Every stable equilibrium in physics is this problem, because every smooth minimum is a parabola once you look closely enough at it.

Chapter 5 solved potentials made of flat pieces by joining the pieces. This chapter takes the first smooth potential in the book — the harmonic oscillator , $V = \frac12kx^2$ — and gets a spectrum of an entirely different character: not a handful of levels at awkward spacings, but an infinite ladder with exactly equal rungs.

It is also the potential that started quantum mechanics. Planck assumed in 1900 that atoms radiate like oscillators with quantized energy; Einstein assumed the same of light in 1905 and of the vibrations of a solid in 1907. Between them those guesses explained black-body radiation, the photoelectric effect and the specific heats of solids — and none of the three had a theory to justify it. This chapter is that theory.

Physics you need first — Hooke's law, and what makes a motion 'simple harmonic'

Hooke’s law. Stretch a spring by $x$ and it pulls back with force $F = -kx$ : proportional to the displacement, opposite in direction. The constant $k$ is the spring or elastic constant , measured in $\mathrm{N\,m^{-1}}$ — a stiff spring has a large $k$ .

Why that one force law matters so much. It is the only force whose equation of motion is linear with constant coefficients, so it is the only one with a clean general solution — and, as this page shows, it is what every smooth force looks like near a point of stable equilibrium. That is why it is worth an entire chapter.

Work and potential energy. Moving the particle from $x$ to $x + \mathrm dx$ against that force costs $kx\,\mathrm dx$ of work, and the potential energy is the accumulated total — the integral in Eq. (6.1) below. This is the standard “work done against a force is stored as potential energy” bookkeeping, and the only calculus on this page.

Simple harmonic motion is the resulting motion: a pure cosine of fixed frequency, whose frequency does not depend on how hard you started it. A pendulum keeps time for exactly this reason.

Carried over from earlier chapters: §4.3 for $\hat H\psi = E\psi$ and $\Psi = \psi e^{-iEt/\hbar}$ , and §3.5 for the operators $\hat x$ and $\hat p$ .

6.1 The classical oscillator

A particle on a spring of elastic constant $k$ , displaced by $x$ , feels $F = -kx$ . The work needed to move it from $x$ to $x + \mathrm dx$ is $kx\,\mathrm dx$ , so the stored potential energy is

V(x) = \int_0^x kx'\,\mathrm dx' = \tfrac12kx^2\tag{6.1}

Newton’s second law gives the equation of motion,

m\frac{\mathrm d^2x}{\mathrm dt^2} = -kx\tag{6.2}

which is conventionally rewritten as

\ddot x = -\omega^2 x,\qquad \omega = \sqrt{k/m}\tag{6.3}

with general solution

x = A\cos(\omega t + \alpha)\tag{6.4}

During the motion the potential and kinetic energies trade places, but their sum is fixed:

E = \tfrac12m\dot x^2 + \tfrac12kx^2 = \tfrac12m\omega^2A^2\tag{6.5}

kinetic ½mẋ²
potential ½kx²
total E — constant, Eq. (6.5)

ωt0.00

amplitude A1.00phase α / π0.00

x 1.000

KE 0.000

PE 0.500

KE + PE 0.500

Drag the phase α and the curve slides sideways without changing shape; drag the amplitude A and it scales without changing period. That independence is the whole content of Eq. (6.4): a classical oscillator can carry any energy, set by A alone, and the frequency does not care. Both facts fail in quantum mechanics — the energies come in a discrete ladder, and the lowest rung is not zero.

The lower panel is Eq. (6.5). Kinetic and potential energy are sinusoids in antiphase, each swinging between 0 and E at frequency 2ω, not ω — the particle passes through the middle twice per period — and their sum is the flat gold line. Watch the KE + PE readout stay fixed while its two parts move.

What this really says, and what it costs quantum mechanics

Three things in Eq. (6.4) are worth naming as promises, because the rest of the chapter breaks all three.

The energy can be anything. $E = \frac12m\omega^2A^2$ , and $A$ is a free constant, so a classical oscillator has a continuous energy spectrum starting at zero. Quantum mechanically the energies are $\left(n + \frac12\right)\hbar\omega$ — discrete, and the lowest is not zero.

