At a glance
A quantum particle in a parabolic well is described by a wave packet. This sim releases a Gaussian packet from rest and draws its density |Ψ|² beside a classical particle released from the same point, so you can see where the two motions agree and where they do not.
What you'll learn
- How the packet's centre compares with the classical particle.
- How its width depends on the starting width σ.
- Which starting width the well singles out.
- How its energy depends on x₀ and σ.
The packet |Ψ|² (teal) starts centred at x₀ with width σ. The pink dot is a classical particle that starts at the same x₀.
What's on screen
- Shaded curve (teal): the density |Ψ(x,t)|², drawn over the well.
- Parabola (slate): the well V(x) = ½x².
- Pink dot: the classical particle. In the phase-space panel (right) it goes round the pink circle.
- Green ellipse (phase space): the spread in position x and momentum p. The Δx projection bracket under it is the width in x.
- Statistics and state banner (sidebar): the mean position ⟨x⟩, and a name for the state.
See itRelease the packet at the opening settings and let it run for one lap. Watch for: ⟨x⟩ swinging from +x₀ through 0 to −x₀ and back (2.00, 0, −2.00 at x₀ = 2), once every 6.3 s.
The controls
Press ⌖ on a row to see where that control sits in the sim.
| Control | What it changes | Range | |
|---|---|---|---|
| Displacement x₀ | Where the packet and the classical particle start, from rest. | 0–3 opens at 2.0 | |
| Width σ | The width of the packet at t = 0. The label next to it names the state. | 0.20–1.50 opens at 0.71 | |
| ⚡ Snap to Coherent | Sets σ to σ₀ = 1/√2 ≈ 0.707 exactly, a value the slider's 0.01 steps cannot reach. | button | |
| Show Classical Particle | Shows or hides the pink dot and its phase-space circle. | on / off opens on | |
| State banner | Says whether σ equals σ₀ (a coherent state) or not (a squeezed state). | readout | |
| Statistics ⟨x⟩ | The mean position of the packet right now. | readout | |
| ⤢ Maximize | Enlarges the phase-space plot. The left plot has its own ⤢ button. | button | |
| Play / Pause | Starts or freezes the motion. | top bar | |
| Reset | Returns to t = 0 and pauses. It keeps your x₀ and σ. | top bar |
Fixed: ħ = m = ω = 1. Lengths are in √(ħ/mω), energies in ħω, time in 1/ω, and one full orbit (2π/ω) takes about 6.3 s. There is no speed control. Moving x₀ or σ while it runs changes the packet at once and does not restart the clock.
What the curve shows
The shaded curve is the probability density |Ψ(x,t)|². The packet starts as a Gaussian of width σ centred on x₀, released from rest in the well V(x) = ½mω²x².
The oscillator's Hamiltonian, and the packet at t = 0.
- Two numbers describe it. The centre ⟨x⟩ is the mean position, and the width Δx is the standard deviation of x. At t = 0 they equal x₀ and σ.
- It is a density, not a probability. The area under the whole curve is 1. The probability of finding the particle between a and b is the area between them.
- The pink dot is one classical particle with a single position. The packet has a spread of positions, so the comparison is between the dot and the packet's centre ⟨x⟩.
- σ₀ = 1/√2 ≈ 0.71 is the reference width the sim marks. The state banner reports whether σ equals it.
See itFreeze the packet at t = 0 with σ = 0.30, then σ = 1.20. Watch for: the same shaded curve drawn narrow, then wide, both centred over the pink dot.
The centre moves classically
Released from rest at x₀, the centre of the packet follows the classical orbit, whatever the width σ:
Ehrenfest's theorem (top), and its solution for this well (bottom).
The theorem gives the rates of change of the average position and momentum. In a parabola V′(x) is a straight line, so its average over the packet equals V′ at the centre, and the equations are the classical ones exactly. In any other well V′ bends and the centre can leave the classical path.
The packet averages V′ over its width (σ = 0.71, ⟨x⟩ = 1). Parabola: ⟨V′⟩ = V′(⟨x⟩) = 1. A quartic well V = x⁴/4, which this sim cannot show: ⟨V′⟩ = 2.5.
Key idea. In a parabolic well the centre of the packet follows the classical orbit for every σ. The sim draws this exact solution, so the pink dot and the centre of the hump agree by construction. The width is where the quantum behaviour appears.
See itRelease the narrowest packet (σ = 0.20) and let it swing. Watch for: the peak of the hump staying over the pink particle, lap after lap.
