The motion of two atoms in a diatomic molecule can be separated into translational, vibrational, and rotational motions. Both rotation and vibrational motions are internal motions that do not change the center of mass for the molecule (Figure \(\PageIndex{1}\)), which is described by translational motion. Quantum translational motions can be modeled with the particle in a box model discussed previously and rotation and vibration can be modeled via the rigid rotor and harmonic oscillator models, respectively.
Figure \(\PageIndex{1}\): Two atoms connected by a vibrating bond. There is a rotation around the common center of mass, and oscillation in bond distance. (CC SA-BY, 3.0; www.cleonis.nl).
Before delving into the quantum mechanical harmonic oscillator, we will introduce the classical harmonic oscillator (i.e., involving classical mechanics) to build an intuition that we will extend to the quantum world. A classical description of the vibration of a diatomic molecule is needed because the quantum mechanical description begins with replacing the classical energy with the Hamiltonian operator in the Schrödinger equation. It also is interesting to compare and contrast the classical description with the quantum mechanical picture.
The Classical Harmonic Oscillator
Simple harmonic oscillators about a potential energy minimum can be thought of as a ball rolling frictionlessly in a curved dish or a pendulum swinging frictionlessly back and forth (Figure \(\PageIndex{2}\)). The restoring forces are precisely the same in either horizontal direction.
Figure \(\PageIndex{2}\): A mass connected to a spring that follows Hooke's law is a system that, when displaced from its equilibrium position (\(x_o=0\)), experiences a restoring force, \(F\) that is proportional to the displacement, \(x\) from the equilibrium length (\(x_o\)). (Public Domain; LucasVB).
If we consider the bond to behave like a mass on a spring (Figure \(\PageIndex{2}\)), then this restoring force (\(F\)) is proportional to the displacement (\(x\)) from the equilibrium length (\(x_o\)) - this is Hooke's Law:
\[ F = - kx \label {5.1.2}\]
where \(k\) is the force constant. Hooke's Law says that the force is proportional to, but in opposite direction to, the displacement (\(x\)). The force constant reflects the stiffness of the spring. The idea incorporated into the application of Hooke's Law to a diatomic molecule is that when the atoms move away from their equilibrium positions, a restoring force is produced that increases proportionally with the displacement from equilibrium. The potential energy for such a system increases quadratically with the displacement.
\[ V (x) = \dfrac {1}{2} k x^2 \label {5.1.3}\]
Hooke's Law or the harmonic (i.e. quadratic) potential given by Equation \(\ref{5.1.3}\) is an excellent approximation for the vibrational oscillations of molecules. The magnitude of the force constant \(k\) depends upon the nature of the chemical bond in molecular systems just as it depends on the nature of the spring in mechanical systems. The larger the force constant, the stiffer the spring or the stiffer the bond. Since it is the electron distribution between the two positively charged nuclei that holds them together, a double bond with more electrons has a larger force constant than a single bond, and the nuclei are held together more tightly.
Solving the Harmonic Oscillator Model
The classical equation of motion for a one-dimensional simple harmonic oscillator with a particle of mass \(m\) attached to a spring having spring constant \(k\) is
\[ m \dfrac{d^2x(t)}{dt^2} = -kx(t) \label{5.1.4a}\]
which can be written in the standard wave equation form:
Equation \(\ref{5.1.4a}\) is a linear second-order differential equation that can be solved by the standard method of factoring and integrating. The resulting solution to Equation \(\ref{5.1.4a}\) is
\[ x(t) = x_o \sin (\omega t + \phi) \label{5.1.5}\]
with
\[\omega = \sqrt{\dfrac{k}{m}} \label{5.1.6}\]
and the momentum \(\) has time dependence
\[\begin{align} p &= mv \\[4pt] &=mx_o \omega \cos (\omega t + \phi) \label{5.1.7} \end{align}\]
Figure \(\PageIndex{4}\) show the displacement of the bond from its equilibrium length as a function of time. Such motion is called harmonic.
Figure \(\PageIndex{4}\): Solution to the Harmonic Oscillator. Displacement (y-axis) is plotted as a function of time.
Harmonic Oscillator Energies
The energy of the vibration is the sum of the kinetic energy and the potential energy. The momentum associated with the harmonic oscillator is
\[p = m \dfrac {dx}{dt} \label {5.1.8}\]
so combining Equations \ref{5.1.8} and \ref{5.1.3}, the total energy can be written as
\[ \begin{align} E &= T + V \\[4pt] &= \dfrac {p^2}{2 m} + \dfrac {k}{2} x^2 \label {5.1.9} \end{align}\]
The total energy of the harmonic oscillator is equal to the maximum potential energy stored in the spring when \(x = \pm A\), called the turning points (Figure \(\PageIndex{5}\)). The total energy (Equation \(\ref{5.1.9}\)) is continuously being shifted between potential energy stored in the spring and kinetic energy of the mass.
Figure \(\PageIndex{5}\): The potential \(V(x)\) for a harmonic oscillator. Vertical axis energy; horizontal axis displacement \(x\). The potential energy \(V(x) = ½ kx^2\) is shown in red. (Paul Wormer Wikipedia).
The motion of a classical oscillator is confined to the region where its kinetic energy is nonnegative, which is what the energy relation Equation \ref{5.1.9} says. Physically, it means that a classical oscillator can never be found beyond its turning points, and its energy depends only on how far the turning points are from its equilibrium position. The energy of a classical oscillator changes in a continuous way. The lowest energy that a classical oscillator may have is zero, which corresponds to a situation where an object is at rest at its equilibrium position. The zero-energy state of a classical oscillator simply means no oscillations and no motion at all (a classical particle sitting at the bottom of the potential well in Figure \(\PageIndex{5}\)). When an object oscillates, no matter how big or small its energy may be, it spends the longest time near the turning points, because this is where it slows down and reverses its direction of motion. Therefore, the probability of finding a classical oscillator between the turning points is highest near the turning points and lowest at the equilibrium position. (Note that this is not a statement of preference of the object to go to lower energy. It is a statement about how quickly the object moves through various regions.)