Unit 3 · The Mathematical MachineryChapter 12 of 25

Quantum Tunneling

Walking through walls you don't have the energy to climb

  • 22 min read
  • 30 min listen
  • 25 videos
01Video briefing

Start with the preview

Start here — today's preview. Watch “What is Quantum Tunneling, Exactly?” from Up and Atom before (or after) this session. The lab will wait. Mostly.

Today's preview

What is Quantum Tunneling, Exactly?

Up and Atom10 min

A friendly ten-minute tour that builds tunnelling from light: an evanescent wave leaks a short way past a surface that totally reflects light, and an electron's wavefunction fades into a barrier in the same way, leaving a small chance of finding it on the far side. As you watch, ask what would happen to that chance if the barrier were twice as thick.

Watch on YouTube

More from the video vault

Deep dive

What is Quantum Tunnelling?

Physics Explained40 min

A forty-minute deep dive that starts from the alpha-decay paradox Rutherford faced, matches the wavefunction and its slope at the barrier edges to reach an explicit tunnelling probability, then approximates the real Coulomb barrier to estimate half-lives. Watch it after Parts 2 and 3, with our transmission formulas and the worksheet's Gamow challenge beside you.

Watch on YouTube
Bonus

The 2025 Nobel Prize in Physics (quantum tunnelling) - Sixty Symbols

Sixty Symbols13 min

Physicists from the University of Nottingham unpack the 2025 Nobel Prize in Physics, awarded for macroscopic quantum tunnelling and energy quantisation in a superconducting circuit. Watch it after Part 6 to see how an entire circuit can tunnel.

Watch on YouTube
  • What is Quantum Tunneling, Exactly? — Up and Atom — https://www.youtube.com/watch?v=WPZLRtyvEqo
  • What is Quantum Tunnelling? — Physics Explained — https://www.youtube.com/watch?v=_f8zeEI0oys
  • The 2025 Nobel Prize in Physics (quantum tunnelling) - Sixty Symbols — Sixty Symbols — https://www.youtube.com/watch?v=wwJf9ZUuKi8
02Cold open

Today's mission

Dr. Q · cold open

Psigor! Roll the bowling ball at the hill. Gently, you magnificent heap of bolts. Up it goes, slower, slower, it stops, and back it comes. Roll it a billion times and it comes back a billion times: that is what lacking the energy to climb means. Now shrink the ball to an electron and the hill to a sliver of insulator one nanometre thick, and listen to the detector on the far side. Click. Click. Great Planck's ghost, it got through! No hole, no ladder, no borrowed energy, and the wall is untouched. Last session our walls were infinitely high. Today they are merely enormous, and the wavefunction learns that a wall is not an ending. It is a fade. That fade lights the Sun, ages uranium, remembers your photos and runs quantum computers.

By the end of this session you will be able to:

  1. Explain why a wavefunction decays exponentially, rather than stopping, inside a classically forbidden region where V exceeds E, and compute its decay length 1/κ.
  2. Estimate transmission probabilities with T ≈ e^(−2κL), compare them with the exact rectangular-barrier result, and predict how T responds to changes in barrier width, energy deficit and particle mass.
  3. Describe how Gamow's 1928 tunnelling model explains alpha decay and its enormous range of half-lives, and why fusion in the Sun's core depends on tunnelling.
  4. Explain how the scanning tunnelling microscope turns an exponentially sensitive current into images of individual atoms.
  5. Connect tunnelling to flash memory, tunnel diodes and Josephson junctions, including the macroscopic tunnelling recognized by the 2025 Nobel Prize in Physics.
  6. Explain why tunnelling drills no holes, lets no person walk through walls, and sends no signal faster than light.
03The lesson

Read it — or listen along

7 parts. Press play on any part and the paragraph being read lights up as Dr. Q speaks.

Part 1

When the Wall Is Merely Enormous

Last session our electron lived behind walls of infinite height, and the rule was absolute: ψ = 0 at the wall. Real walls are never infinite. The oxide in a transistor, the vacuum gap between two metals, the electric repulsion around a nucleus: each is a potential barrier, a region where the potential energy V rises above the particle's total energy E by a finite amount over a finite distance. Classically, the kinetic energy E − V can never be negative, so a ball rolling toward a hill it cannot climb slows, halts at the turning point where E = V, and rolls back. Everything beyond is a classically forbidden region.

Quantum mechanics reads the same landscape through the Schrödinger equation of chapter 10. Where E exceeds V, the curvature of ψ points back toward the axis and the wavefunction oscillates. Where V exceeds E, the curvature points away from the axis, so ψ grows or decays exponentially. A growing solution could never be normalized in an endless barrier, and even in a thick finite one the decaying part dominates. So the wavefunction does not slam to zero at the edge. It enters, and it fades: ψ(x) ≈ ψ(0)e^(−κx), where x is the distance into the barrier and the Greek letter κ stands for the decay constant.

The decay constant is set by the energy deficit V − E, the particle's shortfall below the top. In words, κ is the square root of twice the mass times the deficit, divided by ħ; in symbols, ħκ = √(2m(V − E)). Its inverse, 1/κ, is the decay length, over which ψ shrinks by a factor of e, about 2.718. For an electron one electron-volt short of the top, κ ≈ 5.12 per nanometre and 1/κ ≈ 0.195 nm, comparable to the radius of an atom. The probability density, |ψ|², fades twice as fast, as e^(−2κx), halving every 0.068 nm. The infinite wall of chapter 11 is the limit in which κ is infinite and the fade is instantaneous.

Watch this part

Part 2

Through the Wall: Tunnelling and Its Exponential Toll

Now give the barrier a far side: a rectangle of height V and width L, with free space on either side. An electron wave arrives from the left. Part reflects, part enters and fades, and if the barrier is thin enough the fading tail is still alive at the far edge. Beyond it ψ oscillates again with its original wavelength, and a transmitted wave travels away. Its amplitude was cut by roughly e^(−κL) during the crossing, and probability goes as amplitude squared, so the transmission probability, the chance that a particle sent at the barrier is found beyond it, obeys T ≈ e^(−2κL). Passing through a barrier you lack the energy to climb is called quantum tunnelling.

Matching ψ and its slope at both edges gives the exact result for a rectangle. Once the exponent 2κL exceeds about 4, it simplifies to T ≈ 16(E/V)(1 − E/V)e^(−2κL). The prefactor never exceeds 4, so nearly all the action lives in the exponential. Take an electron with E = 1.0 eV meeting a barrier of height V = 2.0 eV. A barrier 0.5 nm wide transmits about 2.4 percent of electrons; double the width to 1.0 nm and T falls to 0.014 percent, 166 times smaller. Doubling the width squares the exponential factor, and the prefactor barely changes. Whatever is not transmitted is reflected: R + T = 1, where R is the reflection probability.