The particle is somewhere definite. $x(t)$ is a function; ask where the particle is at time $t$ and there is one answer. Quantum mechanically there is a probability distribution, and it reaches beyond the classical turning points.

It oscillates. That is the whole point of the word. Yet Phillips warns on this very page: “oscillators either have definite energy and do not oscillate, or they oscillate with uncertain energy.” A state of definite energy is a stationary state, and nothing about it changes with time — §6.3 shows that getting anything to move requires mixing energies, and getting it to move like Eq. (6.4) requires mixing a great many of them.

Engineer's bridge — the quadratic form, and why this one potential is everywhere

Every stable equilibrium is this problem. Expand any smooth $V(x)$ about a minimum at $x = 0$ :

V(x) = V(0) + V'(0)\,x + \tfrac12V''(0)\,x^2 + \dots

$V(0)$ is a choice of where to put zero. $V'(0) = 0$ because it is a minimum. So the leading behaviour is $\frac12V''(0)x^2$ — Eq. (6.1) with $k = V''(0)$ . There is nothing special about springs; the harmonic oscillator is just the second-order term of everything, and $k$ is a curvature. The widget’s second view is this statement with a real molecule in it.

In more than one variable it is a Hessian. Near a minimum, $V(\mathbf x) \approx \frac12\mathbf x^\mathsf{T}\mathbf H\mathbf x$ with $\mathbf H$ the matrix of second derivatives — positive definite exactly when the equilibrium is stable. Diagonalize it and you get normal modes: $N$ independent oscillators, one per eigenvector, each with $\omega_i = \sqrt{\lambda_i/m}$ . That is the same eig() that §4.2 used on a string, and it is how every vibrational spectrum, every modal analysis and every phonon calculation is actually done.

And it is why quadratic cost functions are the ones you can solve. Least squares, LQR control, Kalman filtering, Newton’s method — all of them are “approximate the thing by a quadratic near its optimum, because a quadratic is the one shape with an exact closed-form answer”. Physics gets the harmonic oscillator for the identical reason.

Where it breaks. The approximation is local, and how local is a real question, not a formality — the widget shows the parabola running 21% high just one oscillator length from the bottom of a hydrogen bond. Worse, it fails qualitatively at large displacement: a parabola rises forever, so a harmonic oscillator can never dissociate, ionize or break, and every real system can. It also silently assumes the equilibrium is stable; at a maximum $V''(0) < 0$ , the cosine becomes a hyperbolic cosine, and the motion runs away instead of oscillating.

Reproduce it in NumPy

The molecule is H₂ with its real Morse parameters, so the k and ħω here can be checked against measured values rather than taken on trust.

import numpy as np

eV, u, hbar = 1.602177e-19, 1.660539e-27, 1.054571e-34

# H2, as a Morse potential: V(r) = De [1 - exp(-beta (r - re))]^2
De, beta, re = 4.7446*eV, 1.9426e10, 0.7414e-10
mu = 0.50391*u

V = lambda x: De*(1 - np.exp(-beta*x))**2          # x = r - re
k = 2*De*beta**2                                    # V''(0), the exact curvature
Vh = lambda x: 0.5*k*x**2

print("why EVERY smooth minimum is a harmonic oscillator")
print("=" * 68)
print("  Taylor: V(x) = V(0) + V'(0) x + (1/2) V''(0) x^2 + ...")
print("  At a minimum V'(0) = 0 and V(0) can be set to zero, so the FIRST")
print("  surviving term is quadratic.  That is Eq. (6.1) with k = V''(0).\n")
print(f"  H2 as a Morse well: De = {De/eV:.3f} eV, beta = {beta:.4e} /m")
print(f"  k = 2 De beta^2 = {k:.1f} N/m      (measured value for H2: ~575 N/m)")
w = np.sqrt(k/mu)
print(f"  omega = sqrt(k/mu) = {w:.4e} /s   ->  hbar omega = {hbar*w/eV:.4f} eV")
print(f"  (measured H2 vibrational quantum: 0.516 eV -- 6 % from a 2-term Taylor)\n")

a = np.sqrt(hbar/(mu*w))
print(f"  the oscillator length a = sqrt(hbar/mu omega) = {a*1e12:.2f} pm")
print(f"  = {100*a/re:.1f} % of the bond length {re*1e12:.1f} pm\n")

