The ground state is not the bottom of the well. It sits above it by exactly the amount the uncertainty principle demands, and no state of this potential does better.

§6.2 wrote down the eigenvalue equation and stopped. This section takes its answer as given — the derivation waits for §6.6 — and spends its length on what the answer means.

It divides in two. First the stationary states, which turn out to carry an unusual number of exact results: the levels are evenly spaced, the eigenfunctions have a definite parity, and $\Delta x\,\Delta p$ comes out to $(n+\frac12)\hbar$ — hitting the uncertainty principle’s floor precisely in the ground state. Then the non-stationary states, and the question §6.1 left hanging: if a state of definite energy does not move, how does a quantum oscillator ever manage to look like $A\cos(\omega t + \alpha)$ ?

Physics you need first — what an expectation value is, and what it is not

$\langle x\rangle$ is the mean of the position over many measurements on identically prepared systems — not “where the particle is”, and not a time-average of one particle’s trajectory. Formally $\langle x\rangle = \int\Psi^*x\Psi\,\mathrm dx$ , from §3.5.

Two consequences matter on this page. A quantity can have a perfectly definite expectation value while being wildly uncertain — $\langle x\rangle = 0$ for every oscillator eigenstate, including highly excited ones spread over many oscillator lengths. And $\Delta x = \sqrt{\langle x^2\rangle - \langle x\rangle^2}$ is the spread, which is the quantity the uncertainty principle constrains.

Also carried over: §4.6 for stationary versus non-stationary states and beats at $|E_m - E_n|/\hbar$ ; §4.5 for expanding a state in a complete orthonormal basis; and §1.4 for $\Delta x\,\Delta p \ge \hbar/2$ itself.

Stationary states

The eigenvalues, derived in §6.6 and quoted here, are

E_n = \left(n + \tfrac12\right)\hbar\omega,\qquad n = 0, 1, 2, 3, \dots\tag{6.12}

A particle with energy $E_n$ has the wave function

\Psi(x,t) = \psi_n(x)\,e^{-iE_nt/\hbar}\tag{6.13}

and the four lowest eigenfunctions are those of Table 6.1: a Gaussian times a polynomial, with the oscillator length $a = \sqrt{\hbar/m\omega}$ setting the width throughout. The polynomials are the Hermite polynomials $H_n(x/a)$ — degree $n$ , and their $n$ roots are precisely the $n$ nodes of $\psi_n$ .

n⇅	energy⇅	eigenfunction⇅
$0$ ⓘ	$E_{0} = \frac{1}{2} ℏ ω$ ⓘ	$(\frac{1}{a π})^{1/2} e^{- x^{2} /2 a^{2}}$ ⓘ
$1$ ⓘ	$E_{1} = \frac{3}{2} ℏ ω$ ⓘ	$(\frac{1}{2 a π})^{1/2} 2 (\frac{x}{a}) e^{- x^{2} /2 a^{2}}$ ⓘ
$2$ ⓘ	$E_{2} = \frac{5}{2} ℏ ω$ ⓘ	$(\frac{1}{8 a π})^{1/2} [2 - 4 (\frac{x}{a})^{2}] e^{- x^{2} /2 a^{2}}$ ⓘ
$3$ ⓘ	$E_{3} = \frac{7}{2} ℏ ω$ ⓘ	$(\frac{1}{48 a π})^{1/2} [12 (\frac{x}{a}) - 8 (\frac{x}{a})^{3}] e^{- x^{2} /2 a^{2}}$ ⓘ

What is observable about a state of definite energy

Parity. Because $\hat H$ is unchanged under $x \to -x$ , each eigenfunction has a definite symmetry:

\psi_n(-x) = +\psi_n(x)\ \ (n\text{ even}),\qquad \psi_n(-x) = -\psi_n(x)\ \ (n\text{ odd})

Even $n$ has positive parity , odd $n$ negative. This is the same property problem 1 of chapter 5 introduced for a symmetric well, and for the same reason: a symmetry of the Hamiltonian becomes a label on its eigenstates.

The density does not move.

|\Psi_n(x,t)|^2 = |\psi_n(x)|^2\tag{6.14}

The time factor $e^{-iE_nt/\hbar}$ has modulus 1, so it cancels — exactly as in §4.6. And unlike a classical particle, which is confined to $-A < x < A$ , the quantum particle has non-zero probability everywhere.

A cross-reference that goes nowhere — the second and last in this book

For that result the book sends you to “Section 4.8”. Chapter 4 ends at §4.6, and the section meant is §4.6, Time Dependence, which is where it showed that a state of definite energy has no observable time dependence. Verified from page-131.png, so it is the book’s error rather than a text-extraction artefact.

