
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.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.
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.
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.Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
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.
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
This experiment is kept only while this page is open. It does not change saved course progress.
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.
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.
- E₁ = π²ℏ²/(2mL²). The numerator is 9.8696 × (1.0546 × 10⁻³⁴ J s)² = 1.0976 × 10⁻⁶⁷ J² s² and 2mL² = 1.8219 × 10⁻⁴⁸ kg m², so E₁ = 6.025 × 10⁻²⁰ J = 0.376 eV, and E₂ = 4E₁ = 1.504 eV.
- Only two levels are occupied, so |ψ(x, t)|² carries one Bohr frequency: ω₂₁ = (E₂ − E₁)/ℏ = 3E₁/ℏ = 3 × 6.025 × 10⁻²⁰ ÷ 1.0546 × 10⁻³⁴ = 1.71 × 10¹⁵ rad s⁻¹.
- Its period is T₂₁ = 2π/ω₂₁ = 6.2832 ÷ 1.7139 × 10¹⁵ = 3.67 × 10⁻¹⁵ s = 3.67 fs, which is just h/(3E₁).
- Tᵣₑᵥ = h/E₁ = 6.626 × 10⁻³⁴ ÷ 6.025 × 10⁻²⁰ = 1.100 × 10⁻¹⁴ s = 11.0 fs, and 4mL²/(πℏ) returns the same number.
- 11.0 ÷ 3.67 = 3.00, as it must. The only gap present is 3E₁ and every gap in this box is an integer multiple of E₁, so Tᵣₑᵥ holds a whole number of every beat period: it is the period of the fundamental E₁/ℏ, although no pair of levels beats at E₁/ℏ itself.
Answer. E₁ = 0.376 eV, ω₂₁ = 1.71 × 10¹⁵ rad s⁻¹, beat period 3.67 fs and Tᵣₑᵥ = 11.0 fs — exactly three beats to the revival.
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.
- ⟨x⟩(t) = Σ over m, n of c*ₘcₙ xₘₙ e(−i(Eₙ − Eₘ)t/ℏ). The diagonal gives ½(L/2) + ½(L/2) = L/2, and the one off-diagonal pair contributes 2c₁c₂x₁₂cos(ω₂₁t) = x₁₂cos(ω₂₁t), since 2c₁c₂ = 1.
- x₁₂ = −16L/(9π²) = −16L/88.826 = −0.18013L = −0.180 nm, so ⟨x⟩(t) = 0.500 − 0.180 cos(ω₂₁t) nm with ω₂₁ = 1.7139 × 10¹⁵ rad s⁻¹. The mean swings between 0.320 nm and 0.680 nm.
- ⟨x⟩ is maximal when cos(ω₂₁t) = −1, that is ω₂₁t = π: t = π/ω₂₁ = 1.83 fs, half a beat and one sixth of Tᵣₑᵥ.
- Ehrenfest: ⟨p⟩ = mₑ d⟨x⟩/dt = mₑ(0.18013L)ω₂₁ sin(ω₂₁t), whose peak is 9.109 × 10⁻³¹ × 1.8013 × 10⁻¹⁰ × 1.7139 × 10¹⁵ = 2.81 × 10⁻²⁵ kg m s⁻¹.
- Directly: ⟨p⟩ = 2Re[c₁c₂p₁₂e(−iω₂₁t)] = Re[(8iℏ/3L)(cos ω₂₁t − i sin ω₂₁t)] = (8ℏ/3L)sin(ω₂₁t), and 8 × 1.0546 × 10⁻³⁴ ÷ (3 × 10⁻⁹) = 2.81 × 10⁻²⁵ kg m s⁻¹. The two agree exactly.
Answer. ⟨x⟩(t) = 0.500 − 0.180 cos(ω₂₁t) nm, running between 0.320 nm and 0.680 nm and first maximal at 1.83 fs = Tᵣₑᵥ/6. Both routes give a peak ⟨p⟩ of 2.81 × 10⁻²⁵ kg m s⁻¹.
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.
- Phase the coefficients: cₙ(t) = (1/√3)e(−in²E₁t/ℏ). Writing τ = E₁t/ℏ, the pairs (1,2), (2,3) and (1,3) beat at 3, 5 and 8 in units of E₁/ℏ, and all three appear in |ψ(x, t)|².
- ⟨x⟩ filters that list. Off the diagonal, xₘₙ = 0 unless m − n is odd, so the (1,3) term dies and only 3E₁/ℏ and 5E₁/ℏ survive. Since 3 and 5 share no common factor, ⟨x⟩ repeats only at τ = 2π, the full Tᵣₑᵥ = h/E₁ = 11.0 fs.
- Assemble it. The diagonal gives L/2 and each surviving pair adds 2cₘcₙxₘₙ cos, with cₘcₙ = 1/3: ⟨x⟩/L = 0.5 − (2/3)(0.18013)cos 3τ − (2/3)(0.19454)cos 5τ = 0.5 − 0.12008 cos 3τ − 0.12969 cos 5τ.
- At t = 0 both cosines equal 1, so ⟨x⟩(0) = (0.5 − 0.12008 − 0.12969)L = 0.2502L = 0.250 nm: the packet starts a quarter of the way along the box.
- t = 5.50 fs is Tᵣₑᵥ/2, so τ = π and cos 3π = cos 5π = −1: ⟨x⟩ = (0.5 + 0.12008 + 0.12969)L = 0.7498L = 0.750 nm.
- Cross-check by symmetry: at half a revival every phase is (−1)ⁿ, which sends ψ(x) to −ψ(L − x), so ⟨x⟩(Tᵣₑᵥ/2) = L − ⟨x⟩(0) = (1 − 0.2502)L = 0.7498L. The two routes agree.
Answer. |ψ|² carries 3E₁/ℏ, 5E₁/ℏ and 8E₁/ℏ; ⟨x⟩ carries only 3E₁/ℏ and 5E₁/ℏ and repeats after the full 11.0 fs. ⟨x⟩(0) = 0.250 nm and ⟨x⟩(5.50 fs) = 0.750 nm, the mirror image the half-revival demands.
Quick check
Select an answer and explain it before checking.
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.