print(f"  {'displacement x':>16} {'x/a':>7} {'V exact/eV':>12} {'(1/2)kx^2/eV':>14} {'error':>9}")
for xp in (1e-12, 5e-12, 1e-11, 2e-11, 4e-11):
    e, h = V(xp), Vh(xp)
    print(f"  {xp*1e12:13.1f} pm {xp/a:7.2f} {e/eV:12.4f} {h/eV:14.4f} {100*(h-e)/e:8.1f} %")
print("\n  Note how fast that error grows: the parabola is within a few percent")
print("  only inside a fraction of an oscillator length, and by x = a it is")
print("  already 21 % high.  Yet the hbar*omega it predicts is within 6 % of")
print("  the measured value -- because omega depends only on the CURVATURE AT")
print("  THE BOTTOM, and the ground state weights small displacements most.")
print("  A crude fit to the well shape still gets the level spacing nearly")
print("  right.  What it cannot get right is the top of the well: a parabola")
print("  rises forever, so a true harmonic oscillator can never dissociate,")
print("  and that failure is exactly what Fig. 6.4 in section 6.4 shows.")

printswhy EVERY smooth minimum is a harmonic oscillator ==================================================================== Taylor: V(x) = V(0) + V'(0) x + (1/2) V''(0) x^2 + ... At a minimum V'(0) = 0 and V(0) can be set to zero, so the FIRST surviving term is quadratic. That is Eq. (6.1) with k = V''(0). H2 as a Morse well: De = 4.745 eV, beta = 1.9426e+10 /m k = 2 De beta^2 = 573.7 N/m (measured value for H2: ~575 N/m) omega = sqrt(k/mu) = 8.2804e+14 /s -> hbar omega = 0.5450 eV (measured H2 vibrational quantum: 0.516 eV -- 6 % from a 2-term Taylor) the oscillator length a = sqrt(hbar/mu omega) = 12.34 pm = 16.6 % of the bond length 74.1 pm displacement x x/a V exact/eV (1/2)kx^2/eV error 1.0 pm 0.08 0.0018 0.0018 2.0 % 5.0 pm 0.41 0.0407 0.0448 10.1 % 10.0 pm 0.81 0.1479 0.1790 21.1 % 20.0 pm 1.62 0.4918 0.7162 45.6 % 40.0 pm 3.24 1.3847 2.8647 106.9 % Note how fast that error grows: the parabola is within a few percent only inside a fraction of an oscillator length, and by x = a it is already 21 % high. Yet the hbar*omega it predicts is within 6 % of the measured value -- because omega depends only on the CURVATURE AT THE BOTTOM, and the ground state weights small displacements most. A crude fit to the well shape still gets the level spacing nearly right. What it cannot get right is the top of the well: a parabola rises forever, so a true harmonic oscillator can never dissociate, and that failure is exactly what Fig. 6.4 in section 6.4 shows.

6.2 The quantum oscillator

The defining property of a quantum system is its Hamiltonian. Take Eq. (6.1), promote position and momentum to operators, and read off

\hat H = \frac{\hat p^2}{2m} + \frac12m\omega^2\hat x^2\tag{6.6}

or, substituting $\hat p = -i\hbar\,\partial/\partial x$ from Eq. (3.30),

\hat H = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + \frac12m\omega^2x^2\tag{6.7}

Every state, stationary or not, obeys the Schrödinger equation with this Hamiltonian:

i\hbar\frac{\partial\Psi}{\partial t} = \hat H\Psi\tag{6.8}

States of definite energy take the form established in §4.3,

\Psi(x,t) = \psi(x)\,e^{-iEt/\hbar}\tag{6.9}

and substituting Eq. (6.9) into Eq. (6.8) leaves the energy eigenvalue equation:

\left[-\frac{\hbar^2}{2m}\frac{\mathrm d^2}{\mathrm dx^2} + \frac12m\omega^2x^2\right]\psi(x) = E\psi(x)\tag{6.10}

step 1 of 4

Four lines, and the same four that produced every eigenvalue equation since §4.3. Worth watching once more, because the reason the time cancels is the reason stationary states are stationary.

