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Quantum core · institution-dependent · Interactive lesson

Unitary evolution and exact revivals

Diagonalising Ĥ looked like the end of a calculation. It was really the purchase of a clock. This is where you spend it: turn a spectrum into a film, learn which parts of a state are free to move and which are frozen for ever, and meet the one box in physics that forgets nothing.

UPV-05.08 · Lesson 8 of 9 in this unit. Position is not a completion score.
Starting model: Unitary evolution and exact revivalsProbability density |ψ(x, t)|² in the box for cos θ |1⟩ + sin θ |q⟩ at θ = 45°, q = 2. Dashed curve: the state at t = 0. Solid curve: the same state at Δφ = 0.72 π rad. Filled dot: ⟨x⟩ = 0.615 L now. Open dot: ⟨x⟩ at t = 0. Dashed vertical: the box centre L/2.|ψ(x, t)|² · levels 1 and 2⟨x⟩ = 0.615 L (filled dot)dashed: t = 0solid: t = 0.12 Tᵣₑᵥx = 0L/2x = L
Start with a picture. Change the model in the lab below.

Quantum core · institution-dependent. Your university determines the required depth of these quantum topics, including the use of operators, differential equations and computation. Institution-dependent does not automatically mean optional.

1

Build the model

Nothing but the relative phases moves — and in this box they all return at once.

The key idea

Unitary evolution in the box means every energy amplitude keeps its magnitude and only turns in phase, cₙ(t) = cₙ(0)exp(−iEₙt/ℏ), so a state changes shape solely through the phase differences between the levels it occupies.

Example. For an electron in L = 1.00 nm, E₁ = 0.376 eV, so in (|1⟩ + |2⟩)/√2 the relative phase runs at 3E₁/ℏ = 1.71 × 10¹⁵ rad s⁻¹ and ⟨x⟩ oscillates between 0.320 nm and 0.680 nm with a period of 3.67 fs.

The complete physical model and assumptions

Time evolution is generated by the Hamiltonian: |ψ(t)⟩ = Û(t)|ψ(0)⟩ with Û(t) = exp(−iĤt/ℏ), and Û†Û = 1̂ is not an extra postulate but the statement that total probability is conserved. Exponentiating an operator is hard in a general basis and trivial in an eigenbasis of Ĥ, where Û is diagonal and its entries are ordinary complex numbers of modulus one. The box therefore hands over its entire dynamics for the price of one expansion: take cₙ = ⟨n|ψ(0)⟩, multiply each by exp(−iEₙt/ℏ), resum. No |cₙ| ever changes, so ⟨Ĥ⟩, ΔE and every function of Ĥ are constants of the motion; all motion lives in phase differences, and it surfaces in the cross terms of |ψ(x, t)|² as beats at the Bohr frequencies ωₘₙ = (Eₘ − Eₙ)/ℏ. Here those frequencies are (m² − n²)E₁/ℏ, integer multiples of a single fundamental, so the beats form a harmonic comb and the state returns exactly — not approximately — at Tᵣₑᵥ = 2πℏ/E₁ = 4mL²/(πℏ). That exactness is the result, and it is also the bill: it is bought entirely by the n² spectrum, which belongs to hard walls around an isolated system. Soften the walls, add a perturbation, or let one environmental photon record which level the system is in, and either the gaps stop being commensurate or the off-diagonal terms decay. The beats survive for a while. The exact revival does not.

1Û(t) is diagonal in the basis you have already built

Û(t) = exp(−iĤt/ℏ) is defined by its power series, and for a general Ĥ that series is real work. In an eigenbasis it is not: Ĥ|n⟩ = Eₙ|n⟩ gives Ĥᵏ|n⟩ = Eₙᵏ|n⟩, the series resums term by term, and Û(t)|n⟩ = exp(−iEₙt/ℏ)|n⟩. So Û is diagonal with entries of modulus one, and a diagonal matrix in an orthonormal basis is unitary precisely when every entry has modulus one. The recipe is three lines and never changes. Expand once, cₙ = ⟨n|ψ(0)⟩ = √(2/L) ∫₀ᴸ sin(nπx/L) ψ(x, 0) dx. Phase each coefficient, cₙ(t) = cₙ e(−iEₙt/ℏ). Resum, ψ(x, t) = Σ cₙ(t) √(2/L) sin(nπx/L). Numerically that is a truncation, not a time-stepper: hold the cₙ in a NumPy array of length N, build phase = numpy.exp(−1j * n * n * E1 * t / hb), and take one matrix product against the mode matrix. There is no timestep error to control, because you never take a step — t = 10⁶ Tᵣₑᵥ costs exactly what t = 0 costs. The only error is the tail you truncated, and since the cₙ of a smooth packet fall away fast, N of order 40 is usually plenty.