This is the second such slip, and the two are related. §5.2a flags a reference to “Section 4.7” that should be §4.5 — also into chapter 4, also overshooting by exactly 2. Chapter 4 evidently lost two section boundaries late in production, and only the references from later chapters were left stale.

It is worth knowing that these are the only two. Every Section N.M reference in all eleven chapters was checked against the real section list while this chapter was being summarized: 3.1 through 11.3 all resolve, and exactly these two dangle. So no later chapter needs this particular suspicion — though a reference can still point at a real section and mean a different one, which §4.1 records happening once.

Position and momentum. For the $n$ -th state,

\langle x\rangle = 0,\qquad \langle x^2\rangle = \left(n+\tfrac12\right)a^2,\qquad \Delta x = \sqrt{n+\tfrac12}\,a

\langle p\rangle = 0,\qquad \langle p^2\rangle = \left(n+\tfrac12\right)\frac{\hbar^2}{a^2},\qquad \Delta p = \sqrt{n+\tfrac12}\,\frac{\hbar}{a}

and therefore

\Delta x\,\Delta p = \left(n + \tfrac12\right)\hbar

The ground state sits exactly on the uncertainty principle's floor

§1.4 says $\Delta x\,\Delta p \ge \hbar/2$ for any state. Put $n = 0$ into the result above and it gives $\Delta x\,\Delta p = \hbar/2$ — equality. The oscillator ground state is not merely allowed by the uncertainty principle; it is the tightest state the principle permits.

That is why $\psi_0$ is a Gaussian. Among all normalized functions, the Gaussian is the unique minimizer of the uncertainty product — a fact §3.6 met from the Fourier side, since a Gaussian is its own transform. The oscillator’s ground state is the same theorem arriving from the other direction.

And it explains the zero-point energy without any algebra. Localize the particle at the bottom of the well and $\Delta x \to 0$ forces $\Delta p \to \infty$ , costing unbounded kinetic energy. Give it definite momentum and $\Delta x$ explodes, costing unbounded potential energy. The zero-point energy is the minimum of that trade-off, and problem 1 turns this paragraph into an inequality that produces $\frac12\hbar\omega$ from the uncertainty principle alone.

The energy divides equally. Because both $x$ and $p$ are uncertain, so are the potential and kinetic energies, and their expectation values are

\tfrac12m\omega^2\langle x^2\rangle = \tfrac12E_n\qquad\text{and}\qquad \frac{\langle p^2\rangle}{2m} = \tfrac12E_n

Half the energy is potential, half kinetic — in every state, exactly. Their sum is $E_n$ , which is sharp even though neither part is.

Engineer's bridge — equipartition, and a Gaussian that minimizes a time–bandwidth product

The even split is the virial theorem, and you have met it as equipartition. For a quadratic potential the time-averaged kinetic and potential energies of a classical oscillator are equal too — which is why classical statistical mechanics assigns $\frac12k_BT$ to each and gets $k_BT$ per oscillator. Here the same split holds state by state, exactly, with no averaging over time and no thermodynamics involved.

$\Delta x\,\Delta p = (n+\frac12)\hbar$ is a time–bandwidth product, and the $n$ -dependence is the interesting part. In DSP, $\Delta t\,\Delta f \ge 1/4\pi$ with equality only for a Gaussian window; every other window — Hamming, Kaiser, rectangular — pays a penalty above the bound. The oscillator states are literally that hierarchy: $\psi_0$ is the Gaussian window sitting on the bound, and $\psi_n$ is a Hermite–Gaussian, the $n$ -th member of a family that is $(2n+1)$ times worse. Hermite–Gaussians are eigenfunctions of the Fourier transform, which is exactly why $x$ and $p$ enter these formulas symmetrically.

And if you have done any optics, you have already drawn Fig. 6.2. The transverse modes of a laser cavity are Hermite–Gaussian beams — TEM₀₀ is $\psi_0$ , TEM₁₀ is $\psi_1$ — for the identical mathematical reason: a paraxial beam in a quadratic index profile obeys the same equation as this one, with $z$ in place of $t$ .

Where it breaks. The DSP analogy is about a window, a passive shape you choose; these are states, and which one the system occupies is decided by energy and dynamics, not by design. And the equipartition analogy fails at the place that matters most: classically the split is $\frac12k_BT$ each and both vanish at $T = 0$ , whereas here both halves stay finite at $\frac14\hbar\omega$ in the ground state. Zero-point energy has no classical counterpart at all.