The exponent also hides the mass, because κ grows as the square root of m. A proton is 1,836 times heavier than an electron, so its κ is larger by a factor of √1836, about 42.85, and through the barrier 0.5 nm wide its transmission is about 10⁻⁹⁵: effectively never. Tunnelling rewards thin barriers, light particles and energies close to the top. Electrons tunnel constantly in electronics and chemistry, hydrogen nuclei tunnel measurably in some chemical reactions, and heavier things almost never do. For smooth barriers, the WKB approximation, introduced in 1926 by Gregor Wentzel, Hendrik Kramers and Léon Brillouin, replaces 2κL by twice the area under the curve of κ(x) across the forbidden region.

Watch this part

Part 3

Gamow's Great Escape: Alpha Decay

Tunnelling's first triumph came from the nucleus. Many heavy nuclei decay by emitting an alpha particle, two protons and two neutrons bound together, identical to a helium-4 nucleus. Inside the nucleus the strong nuclear force holds the alpha in; just outside, the electric repulsion of the remaining protons shoves it away. Together they form the Coulomb barrier, roughly 30 MeV high around uranium. Yet uranium-238 emits alpha particles carrying only about 4.2 MeV, while scattering experiments analysed by Ernest Rutherford in 1927 showed that alpha particles with about twice that energy, fired at uranium, were turned back by electric repulsion without ever getting in. How can a particle escape a wall that a projectile with twice its energy cannot climb?

In 1928 George Gamow, and independently Ronald Gurney and Edward Condon, answered: it does not go over, it goes through. The alpha rattles inside the nucleus, striking the barrier about 10²¹ times per second, each strike with a minuscule, fixed chance of tunnelling out. A nucleus therefore has no memory and no clock: decay times are random, and a large sample decays exponentially with a definite half-life, the time in which half the nuclei decay. Because electric repulsion fades as 1/r with distance r from the centre, a uranium-238 alpha must tunnel from the nuclear surface, about 9 fm out, to about 61 fm, where the hill sinks below its energy (a femtometre is 10⁻¹⁵ metres). The WKB recipe gives odds of a few times 10⁻³⁹ per strike: billions of years.

The model's glory is the Geiger–Nuttall law, found by Hans Geiger and John Nuttall in 1911: the more energetic an emitter's alpha particles, the shorter its half-life. Polonium-212 emits alphas of 8.78 MeV and has a half-life of 0.29 μs; uranium-238 emits alphas of 4.20 MeV and lasts 4.47 billion years. Barely a factor of two in energy, a factor of about 5×10²³ in lifetime! The more energetic alpha meets the 1/r wall where it is thinner, needing to tunnel only out to about 26 fm, and because width sits in an exponent, a modest change in energy becomes an astronomical change in lifetime. Bismuth-209, long thought stable, was caught alpha-decaying in 2003; its half-life, 2.0×10¹⁹ years, is over a billion times the age of the universe.

Watch this part

Part 4

Why the Sun Can Shine

The Sun fuses hydrogen into helium in its core, and the first step is the hardest: two protons must come within a femtometre or so before the strong force can grab them, and their electric repulsion at that range is a barrier of roughly 1.0 MeV. At a core temperature of about 15.7 million kelvin, the matching thermal energy scale kT, where k is Boltzmann's constant, is just 1.35 keV, and a typical proton carries about 2 keV, some five hundred times too little. The fraction of protons with enough energy to climb the barrier is of order e^(−739), about 10⁻³²¹. Even with about 10⁵⁷ protons in the Sun, classically not one pair would fuse in the age of the universe.

A proton with just a few thousand electron-volts has a tiny but real chance of slipping through, a chance that climbs steeply with energy while the number of protons at each energy falls steeply. Their product peaks in a narrow window, the Gamow peak, near 6 keV in the solar core: most fusing pairs tunnel through a barrier more than a hundred times higher than their energy. Arthur Eddington, who had argued in 1920 that stars burn hydrogen, told critics who said the stars were not hot enough to go and find a hotter place. In 1929 Robert Atkinson and Fritz Houtermans vindicated him with Gamow's tunnelling, and in 1938 and 1939 Hans Bethe, partly with Charles Critchfield, worked out the reaction chains that power the stars, work that won Bethe the 1967 Nobel Prize in Physics.

Even with tunnelling the first step is extraordinarily slow, because it also needs the weak nuclear force to turn a proton into a neutron. That bottleneck is a blessing: the Sun shines steadily for about ten billion years instead of burning out before life could evolve.

Watch this part

Part 5

Feeling Atoms with a Tunnelling Current

Exponential sensitivity, a curse for anyone predicting lifetimes, is a gift for anyone measuring distances. In March 1981, at IBM's research laboratory near Zurich, Gerd Binnig and Heinrich Rohrer first measured a controlled tunnelling current across a vacuum gap, the decisive step toward their scanning tunnelling microscope, or STM. A sharp metal tip is held within about a nanometre of a conducting surface, never touching it, and a small voltage is applied. The vacuum gap is a barrier whose height is roughly the work function, the minimum energy needed to pull an electron out of a metal, typically 4 to 5 eV. Electrons tunnel across, giving a current of around a nanoampere that falls as e^(−2κd) with the gap d.

For a barrier of 4.5 eV, κ ≈ 10.9 per nanometre, so retracting the tip by just 0.1 nm cuts the current about ninefold. A feedback loop moves the tip up and down to hold the current constant as it scans, and the record of those movements is a contour map of the surface with height resolution at the picometre level. The same exponential delivers sideways resolution: if the tip ends in a single atom, that atom sits a fraction of a nanometre closer than its neighbours and carries almost all the current. Binnig and Rohrer shared the 1986 Nobel Prize in Physics with Ernst Ruska, the pioneer of the electron microscope.

The STM soon did more than look. In 1989 Donald Eigler and Erhard Schweizer at IBM's Almaden laboratory in California used the tip to slide 35 xenon atoms across a nickel crystal until they spelled the letters IBM, a feat published in 1990. Building with single atoms had begun.

