University Physics V · Particle in a Box · 5.8
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.
Build the model
Connect the measurement to the mechanism.
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 gaps 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.
- Simple definition
- 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⟩ swings between 0.320 nm and 0.680 nm every 3.67 fs.
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.
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.
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.
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.
Halfway through, any packet is its own reflection about L/2, so ⟨x⟩(Tᵣₑᵥ/2) = L − ⟨x⟩(0) whatever the cₙ happen to be.
From e(−iπn²) = (−1)ⁿ together with ψₙ(L − x) = (−1)ⁿ⁺¹ψₙ(x); the overall minus is unobservable.
⟨x⟩ beats only when levels of opposite parity are both occupied; same-parity pairs stir the density but leave its mean at L/2.
xₙₙ = L/2; x₁₂ = −16L/(9π²) = −0.1801L, while x₁₄ = −32L/(225π²) = −0.0144L.
Û(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, which is exactly what unitary means in an orthonormal basis. 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**2 * 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.
What 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.
The 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. An anharmonic potential has gaps in no integer ratio, so its beats never re-phase and the motion is merely quasi-periodic; here every beat is a harmonic of one frequency, and a harmonic series has a period.
Why 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 is periodic with the same period. 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.
Which 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, 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: m 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.
What the exact revival costs
The revival is a property of this spectrum, not of quantum mechanics. It needs every 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.
Change one variable at a time
Make the relationship visible.
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.
RELATIVE PHASE Δφ0.72 π rad
⟨x⟩ / L0.615
⟨H⟩ / E₁, CONSTANT2.50
BEAT PERIOD, 1 nm3.67 fs
Live interpretationRELATIVE PHASE Δφ: 0.72 π rad. ⟨x⟩ / L: 0.615. ⟨H⟩ / E₁, CONSTANT: 2.50. BEAT PERIOD, 1 nm: 3.67 fs
Catch the common trap
Explain before calculating.
An electron in a rigid box of width L is prepared in (|1⟩ + |3⟩)/√2 and then left undisturbed. What does ⟨x⟩ do?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn 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.
- E₁ = π²ℏ²/(2mL²). The numerator is 9.8696 × (1.0546 × 10⁻³⁴ J s)² = 1.0976 × 10⁻⁶⁷ 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, every gap in this box is an integer multiple of E₁, and the revival is therefore the slowest beat of them all.
AnswerE₁ = 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.
MediumThe 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 m d⟨x⟩/dt and from the matrix element directly — and check that Ehrenfest's theorem holds.
- ⟨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⟩ = m d⟨x⟩/dt = m(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.
HardThe 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.
- 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. 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.