The harmonic oscillator
The two atoms of a carbon monoxide molecule are held together by a bond that behaves like a stiff spring. Cool a gas of these molecules towards absolute zero and each one settles into its lowest vibrational state. You might expect that state to be no vibration at all: the two atoms fixed at exactly the right distance. It is not. Why can't a molecule's vibration be removed completely?
Almost anything held near a stable position is pulled back with a force proportional to how far it has strayed: atoms in molecules and crystals, the electromagnetic field in a cavity, a pendulum swinging gently. That is a harmonic oscillator, and in quantum mechanics it has a result simple enough to state in one line and important enough to recur throughout physics: its energy levels are evenly spaced, and the lowest is not zero.
The levels
Put the parabola $V = \tfrac12 m\omega^2 x^2$ into the stationary-state equation. Like the box of the previous essay, the parabola holds the particle in, so only certain energies give wavefunctions that die away far out and can be normalised. Unlike the box, its walls get steadily steeper rather than being vertical, and the result is different. The allowed energies are
$$E_n = \left(n + \tfrac12\right)\hbar\omega,\qquad n = 0, 1, 2, \dots$$Two things stand out. The levels are evenly spaced, each one $\hbar\omega$ above the last, where the box's spread out as $n^2$. So a quantum oscillator can only gain or lose energy in units of $\hbar\omega = h\nu$, where $\nu$ is its classical frequency. This is exactly the assumption Planck made in 1900 about the oscillators in the walls of a hot box, found a quarter of a century later to follow from the wave equation. And the lowest level, $n = 0$, has energy $\tfrac12\hbar\omega$, not zero. That is the zero-point energy.
Checking the ground state by hand
The general solution takes some work, but the ground state can be checked in a few lines. Try a bell-shaped wavefunction of width $a$,
$$\varphi_0(x) = e^{-x^2/2a^2}.$$Differentiating twice gives $\varphi_0'' = \left(x^2/a^4 - 1/a^2\right)\varphi_0$. Put that into the stationary-state equation and divide by $\varphi_0$:
$$-\frac{\hbar^2}{2m}\left(\frac{x^2}{a^4} - \frac{1}{a^2}\right) + \frac12 m\omega^2 x^2 = E .$$The right side is a constant, so the $x^2$ terms on the left must cancel. That happens only if $\hbar^2/(2ma^4) = \tfrac12 m\omega^2$, which fixes the width, $a^2 = \hbar/m\omega$. What is left is the energy, $E = \hbar^2/2ma^2 = \tfrac12\hbar\omega$. The bell curve works, but only with one particular width and only with the zero-point energy.
This ground state is a Gaussian, which the uncertainty essay singled out as the shape with the smallest possible product $\Delta x\,\Delta p = \hbar/2$. The oscillator settles into the best compromise between being localised, which costs kinetic energy, and being spread out, which costs potential energy in the spring. The zero-point energy is the price of that compromise, and no amount of cooling removes it.
It is worth being careful about what this state is. It is stationary: its probability distribution for the bond length is the same at every moment, so nothing in it swings back and forth. What is not zero is the spread of that distribution, the spread of momentum that goes with it, and the average kinetic energy, which is a quarter of $\hbar\omega$. The zero-point energy is not a picture of atoms wobbling on a trajectory; it is the energy of a state that cannot be sharper than this.
Worked example: carbon monoxide
The spacing. Carbon monoxide absorbs infrared light strongly at a wavelength of 4.67 μm, as its vibration jumps from $n = 0$ to $n = 1$. The photon energy is $1240/4670 = 0.266$ eV, so $\hbar\omega = 0.266$ eV. The zero-point energy is half that, 0.133 eV.
The spring. The frequency is $\omega = 0.266\ \text{eV}/\hbar = 4.04\times10^{14}$ rad/s. Both atoms move, each about the molecule's centre of mass, and two bodies moving that way behave exactly like a single body of the reduced mass $\mu = m_Cm_O/(m_C + m_O)$ on the same spring. For carbon monoxide $\mu = 6.86$ atomic mass units, or $1.14\times10^{-26}$ kg. The bond's stiffness is then $k = \mu\omega^2 = 1860$ N/m. Stretching such a spring by 5 cm would take the weight of about 9 kg: an ordinary stiff spring, carried by a single bond between two atoms.
How uncertain the bond length is. The ground-state width is $a = \sqrt{\hbar/\mu\omega} = 4.8$ pm, and the standard deviation of the bond length is $a/\sqrt2 = 3.4$ pm. The bond is 113 pm long, so even at absolute zero its length is uncertain by about 3 percent.
Check: why vibration is frozen at room temperature. At 300 K, $k_BT = 0.026$ eV, a tenth of $\hbar\omega$. By the Boltzmann factor from the blackbody essay, the fraction of molecules in $n = 1$ compared with $n = 0$ is $e^{-0.266/0.026} \approx 3\times10^{-5}$. Almost every molecule sits in its ground state. This is Planck's freezing-out again, now for a molecule instead of a mode of light.