Watch this part

Part 6

Tunnelling in Your Pocket and in Quantum Computers

The photos on your phone are stored by tunnelling. Flash memory, invented by Fujio Masuoka's team at Toshiba in the 1980s, keeps each bit as charge trapped on an isolated layer walled in by oxide only several nanometres thick. To write or erase the NAND flash used for mass storage, the chip applies roughly 15 to 20 volts, which tilts the barrier into a thin triangle that electrons tunnel through: Fowler–Nordheim tunnelling. Remove the voltage, the barrier widens again, and the charge stays put for years without power. Tunnelling can also be a nuisance: by the mid-2000s the silicon dioxide layer insulating each transistor's gate was only about 1.2 nm thick, and leakage through it pushed Intel, in 2007, to a physically thicker hafnium-based insulator.

In 1957 Leo Esaki, at the company soon renamed Sony, doped a germanium junction so heavily that the junction was only some ten to twenty nanometres wide. Electrons tunnelled straight across, and over a range of voltages the current fell as the voltage rose, a negative resistance that let tunnel diodes switch extremely fast. In 1962 Brian Josephson, a 22-year-old Cambridge graduate student, predicted that Cooper pairs, the bound electron pairs that carry a superconductor's resistance-free current, would tunnel through a thin insulator between two superconductors, a device now called a Josephson junction. A current then flows with no voltage at all, and a steady voltage makes it oscillate at 483.6 GHz per millivolt, the basis of modern voltage standards. Esaki, Josephson and Ivar Giaever, who used tunnelling to probe superconductors, shared the 1973 Nobel Prize in Physics.

Can something big tunnel? In 1984 and 1985 at the University of California, Berkeley, John Clarke, Michel Devoret and John Martinis tested exactly that with a Josephson junction on a chip about a centimetre across. A current flowed with zero voltage, a state trapped behind an energy barrier. What tunnelled out was not one electron but a single collective variable describing billions of Cooper pairs, and a voltage suddenly appeared. Below a crossover temperature the escape rate stopped depending on temperature, the fingerprint of tunnelling rather than thermal hopping, and microwaves revealed quantized energy levels. The work earned the 2025 Nobel Prize in Physics; its descendants, superconducting qubits built around Josephson junctions, are as of 2026 the heart of quantum processors at Google, IBM and elsewhere.

Watch this part

Part 7

What Tunnelling Is Not

A tempting thought: perhaps a tunnelling particle crosses the barrier instantly. In 1962 Thomas Hartman calculated that the delay of a transmitted wave packet stops growing once a barrier is thick, implying ever-faster apparent speeds. In 1993 Aephraim Steinberg, Paul Kwiat and Raymond Chiao at Berkeley timed single photons tunnelling through a multilayer mirror 1.1 μm thick: the transmitted peaks arrived 1.47 fs earlier than those of twin photons crossing the same distance in air, an apparent speed of about 1.7 times the speed of light.

No signal outran light. The barrier reshapes the pulse, transmitting a faint copy built mostly from its leading edge, which was already ahead; the peak shifts forward, but the front, the earliest moment anything can be detected, never outruns light. Nor is tunnelling instantaneous: in 2020 Steinberg's group in Toronto sent rubidium atoms cooled to about two nanokelvin through a barrier of laser light 1.3 μm thick and found that transmitted atoms spent about 0.6 ms inside. What happens in between is interpretation-dependent: standard quantum mechanics assigns no path, while the pilot-wave theory of chapter 22 gives every particle a definite trajectory.

Finally, strolling through your bedroom wall. Treat a person of 70 kg as a single particle, grant a generous energy deficit of one joule and a wall 10 cm thick: 2κL ≈ 2.2×10³⁴, so T is about 1 divided by a number with 10³⁴ digits. Trying once a second for the age of the universe, about 4×10¹⁷ seconds, changes nothing. Worse, you are not one isolated quantum object but about 7×10²⁷ atoms, jostled by air and light, which destroys the coherence any collective tunnelling would need. Mwahaha, measurably: the wall wins.

Watch this part

04Video shelf

Go deeper on video

22 more videos for this chapter, 16 of them from universities, research institutes and Nobel lectures. The clips under “Watch this part” in the lesson start at the moment that matters.

22 videos · 8 h 10 min

  • Lecture · Stanford

    Barriers and boundary conditions – David Miller

    David Miller Quantum Mechanics Courses · David A. B. Miller

    Stanford's David Miller shows why ψ and dψ/dz must both be continuous, then sends a wave at an infinitely thick barrier: inside, ψ falls off as e^(−κz) instead of stopping dead, and the barrier still reflects 100% but with a phase shift that has no classical analogue. His worked example is section 1's own: an electron of 1 eV meeting a 2 eV barrier, giving an amplitude decay length 1/κ of about 0.2 nm and a probability decay length of about 0.1 nm.

  • Explainer · YouTube

    The Quantum Barrier Potential Part 2: Defining the Transmission and Reflection Coefficients

    Professor Dave Explains

    The full barrier calculation behind section 2's formula, worked step by step: reflection and transmission coefficients (from 2:54), the transmission factor T (5:33), hyperbolic sines and cosines (10:29) and an application (18:48). Once 2κL is large, the exact result shrinks to the lesson's T ≈ 16(E/V)(1 − E/V)e^(−2κL); plug in E = 1 eV, V = 2 eV and L = 0.5 nm and check that you get about 2.4%.

  • Lecture · Stanford

    Penetration factor for slowly varying barriers – David Miller

    David Miller Quantum Mechanics Courses · David A. B. Miller

    Miller slices a smoothly varying barrier into thin layers whose internal boundary matrices become the identity, leaving an amplitude factor exp(−∫κ(z)dz), with κ = √(2m(V(z) − E))/ħ, and a tunnelling probability, his "penetration factor", of an edge-dependent prefactor times exp(−2∫κ(z)dz). That is section 2's WKB rule (twice the area under κ in the exponent) and the engine of Gamow's calculation in section 3; it builds on his transfer-matrix lecture (33c), so treat it as a stretch goal.

  • Explainer · YouTube

    We finally understood why radium is radioactive! (My mind is blown)

    FloatHeadPhysics

    A story-driven tour of section 3's puzzle, with chapters running from "The alpha decay paradox" (2:30) through how quantum mechanics solves it (9:44) with tunnelling (11:36), where half-lives come from (12:50) and why not every heavy isotope alpha-decays (14:43), then a look ahead to section 4 ("The sun is too cold to be alive", 19:12). Keep the Geiger–Nuttall numbers beside you: a factor of two in alpha energy, a factor of about 5×10²³ in lifetime.