2What is frozen, and what is left free to move

Read the propagator twice. First on the amplitudes: |cₙ(t)| = |cₙ(0)| for every n, because the phase has modulus one. So the energy distribution of the state is a constant of the motion, and with it ⟨Ĥ⟩ = Σ|cₙ|²Eₙ, the spread ΔE, and the expectation of any operator that commutes with Ĥ. For an electron in a 1.00 nm box prepared in (|1⟩ + |2⟩)/√2, ⟨Ĥ⟩ = 2.5E₁ = 0.940 eV and ΔE = 1.5E₁ = 0.564 eV, now and for ever. Second on the phases. The density matrix in the energy basis has elements ρₘₙ(t) = cₘ(t)c*ₙ(t) = ρₘₙ(0)e(−i(Eₘ − Eₙ)t/ℏ): the diagonal is untouched and every off-diagonal element rotates. That is the whole of the dynamics of a closed system with a known spectrum. If an observable's matrix is diagonal in the energy basis, its expectation cannot move; if it carries off-diagonal elements, they are the only route by which it can.

3The beat comb: every gap is a multiple of E₁

Square the evolved state and the phases stop hiding: |ψ(x, t)|² = Σ over m, n of c*ₘcₙψₘ(x)ψₙ(x)e(−i(Eₙ − Eₘ)t/ℏ). The m = n terms are static; every other term oscillates at a Bohr frequency ωₘₙ = (Eₘ − Eₙ)/ℏ. Because Eₙ = n²E₁ those frequencies are (m² − n²)E₁/ℏ, and m² − n² is an integer for every pair. With E₁ = 0.376 eV at L = 1.00 nm the fundamental is E₁/ℏ = 5.71 × 10¹⁴ rad s⁻¹, and the low pairs sit at ω₂₁ = 3E₁/ℏ = 1.71 × 10¹⁵, ω₃₂ = 5E₁/ℏ and ω₃₁ = 8E₁/ℏ rad s⁻¹. The same numbers are spectroscopy read the other way round: 3E₁ = 1.128 eV is the 2 → 1 photon, wavelength 1.10 μm. This is what is special about the box. A generic potential has gaps in no rational ratio, so its beats never all re-phase at once and the motion is merely quasi-periodic; here every beat is a harmonic of one frequency, and a harmonic series has a period.

4Why the state comes back exactly, and the mirror halfway

Ask when every phase returns to 1 at once. Requiring exp(−in²E₁T/ℏ) = 1 for all n means n²E₁T/ℏ must be a multiple of 2π, and since n² is already an integer the single condition E₁T/ℏ = 2π suffices: Tᵣₑᵥ = 2πℏ/E₁ = h/E₁ = 4mL²/(πℏ). For an electron in a 1.00 nm box that is 11.0 fs; widen the box to 1.00 μm and the L² scaling stretches it to 11.0 ns. Nothing was approximated and no particular state was assumed — every state in this box returns exactly after the same Tᵣₑᵥ, even one such as (|1⟩ + |2⟩)/√2 whose density also repeats sooner. Half a revival is the more interesting instant. At t = Tᵣₑᵥ/2 each phase is exp(−iπn²) = (−1)ⁿ, and since ψₙ(L − x) = (−1)ⁿ⁺¹ψₙ(x), that combination is precisely a reflection: ψ(x, Tᵣₑᵥ/2) = −ψ(L − x, 0). A packet released near the left wall arrives, exactly mirrored, near the right wall at 5.50 fs. Rational fractions Tᵣₑᵥ p/q in between give fractional revivals, where the state is a finite superposition of shifted copies of itself — the fine structure of the quantum carpet.

5Which superpositions actually move ⟨x⟩

A moving density is not the same thing as a moving mean. ⟨x⟩(t) = Σ over m, n of c*ₘcₙ xₘₙ e(−i(Eₙ − Eₘ)t/ℏ), so a frequency reaches ⟨x⟩ only if its matrix element survives. In this box xₘₙ = −8Lmn/(π²(m² − n²)²) when m − n is odd and vanishes when m − n is even with m ≠ n, because ψₘψₙ is then even about L/2 while x − L/2 is odd. Two consequences follow. A packet prepared symmetrically about L/2 excites only odd n, every pair then has m − n even, and ⟨x⟩ sits at L/2 for all time while |ψ(x, t)|² breathes visibly. And in (|1⟩ + |2⟩)/√2 the one surviving element gives ⟨x⟩(t) = L/2 − 0.1801L cos(3E₁t/ℏ), a swing from 0.320 nm to 0.680 nm at L = 1.00 nm. Differentiate and Ehrenfest's theorem becomes checkable: mₑ d⟨x⟩/dt = (8ℏ/3L) sin(3E₁t/ℏ), which is exactly 2Re[c*₁c₂p₁₂e(−iω₂₁t)] built from p₁₂ = 8iℏ/(3L). Peak value 2.81 × 10⁻²⁵ kg m s⁻¹, by either route.

