At a glance
This sim takes the quantum harmonic oscillator far up its energy ladder, to n = 60, and sets its probability density beside what a classical mass on a spring would do. It shows how the quantum picture and the classical one meet when the energy is large.
What you'll learn
- Where the energy ladder Eₙ = (n + ½)ħω comes from, and what sets its rungs.
- How a classical mass's position distribution follows from its speed.
- How the quantum density |ψₙ(x)|² compares with that distribution from n = 0 to n = 60.
- How one rung compares with the total energy as n changes, and what the time dependence of an eigenstate looks like.
The parabola V(x) = ½x² with the allowed energies drawn across it; one rung (gold) ends at its two turning points.
What's on screen
- Top panel: the potential V(x) with every rung Eₙ; the state you picked is the gold one, with dashed turning points.
- Blue: the real part of ψₙ(x,t), in the middle panel.
- Pink: the imaginary part of ψₙ(x,t), in the middle panel.
- Bottom panel: the probability density |ψₙ(x)|² in green.
- Orange dashed curve: the classical distribution Pclassical(x), for a mass with the same energy.
- Sidebar readouts: Classical Limit, Quantum Numbers, Turning Points and Time Evolution.
See itGo to the opening state, n = 40. Watch for: the gold rung and its dashed turning points on the top panel, and the three plots filled in below it.
The controls
Press ⌖ on a row to see where that control sits in the sim.
| Control | What it changes | Range | |
|---|---|---|---|
| Energy level n | The state ψₙ shown in every panel: the gold rung, the curves and all the readouts. | 0–60, whole numbers opens at 40 | |
| Animation speed | How fast time runs on screen, so how fast the blue and pink curves turn. It changes no number in the physics. | 0.1×–1× | |
| Classical Limit | The readout ΔE/Eₙ = 1/(n + ½), the size of one rung next to the energy. | readout | |
| Quantum Numbers | The energy Eₙ in units of ħω, the gap to the next rung, and the number of nodes of ψₙ. | readout | |
| Classical Turning Points | Where a classical mass with energy Eₙ turns round: ±√(2n + 1). | readout | |
| Time Evolution | The classical period Tcl, the phase period Tphase of ψₙ, and the total ∫|Ψ|²dx. | readout | |
| ▶ Play / ⏸ Pause | Runs or freezes the clock that turns the phase. | top bar | |
| ↻ Reset | Returns to n = 40 at speed 1×, restarts the clock and plays. | top bar |
Fixed: ħ = m = ω = 1, so x is measured in x₀ = √(ħ/mω), energy in ħω and time in 1/ω.
The energy ladder
The well is the parabola V(x) = ½mω²x², which in the sim's units is V = ½x². Schrödinger's equation inside it is:
The potential on screen, and the equation each ψₙ must satisfy.
Only certain energies give a ψₙ that fits in the well and dies away outside it. They form a ladder:
The allowed energies, and the state belonging to each one.
- Neighbouring rungs are ħω apart, whatever n is.
- The lowest rung, E₀ = ½ħω, is the zero-point energy: the oscillator cannot be at rest at the bottom of the well.
- Rung n meets the parabola at the turning points x = ±√(2n + 1), the dashed lines in the sim.
See itClimb the first rungs, n = 0 to 4, one at a time. Watch for: the gold rung rising by the same step each time, and the Energy readout going 0.5, 1.5, 2.5, 3.5, 4.5.
Where a classical mass spends its time
A mass on a spring moves fastest through the centre and stops at each end of its swing. The time it spends in a thin strip dx is dx/|v|, so the strip's share of one period Tcl = 2π/ω is:
The classical distribution for a mass of energy Eₙ.
- Pclassical(x) is lowest at the centre and rises toward the turning points, where the mass is slowest.
- It diverges at ±xtp yet its total is still 1, so the sim clips the drawing there.
- It contains no ħ. It is Newton's answer, drawn on the quantum plot for comparison.
See itGo to n = 60, where the turning points are at ±11.00. Watch for: the orange dashed curve lowest at the centre, rising toward both ends.
From the ground state to sixty rungs up
The green curve is a probability density: the probability of finding the particle in a strip dx is |ψₙ(x)|² dx, and the total over all x is 1.