  • Explainer · YouTube

    The Sun can’t work without Quantum Tunneling

    The Science Asylum

    Takes the Sun's 15-million-degree core, hot enough to turn gas into plasma, and explains why thermodynamics alone is still not enough for fusion. Check the argument against section 4's numbers: a typical proton carries about 2 keV against a barrier of roughly 1 MeV, and most fusing pairs tunnel near the Gamow peak at about 6 keV.

  • Interview · Nobel Prize

    Interview with Heinrich Rohrer, Nobel Prize in Physics 1986

    Nobel Prize · Heinrich Rohrer

    Heinrich Rohrer, who built the scanning tunnelling microscope with Gerd Binnig at IBM Zurich, interviewed by Adam Smith at the Nobel Foundation in Stockholm in April 2008. It starts here on the microscope itself and its first surprises (2:25); later chapters cover the partnership with Binnig (18:15), the opposition they met (22:05), the prize (24:45) and the nanoscience revolution that followed (26:05).

  • Demo · YouTube

    Moving Atoms: Making The World's Smallest Movie

    IBM

    IBM's short behind-the-scenes documentary on how its researchers moved individual atoms, frame by frame, to make the world's smallest stop-motion film. It is section 5's last paragraph in action: the tunnelling tip used to build as well as to see, the line that runs from Eigler and Schweizer's 35 xenon atoms in 1989.

  • Documentary · miSci – Museum of Innovation and Science

    A Story of Research: Ivar Giaever - 1982

    miSci - museum of innovation and science · Ivar Giaever

    A 1982 documentary on Ivar Giaever, who shared the 1973 prize with Esaki and Josephson, starting at the chapter "Tunnel Junction" (17:49) and running into "The Energy Gap" (19:47): the experiment in which section 6 says he used tunnelling to probe superconductors. Rewind to 5:02 for his thin-film research and to 10:04 for how he learned quantum mechanics.

  • Explainer · YouTube

    New Results in Quantum Tunneling vs. The Speed of Light

    PBS Space Time

    Space Time takes on the long-running argument over whether a tunnelling particle can beat light, the question section 7 settles. Check its conclusions against the lesson's: the transmitted peak can arrive early, but the front, the first moment anything can be detected, never outruns light, and the atoms in the 2020 Toronto experiment spent about 0.6 ms inside the barrier.

  • Lecture · Instituto de Física

    Measuring the Tunneling Time, and Other Problems in Quantum Archaeology | COLÓQUIOS IFUSP

    Instituto de Física USP · Aephraim Steinberg

    A University of São Paulo physics colloquium by Aephraim Steinberg of Toronto, whose experiments section 7 describes: from "Background" (12:39) he covers the stationary-phase approach (14:15), defines a Larmor time (24:43), introduces the atoms (25:53) and walks through the tunnelling experiment (27:54). Advanced, but it shows how "time spent inside a barrier" can be measured at all.

  • Lecture · MIT

    Lecture 12: The Dirac Well and Scattering off the Finite Step

    MIT OpenCourseWare · Allan Adams · 2013

    The lecture from MIT 8.04 Quantum Physics I (Allan Adams, 2013) that goes with Chapter 12. The whole course is in the Lecture Hall.

  • Lecture · Oxford

    015 Tunnelling and Radioactive Decay

    University of Oxford · James Binney · 2009–10

    The lecture from University of Oxford Quantum Mechanics (James Binney, 2009–10) that goes with Chapter 12. The whole course is in the Lecture Hall.

  • Explainer · Perimeter

    Quantum Tunneling Explained | Perimeter Institute for Theoretical Physics

    Perimeter Institute for Theoretical Physics · Katie Mack · 2023

    A short episode from Perimeter Institute's Quantum 101 series with Katie Mack. Watch it alongside Chapter 12.

  • Lecture · Nobel Prize

    Nobel Prize lecture: John Clarke, Nobel Prize in Physics 2025

    Nobel Prize · John Clarke · 2025

    "From SLUGs to macroscopic quantum phenomena": John Clarke on the 1984–85 Berkeley experiments in which an entire superconducting circuit, big enough to hold in your hand, tunnelled out of a trapped state as a single quantum object.

  • Lecture · Nobel Prize

    Nobel Prize lecture: Michel Devoret, Nobel Prize in Physics 2025

    Nobel Prize · Michel H. Devoret · 2025

    "From macroscopic quantum phenomena to artificial atoms": Michel Devoret on how a circuit's quantized energy levels — measured in the same experiments — became the foundation of today's superconducting qubits.

  • Lecture · Nobel Prize

    Nobel Prize lecture: John Martinis, Nobel Prize in Physics 2025

    Nobel Prize · John M. Martinis · 2025

    "Prehistoric Superconducting Qubits": John Martinis, who later led Google's quantum-hardware team, traces the line from the 1980s tunnelling experiments to the superconducting qubit.

  • Course · MIT

    Quantum Physics I

    MIT OpenCourseWare · Allan Adams · 2013

    Allan Adams' celebrated course starts from superposition and the experimental facts, builds the wavefunction, operators and the Schrödinger equation, then tours scattering, hydrogen, identical particles and band structure before closing on entanglement and Bell's theorem. Calculus and a little linear algebra are enough to keep up.

  • Series · Perimeter

    Quantum 101

    Perimeter Institute for Theoretical Physics · Katie Mack · 2023

    Ten five-minute episodes hosted by the cosmologist Katie Mack: duality, quantization, uncertainty, superposition, entanglement, probability, tunnelling and quantum computing. The friendliest possible on-ramp, and a quick refresher before any chapter.

  • Course · MIT

    Quantum Physics I

    MIT OpenCourseWare · Barton Zwiebach · 2016

    The same course rebuilt as 115 short segments of roughly ten to twenty minutes, each devoted to one idea, so you can watch exactly the concept a chapter needs. Barton Zwiebach is meticulous: every step on the board is justified.

  • Course · Oxford

    Quantum Mechanics

    University of Oxford · James Binney · 2009–10

    James Binney's Oxford course, given over Michaelmas Term 2009 and Hilary Term 2010 and built on probability amplitudes and Dirac notation from the very first lecture. Highlights: tunnelling and radioactive decay, the EPR argument and Bell's inequality, spin and the Stern–Gerlach experiment, and a full treatment of hydrogen.

  • Course · IIT Madras

    Quantum Physics

    nptelhrd · V. Balakrishnan · 2008

    V. Balakrishnan of IIT Madras is one of the most admired physics lecturers anywhere: unhurried, exact and full of physical insight. Thirty-one lectures that build the subject from the uncertainty principle and the failures of classical physics to the full formalism.

