Building a wave packet and evolving it

1. Each component is multiplied by e^(−iω(k)t) with ω = ħk²/(2m). Dimensions: [ħk²t/m] = (kg m² s⁻¹)(m⁻²)(s)/(kg) = 1, a pure number, as a phase must be.
2. ħ/(2m) = 1.0546 × 10⁻³⁴ J s ÷ (2 × 9.109 × 10⁻³¹ kg) = 5.789 × 10⁻⁵ m² s⁻¹. So ω = 9.26 × 10¹⁴ rad s⁻¹ at 4.0 × 10⁹ m⁻¹, 1.447 × 10¹⁵ rad s⁻¹ at k₀ and 2.084 × 10¹⁵ rad s⁻¹ at 6.0 × 10⁹ m⁻¹.
3. After t = 1.0 fs the three factors have turned through ωt = 0.926, 1.447 and 2.084 rad. Measured from the centre, the lower edge has turned 0.521 rad less and the upper edge 0.637 rad more.
4. Each factor has modulus 1, so |φ(k, t)|² = |φ(k)|² at every k: every momentum probability, and ⟨p⟩ = ħk₀ = 5.3 × 10⁻²⁵ kg m s⁻¹, are exactly as they were. A free particle feels no force, so classically its momentum is conserved too.
5. The phase differences are the dynamics. Their part odd in k − k₀, (0.637 + 0.521)/2 = 0.579 rad at the edges, equals (ħk₀t/m)(k − k₀) and slides the packet by ħk₀t/m = 0.58 nm, the classical vt with v = 5.8 × 10⁵ m s⁻¹. The even part, (0.637 − 0.521)/2 = 0.058 rad, reshapes it and is still small after 1 fs.

Answer. Phases 0.926, 1.447 and 2.084 rad; |φ(k)|² unchanged; the relative phases slide the packet 0.58 nm, at v = 5.8 × 10⁵ m s⁻¹, and barely reshape it.

1. |ψ(x,0)|² = (1/2L)|e^(ik₃x) + e^(ik₅x)|² = (1/L)[1 + cos((k₅ − k₃)x)]. Over one circumference the cosine makes (k₅ − k₃)L/2π = 2 whole turns and integrates to zero, so ∫₀ᴸ |ψ|² dx = 1, and the density carries m⁻¹ as it must.
2. k₃ = 9.42 × 10⁹ m⁻¹ and k₅ = 1.571 × 10¹⁰ m⁻¹, so ω = ħk²/(2m) gives ω₃ = 5.14 × 10¹⁵ and ω₅ = 1.428 × 10¹⁶ rad s⁻¹. Each term takes its own factor: ψ(x, t) = [e^{(i(k₃x − ω₃t))} + e^{(i(k₅x − ω₅t))}]/√(2L), and the weights |c₃|² = |c₅|² = 1/2 never change.
3. |ψ(x, t)|² = (1/L)[1 + cos(Δk x − Δω t)] with Δk = 6.28 × 10⁹ m⁻¹ and Δω = 9.14 × 10¹⁵ rad s⁻¹: two maxima of height 2/L = 1.0 × 10⁹ m⁻¹ that slide rigidly round the ring at Δω/Δk = 1.45 × 10⁶ m s⁻¹.
4. Classical check: ⟨p⟩ = ħ(k₃ + k₅)/2 = 1.33 × 10⁻²⁴ kg m s⁻¹, so ⟨p⟩/m = 1.45 × 10⁶ m s⁻¹, the same speed. The match is exact, because ω = ħk²/(2m) makes (ω₅ − ω₃)/(k₅ − k₃) = ħ(k₃ + k₅)/(2m).
5. The density repeats when Δω t = 2π, after 0.687 fs, having moved 1.00 nm: exactly one fringe spacing, 2π/Δk = L/2.
6. With k₄ added and all three weights 1/3, the pairs (3, 4) and (4, 5) each make fringes of spacing L, moving at ħ(k₃ + k₄)/(2m) = 1.27 × 10⁶ and ħ(k₄ + k₅)/(2m) = 1.64 × 10⁶ m s⁻¹, so the density changes shape as it goes. Two components cannot show this: a straight line fits any two points of ω(k). The whole state returns only after T = mL²/(πħ) = 11.0 fs.