1Start from the Schrödinger equation with the oscillator Hamiltonian, and propose a state of definite energy — Eq. (6.9). This is a GUESS about the form of the solution, justified by §4.3, not a derivation.
$Ψ (x, t) = ψ (x) e^{- i E t /ℏ}$
2Differentiate the left-hand side in time. Only the exponential depends on t, and differentiating it brings down −iE/ħ.
$i ℏ \frac{\partial Ψ}{\partial t} = i ℏ (- \frac{i E}{ℏ}) ψ (x) e^{- i E t /ℏ} = E ψ (x) e^{- i E t /ℏ}$
The i and the ħ both cancel, leaving a bare E. This is why E ends up as the eigenvalue.
3Apply Ĥ on the right. It differentiates in x and multiplies by x², so it never touches the time factor at all — the exponential rides along untouched.
$\hat{H} Ψ = [- \frac{ℏ ^{2}}{2 m} \frac{d ^{2}}{d x ^{2}} + \frac{1}{2} m ω^{2} x^{2}] ψ (x) e^{- i E t /ℏ}$
4Both sides now carry the same factor e^{−iEt/ħ}, which is never zero, so divide it out. The time dependence has vanished from the problem entirely — which is exactly why a state of definite energy has no observable dynamics.
$[- \frac{ℏ ^{2}}{2 m} \frac{d ^{2}}{d x ^{2}} + \frac{1}{2} m ω^{2} x^{2}] ψ (x) = E ψ (x)$
Equation (6.10). An eigenvalue problem: find the ψ that Ĥ merely rescales.

Solutions must be normalizable, which here means

\psi(x) \to 0 \quad\text{as}\quad x \to \pm\infty\tag{6.11}

Why that boundary condition is the entire source of quantization

Equation (6.10) has a solution for every value of $E$ — it is a second-order differential equation, and they always do. Almost all of those solutions blow up as $x \to \pm\infty$ , which makes $\int|\psi|^2\,\mathrm dx$ infinite and the state meaningless.

Equation (6.11) throws them away. Only for a discrete set of $E$ does the solution decay at both ends, and those are the energy levels. This is the same mechanism as every quantized spectrum so far — the infinite well’s $\psi(0) = \psi(a) = 0$ , the finite well’s smooth join — dressed differently: a boundary condition selects, and what survives the selection is discrete.

Because the sides of the oscillator rise without limit, no state can escape and there is no continuum. Contrast §5.1b, where a finite well let every positive energy through: there the walls ran out, here they never do. So this potential has infinitely many bound states and nothing else.

One more thing falls out of Eq. (6.10) before it is solved at all. It contains exactly three dimensional constants — $\hbar$ , $m$ and $\omega$ — and there is only one way to combine them into a length:

a = \sqrt{\hbar/m\omega}

This is the oscillator length , and it is the size the problem gives itself. A classical oscillator has no such scale — its amplitude is whatever you chose — so the mere existence of $a$ is already a quantum statement. Every eigenfunction on the next page turns out to be this wide, to within a factor of $\sqrt{n+\frac12}$ , and the $12.3\ \mathrm{pm}$ that the widget above reports for a hydrogen bond is this quantity.

V(x)
energy level Eₙ
ψₙ(x)
classically forbidden (V > E)

ω1.0

selected level: E₀ = 0.5000
nodes in ψ: 0
bound states: infinitely many
E₁ − E₀: 1.0000

The one potential with an exact, elegant solution — and the reason it matters is that the bottom of ANY smooth well looks like this. The levels are evenly spaced at ħω, and the lowest sits at ½ħω, not zero.

The quantum number starts at n = 0 here, and at n = 1 for the box

The book labels the oscillator’s ground state $n = 0$ and its excited states $n = 1, 2, 3, \dots$ , and flags in a footnote that the infinite square well of §4.4 uses $n = 1$ for its ground state.

The reason is that each convention makes its own formula tidy — $E_n = (n+\frac12)\hbar\omega$ wants to start at zero, while $E_n = n^2\pi^2\hbar^2/2ma^2$ would give zero energy for $n = 0$ and no state at all. There is no principle here, only two habits that have both stuck.

Where it bites: the number of nodes. The box’s $\psi_n$ has $n - 1$ interior nodes; the oscillator’s $\psi_n$ has $n$ . Both amount to “the ground state has none”, but the arithmetic differs, and comparing the two spectra without noticing the offset is a reliable way to be off by one.