The width runs between two values
The width of |Ψ|² does not stay at σ. For a packet released from rest it follows:
σ₀ is the width of the oscillator's ground state.
- σ = σ₀. The two values coincide. Δx stays at 0.71 and the packet moves without changing shape: a coherent state.
- Any other σ. Δx runs between σ and σ₀²/σ: a squeezed state. At σ = 0.40 that is 0.40 (at the turning points, ⟨x⟩ = ±x₀) and 1.25 (crossing x = 0). At σ = 1.50 the roles swap: 1.50 at the turning points, 0.33 at the centre.
The state banner and the label next to the σ slider tell you which case you are in. At σ = σ₀ the phase-space shape is a circle.
See itRelease a σ = 0.40 packet from x₀ = 2. Watch for: the hump narrowest (0.40) at the turning points and widest (1.25) as it crosses the centre, with the Δx bracket opening and closing.
Two breaths per orbit
Square Δx(t) and write cos² and sin² with the double angle:
The width depends on time only through cos 2ωt.
So Δx repeats every π/ω, half the orbit period 2π/ω of ⟨x⟩. The packet "breathes" at 2ω: one breath is about 3.1 s on screen, one orbit about 6.3 s.
Δx over one orbit: flat at σ₀, and two breaths at σ = 0.40.
- At σ = 0.40 the packet is narrowest twice per orbit: at x = +x₀ and again at x = −x₀.
- It is widest twice: each time it crosses x = 0.
See itRun the σ = 0.40 packet for one lap of ⟨x⟩. Watch for: the Δx bracket opening and closing twice while ⟨x⟩ goes 2.00 → −2.00 → 2.00 once.
The energy of the packet
The mean energy of the packet is the sum of a swing term and an internal term:
The sim has no energy readout. These numbers are computed from this formula.
x₀
σ
These sliders move the sim too.
- The internal term is smallest, ½ħω, at σ = σ₀: the zero-point energy. At x₀ = 2, σ = σ₀ gives ⟨H⟩ = 2.50 ħω and σ = 0.40 gives 2.86 ħω.
- ⟨H⟩ does not depend on time: the well is fixed, so breathing adds no energy.
See itSet x₀ = 0 and σ = σ₀, the lowest energy the well allows (½ħω). Watch for: the hump staying put and ⟨x⟩ reading 0.00.
Where it matters
- Lasers and trapped ions. The light of a laser is described well by a coherent state of the electromagnetic field. A trapped ion's motion can be prepared in a coherent state of its own oscillator: a packet that rings in its well without changing shape.
- Squeezed light. Gravitational-wave detectors such as LIGO inject squeezed light to lower their quantum noise. Noise is moved out of the quantity being measured and into the one that is not.
- Molecular vibrations and zero-point energy. A bond near its energy minimum behaves like this oscillator, with level spacing ħω of a few tenths of an eV (HCl: about 0.36 eV). The zero-point energy ½ħω is part of why helium stays liquid at atmospheric pressure down to absolute zero.
Go deeper: the coherent-state amplitude
The amplitude α, and the mean number of quanta ⟨n⟩ = |α|².
- A coherent state is labelled by one number α, here real because the packet starts at rest. Its mean quantum number is |α|² and the spread in n is √⟨n⟩.
- The packet's ⟨x⟩ is the classical orbit with amplitude x₀; α measures how large that orbit is in units of the zero-point width.
Summary
In one line. In a parabolic well the centre of a packet released from rest follows the classical orbit. Its width stays at σ₀ if it starts there, and otherwise breathes twice per orbit.
Key points
- ⟨x⟩ = x₀ cos ωt for every σ, because V′ is linear (Ehrenfest).
- σ₀ = √(ħ/2mω) ≈ 0.71 is the matched width: the state is coherent.
- Otherwise Δx runs between σ and σ₀²/σ, repeating every π/ω.
- ⟨H⟩ = ½x₀² + 1/(8σ²) + ½σ² (in ħω) is constant, with minimum ½ħω.
Common mix-ups
- "A packet spreads out for ever." A free one does. In this well the width repeats every half orbit.
- "Breathing means gaining and losing energy." ⟨H⟩ is constant.
- "A coherent state is a classical particle." It is still a spread-out packet. Its centre moves classically and its shape stays fixed.
Things to try on your own
- Set x₀ = 3 and σ = 0.30. Predict the range of Δx from section 05, then check the bracket.
- Set σ = 1.50 and find when the packet is narrowest. Compare with σ₀²/σ.
- With x₀ = 2, use the section 07 sliders to find two values of σ that give ⟨H⟩ = 3 ħω.