Modern Physics · Chapter 20

Quantum Mechanics

At the atomic scale, nature is neither wave nor particle but something fundamentally stranger — a probability amplitude that collapses to a definite outcome only upon measurement.

PrerequisitesWaveproperties(Ch.8)⋅Specialrelativity(Ch.19)⋅BasiccalculusWave properties (Ch. 8) \cdot Special relativity (Ch. 19) \cdot Basic calculus
Learning Goals
  • Explain the photoelectric effect and blackbody radiation as evidence for energy quantization.
  • Calculate the de Broglie wavelength of a particle from its momentum.
  • Apply the Heisenberg uncertainty principle to estimate minimum kinetic energies of confined particles.
  • Interpretthewavefunctionasaprobabilityamplitudeand∣ψ∣2asaprobabilitydensityInterpret the wavefunction as a probability amplitude and |\psi|^{2} as a probability density.
  • Find the quantized energy levels of a particle in an infinite square well.

20.1 The Failure of Classical Physics

By 1900, several experiments could not be explained by classical mechanics and electromagnetism. Three were decisive:

Blackbody radiation.A hot object emits light across a spectrum of wavelengths. Classical theory (the Rayleigh–Jeans law) predicted infinite energy emission at short wavelengths — the "ultraviolet catastrophe." In 1900, Planck resolved this by assuming energy is emitted in discrete quanta E = hf, where h = 6.626×10⁻³⁴ J·s is Planck's constant.

Photoelectric effect. Light shining on a metal ejects electrons, but only if the frequency exceeds a threshold — more intensity at low frequency does nothing. In 1905, Einstein explained this by treating light as particles (photons) each carrying energy E = hf.

Atomic spectra.Hydrogen emits light at only discrete wavelengths — a spectrum of sharp lines. Classical orbiting electrons should radiate continuously and spiral into the nucleus in nanoseconds. Bohr's 1913 model imposed quantization by fiat; the explanation had to wait for Schrödinger.

20.2 Wave-Particle Duality

Definition 20.1 — De Broglie Hypothesis (1924)
Every particle with momentum p has an associated wavelength:λ=h/p=h/(mv)(deBrogliewavelength\lambda = h/p = h/(mv) \qquad (de Broglie wavelengthThis applies to electrons,protons,neutrons—and,inprinciple,baseballs,thoughtheirwavelengths( 10−lectrons, protons, neutrons — and, in principle, baseballs, though their wavelengths (~10^{-}rablysmall.Forelectronsatatomicscales,λ≈0.1–1nm,comparabletoatomicspacingsrably small. For electrons at atomic scales, \lambda \approx 0.1–1 nm, comparable to atomic spacings, and diffraction effects are observable.

The double-slit experiment with electrons (Davisson–Germer, 1927; Jönsson, 1961) shows interference fringes identical to light — even when electrons are sent one at a time. Each electron passes through both slits simultaneously (as a wave), then lands at a definite spot (as a particle). No classical picture explains this.

Example 20.1 — De Broglie Wavelength of an Electron

AnelectronisacceleratedthroughV=100V.FinditsdeBrogliewavelengthAn electron is accelerated through V = 100 V. Find its de Broglie wavelength.

Kinetic energy:K=eV=1.6×10−19×100=1.6×10−17JK = eV = 1.6\times10^{-19} \times 100 = 1.6\times10^{-17} J
Momentum:K=p2/2m→p=(2mK)=(2×9.11×10−31×1.6×10−17)=1.71×10−24kg\cdotm/sK = p^{2}/2m \to p = \sqrt(2mK) = \sqrt(2 \times 9.11\times10^{-31} \times 1.6\times10^{-17}) = 1.71\times10^{-24} kg\cdotm/s
Wavelength:λ=h/p=6.626×10−34/1.71×10−24=\lambda = h/p = 6.626\times10^{-34} / 1.71\times10^{-24} = 3.88×10−10m=0.388nm3.88\times10^{-10} m = 0.388 nm
Context:This is comparable to atomic spacings — explaining why electron microscopes achieve atomic resolution.

20.3 The Heisenberg Uncertainty Principle

Definition 20.2 — Heisenberg Uncertainty Principle (1927)
It is impossible to simultaneously know a particle's position and momentum with arbitrary precision. The product of their uncertainties is bounded:\Deltax⋅\Deltap≥ℏ/2(ℏ=h/2π=1.055×10−34J⋯\Deltax \cdot \Deltap \ge \hbar/2 \qquad (\hbar = h/2\pi = 1.055\times10^{-34} J\cdotsSimilarlyforenergyandtime:\DeltaE⋅\Deltat≥ℏ/2Similarly for energy and time: \DeltaE \cdot \Deltat \ge \hbar/2.

This is not a statement about measurement clumsiness — it is a fundamental property of nature. A particle with a precisely defined momentum has a perfectly defined wavelength (λ = h/p) and therefore a completely delocalized position (a pure sine wave extends to infinity). Localizing a particle requires superposing many wavelengths (many momenta), so Δp grows.

The uncertainty principle explains atomic stability: an electron cannot collapse into the nucleus because confining it to Δx ≈ 10⁻¹⁵ m would require Δp ≥ ℏ/(2Δx) — enormous momentum and kinetic energy that blows it back out. The hydrogen atom sits at the radius where kinetic and potential energies balance.

Example 20.2 — Uncertainty Principle Applied

Anelectronisconfinedtoaregionofsize\Deltax=0.1nm(atomicscale).EstimatetheminimAn electron is confined to a region of size \Deltax = 0.1 nm (atomic scale). Estimate the minimum uncertainty in its momentum and kinetic energy.

