Start with the preview
Start here — today's preview. Watch “Particle in a Box Part 1: Solving the Schrödinger Equation” from Professor Dave Explains before (or after) this session. The lab will wait. Mostly.
Particle in a Box Part 1: Solving the Schrödinger Equation
A clear, beginner-paced walk through today's central calculation: set ψ to zero at both walls and watch the allowed energies appear. Pause when the n² shows up and predict what happens to the levels if the box shrinks.
Watch on YouTubeMore from the video vault
23. Quantum Mechanics V: Particle in a Box
Ramamurti Shankar's Yale lecture (PHYS 201, recorded 2010) opens by reviewing how to predict energy measurements from a superposition, revisits the particle on a ring, and gives the particle in a box its full treatment from about the 19-minute mark to the 54-minute mark. The last quarter hour studies scattering, where a particle enters a classically forbidden region, a perfect bridge to the next chapter.
Watch on YouTubeQuantum Dots (Nobel Prize 2023) - Periodic Table of Videos
A chemist's-eye view of the 2023 Nobel Prize in Chemistry: quantum dots, the particle in a box you can hold in a vial. Watch for how size alone, not composition, changes the colour.
Watch on YouTube- Particle in a Box Part 1: Solving the Schrödinger Equation — Professor Dave Explains — https://www.youtube.com/watch?v=LBB39u8dNw0
- 23. Quantum Mechanics V: Particle in a Box — YaleCourses — https://www.youtube.com/watch?v=lGKKdLmRNQw
- Quantum Dots (Nobel Prize 2023) - Periodic Table of Videos — Periodic Videos — https://www.youtube.com/watch?v=A7fDBnjyEgk
Today's mission
Dr. Q · cold openPsigor! Fetch me the smallest box in the laboratory. No, smaller. Smaller than that! I want a box one billionth of a metre wide, with walls so tall that nothing, not even a rumour, can climb over them. Into it we drop a single electron, and then we ask the most innocent question in physics: how much energy can it have? Any amount you like, says the classical physicist, polishing a billiard ball. Nonsense! I'm mad, not wrong: the electron is a wave, and a wave trapped between two walls may only take shapes that fit. Today nobody postulates quantization. It falls out of the box all by itself, and, as you will see, it is lighting up television screens around the world.
By the end of this session you will be able to:
- Explain why requiring ψ = 0 at two walls allows only whole numbers of half-wavelengths and forces Eₙ = n²h²/(8·m·L²).
- Sketch ψₙ and |ψₙ|² for the first few states, count their nodes, and explain why the ground-state energy can never be zero.
- Calculate energy levels and photon wavelengths for an electron in a nanometre-scale box, and predict how they change when the box shrinks or the mass changes.
- Describe how a superposition of n = 1 and n = 2 sloshes back and forth, and compute its period from T = h/(E₂ − E₁).
- Connect the model to quantum dots and conjugated dye molecules, and state where the simple box is too simple.
- Use the correspondence principle to explain why a marble in a shoebox looks perfectly classical.
Read it — or listen along
7 parts. Press play on any part and the paragraph being read lights up as Dr. Q speaks.
Building the Simplest Prison in Physics
Last session we met the Schrödinger equation, the engine that moves wavefunctions through time. Today we bolt it to the simplest problem that still behaves in a genuinely quantum way: the infinite square well, nicknamed the particle in a box. Inside a region of width L the potential energy V(x) is zero, so the particle feels no force; outside, V is infinite. No real wall is infinitely high, but the idealization captures any tightly confined particle: an electron in a semiconductor nanocrystal, an electron roaming along a dye molecule, a proton rattling inside a nucleus.
What do the walls do to the wavefunction? No finite energy can pay an infinite potential, so the particle is never found outside and ψ = 0 there. In chapter 10 we demanded that acceptable wavefunctions be continuous, so ψ(0) = 0 and ψ(L) = 0. Requirements imposed at the edges of a region are called boundary conditions, and they are the entire secret of this chapter: they alone manufacture quantization.
Inside the box V = 0, so the time-independent Schrödinger equation keeps only its kinetic term: −ħ²/(2·m) times the curvature of ψ equals E times ψ. In words, the curvature of ψ is proportional to minus ψ itself, so the wavefunction always bends back toward the axis. Functions that behave like that are sines and cosines, with wavenumber k = √(2·m·E)/ħ, where k is 2π divided by the wavelength. A cosine equals one at x = 0 and breaks the left-hand condition, so only a sine survives: ψ is A times the sine of kx, where A is a constant amplitude. The right-hand wall is where the drama happens.
