40.8 The Quantum Harmonic Oscillator

Simple harmonic motion is exceptionally important in classical physics, where it serves as a prototype for more complex oscillations. As you might expect, a microscopic oscillator—the quantum harmonic oscillator—is equally important as a model of oscillations at the atomic level.

The defining characteristic of simple harmonic motion is a linear restoring force: F=−kx, where k is the spring constant. The corresponding potential-energy function, as you learned in Chapter 10, is

U(x)=12 kx2 (40.42)

where we’ll assume that the equilibrium position is xe=0. The potential energy of a harmonic oscillator is shown in FIGURE 40.20. It is a potential-energy well with curved sides.

FIGURE 40.20 The potential energy of a harmonic oscillator.

The figure illustrates the potential energy of a harmonic oscillator graphically.
FIGURE 40.20 Full Alternative Text

A classical particle of mass m oscillates with angular frequency

ω=km (40.43)

between the two turning points where the energy line crosses the parabolic potential-energy curve. As you’ve learned, this classical description fails if m represents an atomic particle, such as an electron or an atom. In that case, we need to solve the Schrödinger equation to find the wave functions.

The Schrödinger equation for a quantum harmonic oscillator is

d2ψdx2=−2mℏ2(E−12 kx2)ψ(x) (40.44)

where we used U(x)=12 kx2. We will assert, without deriving them, that the wave functions of the first three states are

ψ1(x)=A1e−x2/2b2ψ2(x)=A2xb e−x2/2b2ψ3(x)=A3(1−2x2b2)e−x2/2b2 (40.45)

where

b=ℏmω (40.46)

The constant b has dimensions of length. We will leave it as a homework problem for you to show that b is the classical turning point of an oscillator in the n=1 ground state. The constants A1, A2, and A3 are normalization constants. For example, A1 can be found by requiring

∫−∞∞| ψ1(x)|2 dx=A12∫−∞∞e−x2/b2 dx=1 (40.47)

The completion of this calculation also will be left as a homework problem.

As expected, stationary states of a quantum harmonic oscillator exist only for certain discrete energy levels, the quantum states of the oscillator. The allowed energies are given by the simple equation

En=(n−12)ℏω  n=1, 2, 3,  … (40.48)

where ω is the classical angular frequency, Equation 40.43, and n is the quantum number.

NOTE   The ground-state energy of the quantum harmonic oscillator is E1=12 ℏω. An atomic mass on a spring can not be brought to rest. This is a consequence of the uncertainty principle.

FIGURE 40.21 shows the first three energy levels and wave functions of a quantum harmonic oscillator. Notice that the energy levels are equally spaced by ΔE=ℏω. This result differs from the particle in a box, where the energy levels get increasingly farther apart. Also notice that the wave functions, like those of the finite potential well, extend beyond the turning points into the classically forbidden region.

FIGURE 40.21 The first three energy levels and wave functions of a quantum harmonic oscillator.

The figure illustrates the graph of the first three energy levels and wave functions of a quantum harmonic oscillator.
FIGURE 40.21 Full Alternative Text

FIGURE 40.22 shows the probability density |ψ(x)|2 for the n=11 state of a quantum harmonic oscillator. Notice how the node spacing and the amplitude both increase as the particle moves away from the equilibrium position at x=0. This is consistent with item 3 of Tactics Box 40.1. The particle slows down as it moves away from the origin, causing its de Broglie wavelength and the probability of finding it to increase.

FIGURE 40.22 The quantum and classical probability densities for the n=11 state of a quantum harmonic oscillator.

The figure illustrates the graph of probability densities for the n equals 11 state of a quantum harmonic oscillator.
FIGURE 40.22 Full Alternative Text

Section 40.5 introduced the classical probability density Pclass(x) and noted that a classical particle is most likely to be found where it is moving the slowest. Figure 40.22 shows Pclass(x) for a classical particle with the same total energy as the n=11 quantum state. You can see that on average the quantum probability density |ψ(x)|2 mimics the classical probability density. This is just what the correspondence principle leads us to expect.

EXAMPLE 40.9 Light emission by an oscillating electron

An electron in a harmonic-oscillator potential well emits light of wavelength 600 nm as it jumps from one level to the next lowest level. What is the “spring constant” of the restoring force?

MODEL The electron is a quantum harmonic oscillator.

SOLVE A photon is emitted as the electron undergoes the quantum jump n→n−1. We can use Equation 40.48 for the energy levels to find that the electron loses energy

ΔE=En−En−1=(n−12)ℏωe−(n−1−12)ℏωe=ℏωe

ΔE=ℏωe for all transitions, independent of n, because the energy levels of the quantum harmonic oscillator are equally spaced. We need to distinguish the harmonic oscillations of the electron from the oscillations of the light wave, hence the subscript e on ωe.

The emitted photon has energy Ephoton=hfph=ΔE. Thus

ℏωe=h2πωe=hfph=hcλ

The wavelength of the light is λ=600 nm, so the classical angular frequency of the oscillating electron is

ωe=2πcλ=3.14×1015 rad/s

The electron’s angular frequency is related to the spring constant of the restoring force by

ωe=km

Thus k=mωe2=9.0 N/m.