Quantum Harmonic Oscillator

In one sentence: A tiny particle on a spring can have only certain energies, spaced evenly like the steps of a staircase, and even its lowest step is above zero.

$E_n = \hbar \omega \left(n + \frac{1}{2}\right), \qquad n = 0, 1, 2, \dots$

Read it out loud: “E sub n equals h-bar times omega times, in brackets, n plus one half, with n equal to 0, 1, 2 and so on. E sub n is the energy of level n, in joules. H-bar (said AITCH-bar) is the reduced Planck constant. Omega (OH-may-guh) is the angular frequency, in radians per second. n is the level number, starting at zero. One half is one divided by two.”

Before you start

The ideas and signs this page uses, explained from zero. For powers of ten, Greek letters and other notation in general, see Math basics.

Powers of ten, like $10^{-34}$ and $10^{14}$: Some numbers here are very large and some are very small, so they use powers of ten. The exponent $-34$ means move the decimal point 34 places to the left, so $10^{-34}$ is a decimal point followed by 33 zeros and then a 1. The exponent 14 means move the point 14 places to the right. To multiply powers of ten, add the exponents: $10^{-34} \times 10^{14} = 10^{-20}$ .
Greek letters $\hbar$ (h-bar), $\omega$ (omega) and $\pi$ (pi): The symbol $\hbar$ is read ‘h-bar’. It is Planck’s constant $h$ divided by $2\pi$ , and its value is $1.054571817 \times 10^{-34}$ J s. The letter $\omega$ is read ‘OH-may-guh’ and is used for angular frequency. The letter $\pi$ is read ‘pie’ and equals 3.14159…, the number that relates a circle’s distance around to its width across.
Frequency, in hertz: Frequency counts how many complete vibrations or wiggles happen each second. Its unit is the hertz, written Hz, and 1 Hz means one vibration per second. A frequency of $9.0 \times 10^{13}$ Hz means 90 trillion vibrations every second. The letter $f$ is the usual symbol for ordinary frequency.
Radians and angular frequency $\omega$: A radian is the angle that cuts off an arc on a circle as long as the circle’s radius. A full turn is $2\pi$ radians, about 6.283 radians. Angular frequency $\omega$ counts radians per second instead of turns per second. Since one full vibration is $2\pi$ radians, $\omega = 2\pi f$ . The formula uses $\omega$ directly, because it sets the size of the energy steps.
Springs and the harmonic oscillator: A harmonic oscillator is something pulled back toward a middle point by a force that grows with its distance from that point. A mass on an ideal spring is the standard example. Pull it farther and the spring pulls harder. Atoms in a molecule act a little like masses on springs when they vibrate gently, which is why this model is so useful in chemistry.
The square root sign, $\sqrt{x}$: The symbol $\sqrt{x}$ means the number that, multiplied by itself, gives $x$ . So $\sqrt{4} = 2$ and $\sqrt{9} = 3$ . A square root undoes squaring. In this topic, $\omega = \sqrt{k/m}$ uses a square root for a spring with stiffness $k$ and mass $m$ .
Joules, electron volts and photons: A joule (J) is a unit of energy. A photon is one packet of light, and its energy is $hf$ . One electron volt (eV) is the energy an electron gains when pushed through one volt, equal to $1.602 \times 10^{-19}$ J. Vibration energies in molecules are small, so chemists often use electron volts or wavenumbers, which count waves per centimeter.

What each symbol means

Symbol	Meaning
$E_n$	The energy of level $n$ , in joules (J). For molecules, electron volts (eV) are more convenient, where $1\ \text{eV} = 1.602 \times 10^{-19}$ J. The energy is never less than half of $\hbar\omega$ , and each higher level adds one more step.
$n$	The level number, a whole number that starts at zero: $n = 0, 1, 2, \dots$ . Level 0 is the ground state, the lowest energy the oscillator can have. The level number has no unit.
$\hbar$	The reduced Planck constant, $1.054571817 \times 10^{-34}$ J s, read ‘h-bar’. It equals Planck’s constant divided by $2\pi$ . Its unit is energy times time, and it sets how big the energy steps are in a quantum system.
$\omega$	The angular frequency, in radians per second (rad/s). It equals $2\pi f$ , where $f$ is the ordinary frequency in hertz. A higher $\omega$ means a faster vibration, and a stiffer spring or a lighter mass gives a higher $\omega$ .