Reproduce it in NumPy

The recurrence is the one the widget above runs, so this checks the widget as well as the book: ⟨x²⟩ = (n+½)a² and the node count come out right at n = 30, where a factorial-based formula would already have overflowed.

import numpy as np
from math import factorial

# hbar = m = omega = 1, so a = sqrt(hbar/m omega) = 1 and E_n = n + 1/2.
NMAX, x = 60, np.linspace(-14, 14, 3001)

def hermite_functions(nmax, x):
    """Normalized oscillator eigenfunctions by a STABLE recurrence:
    psi_n = sqrt(2/n) x psi_{n-1} - sqrt((n-1)/n) psi_{n-2}."""
    p = np.zeros((nmax+1, x.size))
    p[0] = np.pi**-0.25*np.exp(-x*x/2)
    if nmax: p[1] = np.sqrt(2.0)*x*p[0]
    for n in range(2, nmax+1):
        p[n] = np.sqrt(2.0/n)*x*p[n-1] - np.sqrt((n-1)/n)*p[n-2]
    return p

psi = hermite_functions(NMAX, x)
print("the eigenfunctions, from the recurrence the widget uses")
print("=" * 66)
print(f"  {'n':>3} {'norm':>9} {'<x^2>':>9} {'(n+1/2)':>9} {'nodes':>6}")
for n in (0, 1, 2, 3, 10, 30):
    nrm = np.trapezoid(psi[n]**2, x)
    x2 = np.trapezoid(psi[n]*x*x*psi[n], x)
    nodes = int((np.diff(np.sign(psi[n][(x>-9)&(x<9)])) != 0).sum())
    print(f"  {n:3d} {nrm:9.6f} {x2:9.4f} {n+0.5:9.4f} {nodes:6d}")

print("\nEq. (6.17): a quasi-classical state really does trace A cos(omega t)")
print("  |c_n|^2 = Poisson(nbar), and A should come out as sqrt(2 nbar) a")
print(f"  {'nbar':>6} {'<E>/hw':>9} {'(nbar+1/2)':>11} {'dE/hw':>8} {'sqrt(nbar)':>11} {'A meas':>8} {'sqrt(2nbar)':>12}")
for nbar in (1.0, 4.0, 9.0, 16.0):
    c = np.array([np.exp(-nbar/2)*np.sqrt(nbar**n/factorial(n)) for n in range(NMAX+1)])
    c /= np.linalg.norm(c)
    n_ = np.arange(NMAX+1)
    Emean = float((c**2*(n_+0.5)).sum())
    dE = float(np.sqrt((c**2*(n_+0.5)**2).sum() - Emean**2))
    # <x>(t) = sum_mn c_m c_n x_mn cos((m-n) t); only |m-n|=1 survives
    ts = np.linspace(0, 2*np.pi, 400)
    xt = np.zeros_like(ts)
    for n in range(NMAX):
        xt += 2*c[n]*c[n+1]*np.sqrt((n+1)/2)*np.cos(ts)
    print(f"  {nbar:6.1f} {Emean:9.4f} {nbar+0.5:11.4f} {dE:8.4f} {np.sqrt(nbar):11.4f} "
          f"{xt.max():8.4f} {np.sqrt(2*nbar):12.4f}")
print("  dE = sqrt(nbar) hbar omega exactly, so dE/<E> -> 1/sqrt(nbar): the")
print("  energy becomes relatively sharper as the state becomes more excited.")
print("\n  and <x>(t) is a PURE cosine at omega, not a sum of harmonics, because")
print("  x_mn vanishes unless |m-n| = 1 -- the selection rule of problem 11.")

printsthe eigenfunctions, from the recurrence the widget uses ================================================================== n norm <x^2> (n+1/2) nodes 0 1.000000 0.5000 0.5000 0 1 1.000000 1.5000 1.5000 1 2 1.000000 2.5000 2.5000 2 3 1.000000 3.5000 3.5000 3 10 1.000000 10.5000 10.5000 10 30 1.000000 30.5000 30.5000 30 Eq. (6.17): a quasi-classical state really does trace A cos(omega t) |c_n|^2 = Poisson(nbar), and A should come out as sqrt(2 nbar) a nbar <E>/hw (nbar+1/2) dE/hw sqrt(nbar) A meas sqrt(2nbar) 1.0 1.5000 1.5000 1.0000 1.0000 1.4142 1.4142 4.0 4.5000 4.5000 2.0000 2.0000 2.8284 2.8284 9.0 9.5000 9.5000 3.0000 3.0000 4.2426 4.2426 16.0 16.5000 16.5000 4.0000 4.0000 5.6569 5.6569 dE = sqrt(nbar) hbar omega exactly, so dE/<E> -> 1/sqrt(nbar): the energy becomes relatively sharper as the state becomes more excited. and <x>(t) is a PURE cosine at omega, not a sum of harmonics, because x_mn vanishes unless |m-n| = 1 -- the selection rule of problem 11.