  • Course · MIT

    Quantum Physics III

    MIT OpenCourseWare · Barton Zwiebach · 2018

    For the ambitious: perturbation theory and the fine structure of hydrogen, the semiclassical WKB approximation behind tunnelling rates, the adiabatic theorem, scattering, and identical particles, in 100 compact segments.

Every video for Chapter 12 in the Video Library

05To the blackboard

Key equations

Equation 1

Decay constant

κ=2m (V−E)ℏ\kappa = \frac{\sqrt{2m\,(V - E)}}{\hbar}

How fast the wavefunction fades inside a barrier. A bigger mass or a bigger energy deficit means a larger κ and a faster fade. For an electron one electron-volt short of the top, κ ≈ 5.12 per nanometre.

What each symbol means
SymbolMeaningUnit
κdecay constant; 1/κ is the decay lengthm⁻¹ (often quoted per nanometre)
mmass of the particlekg
V − Eenergy deficit: barrier height minus particle energyJ (often quoted in eV)
ħreduced Planck constant, 1.055×10⁻³⁴J·s

Say it: “kappa equals the square root of two m times the quantity V minus E, all divided by h-bar”

Equation 2

Wavefunction inside the barrier

ψ(x)≈ψ(0) e−κx,∣ψ(x)∣2≈∣ψ(0)∣2 e−2κx\psi(x) \approx \psi(0)\,e^{-\kappa x}, \qquad |\psi(x)|^2 \approx |\psi(0)|^2\,e^{-2\kappa x}

In a classically forbidden region the wavefunction does not stop; it decays exponentially with distance x into the barrier, and the probability density decays twice as fast.

What each symbol means
SymbolMeaningUnit
ψ(x)wavefunction at depth x inside the barrierm⁻¹ᐟ²
xdistance into the barrier from its edgem

Say it: “psi of x is approximately psi of zero times e to the minus kappa x, so the probability density falls as e to the minus two kappa x”

Equation 3

Thick-barrier transmission

T≈e−2κLT \approx e^{-2\kappa L}

The chance of getting through a barrier of width L falls exponentially with the width, and with the square roots of the mass and of the energy deficit hidden inside κ. Doubling L squares this exponential factor.

What each symbol means
SymbolMeaningUnit
Ttransmission probabilitydimensionless, between 0 and 1
Lbarrier widthm

Say it: “T is approximately e to the minus two kappa L”

Equation 4

Exact rectangular barrier

T=[1+V2sinh⁡2(κL)4E (V−E)]−1≈16 EV(1−EV)e−2κLT = \left[1 + \frac{V^2 \sinh^2(\kappa L)}{4E\,(V - E)}\right]^{-1} \approx 16\,\frac{E}{V}\left(1 - \frac{E}{V}\right) e^{-2\kappa L}

The full answer for a rectangular barrier when E is below V, found by matching ψ and its slope at both edges. Once 2κL exceeds about 4 the simple form on the right is accurate to a few percent; its prefactor is never larger than 4.

What each symbol means
SymbolMeaningUnit
Vbarrier heightJ or eV
Eparticle energy, below VJ or eV
sinhhyperbolic sine, sinh(y) = (eʸ − e⁻ʸ)/2dimensionless

Say it: “T equals one over the quantity one plus V squared times the hyperbolic sine squared of kappa L, divided by four E times the quantity V minus E. For thick barriers this is approximately sixteen times E over V, times one minus E over V, times e to the minus two kappa L”

Equation 5

Barriers of any shape (WKB)

T≈exp⁡ ⁣(−2ℏ∫x1x22m [V(x)−E]  dx)T \approx \exp\!\left(-\frac{2}{\hbar}\int_{x_1}^{x_2} \sqrt{2m\,[V(x) - E]}\;dx\right)

For a smooth barrier, add up κ(x) slice by slice between the classical turning points x₁ and x₂. This is the recipe Gamow used for the Coulomb barrier in alpha decay.

What each symbol means
SymbolMeaningUnit
V(x)potential energy, varying with positionJ
x₁, x₂turning points where V(x) = E, the edges of the forbidden regionm

Say it: “T is approximately e to the power minus two over h-bar times the integral, from x one to x two, of the square root of two m times V of x minus E, d x”

Equation 6

Scanning tunnelling microscope current

I∝Vbias e−2κd,κ=2me ϕℏI \propto V_{\mathrm{bias}}\, e^{-2\kappa d}, \qquad \kappa = \frac{\sqrt{2m_e\,\phi}}{\hbar}

The current between tip and surface falls exponentially with the gap d. With a work function near 4.5 eV it drops about ninefold for every 0.1 nm of extra gap, which is what lets an STM feel single atoms.

What each symbol means
SymbolMeaningUnit
Itunnelling currentA (typically about 1 nA)
V_biassmall voltage applied between tip and sampleV
dtip-to-surface gapm
φwork function, the effective barrier heighteV (typically 4 to 5)

Say it: “The tunnelling current I is proportional to the bias voltage times e to the minus two kappa d, where kappa is the square root of two times the electron mass times phi, divided by h-bar”

06Worked examples

Real numbers, step by step

Cover the steps, try the problem yourself, then reveal the solution one step at a time.

Example 1

An electron meets a thin barrier

Problem

An electron with energy E = 1.00 eV meets a rectangular barrier of height V = 2.00 eV. Find κ, then estimate T for barrier widths of 0.50 nm and 1.00 nm, first with e^(−2κL) alone and then including the thick-barrier prefactor.

  1. Step 1: The deficit is V − E = 1.00 eV = 1.602×10⁻¹⁹ joules. With m = 9.109×10⁻³¹ kilograms, the product 2m(V − E) is 2.919×10⁻⁴⁹ in SI units, and its square root is 5.403×10⁻²⁵. Dividing by ħ = 1.0546×10⁻³⁴ J·s gives κ = 5.123×10⁹ per metre, which is 5.123 per nanometre. Keep the extra digit: in an exponent, small rounding errors grow.

  2. Step 2: For L = 0.50 nm the exponent is 2κL = 2 × 5.123 × 0.50 = 5.123, and e^(−5.123) = 5.96×10⁻³.

  3. Step 3: The prefactor is 16(E/V)(1 − E/V) = 16 × 0.5 × 0.5 = 4, so T ≈ 4 × 5.96×10⁻³ = 0.0238. The exact rectangular-barrier formula gives 0.0235.

