University Quantum Mechanics II · Semiclassical Methods: WKB and the Path Integral · 11.2
Time slicing and Feynman's path integral
Feynman's sum over histories is usually waved into existence. Here it is built: cut e(−iHt/ħ) into N slices, pay one commutator per slice to split kinetic from potential energy, and a free kernel per slice turns into exp(iSN/ħ) over zigzag paths. You leave knowing what ∫D[x] means, and what it does not.
Build the model
Connect the measurement to the mechanism.
For a free particle the propagator K(x, t; x₀, 0) = ⟨x|e(−iHt/ħ)|x₀⟩ has a closed form; for H = p²/2m + V(x) it usually has none, because kinetic and potential energy do not commute. Time slicing sidesteps that instead of solving it. Write e(−iHt/ħ) exactly as N factors e(−iHε/ħ) with ε = t/N, split each into e(−iTε/ħ) e(−iVε/ħ) at the price of a term (ε²/2ħ²)[T, V], and insert a position completeness relation between neighbouring factors.
Each factor becomes a free kernel over the time ε times the phase exp(−iεV/ħ), and their product is (m/(2πiħε))(N/2) exp(iSN/ħ), with SN the action of a zigzag path pinned at x₀ and x and free at the N − 1 times between. Integrating over those positions is the sum over paths. Two checks come built in: with V = 0 nothing is dropped and the integrals return the free kernel at every N, and one slice expanded to first order in ε returns the Schrödinger equation.
The literal picture fails in the limit. The N dropped terms total of order ε, and the Trotter product formula, quoted, sends them to zero; but the prefactor diverges and the steps that matter shrink only as √(ħε/m), so the paths are nowhere differentiable, and ∫D[x] e(iS/ħ) is shorthand for the limit of the whole N-fold integral, never an integral over smooth paths.
- Simple definition
- Feynman's path integral is the propagator written as the N → ∞ limit of (m/(2πiħε))(N/2) times an integral over the N − 1 intermediate positions of a path cut into slices of time ε = t/N, each path weighted by exp(iSN/ħ).
- Example
- An electron followed for t = 1.0 fs in N = 100 slices has ε = 10 as and 99 integrated positions. Each slice spreads the path over a Fresnel length √(ħε/m) = 34 pm, a step speed of 3.4 × 10⁶ m s⁻¹; a hundred times as many slices make that speed ten times larger.
Derived by expanding both sides to second order in ε. U(t) = [e(−iHε/ħ)]N is exact, so this term is all that slicing gives up, and it vanishes for a uniform V.
ε = t/N in s; T = p²/2m and V(x) in J; [T, V] = −(iħ/2m)(pV′ + V′p), with V′ = dV/dx
Derived: V is diagonal in position and acts as a phase at xₖ, and momentum completeness turns the kinetic factor into a Fresnel integral. Each slice is an explicit Gaussian.
Kernel in m⁻¹; mass m in kg, xₖ in m, ε in s. Root and kinetic phase: the free kernel at time ε, quoted from the last lesson
Derived by inserting position completeness between the N factors. For V ≠ 0 only the limit is K; each finite N drops N commutator terms of order ε².
K in m⁻¹: the prefactor carries (length)(−N) and the N − 1 integrals (length)(N−1); x₀ and xN stay fixed
Derived: the Lagrangian comes out of the operator algebra, nobody put it in. V at xₖ₊₁ gives the same limit; a vector potential needs the midpoint rule (quoted).
SN in J s: kinetic minus potential energy times ε on each straight leg, with V taken at the leg's start
Derived as a Gaussian moment. Steps of size √(ħε/m) dominate, so expanding ψ to η² is expanding to first order in ε, and that returns the Schrödinger equation.
η = xₖ − xₖ₊₁ in m. The same weight gives ⟨1⟩ = 1 and ⟨η⟩ = 0, and ⟨η⁴⟩ = −3(ħε/m)² is of order ε²
Counting N errors of order ε² gives a total of order ε = t/N, but [T, V] is unbounded for most V, so the count is only a heuristic. This theorem licenses the limit.