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Only Whole Half-Waves Allowed
For ψ to vanish at x = L as well, the sine of kL must be zero, which happens only when kL is a whole-number multiple of π: kL = nπ, with n = 1, 2, 3 and so on. In terms of wavelength, a whole number of half-wavelengths must fit exactly across the box, λₙ = 2L/n, the same rule that fixes the notes of a guitar string. Two candidates are struck off the list. The value n = 0 gives a wavefunction that is zero everywhere, which describes no particle at all. Negative n merely flips the sign of ψ, an unobservable overall phase, so it adds no new state.
Now chain together two ideas you already own. De Broglie's relation p = h/λ turns each allowed wavelength into an allowed momentum, pₙ = n·h/(2L). Inside the box all the energy is kinetic, p²/(2·m), so Eₙ = n²h²/(8·m·L²). Stop and admire this. Bohr, in 1913, had to decree that angular momentum comes in fixed units; here nobody decrees anything. Quantization falls out of a wave equation plus two boundary conditions, the very point Erwin Schrödinger made in 1926 when he titled his papers Quantization as an Eigenvalue Problem. An eigenvalue problem is an equation that has acceptable solutions only for special values of a parameter, here the energy.
One loose end remains: the amplitude A. Born's rule from chapter 9 requires the total probability of finding the particle somewhere in the box to be one, a condition called normalization. A squared sine averages to one half, so A² times L/2 must equal one, giving A = √(2/L). The complete solution, ψₙ(x), is √(2/L) times the sine of nπx/L inside the box and zero outside. Each ψₙ is a stationary state, or energy eigenstate: a state of definite energy whose probability density, |ψₙ|², never changes in time. The integer n is a quantum number, a label that counts the allowed states.
Nodes, the Ground State, and Energy You Cannot Drain
Sketch the first three states. ψ₁ is a single arch, largest at the centre. ψ₂ is one full wave, positive on the left, negative on the right, and zero exactly in the middle. ψ₃ is three half-arches. A point inside the box where ψ = 0 is called a node, and there the probability density is exactly zero. The state ψₙ has n − 1 interior nodes, and the count is no accident: as chapter 10 showed, kinetic energy is tied to curvature, so more nodes in the same width mean more wiggle and more energy. Each probability density is a row of n equal bumps.
The lowest-energy state, the ground state, is n = 1, with E₁ = h²/(8·m·L²). That is not zero, and it never can be. This irreducible minimum is the zero-point energy: the energy a confined quantum system keeps even at absolute zero temperature. The bookkeeping reason is that n = 0 is not a state. The physical reason is the uncertainty principle of chapter 8: pinning a particle within a width L forces a spread in its momentum, and a momentum spread means kinetic energy. For an electron in a box one nanometre wide, E₁ ≈ 0.376 eV, about fifteen times the room-temperature thermal energy scale of 0.026 eV. Cooling cannot remove it: it is not heat but the price of confinement.
Now resist the most tempting picture in the subject: a tiny ball rebounding between the walls. In a stationary state the probability density is frozen: |ψ₂|² is zero at the centre now and forever, yet a shuttling ball would cross the middle on every trip. The standing wave is better read as a right-moving and a left-moving wave present simultaneously, not alternately. A momentum measurement on a highly excited state gives values near plus or minus n·h/(2L) with equal odds, but between measurements standard quantum mechanics assigns no trajectory. Interpretations differ on what lies underneath: in the de Broglie and Bohm pilot-wave picture of chapter 22, the particle in a stationary box state is actually at rest. Every interpretation agrees that |ψₙ|² is frozen.
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Shrink the Box, Stretch the Ladder
Two features of Eₙ = n²h²/(8·m·L²) govern everything that follows. First, the energies climb as n², in the ratio 1, 4, 9, 16, so the rungs spread apart as you climb: the gap between level n and level n + 1 is (2n + 1)E₁. Second, every energy is inversely proportional to the mass and to the square of the width. Halve the box and every level quadruples; double it and every level drops to a quarter. Light particles in small boxes have huge level spacings; heavy particles in big boxes, tiny ones. That is the quantitative reason the quantum world is the world of the small.
