At a glance
A quantum particle held by a spring-like force can only have certain energies, and each one comes with its own wave shape. This sim shows one such state at a time, so you can compare the quantum picture with a classical mass on a spring.
What you'll learn
- How the allowed energies En depend on n, m and ω.
- How the shape of ψ and |ψ|² depends on n, and on the length scale that m and ω set.
- How the quantum density compares with the classical motion between the turning points.
- What fixes the lowest allowed energy, and how a state's time dependence enters ψ(x, t) and |ψ|².
The model: one particle in the parabola V(x) = ½mω²x². The sim draws one allowed energy at a time, with its two turning points.
What's on screen
- Top panel, Harmonic Potential & Energy Levels: the purple parabola is V(x). The horizontal lines, labelled n = 0, 1, 2 …, are the allowed energies. The state you chose is blue, with dotted arrows ending at its turning points.
- Middle panel, Wavefunction ψ(x, t): Re ψ in blue and Im ψ in pink.
- Bottom panel, Probability density |ψ|²: the green curve.
- Parameters box: the clock, the state's energy, its turning points ±xtp, and the oscillator length √(ħ/mω).
See itStart from n = 3, with m = ω = 1. Watch for: the blue n = 3 line with its dotted arrows in the top panel, and one curve in each panel below.
The controls
Press ⌖ on a row to see where that control sits in the sim.
| Control | What it changes | Range | |
|---|---|---|---|
| Quantum state (n) | Which allowed state is shown. The top panel highlights its energy line; the lower panels show its ψ and |ψ|². | 0–10, steps of 1 opens at 3; the inquiry sets n itself | |
| Mass (m) | The particle's mass, in the sim's units. | 0.7–1.2 opens at 1.00 | |
| Angular freq (ω) | The stiffness of the well in V = ½mω²x². It is the angular frequency of the matching classical oscillator. | 0.7–1.2 opens at 1.00 | |
| Animation speed | How fast the on-screen phase advances. It is a viewing rate only. | 0.1×–1.0× opens at 1.0× | |
| Parameters | Time (the on-screen clock, in seconds), the energy En, the turning points ±xtp, and the oscillator length √(ħ/mω). | readout in the inquiry, Energy appears once step 6 is answered | |
| Equations | The Schrödinger equation for this well, and the formula for its allowed energies. | readout in the inquiry, it appears once step 5 is answered | |
| Play / Pause | Runs or freezes the clock, and with it the on-screen phase. | top bar | |
| Reset | Sets the clock back to 0 and resumes playing. The sliders keep their values. In the Guided Inquiry or Controls Guide it restarts that mode. | top bar |
Fixed: one particle in one dimension, with ħ = 1, so every number on screen is dimensionless. The sim shows one state at a time. The axes never rescale, so changing m or ω shows up as a change in the curves.
The well fixes the allowed energies
Inside the well, ψ obeys the Schrödinger equation with V(x) = ½mω²x². There is no fixed width as in a box; instead ψ must die away at large |x|, and that works only at certain energies.
The Equations box: the Schrödinger equation for the parabola.
The allowed energies, with n = 0, 1, 2, … (ħ = 1 in the sim).
- Equal gaps. En is linear in n, so every neighbouring pair of rungs is ħω apart. In a box, E grows as n² and the gaps widen.
- The lowest rung is E₀ = ½ħω, not 0. Section 06 shows why.
- Only ω sets the spacing. The mass m does not appear in En. At ω = 1 the readout is n + ½.
Computed at m = ω = ħ = 1: the parabola, all eleven rungs the n slider reaches (n = 0 to 10), and n = 3 with the turning-point arrows the sim draws in blue.
See itStep n from 0 to 4 and read the Energy each time. Watch for: the Energy readout going up by the same amount at every step, and the blue line climbing one rung at a time.
One length sets every state's shape
What the sim's ψn evaluates: a constant, a Hermite polynomial Hn, and a Gaussian.
- Same Gaussian, different polynomial. The Gaussian e−x²/2x₀² sets the size; Hn adds the wiggles.
- n zero crossings, n + 1 peaks. ψn crosses zero n times, so |ψn|² has n + 1 peaks. n = 0 is a single hump. Even n gives a symmetric ψ, odd n an antisymmetric one.
- One length, x₀ = √(ħ/mω). It is the sim's Osc. length readout. Only the product mω enters it, so raising m or ω narrows every curve by the same factor, 1/√(mω).
- Energy is different. En depends on ω alone. Raising m narrows the wave and leaves E unchanged; raising ω narrows it and raises E.
See itGo to n = 5. Watch for: five zero crossings in the middle panel, and six peaks in the green curve.
See itAt n = 3, raise m to 1.2, then return m to 1 and raise ω to 1.2. Watch for: Osc. length falling to 0.913 both times, while Energy stays 3.50 for m and rises to 4.20 for ω.
The tails reach past the turning points
A classical mass with energy En cannot go where V(x) > En. Its turning points ±xtp are where En = ½mω²xtp², the ends of the blue arrows: xtp = √(2n + 1)·x₀.
The sim's ψ and |ψ|² panels at m = ω = 1 on its fixed axes: n = 0 on the left, n = 10 on the right. The amber dashed curve is the classical time-average density for the same energy. It is added here for comparison, and the sim never draws it.