A strong operator limit, valid when T + V is essentially self-adjoint on the common domain of T and V; quoted, not derived
Slicing is exact; splitting costs one commutator
Because H commutes with itself, e(−iHt/ħ) = [e(−iHε/ħ)]N with ε = t/N holds exactly for every N. Nothing is gained yet: e(−iHε/ħ) is diagonal in neither position nor momentum. The gain comes from splitting each slice into e(−iTε/ħ), diagonal in momentum, times e(−iVε/ħ), diagonal in position. Expanding both sides to second order in ε shows the price: e(−iHε/ħ) − e(−iTε/ħ)e(−iVε/ħ) = (ε²/2ħ²)[T, V] + O(ε³), with [T, V] = −(iħ/2m)(pV′ + V′p). It vanishes for a uniform potential and grows with the force. For an electron on the atomic scale, where T and V are tens of eV and ħ/(27.2 eV) = 24.2 as, it is of order ½(ε/24.2 as)², about 10⁻³ per slice at ε = 1 as. With V = 0 the split is exact, which the free-particle check relies on.
Insert positions, and every slice becomes a Gaussian
Write K(xN, t; x₀, 0) = ⟨xN|U(t)|x₀⟩ and put a completeness relation ∫dxₖ|xₖ⟩⟨xₖ| between each pair of neighbouring slices, N − 1 of them. Each factor is now ⟨xₖ₊₁|e(−iTε/ħ)e(−iVε/ħ)|xₖ⟩. The potential acts first, on a position eigenstate, so it contributes only the phase exp[−iεV(xₖ)/ħ]. The kinetic factor needs momentum completeness: with ⟨x|p⟩ = e(ipx/ħ)/√(2πħ) from Quantum Mechanics I it becomes ∫dp exp[ip(xₖ₊₁ − xₖ)/ħ − ip²ε/(2mħ)]/(2πħ), a Fresnel integral equal to the last lesson's free kernel at time ε, √(m/(2πiħε)) exp[im(xₖ₊₁ − xₖ)²/(2ħε)]. Multiply the N factors: the roots give (m/(2πiħε))(N/2) and the phases add. Units check: each root is a length⁻¹ and each integral a length, so K ends as length⁻¹, as a one-dimensional position kernel must.
The phases add up to a discretised action
The exponent collected from all N slices is i/ħ times SN = Σ ε[½m((xₖ₊₁ − xₖ)/ε)² − V(xₖ)]. Nobody put a Lagrangian in: kinetic minus potential energy, summed over time, came out of splitting an operator. SN is an ordinary function of the N − 1 intermediate positions, and reading them as a path through the slice times is Feynman's picture: every zigzag from x₀ to xN enters with exp(iSN/ħ), whose modulus is 1. For a free electron taken 1.00 nm in 1.00 fs, the straight path has SN/ħ = m(xN − x₀)²/(2ħt) = 4.32 at every N. Taking V at the end of each leg instead of its start changes SN by exactly ε[V(x₀) − V(xN)], which vanishes in the limit, so that choice is free. For a vector potential, where the term is qA⋅ẋ, it is not, and the midpoint rule is required (quoted).
Two checks: the free particle, and one slice
Set V = 0. Nothing was dropped in the split, and each xₖ integral is a Gaussian that composes free kernels over ε₁ and ε₂ into one over ε₁ + ε₂, prefactor included, so the N − 1 integrals return √(m/(2πiħt)) exp[im(xN − x₀)²/(2ħt)] exactly, at every N. Now keep V and take one slice: ψ(x, t + ε) = √(m/(2πiħε)) ∫dη exp(imη²/(2ħε)) exp(−iεV/ħ) ψ(x + η, t). The weight has ⟨1⟩ = 1, ⟨η⟩ = 0 and ⟨η²⟩ = iħε/m, so the displacements that matter are of size √(ħε/m), and ψ must be expanded to second order in η to reach first order in ε: ψ(x, t + ε) = ψ − (iε/ħ)[−(ħ²/2m)ψ″ + Vψ] + O(ε²). That is iħ ∂ψ/∂t = −(ħ²/2m) ∂²ψ/∂x² + Vψ. The prefactor is no free normalisation: ⟨1⟩ = 1 says the kernel tends to δ(x − x′) as ε → 0.