Take the formula on a tour of scales. An electron in a box 0.1 nm wide, roughly the size of an atom, has E₁ of about 37.6 eV, the same order of magnitude as real atomic energies: hydrogen's ionization energy, from chapter 5, is 13.6 eV. A one-dimensional box is a crude caricature of an atom, yet it lands on the right scale. A proton, about 1,836 times heavier than an electron, in a box 10 fm across, the size of a nucleus, has E₁ of about 2.0 MeV. The box is ten thousand times narrower, which raises the energy a hundred-million-fold, while the extra mass lowers it 1,836-fold. Same formula, nuclear physics.
Then there are balls in boxes. A one-gram marble in a ten-centimetre box has E₁ of about 5.5×10⁻⁶³ joules, and its levels are far too close for any instrument to tell apart. Electrons in atoms and marbles in shoeboxes both obey quantum mechanics. What differs is whether the level spacing is large compared with every other energy in play, such as thermal jiggling or the resolution of your instrument. When the spacing wins, quantization is unmissable; when it loses, energy looks continuous and classical physics works beautifully.
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Superpositions Slosh
If every stationary state is frozen, where does motion come from? From superposition. The Schrödinger equation is linear, so any sum of solutions is a solution. Mix equal parts of n = 1 and n = 2. Each component carries a phase that rotates in the complex plane at its own rate, e to the power minus i Eₙt/ħ. A single rotating phase is invisible in |ψ|², which is why stationary states stand still. But two phases rotating at different rates drift in and out of step, and the interference term in |Ψ|² oscillates as the cosine of (E₂ − E₁)t/ħ.
Now picture it: ψ₁ is positive everywhere; ψ₂ is positive on the left and negative on the right. When the two are in step they reinforce on the left and cancel on the right, so probability piles up in the left half of the box; half a period later the lump sits in the right half. Mathematically, the average position oscillates about the centre, L/2, as a cosine with amplitude 16L/(9π²): it swings about 18 percent of the box width either side of centre, with period T = h/(E₂ − E₁). For an electron in a box one nanometre wide, T ≈ 3.7 fs. Don't panic — superpose. All motion in quantum mechanics arises this way: a wave packet crossing a box is a superposition of many energy states whose phases march in and out of step.
Two consequences deserve a drumroll. First, the sloshing charge oscillates at a frequency equal to the energy gap divided by h, exactly the frequency of the photon emitted in the jump from n = 2 to n = 1. An oscillating charge radiates at its own frequency, which gave Schrödinger a natural account of Bohr's 1913 rule linking photon frequency to the energy gap. Second, an energy measurement on the sloshing state never returns an average: it gives E₁ or E₂, each with probability one half, and leaves the system in the state found, so the sloshing stops. Measurement means a physical interaction with an apparatus, not a conscious glance. Sloshing is real, too: Ahmed Zewail won the 1999 Nobel Prize in Chemistry for using femtosecond laser pulses to follow wave packets moving inside molecules.
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Boxes You Can Buy: Quantum Dots and Dye Molecules
A quantum dot is a semiconductor nanocrystal a few nanometres across. In a semiconductor, light lifts an electron across the band gap, a forbidden range of energies, leaving behind a positively charged vacancy called a hole. In a nanocrystal both are trapped in a tiny box, so each gains a confinement energy that grows as 1/L², added to the band gap. Shrink the dot and its light shifts toward blue; grow it and the light reddens. One material, a whole rainbow, selected purely by size. Real dots are more complicated: they are three-dimensional, in typical dot materials the electron moves with an effective mass well below its free mass, and electron and hole attract each other. Even so, the confinement term still grows as 1/L² and dominates in the smallest dots.
Glassmakers knew that one additive could tint glass different colours depending on how the molten glass was heated and cooled. In 1981 Aleksey Yekimov, in the Soviet Union, showed that copper chloride crystals inside such glass ranged from about 2 to 30 nm, and the smaller the crystals, the bluer the light they absorbed. In 1983 Louis Brus at Bell Laboratories, unaware of that work, saw the same effect in cadmium sulphide particles in solution: particles 4.5 nm across absorbed bluer light than particles 12.5 nm across. In 1993 Moungi Bawendi's group at MIT learned to grow near-perfect dots of chosen size. The three shared the 2023 Nobel Prize in Chemistry; today quantum dots turn blue light into pure reds and greens in QLED screens.
