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Module 03

Bound states in one dimension: wells and the oscillator

See where energy levels come from, and why a state that is visibly spinning is called stationary.

What you will see

  • The blue ψ tube turns in the complex plane while the purple |ψ|² curtain does not move at all
  • Each increment of the quantum number adds one node; level spacings go as n² or stay constant
  • In a finite well the wavefunction leaks through the wall and decays exponentially — the seed of tunnelling

Assumed background

  • The wavefunction and the Born rule (module 02)
  • Boundary conditions for second-order differential equations

“Energy is quantised” is probably the best-known sentence in quantum mechanics, and the one most often memorised as an incantation.

It is not mysterious. Trap a wave in a box and only certain waves fit. Guitar strings, drum heads and organ pipes all do this. What actually needs explaining is not the quantisation but why it took so long to notice that electrons are waves.

Bound states in one dimension: levels, wavefunctions, and why a stationary state is "stationary"

The blue tube is ψ drawn in the complex plane, and it turns as time passes. The purple curtain is |ψ|², and it never moves. That contrast is the entire meaning of the word "stationary".

Loading 3D scene…

Potential well
4.0
Energy levels
2

Click the level ruler or any curve in the 3D view to switch quantum number.

Display
Time evolution
Phase 0°
Speed
E2=2.7758E_{2} = 2.7758analytic 2.7758 (error 0.00%)
nodes 2⟨x⟩ = 0.000Δx = 1.115bound
  • ψ (complex, turning in time)
  • |ψ|² (time-independent)
  • Potential V(x)
  • Energy levels Eₙ

What to look for

  • Press play, then switch to the "Along x" view. Facing the complex plane you see ψ spinning on the spot. What spins is the phase, not the probability — the purple curtain never moves.
  • Take n from 0 up to 5 and count the zero crossings: there are exactly n of them. Watch the level spacing at the same time — n² for the infinite well, evenly spaced for the oscillator.
  • Switch to the finite well and lower V₀: levels get squeezed out of the well one by one. Notice that the wavefunction is not zero outside the walls — it leaks out with exponential decay, and that leak is the root of tunnelling.
  • Pick the oscillator, push n to its maximum and switch on the classical comparison: the envelope of the quantum distribution hugs the classical one, largest at the turning points and smallest in the middle. That is the correspondence principle.
  • Switch on the other eigenstates and rotate: the level diagram and the wavefunction plot turn out to be two projections of a single picture.

✕ "The wavefunction is the shape of the particle in space"

ψ is not a distribution of stuff. It is a complex amplitude; what carries physical meaning is |ψ|² (a probability density) and the relative phase between states.

✕ "In a stationary state the particle rattles back and forth"

The probability density of a stationary state does not change at all, and there is no probability current. Only superpositions flow — that is module 09.

✕ "ψ vanishes outside an infinite well, so it must vanish outside a finite one too"

The opposite. In a finite well the wavefunction penetrates the classically forbidden region as e^{−κ|x|}. That is a purely quantum effect, and it is where tunnelling comes from.

Think it through

  1. Why is the ground-state energy not zero? What happens to it as you narrow the well? Estimate E ~ ħ²/(2mL²) from the uncertainty principle and compare with the readout.
  2. The infinite well gives E_n ∝ n², the oscillator gives evenly spaced E_n ∝ (n + ½). Explain the difference starting from "an integer number of half-wavelengths must fit" and "the potential is quadratic".
  3. However shallow it is, a one-dimensional finite well always holds at least one bound state. Try V₀ = 0.5. This fails in three dimensions — why should dimensionality matter?

Stationary states: the phase turns, the probability does not

The time-dependent Schrödinger equation

iℏ ∂tΨ=H^Ψ\ii\hbar\,\partial_t\Psi = \hat H \Psi

acting on a solution of H^ψn=Enψn\hat H\psi_n = E_n\psi_n gives

Ψn(x,t)=ψn(x) e−iEnt/ℏ.\Psi_n(x,t) = \psi_n(x)\,\ee^{-\ii E_n t/\hbar}.

The exponential has modulus one, so

∣Ψn(x,t)∣2=∣ψn(x)∣2independent of t.|\Psi_n(x,t)|^2 = |\psi_n(x)|^2 \quad\text{independent of } t.

That is the central image of the scene: the tube spins, the shadow does not. Switch to the “Along x” view and you are facing the complex plane, watching the whole curve rotate like a twisted rope. The rate is proportional to EnE_n.