What comes next, and why the book delays it

Phillips now states the answer and defers the derivation to §6.6 — energies $E_n = (n + \frac12)\hbar\omega$ , eigenfunctions a Gaussian times a polynomial — so that §6.3 can spend its length on physics rather than on differential equations.

That is a defensible order, and this site follows it. But two things are worth knowing now, because they are what make the delay safe.

The answer can be checked without the derivation. The widget above solved Eq. (6.10) by the recipe of §0.3 — discretize $x$ , build $\hat H$ as a matrix, call eigh — and the ladder it draws is the ladder §6.6 will derive. Nothing in this chapter has to be taken on faith.

The derivation is worth more than its result. §6.6 finds the spectrum with raising and lowering operators rather than by solving a differential equation at all: build one state, then step up the ladder algebraically. That method is the rehearsal for angular momentum in chapter 8, and it is the reason the harmonic oscillator is the first thing anyone meets in quantum field theory. The book marks §6.6 optional; it is the most reusable thing in the chapter.

Put a number on it — why nobody noticed the ladder in a pendulum

Take a $1\ \mathrm{kg}$ mass swinging at $1\ \mathrm{Hz}$ with an amplitude of $1\ \mathrm{cm}$ .

Step 1 — its energy, from Eq. (6.5) with $\omega = 2\pi\ \mathrm{rad\,s^{-1}}$ :

E = \tfrac12m\omega^2A^2 = \tfrac12(1)(2\pi)^2(0.01)^2 = 1.97\times10^{-3}\ \mathrm{J}

Step 2 — the size of one quantum:

\hbar\omega = (1.055\times10^{-34})(6.283) = 6.63\times10^{-34}\ \mathrm{J}

Step 3 — divide. $n \approx E/\hbar\omega = 3.0\times10^{30}$ .

The pendulum is on rung three thousand billion billion billion. Adding one quantum changes its energy by 3 parts in $10^{31}$ — far below any conceivable measurement, and far below the thermal noise in the pivot. The ladder is there; the rungs are simply unresolvable.

Now the same arithmetic for a CO molecule, which is a spring with $k = 1907\ \mathrm{N\,m^{-1}}$ and a reduced mass of $6.85$ amu: $\hbar\omega = 0.270\ \mathrm{eV}$ , while $k_BT$ at room temperature is $0.026\ \mathrm{eV}$ . That is $n \approx 0.1$ — less than one quantum of thermal energy available, so the molecule sits in its ground state and stays there.

Same equation, same constants. Only $m$ and $k$ changed, and the answer moved from “continuous beyond any hope of resolution” to “frozen in a single state”.

Reproduce it in NumPy

Three oscillators spanning thirty orders of magnitude in ħω, from the same two-line formula. The elastic constants for H₂ and CO are the measured ones; §6.4 will show how they are measured.

import numpy as np

hbar, eV, u = 1.054571e-34, 1.602177e-19, 1.660539e-27

def osc(m, k):
    """Eq. (6.3): omega = sqrt(k/m), and the period is 2 pi / omega."""
    w = np.sqrt(k/m)
    return w, 2*np.pi/w

print("Eq. (6.5): the classical oscillator carries E = (1/2) m omega^2 A^2")
print("=" * 70)
print(f"  {'system':<26} {'k / N m^-1':>11} {'omega / s^-1':>13} {'period':>12} {'hbar*omega':>13}")
rows = [("a 1 kg mass, 1 Hz", 1.0, 39.478),
        ("H2 molecule (vibration)", 0.50391*u, 575.0),
        ("CO molecule (vibration)", 6.85*u, 1907.3)]
for name, m, k in rows:
    w, T = osc(m, k)
    e = hbar*w
    per = f"{T:.3e} s" if T < 1 else f"{T:.3f} s"
    print(f"  {name:<26} {k:11.3e} {w:13.4e} {per:>12} {e/eV:10.3e} eV")