Chemistry found its boxes even earlier. In a conjugated molecule, a carbon chain with alternating single and double bonds, the π electrons are not tied to any one bond but spread along the chain, like particles in a one-dimensional box about as long as the molecule. In 1948 the Swiss chemist Hans Kuhn built this into the free-electron model of dye colour. Electrons fill the levels two at a time, the Pauli exclusion principle of chapter 18, so with N π electrons the highest filled level is n = N/2, and absorbing light lifts one electron up a level, costing ΔE = (N + 1)h²/(8·m·L²).
Try it on a simple cyanine dye: a three-carbon chain between two nitrogen atoms, with six π electrons spread over about 0.85 nm. The model predicts absorption near 340 nm; the measured peak is about 309 nm, within about ten percent. Longer chains shrink the gap because L² grows faster than N + 1, which is why the fluorescent label Cy5, whose chain is two carbon atoms longer than that of Cy3, absorbs near 650 nm rather than 550 nm. Cyanines suit the model because resonance makes their chain bonds nearly equal in length. Polyenes such as beta-carotene, whose single and double bonds stay unequal, fare much worse, a failure Kuhn traced to that bond alternation.
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Where the Box Forgets It Is Quantum
The correspondence principle, named by Niels Bohr in 1920 for an idea he had used since 1913, is stated in textbooks as the requirement that quantum predictions merge with classical ones at large quantum numbers. A classical ball moving at constant speed spends equal time everywhere, so its probability density is flat, 1/L. At n = 1 quantum mechanics disagrees sharply: about 61 percent of the probability lies in the middle third of the box, against the classical 33 percent, and at n = 2 only about 20 percent. But as n grows, |ψₙ|² becomes a comb of n ever-narrower bumps, and any detector of finite width averages them into the flat classical value.
The energies agree. The fractional gap between neighbouring levels, (2n + 1)/n², shrinks toward zero. Our marble, crawling at one centimetre per second, sits at n of roughly 3×10²⁷, where adjacent levels differ by roughly one part in 10²⁷. Quantum mechanics has not switched off; its graininess is too fine to detect. Yet even at enormous n, a single stationary state is still frozen. A real bouncing ball corresponds to a wave packet built from many neighbouring states, sloshing like our two-state superposition but with countless components forming one compact lump that travels from wall to wall. The bouncing ball is what a well-built superposition becomes in the classical limit.
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Go deeper on video
17 more videos for this chapter, 13 of them from universities, research institutes and Nobel lectures. The clips under “Watch this part” in the lesson start at the moment that matters.
Key equations
Standing-wave condition
Only wavelengths that fit a whole number of half-waves between the walls are allowed, because ψ must vanish at both walls.
| Symbol | Meaning | Unit |
|---|---|---|
| λₙ | wavelength of the nth allowed state | m |
| L | width of the box | m |
| n | quantum number, a positive integer | dimensionless |
Say it: “lambda n equals two L divided by n, where n is one, two, three, and so on”
Box wavefunctions
Each allowed state is a sine wave with n half-arches and n minus 1 interior nodes; the prefactor makes the total probability one. Outside the box the wavefunction is zero.
| Symbol | Meaning | Unit |
|---|---|---|
| ψₙ(x) | wavefunction of the nth state | m⁻¹ᐟ² (so that |ψ|² is a probability per metre) |
| x | position measured from the left wall | m |
| L | width of the box | m |
Say it: “psi n of x equals the square root of two over L, times the sine of n pi x over L, for x between zero and L”
Energy levels
Energies grow as n squared and fall as the mass and the square of the box width grow. The lowest level, E₁, is never zero: that is the zero-point energy.
| Symbol | Meaning | Unit |
|---|---|---|
| Eₙ | energy of the nth level | J (often converted to eV) |
| h | Planck constant, 6.626×10⁻³⁴ | J·s |
| m | mass of the trapped particle | kg |
| L | width of the box | m |
Say it: “E n equals n squared h squared divided by eight m L squared”
Photon from a jump
A particle dropping between levels emits a photon carrying exactly the energy difference. For the jump from 2 to 1 that is three times E₁.
| Symbol | Meaning | Unit |
|---|---|---|
| λ | wavelength of the emitted photon | m |
| c | speed of light, 2.998×10⁸ | m/s |
| nᵢ, n_f | initial and final quantum numbers | dimensionless |
Say it: “h c over lambda equals the energy of the initial level minus the energy of the final level, which is n i squared minus n f squared, times h squared over eight m L squared”
Sloshing superposition
An equal mix of the two lowest states. Each part's phase turns at its own rate, so the two drift in and out of step and the probability density sloshes from side to side.