  4. Step 4: For L = 1.00 nm the exponent doubles to 10.246, and e^(−10.246) = 3.55×10⁻⁵, so T ≈ 4 × 3.55×10⁻⁵ = 1.42×10⁻⁴, matching the exact value of 1.42×10⁻⁴.

Answer

κ ≈ 5.12 per nanometre; T ≈ 2.4 percent for a width of 0.50 nm, and T ≈ 1.4×10⁻⁴, about 0.014 percent, for a width of 1.00 nm

Why it matters: Doubling the width cut the transmission 166-fold. The bare exponential missed the factor of 4 in front, but it captured the dramatic part: how fast T collapses as the barrier thickens.

Try it yourself first — then check each step.
Example 2

Swap the electron for a proton

Problem

Send a proton with the same energy, 1.00 eV, at the same barrier: height 2.00 eV, width 0.50 nm. Estimate its transmission probability.

  1. Step 1: κ is proportional to the square root of the mass, and the proton is 1,836 times heavier than the electron, so the proton's κ is 5.1232 × √1836 = 5.1232 × 42.849 = 219.52 per nanometre.

  2. Step 2: The exponent is 2κL = 2 × 219.52 × 0.50 = 219.52.

  3. Step 3: Convert to a power of ten by dividing the exponent by the natural logarithm of 10, which is 2.3026: 219.52 ÷ 2.3026 = 95.34, so e^(−219.52) = 10^(−95.34) ≈ 4.6×10⁻⁹⁶.

  4. Step 4: Including the prefactor of 4, T ≈ 1.8×10⁻⁹⁵. Even a barrier only 0.10 nm wide gives the proton just 3.4×10⁻¹⁹, while an electron would cross it 78 percent of the time.

Answer

T ≈ 1.8×10⁻⁹⁵ for the proton, against about 2.4×10⁻² for the electron

Why it matters: Mass enters under a square root inside the exponent, so a factor of 1,836 in mass became a factor of about 43 in the exponent, and that turned a few percent into essentially never. Tunnelling is a story about light particles.

Try it yourself first — then check each step.
Example 3

The sensitivity of a tunnelling microscope

Problem

An STM tip hovers above a metal surface whose effective barrier height, roughly its work function, is 4.5 eV. First, by what factor does the tunnelling current fall when the tip retracts by 0.10 nm? Second, if the electronics can hold the current steady to within 2 percent, what is the smallest change in height the microscope can sense?

  1. Step 1: κ grows as the square root of the deficit, so scale the value for a deficit of one electron-volt: κ = 5.123 × √4.5 = 5.123 × 2.121 = 10.87 per nanometre.

  2. Step 2: The current goes as e^(−2κd). Retracting by Δd = 0.10 nm multiplies it by e^(−2 × 10.87 × 0.10) = e^(−2.174) = 0.114, a drop by a factor of about 8.8.

  3. Step 3: A 2 percent change in current means e^(2κΔd) = 1.02, so 2κΔd equals the natural logarithm of 1.02, which is 0.0198. Then Δd = 0.0198 ÷ 21.74 per nanometre = 9.1×10⁻⁴ nanometres.

  4. Step 4: That is 0.91 pm, a few hundred times smaller than the diameter of a typical atom.

Answer

The current falls by a factor of about 8.8, nearly tenfold, and the smallest detectable height change is about 0.9 pm, less than a picometre

Why it matters: No lens is involved: the exponential law itself is the magnifying glass. That is why an STM registers height changes far smaller than an atom.

Try it yourself first — then check each step.
Example 4

Knocking to escape a uranium nucleus

Problem

Uranium-238 has a half-life of 4.47×10⁹ years and polonium-212 a half-life of 294 ns. Estimate how often an alpha particle strikes the nuclear barrier, then the escape probability per strike for each nucleus.

  1. Step 1: Strike rate: an alpha particle with about 4.2 MeV moves at about 1.4×10⁷ m/s. Crossing a nucleus of radius 7.4 fm from wall to wall, a diameter of 1.48×10⁻¹⁴ metres, takes about 10⁻²¹ seconds, so it strikes the wall roughly 10²¹ times per second. This is a crude, standard estimate.

  2. Step 2: The decay rate λ, the probability of decay per second, is the natural logarithm of 2, which is 0.693, divided by the half-life. For uranium-238 the half-life is 4.47×10⁹ years × 3.156×10⁷ seconds per year = 1.41×10¹⁷ seconds, so λ = 0.693 ÷ 1.41×10¹⁷ = 4.9×10⁻¹⁸ per second.

  3. Step 3: Escape probability per strike: T = λ ÷ (strike rate) = 4.9×10⁻¹⁸ ÷ 10²¹ ≈ 5×10⁻³⁹.

  4. Step 4: For polonium-212, λ = 0.693 ÷ 2.94×10⁻⁷ seconds = 2.4×10⁶ per second, so T ≈ 2.4×10⁶ ÷ 10²¹ ≈ 2×10⁻¹⁵.

Answer

About 5×10⁻³⁹ per strike for uranium-238, so a typical alpha knocks about 2×10³⁸ times before escaping, and about 2×10⁻¹⁵ per strike for polonium-212

Why it matters: The alpha energies differ by only a factor of 2.1, 4.20 against 8.78 MeV, yet the escape odds differ by a factor of about 5×10²³. The mad scientist challenge on the worksheet shows that Gamow's WKB integral reproduces both numbers to within a factor of about four.

Try it yourself first — then check each step.
07Myth busters

What everyone gets wrong

Read the myth, decide whether you believed it, then bust it.

  • Myth #1

    A tunnelling particle drills a hole through the barrier, or squeezes through a gap in it.

    Truth

    Nothing is bored, broken or opened. The barrier is identical before and after, ready to transmit or reflect the next particle with exactly the same odds, and for a barrier that does not change in time the particle emerges with the energy it arrived with. Tunnelling is a wave property: the wavefunction fades across the forbidden region without reaching zero, so the probability of finding the particle beyond it is small but not zero.

  • Myth #2

    A person could walk through a wall by tunnelling, if they only tried often enough.

    Truth

    Even treating a person of 70 kg as a single particle with a generous one-joule energy deficit and a wall 10 cm thick, T is about 1 divided by a number with 10³⁴ digits; one attempt per second for the age of the universe, about 4×10¹⁷ attempts, changes nothing. Worse, a person is about 7×10²⁷ atoms constantly disturbed by air and light, not one isolated quantum object. The macroscopic tunnelling of the 2025 Nobel Prize needed a single collective variable in a carefully isolated circuit cooled to a few hundredths of a kelvin.