Non-stationary states

Nothing above moves. To get motion, superpose:

\Psi(x,t) = \sum_{n=0,1,2,\dots} c_n\psi_n(x)\,e^{-iE_nt/\hbar}\tag{6.15}

This is a state of uncertain energy: measuring $E$ gives $\frac12\hbar\omega$ with probability $|c_0|^2$ , $\frac32\hbar\omega$ with probability $|c_1|^2$ , and so on. It is also non-stationary, because $|\Psi|^2$ now carries interference between different eigenfunctions. The interference of $\psi_m$ with $\psi_n$ oscillates at

\omega_{m,n} = \frac{|E_m - E_n|}{\hbar}\tag{6.16}

which for this spectrum is always an integer multiple of $\omega$ . So $|\Psi(x,t)|^2$ can carry $\omega$ , $2\omega$ , $3\omega$ and so on.

⟨x⟩(t), from the full Σ c*ₘcₙ xₘₙ sum
A cos ωt — the classical motion, Eq. (6.4)

ωt0.00

⟨E⟩ / ħω 1.000

ΔE / ħω 0.500

ΔE / ⟨E⟩ 0.500

amplitude A / a 0.707

Two energies, so the density sloshes. Every pair of levels contributes a beat at ω_m,n = |Eₘ − Eₙ|/ħ, which for this potential is always an integer multiple of ω — so |Ψ|² is exactly periodic with period 2π/ω, unlike the box of §4.6, whose n² spectrum gives it a long revival time instead.

Yet ⟨x⟩ below is a pure cosine at ω alone, with no 2ω component even for the three-level mix. That is the selection rule: the matrix xₘₙ is tridiagonal, so only neighbouring levels can contribute to the average position, and neighbours are always exactly ħω apart.

Why this is not §4.6's widget, and why the difference matters

§4.6 built the same kind of superposition for a particle in a box, and the two behave quite differently.

The box has $E_n \propto n^2$ , so the beat frequencies $|E_m - E_n|/\hbar$ are not commensurate with any single frequency. Its wave packet spreads, sloshes irregularly, and only reassembles after a long revival time.

The oscillator has $E_n \propto (n + \frac12)$ , so every beat frequency is an integer multiple of $\omega$ and every superposition is exactly periodic with period $2\pi/\omega$ — no revival structure, no spreading of a coherent state, nothing to wait for. That is a special property of an evenly spaced ladder, and it is the deep reason the oscillator can imitate classical motion at all.

Why the average position oscillates at $\omega$ and nothing else

$|\Psi|^2$ carries harmonics, so one expects $\langle x\rangle$ to carry them too. It does not. Evaluating $\langle x(t)\rangle = \int\Psi^*x\Psi\,\mathrm dx$ produces terms in $c_m^*c_nx_{m,n}e^{i\omega_{m,n}t}$ with

x_{m,n} = \int_{-\infty}^{+\infty}\psi_m^*(x)\,x\,\psi_n(x)\,\mathrm dx

and $x_{m,n}$ vanishes unless $|m - n| = 1$ (problem 11). Adjacent levels are exactly $\hbar\omega$ apart, so the only surviving frequency is $\omega$ :

\langle x(t)\rangle = A\cos(\omega t + \alpha),\qquad \langle p(t)\rangle = -m\omega A\sin(\omega t + \alpha)\tag{6.17}

A selection rule is a sparsity pattern

$x_{m,n}$ is a matrix indexed by two quantum numbers, and the statement “zero unless $|m-n| = 1$ ” says it is tridiagonal with a hollow diagonal. That is all a selection rule ever is: a statement about which entries of an operator’s matrix are non-zero.

Everything follows from the sparsity. Only neighbouring levels are connected, so only their energy difference — $\hbar\omega$ , always — can appear in $\langle x\rangle$ . The same tridiagonality is why a vibrating molecule emits at one wavelength rather than a comb of them (§6.4), and why the ladder operators of §6.6 move exactly one rung: $\hat x$ is built out of those operators, and they are the two off-diagonals.