Minimum Δp:\Deltap≥ℏ/(2\Deltax)=1.055×10−34/(2×10−10)=5.28×10−25kg\cdotm/s\Deltap \ge \hbar/(2\Deltax) = 1.055\times10^{-34} / (2 \times 10^{-10}) = 5.28\times10^{-25} kg\cdotm/s
Minimum K:K=(\Deltap)2/(2m)=(5.28×10−25)2/(2×9.11×10−31)=1.53×10−19J=K = (\Deltap)^{2}/(2m) = (5.28\times10^{-25})^{2} / (2 \times 9.11\times10^{-31}) = 1.53\times10^{-19} J = 0.96 eV
Context:This is comparable to atomic binding energies (~13.6 eV for hydrogen). The electron cannot be confined more tightly without enormous energy cost.

20.4 The Schrödinger Equation

De Broglie's matter waves needed a wave equation. In 1926, Schrödinger provided it. The time-dependent Schrödinger equation governs the quantum state ψ(x,t) — the wavefunction:

iℏ∂ψ/\partialt=[−ℏ2/2m⋅∂2/\partialx2+V(x)]ψi\hbar \partial\psi/\partialt = [-\hbar^{2}/2m \cdot \partial^{2}/\partialx^{2} + V(x)] \psi(20.1)

The wavefunction ψ is complex-valued. Its physical meaning, given by Born (1926): |ψ(x,t)|² is the probability density — the probability of finding the particle between x and x + dx is |ψ(x)|² dx.

For a particle in a box (infinite square well of width L), the allowed energies are:

En=n2π2ℏ2/(2mL2)n=1,2,3,⋯E_{n} = n^{2} \pi^{2} \hbar^{2} / (2mL^{2}) \qquad n = 1, 2, 3, \cdots(20.2)

Energy is quantized — only discrete values are allowed. This is not an assumption; it follows from the boundary conditions on ψ. The wavefunctions are ψ_n(x) = √(2/L) sin(nπx/L), standing waves just like a guitar string — but the "string" is a probability amplitude.

Figure 20.1. Particleinaninfinitesquarewell.Togglebetweenthewavefunctionψ(x)(animated,showiParticle in an infinite square well. Toggle between the wavefunction \psi(x) (animated, showingtheoscillationintime)andtheprobabilitydensity∣ψ∣2.Higherquantumnumbernmeanng the oscillation in time) and the probability density |\psi|^{2}. Higher quantum number n meansmorenodesandhigherenergy.Energyscalesasn2s more nodes and higher energy. Energy scales as n^{2}.
Theorem 20.1 — Quantum Numbers and Ground State Energy
Thelowestenergystate(n=1)hasenergyE1=π2ℏ2/(2mL2The lowest energy state (n=1) has energy E_{1} = \pi^{2}\hbar^{2}/(2mL^{2} > 0. A quantum particle can never be at rest at the bottom of a potential well. This zero-point energyisanotherconsequenceoftheuncertaintyprinciple:zeromomentumwouldmean\Deltap=0,requis another consequence of the uncertainty principle: zero momentum would mean \Deltap = 0, requiring\Deltax=∞iring \Deltax = \infty.
Definition 20.3 — Common Traps
  • The wavefunction is not the probability: probabilitydensityis∣ψ∣2probability density is |\psi|^{2}
  • Uncertainty is not bad equipment: \Deltax\Deltapisapropertyofthestate\Deltax\Deltap is a property of the state
  • Energy levels depend on boundary conditions: changing the well changes the spectrum.
  • Measurement changes the state: a definite outcome generally prepares a new state.
Exercises — 20.1–20.4 Quantum Mechanics
1.
Aphotonhaswavelengthλ=1020nm.FinditsenergyineVA photon has wavelength \lambda = 1020 nm. Find its energy in eV.
eV
Straightforward
2.
An electron is accelerated through 100 V. Find its de Broglie wavelength.
nm
Straightforward
3.
Anelectronisconfinedto\Deltax=0.1nm.EstimateitsminimumkineticenergyfromtheuncerAn electron is confined to \Deltax = 0.1 nm. Estimate its minimum kinetic energy from the uncertainty principle.
eV
Intermediate
4.Anelectronisconfinedina1DboxofwidthL=0.5nm.FindtheenergiesofthefirstthAn electron is confined in a 1D box of width L = 0.5 nm. Find the energies of the first threelevelsandthewavelengthofthephotonemittedinthe3→2transitionree levels and the wavelength of the photon emitted in the 3\to2 transition.
Intermediate
5.Explain quantum tunneling. Why is it allowed by quantum mechanics but forbidden classically? Give three real-world phenomena that depend on tunneling.
Challenging
Key Takeaways
  • Energyisquantized:E=hf(Planck);photonscarryenergyhfregardlessofintensityEnergy is quantized: E = hf (Planck); photons carry energy hf regardless of intensity.
  • Wave−particleduality:matterhaswavelengthλ=h/p(deBroglie);confirmedbyelectrondWave-particle duality: matter has wavelength \lambda = h/p (de Broglie); confirmed by electron diffraction.
  • Uncertaintyprinciple:\Deltax⋅\Deltap≥ℏ/2—afundamentallimit,notameasurementproblemUncertainty principle: \Deltax\cdot\Deltap \ge \hbar/2 — a fundamental limit, not a measurement problem.
  • Wavefunctionψ:∣ψ∣2isprobabilitydensity.Schro¨dinger′sequationgovernsitsevolutionWavefunction \psi: |\psi|^{2} is probability density. Schrödinger's equation governs its evolution.
  • Quantizedenergyinabox:En=n2π2ℏ2/2mL2—onlydiscreteenergiesallowedQuantized energy in a box: E_{n} = n^{2}\pi^{2}\hbar^{2}/2mL^{2} — only discrete energies allowed.
  • Zero-point energy: quantum particles can never be at rest; minimum energy is always nonzero.