The formula piece by piece

$\omega$: Start with the angular frequency $\omega$ . It says how fast the oscillator swings or vibrates, in radians per second. A faster vibration has a bigger $\omega$ , and the energy steps grow with $\omega$ .
$\hbar\omega$: Multiply by $\hbar$ . The product $\hbar\omega$ is the size of one energy step, in joules. Each level sits one step above the level below it, so $\hbar\omega$ is the gap between neighboring levels.
$n + \frac{1}{2}$: Now add the count. $n$ counts how many steps above the ground state we are, and $\frac{1}{2}$ is an extra half step. For $n = 0$ the count is $\frac{1}{2}$ , so even the lowest level sits half a step above zero.
$\hbar\omega\left(n + \frac{1}{2}\right)$: Multiply the step size by the count. Level 0 gives half a step, level 1 gives one and a half steps, and level 2 gives two and a half steps. The levels are evenly spaced because the step size is the same for every level.
$E_n = \hbar \omega \left(n + \frac{1}{2}\right), \qquad n = 0, 1, 2, \dots$: The whole formula in one sentence: the energy of level $n$ equals the reduced Planck constant times the angular frequency times $n$ plus one half, for $n = 0, 1, 2$ and so on.

What the formula is doing, and why

A harmonic oscillator is a particle pulled back toward a fixed point by a force that grows with its distance, like a mass on a spring. What goes in is the angular frequency $\omega$ and the level number $n$ . What comes out is the energy of that level, in joules. In ordinary physics a swinging spring can have any energy. In quantum physics, it can only have the energies in the list.

Why equal steps? For an ideal spring, the time for one swing does not depend on how far it swings. That fixed rhythm means every step up adds the same energy, $\hbar\omega$ , and that energy depends only on the angular frequency. A stiffer spring or a lighter mass gives a higher $\omega$ , so the steps are bigger. For a mass $m$ on a spring of stiffness $k$ , $\omega = \sqrt{k/m}$ .

Why the one half? The lowest level is $n = 0$ , and its energy is $\frac{1}{2}\hbar\omega$ , not zero. This is the zero-point energy. A particle at the very bottom of the bowl would have to sit still at one spot, with a definite position and a definite momentum of zero. The uncertainty principle does not allow both to be exact at once, so the lowest state keeps some energy.

Why it matters for molecules. The bonds between atoms act like springs. A molecule can absorb a photon whose energy equals one step, $\hbar\omega = hf$ , and move up one level. That is why infrared light is absorbed at the frequencies of molecular vibrations. Each kind of bond has its own vibration frequency, so each bond absorbs its own infrared light.

Where does it apply? The formula is an approximation. Real bonds stretch further and eventually break, so at high energy the steps get smaller than the formula predicts. The harmonic model works best for small vibrations near the bottom of the energy well.

An everyday picture

Think of a staircase with evenly spaced steps. You can stand on a step but never between two steps. The quantum version of a bowl has its lowest step not on the floor. It sits half a step up, because the particle can never come to a complete rest at the bottom. A stiffer bond or a lighter atom makes the steps taller, so the staircase rises faster.

Worked example: Vibration of a C-H bond

A carbon-hydrogen (C-H) bond in a molecule vibrates with a frequency of $f = 9.0 \times 10^{13}$ Hz. That is close to the stretching frequency of C-H bonds, which spectroscopists often quote as about 3000 waves per centimeter. We find the energy step, the lowest energy and the first excited energy in electron volts (eV), and the wavelength of a photon with the energy step. The numbers are round values for practice.