6What the exact revival costs

The revival is a property of this spectrum, not of quantum mechanics. It needs every gap Eₘ − Eₙ to be an integer multiple of one energy, and few systems oblige: the infinite box through n², the oscillator through n, the rigid rotor through l(l + 1). A cubic box still qualifies, since Eₙ = (nₓ² + ny² + nz²)E₁ is still an integer multiple, and its revival time is the same h/E₁. Round the walls off, though — a finite well, a real quantum dot — and the upper levels are pulled below n²E₁, the ratios stop being rational, and the exact revival degrades into a partial one that fades with each pass. Isolation is the second cost. Let the environment learn which level the particle occupies and ρₘₙ decays for m ≠ n while the diagonal survives, so the density relaxes to the static incoherent mixture Σ|cₙ|²|ψₙ(x)|² with ⟨x⟩ pinned at L/2. No energy has been absorbed; the state has simply stopped interfering with itself, and that is the failure mode a revival experiment is really testing.

2

Key equations

Read the units, assumptions and limits before substituting.

The propagator, diagonal in the energy basis

Û(t) = e(−iĤt/ℏ) = Σ over n of e(−iEₙt/ℏ) |n⟩⟨n|

The basis that diagonalised Ĥ diagonalises every Û(t) at once, so exponentiating an operator collapses into multiplying numbers.

Eₙ = n²π²ℏ²/(2mL²) in J and t in s. Û†Û = 1̂, so ⟨ψ(t)|ψ(t)⟩ = 1 at every t.

Phasing the coefficients

cₙ(t) = cₙ(0) e(−i n² E₁ t/ℏ), with cₙ(0) = ⟨n|ψ(0)⟩

Every |cₙ| is frozen, so ⟨Ĥ⟩ and ΔE never move; the whole of the dynamics is the phase you attach right here.

cₙ is dimensionless; E₁ = π²ℏ²/(2mL²) = 0.376 eV for an electron at L = 1.00 nm.

Bohr frequencies: a harmonic comb

ωₘₙ = (Eₘ − Eₙ)/ℏ = (m² − n²) E₁/ℏ

Every beat in |ψ(x, t)|² is a harmonic of one fundamental, which is why this box has a single period rather than a quasi-period.

E₁/ℏ = 5.71 × 10¹⁴ rad s⁻¹ at L = 1.00 nm, and m² − n² is always an integer.

Exact revival time

Tᵣₑᵥ = 2πℏ/E₁ = h/E₁ = 4mL²/(πℏ)

All the phases reach 1 together because n² is an integer, so ψ(x, Tᵣₑᵥ) = ψ(x, 0) exactly, not to within some tolerance.

m in kg and L in m give Tᵣₑᵥ in s: 11.0 fs for an electron at L = 1.00 nm, 11.0 ns at L = 1.00 μm.

Mirror image at half a revival

ψ(x, Tᵣₑᵥ/2) = −ψ(L − x, 0)

Halfway through, any packet is its own reflection about L/2, so ⟨x⟩(Tᵣₑᵥ/2) = L − ⟨x⟩(0) whatever the cₙ happen to be.

x and L in m, ψ in m(−1/2). From e(−iπn²) = (−1)ⁿ together with ψₙ(L − x) = (−1)ⁿ⁺¹ψₙ(x); the overall minus is unobservable.

Which pairs of levels make ⟨x⟩ move

xₘₙ = −8Lmn/(π²(m² − n²)²) if m − n is odd, and 0 if m − n is even with m ≠ n

⟨x⟩ beats only when levels of opposite parity are both occupied; same-parity pairs stir the density but leave its mean at L/2.

xₘₙ = ⟨m|x̂|n⟩ carries the units of L, and xₙₙ = L/2; x₁₂ = −16L/(9π²) = −0.1801L, while x₁₄ = −32L/(225π²) = −0.0144L.
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

0.12 Tᵣₑᵥ
0 to 1 Tᵣₑᵥ; steps of 0.01.
45 °
0 to 90 °; steps of 5.
2
2 to 4; steps of 1.

Try this. Drive time to 1.00 and the solid curve lands back exactly on the dashed one: that is the revival. Stop at 0.50 with q = 2 and the packet is the mirror image about L/2. Switch to q = 3 and the filled dot never leaves the centre, because ⟨x⟩ moves only for levels of opposite parity.