The widget above computes $\langle x\rangle$ as the full double sum over every pair $(m,n)$ . The harmonics are absent from the plot because $x_{m,n}$ makes them zero — not because they were left out of the sum.

Quasi-classical states

Equation (6.17) says the average moves like a classical oscillator, but says nothing about the spread around that average, which may be large and may change through the cycle. The states that look most classical are the quasi-classical states , with energy amplitudes following a Poisson distribution:

|c_n|^2 = \frac{\bar n^{\,n}}{n!}e^{-\bar n},\qquad \bar n \gg 1

Using the mean and standard deviation of the Poisson distribution from problem 1 of chapter 3,

\langle E\rangle = \left(\bar n + \tfrac12\right)\hbar\omega,\qquad \Delta E = \sqrt{\bar n}\,\hbar\omega

so the relative energy uncertainty is

\frac{\Delta E}{\langle E\rangle} = \frac{\sqrt{\bar n}}{\bar n + \frac12} \longrightarrow \frac{1}{\sqrt{\bar n}}

Put a number on it — how classical is a real pendulum?

§6.1 found that a $1\ \mathrm{kg}$ pendulum swinging at $1\ \mathrm{Hz}$ with a $1\ \mathrm{cm}$ amplitude sits at $\bar n \approx 3.0\times10^{30}$ . Feed that into the formula above:

\frac{\Delta E}{\langle E\rangle} \approx \frac{1}{\sqrt{\bar n}} = \frac{1}{1.7\times10^{15}} = 5.8\times10^{-16}

Its energy is uncertain by six parts in ten million billion — far below any measurement, and below the fractional energy loss to air resistance in a single swing.

And the amplitude works out too. A quasi-classical state swings with $A = \sqrt{2\bar n}\,a$ , and for this pendulum $a = \sqrt{\hbar/m\omega} = 4.1\times10^{-18}\ \mathrm{m}$ , giving

A = \sqrt{2(3.0\times10^{30})}\times 4.1\times10^{-18}\ \mathrm{m} = 1.0\ \mathrm{cm}

which is the amplitude we started from. The quantum description of a pendulum reproduces the classical one exactly, and it does so through a state that is a superposition of about $10^{30}$ energy eigenstates — none of which, alone, would move at all.

What this really says — Planck and Einstein assumed the ladder; here it is derived

Phillips ends the section by pointing out the direction of the argument. Planck and Einstein assumed that things which visibly oscillate might have quantized energies — a guess that ran from the classical world toward the quantum one.

This chapter runs the other way. It starts from quantized energies and shows that they can, in the right combination, almost imitate simple harmonic motion. “Almost” is the honest word: a quasi-classical state has uncertain energy, its position has a spread of order $a$ at every instant, and it never becomes a point moving along a trajectory. What it becomes is a probability distribution whose shape stays fixed while its centre traces $A\cos(\omega t + \alpha)$ — which is as close to a classical particle as quantum mechanics ever gets, and close enough that nobody noticed for two hundred years.

Where this goes next

§6.4 puts these results to work on a real system: two nuclei vibrating in a chemical bond, where $\hbar\omega$ becomes an infrared photon and Eq. (6.12) becomes a way of weighing the bond. Then §6.5 takes the oscillator into three dimensions, where the evenly spaced ladder acquires degeneracies. §6.6 finally derives Eq. (6.12), by an algebraic method that will reappear in chapter 8 as the whole theory of angular momentum.

Check yourself

0 / 6 answered

Select "ground state alone" in the widget and press play.
1.Nothing moves. What is the wave function doing while nothing moves?
2.Why is $Δ x Δ p = \frac{ℏ}{2}$ exactly, and not merely close to it, in the oscillator's ground state?
Switch to "ψ₀ + ψ₁ + ψ₂" and look at the lower panel.
3. $∣Ψ ∣^{2}$ visibly carries a $2 ω$ component, yet $⟨ x ⟩$ is a clean cosine at $ω$ . Why?
Switch to "quasi-classical" and drag n̄ from 0.5 up to 16.
4. $Δ E$ grows as $\overset{n}{ˉ} ℏ ω$ — it gets *bigger*. In what sense is the state becoming more classical?
5.A superposition in a box eventually reassembles after a long revival time; an oscillator superposition repeats every $2 π / ω$ exactly. What is responsible?
6.In every oscillator eigenstate, $\frac{1}{2} m ω^{2} ⟨ x^{2} ⟩$ and $⟨ p^{2} ⟩ /2 m$ each equal $\frac{1}{2} E_{n}$ . What is the classical counterpart, and where does the analogy fail?