print("\nwhy nobody noticed quantization in a pendulum")
print("  a 1 kg pendulum, 1 Hz, swinging with amplitude 1 cm:")
m, f, A = 1.0, 1.0, 0.01
w = 2*np.pi*f
E = 0.5*m*w**2*A**2
print(f"    E        = {E:.4e} J        (Eq. 6.5)")
print(f"    hbar*w   = {hbar*w:.4e} J")
print(f"    n = E/hbar*w - 1/2 = {E/(hbar*w) - 0.5:.3e}")
print("    Adding one quantum changes the energy by 3 parts in 10^31.  The")
print("    ladder is there; the rungs are unresolvably close.")
print("\n  the same sum for a CO molecule at room temperature:")
w_co, _ = osc(6.85*u, 1907.3)
kT = 1.380649e-23*300
print(f"    hbar*w   = {hbar*w_co/eV:.4f} eV     kT at 300 K = {kT/eV:.4f} eV")
print(f"    n at kT  = {kT/(hbar*w_co):.3f}")
print("    Under one quantum, so the molecule sits in n = 0 and stays there.")
print("    Same equation, same constants: only the mass and stiffness differ.")

printsEq. (6.5): the classical oscillator carries E = (1/2) m omega^2 A^2 ====================================================================== system k / N m^-1 omega / s^-1 period hbar*omega a 1 kg mass, 1 Hz 3.948e+01 6.2832e+00 1.000 s 4.136e-15 eV H2 molecule (vibration) 5.750e+02 8.2896e+14 7.580e-15 s 5.456e-01 eV CO molecule (vibration) 1.907e+03 4.0949e+14 1.534e-14 s 2.695e-01 eV why nobody noticed quantization in a pendulum a 1 kg pendulum, 1 Hz, swinging with amplitude 1 cm: E = 1.9739e-03 J (Eq. 6.5) hbar*w = 6.6261e-34 J n = E/hbar*w - 1/2 = 2.979e+30 Adding one quantum changes the energy by 3 parts in 10^31. The ladder is there; the rungs are unresolvably close. the same sum for a CO molecule at room temperature: hbar*w = 0.2695 eV kT at 300 K = 0.0259 eV n at kT = 0.096 Under one quantum, so the molecule sits in n = 0 and stays there. Same equation, same constants: only the mass and stiffness differ.

question⇅	classical oscillator⇅	quantum oscillator⇅
what energies are allowed?	any $E \geq 0$ , set by $A$ ⓘ	$E_{n} = (n + \frac{1}{2}) ℏ ω$ onlyⓘ
can it be at rest?	yes — $A = 0$ , $E = 0$ ⓘ	no — the lowest energy is $\frac{1}{2} ℏ ω$ ⓘ
where can it be found?	only within $- A < x < A$ ⓘ	anywhere, with probability $∣ ψ_{n} ∣^{2}$ ⓘ
does it oscillate?	always, at $ω$ ⓘ	only if its energy is uncertainⓘ
what sets the length scale?	the amplitude $A$ — you choose itⓘ	$a = ℏ/ mω$ — you do notⓘ

Where this goes next

§6.3 takes the eigenvalues and eigenfunctions as given and works out what they mean: the equally spaced ladder, the Gaussian eigenfunctions and their nodes, parity, and the exact result $\Delta x\,\Delta p = (n + \frac12)\hbar$ — which saturates the uncertainty principle precisely in the ground state. Then it asks the question this section raised and left hanging: how does a quantum oscillator ever manage to look like Eq. (6.4)?

Check yourself

0 / 6 answered

Open the "why a parabola?" view and drag the zoom out to ±120 pm.
1.The dashed parabola and the real molecular bond come apart badly. What does that divergence tell you about the harmonic oscillator as a model?
2.Why does *every* smooth potential minimum look like $\frac{1}{2} k x^{2}$ close enough in?
3.In the lower panel of the motion view, the kinetic and potential energies each complete two full cycles per period of $x (t)$ . Why twice?
4.Equation (6.10) has a solution for every value of $E$ , yet the oscillator's energies are discrete. What does the selecting?
5.A 1 kg pendulum swinging at 1 Hz with a 1 cm amplitude sits at roughly $n = 3 \times 1 0^{30}$ . What is the right conclusion?
6.In $\hat{H} = \overset{p}{^}^{2} /2 m + \frac{1}{2} m ω^{2} \overset{x}{^}^{2}$ , what has become of the classical frequency $ω$ ?