Write down the frequency. The ordinary frequency is $f = 9.0 \times 10^{13}$ Hz, which is 90 trillion vibrations every second. The formula needs the angular frequency instead, so we convert it in the next step.
Convert to angular frequency. Multiply by $2\pi$ , because one vibration is one full turn of $2\pi$ radians: $\omega = 2\pi (9.0 \times 10^{13}\ \text{Hz}) \approx 5.65 \times 10^{14}$ rad/s. Multiply the front numbers first, $2\pi \times 9.0 \approx 56.5$ , and then write $56.5 \times 10^{13}$ , which tidies to $5.65 \times 10^{14}$ .
Find the energy step. Multiply $\hbar$ by $\omega$ : $\hbar\omega = (1.054571817 \times 10^{-34})(5.65 \times 10^{14}) \approx 5.96 \times 10^{-20}$ J. The exponents add, $-34 + 14 = -20$ , and the front numbers multiply to $1.0546 \times 5.65 \approx 5.96$ .
Convert the energy step to electron volts. Divide by $1.602 \times 10^{-19}$ to change joules into electron volts: $\frac{5.96 \times 10^{-20}}{1.602 \times 10^{-19}} \approx 0.372$ eV. The powers of ten differ by one, so the result is a tenth of $5.96 / 1.602 \approx 3.72$ . Every energy in this example is a little more than a third of an electron volt.
Find the zero-point energy. The lowest level is $\frac{1}{2}\hbar\omega = 0.5 \times 0.372 \approx 0.186$ eV. Even in its ground state, the bond keeps this much energy, so it never sits perfectly still.
Find the first excited level. Level 1 has $n + \frac{1}{2} = 1.5$ steps, so $E_1 = 1.5 \times 0.372 \approx 0.558$ eV. The gap between levels 0 and 1 is one full step, 0.372 eV, and that is the energy a photon must carry to lift the bond up one level.
Find the wavelength of the photon. A photon with energy $hf$ has wavelength $\lambda = \frac{c}{f}$ . Using $c = 3.00 \times 10^{8}$ m/s, $\lambda = \frac{3.00 \times 10^{8}}{9.0 \times 10^{13}} \approx 3.33 \times 10^{-6}$ m. The exponents give $8 - 13 = -5$ and the front numbers give $0.333$ , which tidies to $3.33 \times 10^{-6}$ m, or 3.33 micrometers.
Read the answer in plain terms. Three and a third micrometers is infrared light, longer than the red end of the visible spectrum, which is about 0.7 micrometers. Our eyes cannot see it, but this is the kind of light that C-H bonds absorb in infrared spectra.

Answer: The lowest vibrational energy of this bond is 0.186 eV, the energy steps are 0.372 eV, and the first excited level sits at 0.558 eV. A photon with the step energy has a wavelength of about 3.33 micrometers, in the infrared, which is why bonds like this absorb infrared light.

The numbers in this example are recomputed by script every time the site is built.

Common mistakes

Thinking the ground state has zero energy. The lowest level is $n = 0$ , with energy $\frac{1}{2}\hbar\omega$ , which is above zero.
Dropping the one half and writing $E_n = n\hbar\omega$ . That formula puts the ground state at zero and moves every level off by half a step.
Using the ordinary frequency $f$ where the formula needs the angular frequency $\omega$ . Since $\omega = 2\pi f$ , the energies come out about $2\pi$ times too small.
Using the harmonic formula for large vibrations. Real bonds stretch and break, so at high energy the levels are closer together than the formula predicts.

Where you will meet it

Infrared spectroscopy, where molecules absorb infrared light at frequencies that match the spacing between their vibrational levels.
Modeling the vibrations of atoms in crystals as quantized sound waves, called phonons.
Quantum field theory, where each mode of a field is treated as a harmonic oscillator.
Quantum chemistry programs, which often start from the harmonic approximation to estimate molecular vibration frequencies.

Questions people ask

Why is there energy even in the ground state?

A particle at the exact bottom of the bowl, with no motion, would have a definite position and a definite momentum of zero. The uncertainty principle does not allow both to be exact, so the lowest state keeps the zero-point energy $\frac{1}{2}\hbar\omega$ .

Why are the energy steps all the same size?

For an ideal spring-like force, the levels are equally spaced. That is why one photon with energy $hf = \hbar\omega$ moves the oscillator from one level to the next.

Are molecules really harmonic oscillators?

Only approximately. The model is good for small vibrations near the bottom of the energy well. At higher energy, real bonds stretch further and the levels bunch together.

Particle in a Box: A particle trapped between two walls can have only certain energies, like steps on a ladder, and the narrower the box, the higher those steps are.
Planck-Einstein Relation (E = hf): E = hf says each packet of light carries energy in proportion to its frequency: higher frequency means more energy per packet, and brightness does not change it.
Heisenberg Uncertainty Principle: The more tightly you pin down where a tiny particle is, the more its speed must spread out, and no instrument can beat that trade-off.