To describe the rotational motion of a free molecule, we begin by constructing the Hamiltonian operator. For an isolated molecule rotating in space, there is no restoring force and no preferred orientation. As a result, the potential energy is zero everywhere.
The Hamiltonian therefore consists of kinetic energy only.
Hamiltonian in Cartesian coordinates
In Cartesian coordinates, the kinetic energy operator for a particle of mass \(m\) is
\[ \hat{H} = -\frac{\hbar^2}{2m} \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \right) = -\frac{\hbar^2}{2m}\nabla^2. \]
While this form is general, it is not well suited for rotational motion, which depends on angular orientation rather than linear displacement.
Hamiltonian in spherical polar coordinates
Expressing the Laplacian in spherical polar coordinates \((r,\theta,\phi)\) gives
\[ \nabla^2 = \frac{1}{r^2}\frac{\partial}{\partial r} \left( r^2\frac{\partial}{\partial r} \right) + \frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta} \left( \sin\theta\,\frac{\partial}{\partial\theta} \right) + \frac{1}{r^2\sin^2\theta}\frac{\partial^2}{\partial\phi^2}. \]
Substituting this into the Hamiltonian yields a kinetic-energy operator that naturally separates radial and angular motion.
The rigid rotor approximation
In the rigid rotor model, the distance between the atoms is fixed. This means that \(r\) is constant and does not vary with time.
Because of this constraint, all derivatives with respect to \(r\) vanish. The Hamiltonian reduces to an operator involving only angular variables:
\[ \hat{H} = -\frac{\hbar^2}{2I} \left[ \frac{1}{\sin\theta}\frac{\partial}{\partial\theta} \left( \sin\theta\,\frac{\partial}{\partial\theta} \right) + \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\phi^2} \right], \]
where \(I=\mu r^2\) is the moment of inertia.
Separation of variables
The time-independent Schrödinger equation,
\[ \hat{H}\psi(\theta,\phi)=E\psi(\theta,\phi), \]
can be solved by assuming that the wavefunction is separable:
\[ \psi(\theta,\phi)=\Theta(\theta)\Phi(\phi). \]
Substituting this product form into the Schrödinger equation and dividing by \(\Theta(\theta)\Phi(\phi)\) leads to two differential equations:
- one involving derivatives only with respect to \(\phi\),
- one involving derivatives only with respect to \(\theta\).
Although the variables are separated, the equations are not completely independent. The equation in \(\phi\) contains a separation constant that also appears in the \(\theta\) equation.
As a result, the allowed solutions for \(\Phi(\phi)\) are constrained by the requirement that \(\Theta(\theta)\) remain finite and single-valued over the full range of angles.
Separation of variables: the \(\theta\) and \(\phi\) equations
Starting from the rigid-rotor angular Schrödinger equation
\[ \left[ \frac{1}{\sin\theta}\frac{\partial}{\partial\theta} \Bigl(\sin\theta\,\frac{\partial}{\partial\theta}\Bigr) + \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\phi^2} \right]\psi(\theta,\phi) = -\frac{2I E}{\hbar^2}\,\psi(\theta,\phi), \]
assume a separable solution \(\psi(\theta,\phi)=\Theta(\theta)\,\Phi(\phi)\). Dividing through by \(\Theta\Phi\) gives
\[ \frac{1}{\Theta}\frac{1}{\sin\theta}\frac{d}{d\theta}\!\Bigl(\sin\theta\frac{d\Theta}{d\theta}\Bigr) + \frac{1}{\Phi}\frac{1}{\sin^2\theta}\frac{d^2\Phi}{d\phi^2} \;=\; -\frac{2I E}{\hbar^2}. \]
Rearranging, the \(\phi\)-dependent term can be isolated and set equal to a constant. Using your chosen separation constant \(-m_\ell^2\) (note the minus sign), we obtain the \(\phi\)-equation:
\[ \frac{1}{\Phi}\frac{d^2\Phi}{d\phi^2} = -m_\ell^2 \quad\Longrightarrow\quad \frac{d^2\Phi}{d\phi^2} + m_\ell^2\,\Phi = 0. \]
The general solution of the \(\phi\)-equation is
\[ \Phi(\phi) = A e^{+i m_\ell\phi} + B e^{-i m_\ell\phi}. \]
Single-valuedness under \(\phi\mapsto\phi+2\pi\) requires \(m_\ell\) to be an integer: \(m_\ell\in\mathbb{Z}\). A convenient normalized choice is
\[ \Phi_{m_\ell}(\phi)=\frac{1}{\sqrt{2\pi}}e^{i m_\ell\phi}. \]
Substituting the separation constant back into the angular equation yields the \(\theta\)-equation (written here before introducing the usual eigenvalue label):
\[ \frac{1}{\sin\theta}\frac{d}{d\theta}\!\Bigl(\sin\theta\frac{d\Theta}{d\theta}\Bigr) -\frac{m_\ell^2}{\sin^2\theta}\,\Theta +\frac{2I E}{\hbar^2}\,\Theta =0. \]
It is standard to write the energy (or separation constant) as \(\dfrac{2I E}{\hbar^2}= \ell(\ell+1)\) with \(\ell\ge 0\). With that choice the \(\theta\)-equation becomes the associated-Legendre differential equation:
\[ \frac{1}{\sin\theta}\frac{d}{d\theta}\!\Bigl(\sin\theta\frac{d\Theta}{d\theta}\Bigr) +\Bigl[\ell(\ell+1)-\frac{m_\ell^2}{\sin^2\theta}\Bigr]\,\Theta = 0. \]
The regular (finite) solutions on \(0\le\theta\le\pi\) are the associated Legendre functions:
\[ \Theta_{\ell}^{m_\ell}(\theta)\propto P_\ell^{m_\ell}(\cos\theta), \qquad\text{with}\quad |m_\ell|\le \ell,\ \ m_\ell\in\mathbb{Z}. \]
Summary of key points
- \(\Phi(\phi)\) satisfies \(\dfrac{d^2\Phi}{d\phi^2}+m_\ell^2\Phi=0\) and gives \(\Phi\propto e^{i m_\ell\phi}\) with integer \(m_\ell\).
- \(\Theta(\theta)\) satisfies the associated-Legendre equation with eigenvalue \(\ell(\ell+1)\).
- The allowed values must satisfy \(|m_\ell|\le\ell\); this constraint links the \(\phi\)- and \(\theta\)-solutions.
These two results together produce the spherical harmonics \(\displaystyle Y_\ell^{m_\ell}(\theta,\phi)=N\,P_\ell^{m_\ell}(\cos\theta)\,e^{i m_\ell\phi}\), which are the simultaneous eigenfunctions of the rigid-rotor Hamiltonian (and of \(\hat{L}^2,\hat{L}_z\)).
Big idea: rotational motion leads to a purely angular Schrödinger equation. Separating the variables reveals that the allowed solutions in \(\phi\) and \(\theta\) are linked, giving rise to quantized angular momentum and discrete rotational energy levels.