Three wells, three patterns of levels

WellLevelsSpacingIndex
Infinite square wellEn=n2π2ℏ22mL2E_n = \dfrac{n^2\pi^2\hbar^2}{2mL^2}grows with nn=1,2,3…n = 1,2,3\ldots
Harmonic oscillatorEn=ℏω(n+12)E_n = \hbar\omega\left(n+\tfrac12\right)perfectly evenn=0,1,2…n = 0,1,2\ldots
Finite square wellno closed form (transcendental)finitely many, thinning outdepth sets the count

Switch between the three in the scene while watching the level ruler on the left, and the difference is immediate.

The infinite well in three lines

Inside the well V=0V=0, so the stationary equation is −ℏ22mψ′′=Eψ-\frac{\hbar^2}{2m}\psi'' = E\psi with general solution ψ=Asin⁡kx+Bcos⁡kx\psi = A\sin kx + B\cos kx, where k=2mE/ℏk=\sqrt{2mE}/\hbar.

The boundary conditions ψ(0)=ψ(L)=0\psi(0)=\psi(L)=0 force B=0B=0 and kL=nπkL = n\pi, so

kn=nπL,En=ℏ2kn22m=n2π2ℏ22mL2,ψn=2Lsin⁡nπxL.k_n = \frac{n\pi}{L},\qquad E_n = \frac{\hbar^2k_n^2}{2m} = \frac{n^2\pi^2\hbar^2}{2mL^2},\qquad \psi_n = \sqrt{\frac{2}{L}}\sin\frac{n\pi x}{L}.

“An integer number of half-wavelengths must fit” is, mathematically, just kL=nπkL=n\pi.

Zero-point energy: uncertainty, directly

E1≠0E_1 \ne 0, and the oscillator even has 12ℏω\tfrac12\hbar\omega left over. Why can the particle not simply sit still at the bottom?

Because sitting still means Δp=0\Delta p = 0, while Δx≤L\Delta x \le L is finite, and that violates ΔxΔp≥ℏ/2\Delta x\Delta p\ge\hbar/2. A rough estimate:

E∼(Δp)22m∼ℏ22mL2,E \sim \frac{(\Delta p)^2}{2m} \sim \frac{\hbar^2}{2mL^2},

which differs from the exact π2ℏ2/2mL2\pi^2\hbar^2/2mL^2 only by the factor π2≈10\pi^2 \approx 10. Narrow the well in the scene and the ground-state energy shoots up as 1/L21/L^2. That cost of confinement is the same reason atoms do not collapse and white dwarfs resist gravity.

The finite well: the wavefunction leaks

This is the picture to take away from the module.

In the classically forbidden region (E<V(x)E < V(x)) the stationary equation becomes

ψ′′=κ2ψ,κ=2m(V0−E)ℏ,\psi'' = \kappa^2\psi,\qquad \kappa = \frac{\sqrt{2m(V_0-E)}}{\hbar},

whose solutions are e±κx\ee^{\pm\kappa x} — not oscillation, but exponential decay. So outside the wall the wavefunction is not zero; it trails off with an exponential tail.

Join that tail to the far side of the wall and you have tunnelling, module 04.

Key formulas

  • Stationary state

    Ψn(x,t)=ψn(x)e−iEnt/ℏ\Psi_n(x,t)=\psi_n(x)\ee^{-\ii E_n t/\hbar}

    Only the phase moves

  • Infinite well

    En=n2π2ℏ2/2mL2E_n = n^2\pi^2\hbar^2/2mL^2

    ∝ n², ∝ 1/L²

  • Harmonic oscillator

    En=ℏω(n+12)E_n = \hbar\omega(n+\tfrac12)

    Even spacing plus zero-point energy

  • Forbidden region

    ψ∼e−κx, κ=2m(V0−E)/ℏ\psi \sim \ee^{-\kappa x},\ \kappa=\sqrt{2m(V_0-E)}/\hbar

    Exponential decay, not zero

Think it through

  1. Double the width of an infinite well. By what factor does the ground-state energy drop? Measure it in the scene, then check against the formula.
  2. The oscillator has evenly spaced levels. What does that have to do with an electromagnetic field being allowed to contain n photons?
  3. A one-dimensional finite well always holds at least one bound state, however shallow. Try a depth of 0.5. This fails in three dimensions — at which step did dimensionality enter?

Go deeper · matching textbook sections

The 3D scenes build the picture; the full derivations and exercises live in the textbook.

Having finished this module