Another form of microscopic motion that shows interesting quantization effects is rotation. Rotation is a fundamental motion of molecules and atoms; for example, gaseous molecules are in a constant state of relatively free rotation, electrons in atoms orbit around the nucleus, and the nuclear particles themselves exhibit spin. Rotational motion is characterized by an angular momentum vector L. On a microscopic scale the angular momentum is quantitized in both length and direction by units of Panck's constant divided by 2¹

This quantization of angular momentum is a fundamental property of quantum mechanical systems. The central role of angular momentum in quantum mechanics is indicated by the fact that Planck's constant itself has units of angular momentum. In the early days of quantum mechanics Niels Bohr developed a successful model of the hydrogen atom spectrum by imposing a quantization condition on the electron's orbital angular momentum. Indeed many features of molecular and atomic spectra results from the fact that the total angular momentum can only take on certain values. Today, we see that quantization of angular momentum does not need to be imposed on a system but is a natural consequence of Schrödinger's model.

The simplest model of rotational motion that can be described quantum mechanically is a rigid rotor consisting of two masses held a fixed distance apart rotating about its center of mass. The rotation is assumed to be free of any outside potential energy. Angular momentum in quantum systems results from a direct correspondence of classical mechanical equations of rotational motion. In rotational motion the moment of inertia and angular velocity play the same role as the mass and velocity does in a straight moving body. In linear motion momentum and kinetic energy are given by p = mv and E =p²/2m , so a rigid rotor rotating with angular velocity w and having a moment of inertia I = µ R² exhibits angular momentum |L| = I w; and an energy of E = L²/2I. To convert these classical equations to quantum equations we simply need to replace the momentum with its quantum operator and solve Schrödinger's eigenvalue equation.

In most respects the analysis of rotational systems is largely a generalization of the types of coordinates used to describe the system. Schrödinger's eigenvalue equation given above is very hard to solve in Cartesian coordinates because motions in the x,y, and z directions are not independent of each other. Polar coordinates most directly describe rotational motion and allow the Hamiltonian to be separated into independent coordinates. For example, the angular velocity is only dependent on the time derivative of the phi coordinate. To solve Schrödinger's equation we need to convert the Hamiltonian to polar coordinates. Chain rule differentiation provides the means for converting differential operators from Cartesian to polar coordinates.

Angular momentum is a vector quantity that results from the cross-product of the position vector r from the center of rotation with the linear momentum vector p of the particles in motion. Conversion of the angular momentum vector to polar coordinates is given in the following table.

For a rigid rotor r is a constant and the Hamiltonian becomes

The wavefunction for this Hamiltonian will be two-dimensional depending on both coordinates. To solve this problem we need to reduce the two-dimensional problem to two one-dimensional problems. Because the derivatives in the Hamiltonian are in respects to only one coordinate at a time the Hamiltonian is separable and we can look for a solution of the form of a product function:

separating like variables gives

For this equation to be true for all variations of both angles, it must be true that both sides are equal to a constant

It is easy to see by inspection that the first equation has a solution of

Differentiating twice shows the constant beta is equal to m². Furthermore, because polar coordinates are cyclic, the wavefunction must be invariant on a rotation of 2¹.

substituting this condition into the wavefunction gives:

Using this in the cyclic boundary condition gives:

Hence, 2m must be a positive or negative even integer restricting m to values: 0, ±1, ±2, ±3 .....

It is easy to see that this solution is also an eigenfunction of the z component of the angular momentum.

The remarkable implication of this equation is that the orientation of a rotating body is quantitized. A macroscopic object like a child's top starts out spinning rapidly in a vertical direction then begins to precess, its axis pointing in an ever increasing angle to the vertical until it wobbles to the floor. In quantum mechanical systems the angular momentum vector is not allowed to point in any direction. A quantum mechanical top would only precess at fixed angles to the vertical. Angular momentum is space quantitized such that z component of angular momentum has values of only m. The quantum number m is called the magnetic quantum number because it describes how spinning particles such as electrons or protons will align with an external magnetic field.

The second differential equation can now be used to solve for the energy.

In this program we will solve this differential equation numerically varing the energy until the cyclic boundary condition is satisfied. Note: this is not an easy differential equation to solve because the sine function cause singularities at 0 and ¹ . Small numerical errors that keep the solution from being zero at these points will be magnified to infinity. You might guess then that the analytical solutions would contain the sine function to keep these singularities in check.

This is a well known Legendre differential equation whose analytical solution are called associate Legendre functions. The cyclic boundary condition requires that the energy is equal to

The first 4 sets of associate Legendre functions are:

Q(q) = Pl|m|

	|m| = 0	|m| = 1	|m| = 2	|m| = 3
l = 0	P00 = 1
l = 1	P10 = cosq	P11 = sinq
l = 2	P20 = (3cos2q -1)/2	P21 = 3 cosq sinq	P22 = 3 sin2q
l = 3	P30 = cosq(5cos2q -3)/2	P30 = 3sinq(5cos2q -1)/2	P31 = 15 cosq sin2q	P33 = 15 sin3q

The associate Legendre functions are normalized in polar coordinate representation

The normalized product function solution of the full rotational Hamiltonian is called a spherical harmonic (vibrational modes of a sphere of jello) and is usually denoted as Y_lm

The spherical harmonics are eigenfunctions of the L² operator

The implication of this equation is again startling, the magnitude and hence the angular velocity is quantitized. In the microscopic world a child's yo-yo could only be set spinning a certain rates. One of these rates is zero. At first glance this seems to violate the uncertainty principle. Most systems do not have zero point energies that are zero. A zero velocity does not violate the uncertainty principle in this case because the angle and hence its position is completely undermined. Also note that there is a high degree of degeneracy. For a given l quantum number there are 2l+1 different m quantum numbers. For example when l=2 there will be five different m values corresponding to five different orientations of the angular momentum with respects to the z direction. Each will have the same rotation rate and energy because the length of the angular momentum vector has the same magnitude by different orientations.

An atom is also a spherical system where the total orbital angular momentum is quantitized. The spherical harmonics give the atomic orbitals their angular shape. For example, l=2 corresponds to a d orbital. Their are five d orbitals each corresponding to a different m quantum number. These five d orbitals are normally degenerate but in the presence of a magnetic field they will separate into five different energies corresponding to the orientation of the angular momentum vector with the field.