| Symbol | Meaning | Unit |
|---|---|---|
| Ψ(x,t) | time-dependent wavefunction of the superposition | m⁻¹ᐟ² |
| t | time | s |
| ħ | reduced Planck constant h/2π, 1.055×10⁻³⁴ | J·s |
Say it: “capital psi of x and t equals one over root two, times the sum of psi one times e to the minus i E one t over h-bar, and psi two times e to the minus i E two t over h-bar”
Slosh period
The time for one full back-and-forth of the probability. Its inverse, the slosh frequency, equals the frequency of the photon emitted in the jump from 2 to 1.
| Symbol | Meaning | Unit |
|---|---|---|
| T | slosh period | s |
| E₂ − E₁ | energy gap between the two mixed states | J |
Say it: “T equals h divided by E two minus E one”
Real numbers, step by step
Cover the steps, try the problem yourself, then reveal the solution one step at a time.
An electron in a one-nanometre box
An electron is trapped in an infinite square well of width L = 1.00 nm. Find E₁, E₂ and E₃ in electron-volts, and the wavelength of the photon emitted when it drops from n = 2 to n = 1.
Step 1: Compute the energy unit h²/(8·m·L²). The numerator is h² = (6.626×10⁻³⁴)² = 4.390×10⁻⁶⁷ in SI units. The denominator is 8 × 9.109×10⁻³¹ × (1.00×10⁻⁹)² = 7.287×10⁻⁴⁸. The ratio is 6.025×10⁻²⁰ joules.
Step 2: Convert to electron-volts by dividing by 1.602×10⁻¹⁹ joules per electron-volt: E₁ = 0.376 eV.
Step 3: Scale by n²: E₂ = 4 × 0.376 = 1.504 eV and E₃ = 9 × 0.376 = 3.384 eV.
Step 4: The photon carries E₂ − E₁ = 3 × E₁ = 1.128 eV. Using hc = 1239.8 eV·nm, the wavelength is 1239.8 ÷ 1.128 ≈ 1099 nm.
E₁ ≈ 0.376 eV, E₂ ≈ 1.50 eV, E₃ ≈ 3.38 eV; the photon from 2 to 1 has a wavelength of about 1.10 μm, in the near infrared
Why it matters: Invisible to your eye, but the same box has visible lines too: the jump from 3 to 2 gives about 659 nm, which is red, and the jump from 3 to 1 gives about 412 nm, which is violet.
Tuning a toy quantum dot to green
In the lab's bare-electron model, what box width L makes the photon from the jump 2 to 1 green, with a wavelength of 530 nm?
Step 1: Find the required photon energy: ΔE = hc/λ = 1239.8 eV·nm ÷ 530 nm = 2.339 eV.
Step 2: For the jump from 2 to 1, ΔE = (4 − 1) × E₁ = 3 × E₁, so the box must have E₁ = 2.339 ÷ 3 = 0.780 eV.
Step 3: Use the scaling E₁ ∝ 1/L² with the reference value of 0.376 eV for a box one nanometre wide: L in nanometres is √(0.376 ÷ 0.780) = √0.482 ≈ 0.694, so L ≈ 0.694 nm.
Step 4: Check from scratch: L² = 3h²/(8·m·ΔE) with ΔE = 3.748×10⁻¹⁹ joules gives L² = 4.822×10⁻¹⁹ square metres, so L = 6.94×10⁻¹⁰ metres.
L ≈ 0.69 nm; for comparison, red light at 650 nm needs about 0.77 nm and blue light at 460 nm about 0.65 nm
Why it matters: Real quantum dots are a few nanometres across, not under one, because the photon energy is the band gap plus the confinement energy and the electron's effective mass is smaller than its free mass. The rule you just used, smaller box means bluer light, is exactly the same.
A marble in a shoebox
A marble of mass 1.0 gram rolls at 0.010 m/s inside a box 10 cm long. Treat it as a particle in a one-dimensional box. Estimate its quantum number n and the energy gap to the next level.
Step 1: Ground-state energy: E₁ = h²/(8·m·L²) = 4.390×10⁻⁶⁷ ÷ (8 × 1.0×10⁻³ × (0.10)²) = 4.390×10⁻⁶⁷ ÷ 8.0×10⁻⁵ ≈ 5.5×10⁻⁶³ joules.
Step 2: Kinetic energy of the marble: mv²/2 = 1.0×10⁻³ × (0.010)² ÷ 2 = 5.0×10⁻⁸ joules.