  • Myth #3

    Tunnelling is instantaneous, so it can carry signals faster than light.

    Truth

    Tunnelling photons in the 1993 Berkeley experiment had peaks that arrived early, as if moving at 1.7 times the speed of light, but the barrier merely reshaped each pulse from its leading edge; the front of any signal never outruns light, so no message can be sent faster than c. Tunnelling also takes time: in 2020 rubidium atoms were measured spending about 0.6 ms inside a barrier 1.3 μm thick.

  • Myth #4

    A particle tunnels by borrowing energy from the uncertainty principle and paying it back quickly.

    Truth

    This popular story misuses the energy-time relation. In the standard calculation the particle's energy is E throughout, and nothing is borrowed. Tunnelling happens because the wavefunction in a forbidden region decays rather than vanishes. If a measurement does catch the particle inside the barrier, that measurement itself supplies the energy.

08Interactive lab

Your turn at the controls

Welcome to the Tunneling Arcade. In the barrier simulator an electron wave packet races toward a rectangular barrier, and you set the electron's energy E, the barrier height V₀ and the barrier width a, which this chapter calls L; energies are in electron-volts and widths in nanometres. Fire the packet and watch it split: part reflects, a faded remnant leaks through, and the lab measures the fraction that ends up on the far side. Beside that simulated transmission sit the exact rectangular-barrier value and the bare exponential e^(−2κa), which the lab labels the thick-barrier estimate, and a log plot of T against width turns the exponential into a straight line. A second screen holds a scanning tunnelling microscope. The arcade has three levels; clear all three to win.

Your tasks

0/6

How it connects to the theory

Every number on screen comes from this chapter. κ uses the real electron mass and ħ, the exact transmission comes from matching ψ and its slope at both edges of the barrier, and e^(−2κa) is the thick-barrier shortcut whose missing prefactor you can watch for yourself. The packet itself moves by the time-dependent Schrödinger equation of chapter 10, solved step by step. Because a real packet carries a spread of energies and its faster components tunnel more easily, the simulated transmission usually sits a little above the single-energy formula; after each run the lab also shows the formula averaged over the packet's energies, which is what the simulation should match. The STM screen runs on the same law, with the current falling as e^(−2κd) under a 4.5 eV work function. The barrier never changes as packets pass, and in a real experiment each electron would be detected whole on one side or the other. Only the odds are shared.

Tunneling Arcade: Get Through the Barrier loads as you scroll down.

Win the challenge to tick “lab won” on your chapter checklist.
Open full-screen lab
09Memory aids

Make it stick

Memory trick

Tunnelling loves the THIN, the LIGHT and the NEARLY THERE. Thin: a narrow barrier, small L. Light: a small mass m. Nearly there: an energy just below the top, so the deficit V − E is small. All three sit together in the exponent, 2κL = 2L√(2m(V − E)) divided by ħ, so getting any one of them wrong is punished exponentially. For the applications, remember FAST JS: Flash memory, Alpha decay, the Sun, Tunnel diodes, Josephson junctions, Scanning tunnelling microscope. And remember the joke hidden in the name: tunnelling is never faster than light. To use it: before any tunnelling problem, ask whether the barrier is thin, the particle light and the energy nearly there. A loud no to any of them means a vanishing transmission.

Analogy — and where it breaks

Tunnelling is like light escaping a glass prism at an angle where it should be totally reflected. Just beyond the glass a fading evanescent wave reaches a fraction of a wavelength into the air, and if a second prism is brought almost into contact, light crosses the gap; Isaac Newton noticed the effect three centuries ago. The mathematics is the same exponential fade. Where it breaks: a bright beam's energy is visibly shared between two beams, which tempts you to picture each electron split in two, but a tunnelling electron is never split; it is detected whole on one side or the other, and only the odds are shared. And light fades over hundreds of nanometres, an electron over a fraction of one.

The whole chapter in one line

Where a particle lacks the energy to climb, its wave fades but never stops; thin, light, nearly-there particles leak through with odds of about e^(−2κL).

Leitner flashcards

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10Games

Play it into long-term memory

Match the vocabulary, then take the Quiz Show. Score 70% or better to tick it off your checklist.

Vocab match · timed

Vocab Match

Match each term to its definition, against the clock.

  • 12 pairs in 2 rounds
  • 100 points a match, more on a streak
  • −25 per mistake · speed bonus each round
  • Finish with 3 mistakes or fewer to win
Live from the lab

Dr. Q's Quantum Quiz Show

Dr. Q, your host

Welcome, contestants! 8 questions, 30 seconds each, and one mad scientist judging you, lovingly. Answer fast for a speed bonus; answer right in a row and your points multiply.

  • 100 per answer
  • +100 speed bonus
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11Worksheet

Prove it on paper

Answer every question, check your work, and read the explanations — the stretch challenge is for the truly mad.

14 questions12 terms1 challenge

Answer every question with the lesson closed, then check your work. Use ħ = 1.055×10⁻³⁴ J·s, electron mass 9.109×10⁻³¹ kg, 1 eV = 1.602×10⁻¹⁹ J and Boltzmann's constant k = 8.617×10⁻⁵ eV per kelvin. Once you have derived it, you may use the shortcut that an electron's κ is 5.12 per nanometre times the square root of the energy deficit in eV. Show every step in the calculation questions.

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Part A

Vocabulary

Potential barrier
A region where the potential energy rises above a particle's total energy, over a finite width.
Classically forbidden region
A region where V exceeds E, which a classical particle can never enter because its kinetic energy would be negative.
Turning point
The position where E = V, at which a classical particle stops and reverses.
Decay constant
κ, with ħκ = √(2m(V − E)); the wavefunction in a barrier fades as e^(−κx), and 1/κ is the decay length.
Transmission probability
T, the chance that a particle sent at a barrier is found beyond it; with the reflection probability R, R + T = 1.
Quantum tunnelling
Passage of a particle through a classically forbidden barrier, possible because its wavefunction fades inside the barrier rather than vanishing.
WKB approximation
A method, introduced in 1926 by Wentzel, Kramers and Brillouin, that estimates tunnelling through smooth barriers by adding up κ(x) across the forbidden region.
Coulomb barrier
The energy hill created by electric repulsion between positively charged nuclear particles, which an alpha particle tunnels out through and fusing protons tunnel in through.
Gamow peak
The narrow energy window in which fusion reactions in a star actually occur, where the falling number of fast particles balances their rising tunnelling probability; near 6 keV in the Sun's core.
Work function
The minimum energy needed to remove an electron from a metal surface; it sets the barrier height in a scanning tunnelling microscope.
Josephson junction
Two superconductors separated by a thin insulating barrier through which Cooper pairs tunnel; the heart of superconducting qubits and voltage standards.
Macroscopic quantum tunnelling
Tunnelling of a collective variable describing enormous numbers of particles, such as the state of a whole superconducting circuit; recognized by the 2025 Nobel Prize in Physics.