Optional: raising and lowering, and why the spacing is even
Measure position in units of $a$ and momentum in units of $\hbar/a$, and define the combinations $\hat a = (\hat x + i\hat p)/\sqrt2$ and $\hat a^\dagger = (\hat x - i\hat p)/\sqrt2$. Using $[\hat x, \hat p] = i$ in these units, the energy operator becomes
$$\hat H = \hbar\omega\left(\hat a^\dagger\hat a + \tfrac12\right),\qquad [\hat a, \hat a^\dagger] = 1 .$$From the commutator, if $\varphi$ has energy $E$ then $\hat a^\dagger\varphi$ has energy $E + \hbar\omega$ and $\hat a\varphi$ has $E - \hbar\omega$. The operators climb and descend a ladder with rungs $\hbar\omega$ apart. The descent must stop somewhere, because energy in a parabola cannot be negative, and it stops at a state with $\hat a\varphi_0 = 0$, whose energy is then $\tfrac12\hbar\omega$. That is the whole spectrum, obtained without solving a differential equation. The same operators later describe photons: a mode of light in the state $n$ holds $n$ photons.
Large $n$ behaves classically
A classical mass on a spring moves fastest through the middle and slowest near its turning points, so a snapshot taken at a random moment most often catches it near the ends. The right panel of the figure shows that the quantum density for $n = 10$ already follows that classical shape on average, with rapid oscillations on top. As $n$ grows the oscillations get finer and the match better: the correspondence principle again. The ground state is the opposite: its distribution peaks at the centre, where a classical mass at rest would sit, yet its momentum is spread and its kinetic energy is not zero.
Two questions
1. Hydrogen chloride. HCl absorbs at 3.47 μm. What are its vibrational quantum $\hbar\omega$ and its zero-point energy? Compared with carbon monoxide, is its vibration more or less frozen out at room temperature?
Worked answer
$\hbar\omega = 1240/3470 = 0.358$ eV, and the zero-point energy is 0.179 eV. At 300 K the ratio of $n = 1$ to $n = 0$ is $e^{-0.358/0.026} \approx 10^{-6}$, even smaller than carbon monoxide's $3\times10^{-5}$, so HCl is more thoroughly frozen out.
A slip to avoid: reading the higher frequency as a stiffer bond. The bond is in fact less stiff than CO's (about 480 N/m against 1860 N/m). The frequency is high because the oscillating mass, dominated by the light hydrogen atom, is small: $\omega = \sqrt{k/\mu}$ depends on both.
2. Motion at absolute zero. Someone says: "At absolute zero all motion stops, so the bond length of a molecule is then fixed exactly." What is right and what is wrong?
Worked answer
What is right: at absolute zero the molecule is in its lowest energy state, and nothing can be taken out of it. What is wrong: that lowest state is not a state with a fixed bond length. It has energy $\tfrac12\hbar\omega$ and a spread in bond length, 3.4 pm for carbon monoxide, which does not change in time. A bond length fixed exactly would mean $\Delta x = 0$, which by the uncertainty relation would need an infinite spread of momentum and infinite energy.
The tempting wrong turn is to equate "coldest possible" with "at rest". Temperature measures how the energy above the ground state is shared out; it says nothing against the ground state itself having spread and kinetic energy. The opposite wrong turn is to picture the atoms still shaking back and forth at absolute zero. Nothing in a stationary state oscillates; its probabilities are constant in time.
Further reading
- Landau & Lifshitz, Quantum Mechanics (2nd ed.), §23 "The linear oscillator": the full solution, including the polynomial wavefunctions.
- Dirac, Principles of Quantum Mechanics (4th ed.), Ch. VI §34 "The Harmonic Oscillator": the ladder-operator method of the optional section, from its source.
- Born, Atomic Physics (Dover, 8th ed.), Ch. V §4 "Wave Mechanics", with the oscillator's solution worked in Appendix XVI.
Where this leaves us
A molecule's vibration cannot be removed completely because its lowest state has energy $\tfrac12\hbar\omega$: a state with the bond length fixed exactly would need an unlimited spread of momentum, and localisation costs kinetic energy. The ground state is a fixed distribution of bond lengths with a nonzero spread, not a classical wobble. Above that floor the levels are evenly spaced by $\hbar\omega$, which is why oscillators, from molecules to modes of light, trade energy in equal quanta.
Vibration is motion along a line. A molecule can also rotate, and an electron in an atom moves around the nucleus in three dimensions. Rotation brings in a new quantity, angular momentum, which the Bohr model had to postulate and got wrong for the ground state. What values can it take, and how much of it can be known at once? The next essay answers both.