Step 3: Since E = n²E₁, n = √(5.0×10⁻⁸ ÷ 5.5×10⁻⁶³) = √(9.1×10⁵⁴) ≈ 3.0×10²⁷.
Step 4: The gap to the next level is (2n + 1)E₁ ≈ 2 × 3.0×10²⁷ × 5.5×10⁻⁶³ ≈ 3.3×10⁻³⁵ joules, a fraction 2/n ≈ 7×10⁻²⁸ of the marble's energy.
n ≈ 3×10²⁷, with neighbouring levels about 3×10⁻³⁵ joules apart, a fractional step of roughly 7×10⁻²⁸
Why it matters: The marble obeys quantum mechanics perfectly well; its energy ladder is simply far too fine to detect. That is the correspondence principle in one calculation.
Timing the slosh
An electron in a box 1.00 nm wide is in an equal superposition of n = 1 and n = 2. Find the slosh period and the range over which the average position swings.
Step 1: From the first example, E₂ − E₁ = 3 × E₁ = 1.128 eV, which is 1.128 × 1.602×10⁻¹⁹ = 1.807×10⁻¹⁹ joules.
Step 2: Period: T = h/(E₂ − E₁) = 6.626×10⁻³⁴ ÷ 1.807×10⁻¹⁹ = 3.67×10⁻¹⁵ seconds, or 3.67 fs.
Step 3: The average position oscillates about L/2 with amplitude 16L/(9π²). Since 16/(9π²) ≈ 0.180, ⟨x⟩ swings between 0.32L and 0.68L, that is, between 0.32 nm and 0.68 nm.
Step 4: Check the link to light: the slosh frequency 1/T = 2.73×10¹⁴ hertz, and a photon of wavelength 1099 nm has frequency c/λ = 2.998×10⁸ ÷ 1.099×10⁻⁶ = 2.73×10¹⁴ hertz. They match.
T ≈ 3.7 fs, with ⟨x⟩ oscillating between about 0.32 nm and 0.68 nm
Why it matters: The charge sloshes at precisely the frequency of the light this box emits. Double the width of the box and the slosh slows fourfold, because T scales as L².
What everyone gets wrong
Read the myth, decide whether you believed it, then bust it.
- Myth #1
The particle bounces back and forth between the walls like a tiny ball.
TruthIn a stationary state the probability density, |ψₙ|², is frozen in time; for n = 2 it is zero at the centre forever, which a bouncing ball could never manage. Standard quantum mechanics assigns no trajectory between measurements. Motion appears only in superpositions of different energies, where the probability cloud sloshes, and a classical-looking bounce emerges only from a wave packet built from many states at large n.
- Myth #2
The lowest possible energy is zero, when the particle sits perfectly still.
TruthThe value n = 0 would make ψ vanish everywhere, which describes no particle at all. The ground state n = 1 has zero-point energy E₁ = h²/(8·m·L²), demanded by the uncertainty principle: confining a particle to width L forces a spread in its momentum. No amount of cooling removes it, because it is not heat.
- Myth #3
Quantum dots of different colours must be made of different coloured materials.
TruthDots of a single semiconductor glow different colours purely because of their size. The confinement energy grows as 1/L², so smaller dots emit bluer light and larger dots redder light.
Your turn at the controls
Welcome to the Quantum Box. On screen is an infinite square well holding a single electron. A slider sets the width of the box anywhere from 0.2 to 3 nanometres, and every energy is read out in electron-volts. The lab has three modes. Energy levels draws the ladder from n = 1 to n = 8, with each wavefunction or probability density sketched on its rung and a detector that finds the particle at random positions. Superposition lets you mix the three lowest states, watch the probability slosh, and make energy measurements that collapse the mixture. Quantum dot fires the photon from the jump 2 to 1 into a spectrometer. Then comes the game: you are a quantum-dot engineer.
Your tasks
0/5How it connects to the theory
Every number on screen comes from Eₙ = n²h²/(8·m·L²) with the real electron mass and Planck constant, converted to electron-volts, and the photon wavelength is λ = hc/(E₂ − E₁). The slosh period is T = h/(E₂ − E₁), only a few femtoseconds for nanometre boxes, so at one nanometre the animation runs about a million billion times slower than reality. The measure button applies the Born rule: each outcome turns up with probability equal to the squared magnitude of its amplitude. Real quantum dots add a band gap, three dimensions and a smaller effective mass, which is why they are a few nanometres across, but they obey the same law: shrink the box, raise the energy, shift toward blue.
