\ell(\ell+1), \qquad \ell = 0,1,2,\ldots \]

A key feature of this result is that the energy depends only on the quantum number \(\ell\). The magnetic quantum number \(m_\ell\) does not affect the energy.


Degeneracy of the rotational energy levels

For a given value of \(\ell\), the allowed values of \(m_\ell\) range from \(-\ell\) to \(+\ell\) in integer steps.

This gives a total of

\[ 2\ell + 1 \]

distinct wavefunctions with the same energy \(E_\ell\). The rotational energy level labeled by \(\ell\) is therefore \((2\ell+1)\)-fold degenerate.

This degeneracy reflects the fact that, in the absence of external fields, all orientations of the angular momentum vector are energetically equivalent.


Behavior of the rotational level spacings

The spacing between adjacent rotational energy levels is not constant. Using \(E_\ell=\dfrac{\hbar^2}{2I}\ell(\ell+1)\), the difference between successive levels is

\[ \Delta E_{\ell\rightarrow \ell+1} = E_{\ell+1}-E_\ell = \frac{\hbar^2}{2I}\big[(\ell+1)(\ell+2)-\ell(\ell+1)\big] = \frac{\hbar^2}{I}(\ell+1). \]

This result shows that the spacing between rotational levels increases linearly with \(\ell\). As the molecule rotates faster, additional rotational energy levels are spaced progressively farther apart.

This increasing spacing is a direct consequence of the quadratic dependence of the energy on \(\ell\) and contrasts with the equally spaced levels of the harmonic oscillator.


Big idea: the rigid rotor has rotational energy levels that depend only on the total angular momentum quantum number \(\ell\), with each level possessing a \((2\ell+1)\)-fold degeneracy arising from different spatial orientations of the rotation.

Your turn

Problem 1
What is the expression for the rotational energy levels of a rigid rotor?
\(E_\ell = \hbar\omega(\ell+\tfrac{1}{2})\)
\(E_\ell = \dfrac{\hbar^2}{I}\ell^2\)
\(E_\ell = \dfrac{\hbar^2}{2I}\ell(\ell+1)\)
\(E_\ell = B\ell\)
Problem 2
On which quantum number do the rigid rotor energy levels depend?
\(\ell\) only
\(m_\ell\) only
Both \(\ell\) and \(m_\ell\)
The bond length only
Problem 3
What is the degeneracy of the rotational energy level with quantum number \(\ell\)?
\(\ell\)
\(\ell+1\)
\(2\ell\)
\(2\ell+1\)
Problem 4
How does the spacing between adjacent rotational energy levels change as \(\ell\) increases?
It remains constant
It increases linearly with \(\ell\)
It decreases with \(\ell\)
It oscillates with \(\ell\)
Problem 5
What physical feature gives rise to the \(2\ell+1\) degeneracy of each rotational energy level?
All orientations of the angular momentum vector are energetically equivalent
The bond length is fixed
The molecule vibrates while rotating
The potential energy is zero

Key points (one glance)

Big picture: the rigid rotor model predicts quantized rotational energy levels whose structure and degeneracy arise directly from angular momentum quantization and spatial symmetry.