Answer. Normalised; |ψ|² = (1/L)[1 + cos(Δk x − Δω t)] slides rigidly at 1.45 × 10⁶ m s⁻¹ = ⟨p⟩/m and repeats every 0.687 fs; with k₄ added, fringes move at 1.27 and 1.64 × 10⁶ m s⁻¹ and the shape changes.

1. φ(k) = (1/√(2π))(2a)⁻¹⁄² πa sech(πa(k − k₀)/2) = (√(πa)/2) sech(πa(k − k₀)/2), in m¹⁄². With ∫ sech²(bq) dq = 2/b and b = πa/2, ∫|φ|² dk = (πa/4)(4/(πa)) = 1, as Plancherel requires of a normalised ψ.
2. Weight above k₀ + Q: (πa/4) ∫ from Q to ∞ of sech²(bq) dq = ½(1 − tanh(bQ)) = 1/(1 + e^(πaQ)). Below 10⁻⁶ needs πaQ ≥ 13.8, so Q ≥ 4.40 × 10⁹ m⁻¹, π/Δx ≥ 1.44 × 10¹⁰ m⁻¹ and Δx ≤ 0.218 nm. Take Δx = 0.20 nm: π/Δx = 1.571 × 10¹⁰ m⁻¹, and the weight above it is 1.6 × 10⁻⁸.
3. The centre moves at ħk₀/m = 1.16 × 10⁶ m s⁻¹, so 11.6 nm in 10 fs. At t = 0 less than 10⁻⁶ lies beyond |x| = 7.3 nm, since 1 − tanh(X/a) ≈ 2e^(−2X/a). Components at k₀ ± Q gain or lose ħQt/m = 5.1 nm on the centre by 10 fs. Even the slow band edge, k₀ − Q = 5.6 × 10⁹ m⁻¹, moves right, so the packet never reaches left of −7.3 nm, while the right edge reaches about 7.3 + 11.6 + 5.1 ≈ 24 nm: a safe estimate keeps the packet between −7.3 nm and +24 nm.
4. L = 51.2 nm, from −25.6 to +25.6 nm, holds that range, and N = L/Δx = 256 = 2⁸. Each transform costs about N log₂N = 2048 operations against N² = 6.6 × 10⁴ for a direct sum, and there are no time steps: one multiplication by e^{(−iħkₙ²t/(2m))} reaches 10 fs. The run gives ⟨x⟩ = 11.6 nm, with σₓ growing from 0.91 nm to 1.13 nm.
5. With Δx = 0.30 nm the top wavenumber π/Δx = 1.047 × 10¹⁰ m⁻¹ sits only 4.7 × 10⁸ m⁻¹ above k₀, and 1/(1 + e^(1.48)) = 0.185 of the weight lies beyond it. The grid folds that weight to k − 2π/Δx, near −1.0 × 10¹⁰ m⁻¹, so about a fifth of the packet runs backwards at about 1.2 × 10⁶ m s⁻¹. By 10 fs some 18% of the probability sits left of x = −3 nm, where the true packet has essentially none.
6. The norm cannot see this: the discrete transform is unitary, so Σ|ψⱼ|²Δx = 1 on both grids. The classical check does: on the coarse grid ⟨k⟩ comes out near 6 × 10⁹ m⁻¹ instead of 1.0 × 10¹⁰ m⁻¹, and the centroid reaches about 7.1 nm at 10 fs instead of ħk₀t/m = 11.6 nm. So does a convergence test: halving Δx to 0.15 nm puts the centroid back at 11.6 nm and removes the backward lobe.

Answer. φ = (√(πa)/2) sech(πa(k − k₀)/2), normalised; Δx ≤ 0.218 nm, so 0.20 nm, with L = 51.2 nm and N = 256. Δx = 0.30 nm sends about a fifth of the weight backwards; the norm stays 1, but ⟨x⟩ (7.1 nm, not 11.6 nm) and a convergence test expose it.