Quantum Box: Trap a Particle and Read Its Energy loads as you scroll down.
Make it stick
Remember the box by its WALLS. W: Whole half-waves fit, λₙ = 2L/n. A: Always above zero, the ground state keeps E₁ = h²/(8·m·L²). L: Little box, Large gaps, because energies scale as 1/(m·L²). L: Ladder of squares, with levels at 1, 4, 9 and 16 times E₁. S: Superpositions Slosh, with period T = h/(E₂ − E₁). To use it, walk along the WALLS whenever a problem traps something: find the waves that fit, set the zero-point floor, check the size of the box, climb the n² ladder, and ask whether more than one state is present.
A particle in a box is like a guitar string pinned at both ends: only vibrations with nodes at the pins are allowed, the fundamental is a single half-arch, each harmonic adds a node, and a shorter string raises every pitch. Where it breaks: a string's frequencies climb as n while the box energies climb as n², because a particle's energy goes as momentum squared; a string can fall silent with zero energy, but a confined particle never can; and the string's wave is real moving matter, whereas ψ is a probability amplitude whose squared magnitude gives the odds of finding the particle.
The whole chapter in one lineFit whole half-waves between the walls and energy comes in n² steps that never reach zero; shrink the box and the steps grow.
Flashcards
Saved on this device. Knew it moves a card up a box; still learning sends it back to box 1. Boxes 4–5 count as mastered.
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
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
Dr. Q's Quantum Quiz Show
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
- ×2 streak bonus
Prove it on paper
Answer every question, check your work, and read the explanations — the stretch challenge is for the truly mad.
Try every question with the lesson closed, then check your work. Use h = 6.626×10⁻³⁴ J·s, electron mass 9.109×10⁻³¹ kg, proton mass 1.673×10⁻²⁷ kg, 1 eV = 1.602×10⁻¹⁹ J and hc = 1239.8 eV·nm. Once you have derived it, you may use the shortcut that an electron's E₁ is 0.376 eV divided by the square of the box width in nanometres. Show every step in the calculation questions.
Part A
Vocabulary
- Infinite square well
- A potential that is zero inside a region of width L and infinite outside it, so a particle is perfectly confined; also called the particle in a box.
- Boundary condition
- A requirement a solution must satisfy at the edges of its region; for the box, ψ = 0 at both walls.
- Stationary state
- A state of definite energy whose probability density does not change in time; also called an energy eigenstate.
- Quantum number
- An integer label, here n = 1, 2, 3 and so on, that counts the allowed states and fixes their energies.
- Node
- A point where the wavefunction is zero, so the probability of finding the particle exactly there is zero.
- Ground state
- The lowest-energy state of a system; n = 1 for the particle in a box.
- Zero-point energy
- The minimum energy a confined quantum system keeps even at absolute zero; E₁ = h²/(8·m·L²) for the box.
- Level spacing
- The energy gap between neighbouring levels; in the box it is (2n + 1)E₁ and grows as the box shrinks.
- Quantum beat
- The oscillation at frequency (E₂ − E₁)/h that appears when two energy states are superposed, making the probability density slosh.
- Quantum dot
- A semiconductor nanocrystal a few nanometres across that confines electrons in all three directions, so its colour is set by its size.
- Conjugated molecule
- A molecule with alternating single and double bonds whose π electrons spread along the chain as if confined in a one-dimensional box.
- Correspondence principle
- The requirement, in its modern textbook form, that quantum predictions reproduce classical physics in the limit of large quantum numbers.
Part B
Questions
Auto-checked except short answers, which you mark yourself.
- Question 1.Multiple choice
An electron is trapped in an infinite square well. Which single change makes its ground-state energy four times larger?
- Question 2.Multiple choice
How many interior nodes does the wavefunction ψ₅ have?
- Question 3.Multiple choice
For which of these states is the probability density exactly zero at the centre of the box?
- Question 4.True or false
The ground-state energy of a particle in an infinite square well is zero.
- Question 5.True or false
In the particle in a box, the energy gap between neighbouring levels grows as you climb the ladder.
- Question 6.True or false
In the n = 1 stationary state, the probability density slowly sloshes from one side of the box to the other.
- Question 7.Fill in the blank
Requirements that a wavefunction must meet at the edges of a region, such as ψ = 0 at both walls of the box, are called conditions.
- Question 8.Fill in the blank
Making a quantum dot smaller shifts the light it emits toward the end of the spectrum.