How fast the errors die, and which ones matter
Each slice drops a term of order ε² and there are N = t/ε slices, so the errors total at most of order ε = t/N and vanish together as N grows. That count is a heuristic, because [T, V] is unbounded for most potentials; the Trotter product formula, quoted, is what guarantees the limit. The count is also pessimistic for the energy levels. The symmetric split exp(−iVε/(2ħ)) e(−iTε/ħ) exp(−iVε/(2ħ)) cancels the ε² term, as the same expansion shows, and errs at ε³ per slice, and N plain slices equal exp(iVε/(2ħ)) [symmetric]N exp(−iVε/(2ħ)). The order-ε error sits in the two end factors, so energies, which a conjugation cannot change, converge as ε². The oscillator shows it exactly, as the Hard example derives: the sliced levels are spaced by ħω̃ with sin(ω̃ε/2) = ωε/2, only 0.04% above ħω at ωε = 0.1.
What the limit means, and what it does not
K = ∫D[x] e(iS/ħ) is shorthand for the limit of the whole N-fold expression, and neither piece has a limit alone. The prefactor |m/(2πħε)|(N/2) diverges. The paths that matter take steps of order √(ħε/m), so their slopes grow as √(ħ/(mε)): for an electron over 1 fs in 100 slices, 34 pm every 10 as, or 3.4 × 10⁶ m s⁻¹, and ten times that with a hundred times more slices. They are continuous but nowhere differentiable, and ∫ẋ² dt is infinite on them. The region they explore does not shrink: completing the square in one intermediate position shows that forcing a free path through one point at time s, straight on either side, costs one radian at a distance √(2ħs(t − s)/(mt)) from the classical path, whatever N is. No countably additive measure D[x] reproduces these integrals, a theorem of R. H. Cameron (1960), quoted. In imaginary time the weights turn real, exp(−SE/ħ), and the limit is a genuine measure, Wiener's, on Brownian paths, also quoted.
Change one variable at a time
Make the relationship visible.
Set the teeth to one Fresnel length and raise N from 2 to 16: each tooth shrinks as 1/√N but steepens as √N, the step speed passes the classical 1.0 × 10⁶ m s⁻¹ at N = 9, and the extra phase grows as 2N − 3, while the lens, fixed by t alone, never moves.
SLICE ε = t/N167 as
FRESNEL LENGTH √(ħε/m)139 pm
ZIGZAG EXTRA PHASE9.00 rad
STEP SPEED √(ħ/(mε))0.83 × 10⁶ m s⁻¹
Live interpretationSLICE ε = t/N: 167 as. FRESNEL LENGTH √(ħε/m): 139 pm. ZIGZAG EXTRA PHASE: 9.00 rad. STEP SPEED √(ħ/(mε)): 0.83 × 10⁶ m s⁻¹
Catch the common trap
Explain before calculating.
The time-sliced formula is applied to a free electron (V = 0) going from x₀ to xN in time t: N slices of ε = t/N, with all N − 1 intermediate integrals done exactly. How does the result compare with the closed-form free propagator √(m/(2πiħt)) exp[im(xN − x₀)²/(2ħt)]?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA free electron (V = 0) goes from x₀ = 0 to x₄ = 1.00 nm in t = 2.00 fs, with the time cut into N = 4 slices. Evaluate SN/ħ for the path through x₁ = 0.60 nm, x₂ = 0.40 nm and x₃ = 1.20 nm, compare it with the straight path, and measure its detours in units of the slice's Fresnel length √(ħε/m).
