2.6. Rigid rotor

Here, we will derive the solution of the Schrödinger equation for the linear rigid rotor.

(2.99)\[\hat{H}Y(\theta,\phi) = E\cdot Y(\theta,\phi)\]

Expressed in spherical coordinates, the Hamiltonian is

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

\(I\) is the moment of inertia around an axis perpendicular to the molecular axis, and \(\hat{J}^2\) is the norm-squared of the angular-momentum operator (\(\hat{J}^2=\hat{J}_x^2+\hat{J}_y^2+\hat{J}_z^2\) in Cartesian coordinates), here expressed in spherical coordinates.

2.6.1. Separation of variables

Moving the right-hand side of the Schrödinger equation to the left gives \((\hat{H}-E)Y = 0\). Dividing by \(-\hbar^2/2I\) and multiplying by \(\sin^2\theta\) gives

(2.101)\[\sin\theta \frac{\partial}{\partial\theta} \sin\theta \frac{\partial Y}{\partial\theta} + \frac{\partial^2 Y}{\partial\phi^2} + \beta^2 \sin^2\theta\cdot Y = 0\]

where we have introduced the abbreviation \(\beta = 2EI/\hbar^2\). The first and the third term on the left-hand side depend on \(\theta\) only, whereas the second term depends on \(\phi\) only. We can therefore use a separation approach and write \(Y\) as a product of a \(\theta\)-dependent function \(\varTheta\) and a \(\phi\)-dependent function \(\varPhi\): \(Y = \varTheta(\theta)\cdot\varPhi(\phi)\). Inserting this and dividing by both \(\varTheta\) and \(\varPhi\) gives

(2.102)\[\frac{\sin\theta}{\varTheta} \frac{\partial}{\partial\theta} \left( \sin\theta \frac{\partial \varTheta}{\partial\theta} \right) + \frac{1}{\varPhi} \frac{\partial^2 \varPhi}{\partial\phi^2} + \beta^2 \sin^2\theta = 0\]

The sum of the first and third terms depends only on \(\theta\), and the second term is a function of \(\phi\) only. Varying \(\theta\) can change only two of the three terms - but since the overall sum must remain constant (zero), the sum of the \(\theta\) terms must be constant - let’s denote this constant as \(m^2\). The same reasoning holds for the \(\phi\) term, where the constant now has to be the negative of the \(\theta\) constant, since they should add up to zero. This yields two equations, one for \(\varTheta\) and one for \(\varPhi\).

(2.103)\[ \begin{align}\begin{aligned}\frac{\sin\theta}{\varTheta} \frac{\partial}{\partial\theta} \left( \sin\theta \frac{\partial \varTheta}{\partial\theta} \right) + \beta^2 \sin^2\theta = \mathrm{const} = m^2\\\frac{1}{\varPhi} \frac{\partial^2 \varPhi}{\partial\phi^2} = -\mathrm{const} = -m^2\end{aligned}\end{align} \]

We chose to write the constant as a square, \(m^2\), of another constant, \(m\), for future convenience.

Now we have successfully separated the original partial differential equation with two variables into two ordinary differential equations, with one variable each. Next, we solve these two equations.

2.6.2. The \(\varPhi\) equation

The general mathematical solution of the \(\varPhi\) equation is

(2.104)\[\varPhi(\phi) = A\mathrm{e}^{\mathrm{i}m\phi} + B \mathrm{e}^{-\mathrm{i}m\phi}\]

as can be verified by insertion. In this general mathematical solution, there are no constraints on the constants \(A\), \(B\), and \(m\). However, for a solution \(\varPhi\) to be physical, it has to be single-valued. In particular, a full \(2\pi\) turn of \(\phi\) (the angle in the equatorial plane) brings the orientation back to its original value, and therefore the function also should have the same value:

(2.105)\[\varPhi(2\pi) = \varPhi(0)\]

i.e. it must be periodic. This condition yields

(2.106)\[A\cdot\mathrm{e}^{2\pi\mathrm{i}m} + B\cdot\mathrm{e}^{-2\pi\mathrm{i}m} = A + B\]

For arbitrary values of \(A\) and \(B\), this is possible only if the exponentials both are equal to 1: \(\mathrm{e}^{\pm2\pi\mathrm{i}m} = 1\). From this it follows that \(m\) must be an integer:

(2.107)\[m = 0, \pm1, \pm2, \pm3, \dots\]

Since \(m\) can be positive or negative, we can write the wavefunction, now including normalization, simply as

(2.108)\[\varPhi(\phi) = \frac{1}{\sqrt{2\pi}}\mathrm{e}^{+\mathrm{i}m\phi}\]

2.6.3. The \(\varTheta\) equation

Next, we need to solve the equation for \(\varTheta\). It is

(2.109)\[\frac{\sin\theta}{\varTheta} \frac{\mathrm{d}}{\mathrm{d}\theta} \left( \sin\theta \frac{\mathrm{d}\varTheta}{\mathrm{d}\theta} \right) + \beta^2 \sin^2\theta = m^2\]

With the substitution \(\cos\theta\rightarrow z\) (with \(z\) ranging from \(-1\) to \(+1\)) and \(\varTheta(\theta)\rightarrow P(z)\), we obtain

(2.110)\[(1-z^2)\frac{\mathrm{d}^2P}{\mathrm{d}^2z} - 2z\frac{\mathrm{d}P}{\mathrm{d}z} +\left(\beta - \frac{m^2}{1-z^2}\right)P = 0\]

This is a well-known equation, the associated Legendre differential equation, and it can be solved using the Frobenius method. Most of the mathematical solutions are not physical, since they diverge at \(z=\pm1\) (corresponding to \(\theta=0\) and \(\pi\)) and are not square integrable. Divergence-free solutions are obtained only if \(\beta\) is a product of two consecutive non-negative integers. If we denote them \(J\) and \(J+1\):

(2.111)\[\beta = \frac{2EI}{\hbar^2} = J(J+1) \quad \text{with integer}\,J \ge|m|\]

The solutions \(P_J^{|m|}(z)\) with integer \(J\) and \(m\) are called associated Legendre polynomials.

Finally, this gives the following ranges for the quantum numbers \(J\) and \(m\):

(2.112)\[J = 0, 1, 2, \dots \qquad m = 0,\pm1,\dots,\pm J\]

Rearranging the expression for \(\beta\) gives

(2.113)\[E = \frac{\hbar^2}{2I}J(J+1) = B J(J+1)\]

with the rotational constant \(B = \hbar^2/2I\).