Part B

Questions

Auto-checked except short answers, which you mark yourself.

  1. Question 1.Multiple choice

    A thick barrier transmits a tiny fraction of the particles that hit it. Which single change raises the transmission probability the most?

  2. Question 2.Multiple choice

    What does the wavefunction of a particle with E below V do inside a wide barrier?

  3. Question 3.Multiple choice

    An STM operates with κ = 10 per nanometre, and its tunnelling current suddenly doubles. What happened to the tip-surface gap?

  4. Question 4.True or false

    A particle that tunnels through a barrier that does not change in time emerges with less energy than it had before.

  5. Question 5.True or false

    Without quantum tunnelling, hydrogen fusion in the Sun's core would essentially stop.

  6. Question 6.True or false

    The 2020 Toronto experiment found that rubidium atoms cross a tunnelling barrier in zero time.

  7. Question 7.Fill in the blank

    The distance 1/κ over which ψ shrinks by a factor of e inside a barrier is called the length.

  8. Question 8.Fill in the blank

    In 1928 George Gamow explained decay as tunnelling through the Coulomb barrier of the nucleus.

  9. Question 9.Short answer

    Alpha energies of common emitters differ by only about a factor of two, yet their half-lives differ by more than twenty orders of magnitude. Explain why in two or three sentences.

  10. Question 10.Short answer

    A popular explanation says a tunnelling particle briefly borrows the energy it needs from the uncertainty principle. In two or three sentences, explain what is wrong with this and what actually makes tunnelling possible.

  11. Question 11.Calculation

    An electron with E = 2.0 eV meets a rectangular barrier of height 5.0 eV and width 0.40 nm. Find κ, the bare exponential e^(−2κL), and the transmission probability including the thick-barrier prefactor.

    This one asks for more than one value — give each, with its units.

  12. Question 12.Calculation

    Barrier design: an electron with E = 1.0 eV meets a 2.0 eV barrier. Using the thick-barrier formula, what width gives a transmission probability of exactly 1 percent?

    Give a number and units. 6.63×10⁻³⁴, 6.63e-34 and 6.63 x 10^-34 all work.

  13. Question 13.Calculation

    The Sun's core is at 1.571×10⁷ K. Find the thermal energy scale kT in keV, the ratio of a 1 MeV Coulomb barrier to kT, and the Boltzmann factor e^(−1 MeV/kT) as a power of ten.

    This one asks for more than one value — give each, with its units.

  14. Question 14.Calculation

    Treat a 70 kg person as a single particle facing a wall 0.10 m thick with an energy deficit of just 1.0 J. Estimate 2κL and express T as a power of ten.

    Give a number and units. 6.63×10⁻³⁴, 6.63e-34 and 6.63 x 10^-34 all work.

Part C · optional, dangerous, extremely rewarding

Mad scientist challenge

Reproduce Gamow's 1928 triumph. For an alpha particle of energy Q escaping a daughter nucleus of charge number Z, the barrier runs from the nuclear radius R out to the turning point b = 2Z × 1.440 MeV·fm ÷ Q, and the WKB exponent is G = (2b/ħ)√(2μQ) × [arccos√(R/b) − √((R/b)(1 − R/b))], with T = e^(−G). Use an alpha mass of 6.645×10⁻²⁷ kg, reduced mass μ = (alpha mass) × A ÷ (A + 4) for a daughter of mass number A, 1 MeV = 1.602×10⁻¹³ J, and a strike rate of 10²¹ per second. (a) Uranium-238: Q = 4.27 MeV, Z = 90, A = 234, R = 9.3 fm. Find b, G, T and the predicted half-life. (b) Polonium-212: Q = 8.95 MeV, Z = 82, A = 208, R = 9.0 fm. (c) Compare both with the measured half-lives, 4.47×10⁹ years and 294 ns.

12Wrap-up

Recap & sign-off

Key takeaways

  • Where the potential V exceeds a particle's energy E, classical physics forbids entry, but ψ decays exponentially, as e^(−κx) with ħκ = √(2m(V − E)), instead of stopping.
  • If the barrier has a far side, the faded tail emerges as a transmitted wave: T ≈ e^(−2κL), with an exact prefactor never larger than 4, and R + T = 1.
  • The exponent punishes width directly and mass and energy deficit through square roots: doubling the width squares the exponential factor, and a proton facing a barrier that an electron crosses 2.4 percent of the time gets through with a probability of about 10⁻⁹⁵.
  • Gamow, and independently Gurney and Condon, explained alpha decay in 1928 as tunnelling through the Coulomb barrier; because width sits in an exponent, a factor of two in alpha energy spans about 24 orders of magnitude in half-life.
  • Protons in the Sun's core carry about 2 keV against a barrier near 1.0 MeV; fusion happens only because they tunnel, mostly near the Gamow peak at about 6 keV.
  • Exponential sensitivity is a tool: the STM (1981) maps single atoms, flash memory writes by Fowler–Nordheim tunnelling, tunnel diodes switch fast, and Josephson junctions underpin voltage standards and superconducting qubits; macroscopic tunnelling in such a circuit won the 2025 Nobel Prize in Physics.
  • Tunnelling drills no holes, borrows no energy, sends no signal faster than light, and offers no hope of walking through walls.
Dr. Q signs off

And that, my fearless barrier-breachers, is how a wave that refuses to stop at a wall lights the stars, ages rocks, guards your photos and runs quantum chips. Next session we trade the wall for a valley: a particle in a parabolic bowl, the quantum harmonic oscillator, the universe's favourite spring. Its energies form a perfectly even ladder, its lowest rung is never zero, and yes, its wavefunction leaks past the classical turning points too. Psigor, oil the springs. Chapter 13 awaits.

Your chapter checklist

Tick these off in any order. It's a guide, not a law of physics — you can mark the chapter complete whenever you're ready.

  • Listen to the whole lesson (not yet)Or read it all — but the voice is more fun.
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