- n = 0: one Gaussian hump, tallest at the centre, where Pclassical is lowest. About 16% of the probability lies beyond the turning points at ±1.
- n = 60: ψₙ has n nodes, so |ψₙ|² is 61 ripples between the turning points at ±11.00. Averaged over a few ripples it follows the orange curve; the tops reach about twice that height.
- The ripples never go away. At n = 60 they are about 0.36 x₀ apart, and a detector that averages over a few of them sees only the orange curve.
Away from the turning points, |ψₙ|² is a fast cos² riding on the classical curve.
See itStep from n = 0 to n = 12 to n = 60. Watch for: one hump at n = 0, then ripples that crowd along the orange curve as n rises.
How big one rung is next to the energy
The gap Eₙ₊₁ − Eₙ = ħω is the same at every n. What changes is the gap measured against the energy of the state:
The absolute gap stays fixed; the relative gap is 1/(n + ½).
n
This slider moves the sim too.
- ΔE/Eₙ is 2 at n = 0, 0.0247 at n = 40 and 0.0165 at n = 60.
- The rungs do not merge: they stay ħω apart. Only their size next to Eₙ shrinks, until no measurement of the energy can resolve one rung.
See itStep n through 0, 5, 20 and 60. Watch for: the Classical Limit readout falling from 2 to 0.0165 while the gap stays 1.000.
What is moving on screen
An energy eigenstate is a fixed shape ψₙ(x) times a rotating phase:
The only time dependence is the phase factor.
- Blue and pink are Re Ψₙ and Im Ψₙ. They share one shape and trade places as the phase turns.
- Green is |Ψₙ|² = |ψₙ|². The phase cancels, so the density does not change in time. That is what stationary state means.
- One turn of the phase takes Tphase = 2π/Eₙ: 1.80 at n = 3, 0.155 at n = 40. The classical period Tcl = 2π/ω = 6.283 is a separate number and does not depend on n.
See itPlay n = 3 at half speed. Watch for: the blue and pink curves trading places while the green curve does not move.
Where it matters
Two oscillators at opposite ends of the ladder.
- Molecular vibrations. A CO molecule's stretch has ħω ≈ 0.27 eV, about ten times kBT at 300 K, so at room temperature it sits in n = 0 and its rungs show up in infrared spectra.
- Everyday oscillators. A 1 kg pendulum with a 2 s period and a 1 cm amplitude sits near n ≈ 10³⁰. One rung is about 10⁻³⁰ of its energy, far below anything we can measure.
- Near any minimum. The bottom of a smooth well is close to a parabola, so this oscillator is the first model for molecular bonds and for lattice vibrations in a crystal.
Go deeper: where the classical period comes back
- A single eigenstate shows no motion in |Ψ|². Mix two neighbouring rungs and |Ψ|² oscillates at (Eₙ₊₁ − Eₙ)/ħ = ω, with period 2π/ω = Tcl.
- The sim shows one eigenstate at a time, so this beat is not drawn.
Summary
In one line. The oscillator's rungs are always ħω apart, but their size next to Eₙ shrinks as 1/(n + ½), and by n = 60 the local average of the quantum density follows the classical curve.
Key points
- Eₙ = (n + ½)ħω, with turning points at ±√(2n + 1).
- Pclassical(x) is the time a classical mass spends in each strip, so it is largest where the mass is slowest.
- |ψₙ|² has n nodes. At large n its local average follows Pclassical.
- An eigenstate's phase turns at Eₙ/ħ, but |ψₙ|² never changes.
Common mix-ups
- "The probability density sloshes back and forth like the classical mass." For an eigenstate, |ψₙ(x)|² has no t in it. Only the blue and pink curves change.
- "At high n the rungs merge into a continuum." They stay exactly ħω apart. Only ΔE/Eₙ = 1/(n + ½) shrinks.
- "En = nħω." The ½ is the zero-point energy. Without it E₀ would be 0.
Things to try on your own
- Set n = 0, then n = 60. At each, note where the green curve is tallest and where the orange one is lowest.
- Raise n from 0 to 12 and compare the Nodes readout with the number of dips in the green curve.
- Read Tphase at n = 10 and n = 60 and check each against 2π/Eₙ.