- The tails go past xtp. ψ falls off like a Gaussian but is not zero beyond the turning points. For n = 0 about 16% of the probability lies there; for n = 10, about 6%.
- n = 0 is the opposite of classical. A classical mass moves slowest near its ends and lingers there. |ψ₀|² peaks at the centre.
- Large n looks more classical. At n = 10 the outermost peaks of |ψ|² are the tallest, about twice the central ones, and they sit just inside the turning points. The quantum density oscillates fast around the classical average.
See itGo to n = 0. Watch for: the green curve still above the axis beyond ±1.00, where the blue arrows in the top panel end.
See itGo to n = 10. Watch for: the tallest green peaks at the two ends, just inside ±4.58. The left one is behind this card; the right one is clear.
The ground state cannot sit still
Exact rest at the bottom would mean x = 0 and p = 0 together, which Δx Δp ≥ ħ/2 forbids. An estimate shows the cost of squeezing either one:
A kinetic term that grows as Δx shrinks, a potential term that grows as Δx grows, and a minimum between them.
The states themselves satisfy this exactly:
Spreads in x and p for state n. Δx and Δp are standard results; the sim does not display them.
n
m
ω
These sliders move the sim too.
- n = 0 sits on the bound. Δx Δp = ħ/2 exactly, and the energy is the minimum ½ħω.
- Higher rungs are further from it. Δx Δp = (n + ½)ħ, which is 2n + 1 times ħ/2.
See itGo to n = 0. Watch for: the Energy readout showing 0.50, not 0.
Stationary states have a turning phase
The whole time dependence is one phase factor, and it cancels in |Ψ|².
- The middle panel is alive. Re Ψ (blue) and Im Ψ (pink) share the shape ψn(x), and the phase moves amplitude between them.
- The bottom panel is frozen. |Ψ|² = |ψn|² has no t in it. That is what "stationary state" means.
- Higher rungs turn faster. The phase turns at En/ħ, so at the same ω, n = 10 turns 21 times faster than n = 0.
- Animation speed is a viewing rate. It scales the on-screen phase only. The energy, the turning points and |ψ|² do not change.
- A density that really moves, such as a packet sloshing in the well, needs a superposition of several rungs. This sim shows one state at a time.
See itGo to n = 1 at full speed. Watch for: the blue and pink curves trading places in about 4 s per turn, while the green curve stays put.
Where it matters
Near any smooth minimum the first surviving term is a parabola, so this one problem appears everywhere.
- Molecular vibration. The C–O bond has ħω ≈ 0.27 eV, so its stretch absorbs infrared light near 2143 cm⁻¹. Its zero-point energy is about 0.13 eV, and the zero-point spread of the bond length is about 3 pm on a 113 pm bond.
- Frozen vibration. At room temperature kT ≈ 0.026 eV, a tenth of ħω. A Boltzmann estimate puts about 3 molecules in 100 000 in n = 1, so nearly all sit in n = 0.
- Helium. Its large zero-point motion and very weak atom-atom attraction keep it liquid at absolute zero unless about 25 atmospheres are applied.
- Phonons and photons. Each vibration mode of a crystal, and each mode of light in a cavity, has the energies (n + ½)ħω. Here n counts the quanta, and every mode keeps its ½ħω.
Go deeper: a real bond is not exactly a parabola
Carbon monoxide computed from its measured constants (ωe = 2170 cm⁻¹, re = 1.128 Å, De = 11.1 eV): near the minimum the real Morse curve and the parabola agree. The harmonic rungs sit ħω = 0.269 eV apart, with the n = 0 floor at 0.134 eV.
- Real levels creep closer together higher up the well, which is why the measured 2143 cm⁻¹ is a little below the harmonic 2170 cm⁻¹. This sim models the parabola only, so its rungs stay exactly ħω apart.
Summary
In one line. In a parabolic well the allowed energies are En = (n + ½)ħω: evenly spaced, starting half a quantum above zero, with a standing wave of n zero crossings at each rung.
Key points
- Every gap is ħω, and the lowest energy is ½ħω because Δx Δp ≥ ħ/2 forbids exact rest.
- x₀ = √(ħ/mω) sets the width of every state. Energy depends on ω alone.
- ψn has n zero crossings, |ψn|² has n + 1 peaks, and the tails reach past the turning points.
- A stationary state's phase turns at En/ħ, while |ψ|² stays fixed.
Common mix-ups
- "At n = 0 the particle rests at the bottom with zero energy." Rest would fix x and p exactly, which the uncertainty principle forbids. The floor is ½ħω.
- "The bottom panel shows the particle oscillating." In one state |ψ|² never changes. Only Re Ψ and Im Ψ move, and they are not probabilities.
- "The particle is always most likely near the ends, like a classical mass." At n = 0 the density peaks at the centre. The classical pattern appears only as n grows.
Things to try on your own
- Step n from 0 to 10 and write down each Energy. Then set ω = 0.7 and check the gap again.
- At n = 10, count the peaks of |ψ|² and compare with n + 1.
- Set n = 3, m = 1.2 and ω = 0.7. Predict the Energy and the Osc. length, then check the readouts.