- Question 9.Short answer
In two or three sentences, explain why a particle in a box has quantized energy while a free particle does not.
- Question 10.Short answer
Why does an equal superposition of n = 1 and n = 2 slosh back and forth, while either state on its own has a probability density that never changes?
- Question 11.Calculation
An electron is confined to an infinite well 0.50 nm wide. Find its ground-state energy in eV and the wavelength of the photon emitted in the jump from n = 2 to n = 1. In which part of the spectrum does it lie?
This one asks for more than one value — give each, with its units.
- Question 12.Calculation
Model a proton inside a small nucleus as a particle in a one-dimensional box of width 5.0 fm (5.0×10⁻¹⁵ m). Estimate its ground-state energy in MeV.
Give a number and units. 6.63×10⁻³⁴, 6.63e-34 and 6.63 x 10^-34 all work.
- Question 13.Calculation
For the ground state ψ₁ = √(2/L) sin(πx/L), find the probability of finding the particle in the middle third of the box, between L/3 and 2L/3. Compare with a classical particle bouncing at constant speed.
Give a number and units. 6.63×10⁻³⁴, 6.63e-34 and 6.63 x 10^-34 all work.
- Question 14.Calculation
An electron in a 2.0 nm box is placed in an equal superposition of n = 1 and n = 2. How long does one full slosh take?
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
Prove that the box obeys Heisenberg. For ψₙ you are given ⟨x⟩ = L/2, ⟨x²⟩ = L²(1/3 − 1/(2n²π²)), ⟨p⟩ = 0 and ⟨p²⟩ = 2mEₙ. (a) Find Δx and Δp, then the product Δx·Δp for n = 1 and n = 2, and check each against Δx·Δp ≥ ħ/2. (b) Show that as n grows, Δx approaches the value for a classical particle spread evenly across the box. (c) Explain in one sentence how part (a) guarantees zero-point energy.
Recap & sign-off
Key takeaways
- A particle behind infinitely high walls must have ψ = 0 at both walls, so only whole numbers of half-wavelengths fit: λₙ = 2L/n.
- That boundary condition alone forces Eₙ = n²h²/(8·m·L²), with levels in the ratio 1, 4, 9, 16, and normalized wavefunctions equal to √(2/L) times the sine of nπx/L.
- ψₙ has n − 1 interior nodes, and the ground state n = 1 keeps a zero-point energy E₁ = h²/(8·m·L²) that no amount of cooling can remove.
- Energies scale as 1/(m·L²): atom-sized boxes give electron-volts, nucleus-sized boxes give millions of electron-volts, and marbles give level spacings far too small to detect.
- A stationary state is frozen in time, so there is no bouncing ball; motion comes from superpositions, and n = 1 plus n = 2 sloshes with period T = h/(E₂ − E₁).
- Quantum dots and conjugated dyes are real boxes: smaller dots glow bluer, and longer dye chains absorb redder light.
- At large n, |ψₙ|² averages to the flat classical distribution and the fractional level spacing vanishes, which is the correspondence principle at work.
Dr. Q signs offAnd that, my magnificent minions, is how two walls and one wave manufacture quantization from nothing but geometry. But I cheated. I built walls of infinite height, and nature never pays infinite bills. Next session, Psigor and I lower the walls to something merely enormous, and the wavefunction does something scandalous: it seeps into the wall, fades, and with a small but stubborn probability appears on the far side. Bring a helmet. Chapter 12 is Quantum Tunneling, and nothing is as solid as it looks.
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.
- Win the lab challenge (not yet)Quantum Box: Trap a Particle and Read Its EnergyGo to the lab
- Score 70% or more on the Quiz Show (not yet)Six to eight questions, no penalties for guessing.Take the quiz
- Check your worksheet (not yet)Answer, check, and read the explanations.Open the worksheet
Sources & further reading
- HyperPhysics: Particle in a Box · hyperphysics.phy-astr.gsu.edu
- NobelPrize.org: Popular information, Nobel Prize in Chemistry 2023 (quantum dots) · nobelprize.org
- Stanford Encyclopedia of Philosophy: Bohr's Correspondence Principle · plato.stanford.edu
- LibreTexts Chemistry: Cyanine Dyes (particle-in-a-box model and measured absorption peaks) · chem.libretexts.org
- NobelPrize.org: Nobel Prize in Chemistry 1999 (Ahmed Zewail, femtochemistry) · nobelprize.org