Live results

RELATIVE PHASE Δφ
0.72 π rad
⟨x⟩ / L
0.615
⟨H⟩ / E₁, CONSTANT
2.50
BEAT PERIOD, 1 nm
3.67 fs

Results follow this lesson’s model and assumptions. Displayed values are rounded.

Interactive Physics V modelProbability density |ψ(x, t)|² in the box for cos θ |1⟩ + sin θ |q⟩ at θ = 45°, q = 2. Dashed curve: the state at t = 0. Solid curve: the same state at Δφ = 0.72 π rad. Filled dot: ⟨x⟩ = 0.615 L now. Open dot: ⟨x⟩ at t = 0. Dashed vertical: the box centre L/2.|ψ(x, t)|² · levels 1 and 2⟨x⟩ = 0.615 L (filled dot)dashed: t = 0solid: t = 0.12 Tᵣₑᵥx = 0L/2x = L
Probability density |ψ(x, t)|² in the box for cos θ |1⟩ + sin θ |q⟩ at θ = 45°, q = 2. Dashed curve: the state at t = 0. Solid curve: the same state at Δφ = 0.72 π rad. Filled dot: ⟨x⟩ = 0.615 L now. Open dot: ⟨x⟩ at t = 0. Dashed vertical: the box centre L/2.

Predict → change one variable → check

Plan a change before running it. Every other input stays at the saved baseline.

Saved baseline. Time t / Tᵣₑᵥ: 0.12 Tᵣₑᵥ; Mixing angle θ: 45 °; Upper level q: 2

Baseline 0.12; choose 0 to 1, in steps of 0.01.
4. Compared with the saved baseline, this result will…

This experiment is kept only while this page is open. It does not change saved course progress.

4

Catch the common trap

The factor exp(−iEₙt/ℏ) is often dismissed as an unobservable phase, and for a single stationary state that is exactly right: |ψₙ(x, t)|² has no t in it at all. The step that fails is carrying that conclusion into a superposition. Only a phase common to the whole vector is unobservable; here each level turns at its own rate, and the differences exp(−i(Eₙ − Eₘ)t/ℏ) are what the cross terms of |ψ(x, t)|² are built from. Delete them and the density collapses to Σ|cₙ|²|ψₙ(x)|², the incoherent mixture you get after decoherence rather than the state you prepared. The clean test is the zero of energy. Shift the potential by V₀ and every phase picks up the same factor exp(−iV₀t/ℏ), a genuine global phase, and no prediction changes. Shift one level alone and ⟨x⟩(t) changes at once. Global phases are free; relative phases are the physics.

5

Practice & worked examples

3 authored examples. Make an attempt, then reveal the reasoning one step at a time.

EasyExample 1

An electron sits in a rigid box of width L = 1.00 nm, prepared in (|1⟩ + |2⟩)/√2. Find E₁ in eV, the single Bohr frequency this state carries, the period of that beat, and the revival time Tᵣₑᵥ — then confirm that exactly three beats fit into one revival.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

MediumExample 2

The same electron and box, now in (|1⟩ + |2⟩)/√2, with x₁₂ = −16L/(9π²) and p₁₂ = 8iℏ/(3L). Write ⟨x⟩(t), give its extremes in nm and the first time it is maximal, then compute the peak ⟨p⟩ twice — from mₑ d⟨x⟩/dt and from the matrix element directly — and check that Ehrenfest's theorem holds.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

HardExample 3

The same electron in the 1.00 nm box is prepared in (|1⟩ + |2⟩ + |3⟩)/√3. (i) List the Bohr frequencies carried by |ψ(x, t)|² and the smaller set carried by ⟨x⟩(t), and give the period of ⟨x⟩. (ii) Evaluate ⟨x⟩(0) using x₁₂ = −16L/(9π²) and x₂₃ = −48L/(25π²). (iii) Predict ⟨x⟩ at t = 5.50 fs twice — once from your expression, once from the half-revival symmetry — and check the two agree.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 6 steps revealed.

6

Quick check

Select an answer and explain it before checking.

An electron in a rigid box of width L is prepared in (|1⟩ + |3⟩)/√2 and then left undisturbed. What does ⟨x⟩ do?
7

Review & continue

Unitary evolution in the box means every energy amplitude keeps its magnitude and only turns in phase, cₙ(t) = cₙ(0)exp(−iEₙt/ℏ), so a state changes shape solely through the phase differences between the levels it occupies.

  • Explain the relationship without looking at the equation.
  • Check a sign, a unit and a model assumption in your working.
  • Return to the lab or worked example if a step is still unclear.

Extra challenging questions

Connect several ideas, explore supplied data, and compare your reasoning with a worked solution.