- Slice and constant: ε = t/N = 0.500 fs. With V = 0, SN/ħ = (m/(2ħε)) Σ(xₖ₊₁ − xₖ)², and m/(2ħε) = 9.109 × 10⁻³¹ kg ÷ (2 × 1.0546 × 10⁻³⁴ J s × 0.500 × 10⁻¹⁵ s) = 8.64 × 10¹⁸ m⁻². Units: kg J⁻¹ s⁻² = m⁻², so SN/ħ is a pure number, as a phase must be.
- The zigzag's increments are +0.60, −0.20, +0.80 and −0.20 nm, with squares summing to 0.36 + 0.04 + 0.64 + 0.04 = 1.08 nm² = 1.08 × 10⁻¹⁸ m². So SN/ħ = 8.64 × 10¹⁸ m⁻² × 1.08 × 10⁻¹⁸ m² = 9.33.
- The straight path has four increments of 0.25 nm, Σ = 0.25 nm², so SN/ħ = 2.16. That equals m(x₄ − x₀)²/(2ħt), the classical free action, and a straight path gives it at every N. The zigzag costs 7.17 rad more.
- Fresnel length: √(ħε/m) = √(1.0546 × 10⁻³⁴ × 5.00 × 10⁻¹⁶ ÷ 9.109 × 10⁻³¹) m = 0.241 nm. Measured from the straight path, which sits at 0.25, 0.50 and 0.75 nm at the three slice times, the detours are +0.35, −0.10 and +0.45 nm: 1.45, 0.42 and 1.87 Fresnel lengths.
- So the zigzag is an ordinary member of the sum, not an outlier: its weight exp(iSN/ħ) has modulus 1 like every other path's, and its detours are the size each slice's Gaussian integral spreads over.
AnswerSN/ħ = 9.33 for the zigzag against 2.16 for the straight path, the classical free action. The Fresnel length is 0.241 nm, so detours of 0.4 to 1.9 Fresnel lengths are typical of the paths the integral sums.
MediumYou plan to time-slice the evolution of hydrogen's 1s electron. Expanding one slice to first order in ε keeps a kinetic term of relative size about (√(ħε/m)/a₀)² = ħε/(ma₀²) and drops the next one, of about (ħε/(ma₀²))², along with a potential term of about (εV/ħ)². Find the largest ε that keeps each dropped term at 0.01, using a₀ = 52.9 pm and V(a₀) = −27.2 eV, the Fresnel length √(ħε/m) at that ε, and the number of slices in one period h/|E₁| of the 1s phase, with E₁ = −13.6 eV.
- Kinetic condition: (ħε/(ma₀²))² ≤ 0.01, so ε ≤ 0.1 ma₀²/ħ. Here ma₀²/ħ = 9.109 × 10⁻³¹ kg × (5.29 × 10⁻¹¹ m)² ÷ 1.0546 × 10⁻³⁴ J s = 2.42 × 10⁻¹⁷ s = 24.2 as, the atomic unit of time; kg m² ÷ J s is indeed s. So ε ≤ 2.42 as.
- Potential condition: (ε|V|/ħ)² ≤ 0.01, so ε|V|/ħ ≤ 0.1 with |V| = 27.2 eV, and ε ≤ 0.1 × 6.582 × 10⁻¹⁶ eV s ÷ 27.2 eV = 2.42 × 10⁻¹⁸ s = 2.42 as.
- The two limits coincide because |V(a₀)| = e²/(4πε₀a₀) = ħ²/(ma₀²) is hydrogen's own energy scale: per slice the kept kinetic and potential terms are both about ε/(24.2 as), and each dropped term is the square of that. At ε = 2.42 as the Fresnel length is √(ħε/m) = √0.1 × a₀ = 0.316a₀ = 16.7 pm. The steps need not be tiny against a₀, because the kept kinetic term is second order in the step and the dropped one fourth order.
- Period of the 1s phase: h/|E₁| = 4.136 × 10⁻¹⁵ eV s ÷ 13.6 eV = 3.04 × 10⁻¹⁶ s = 304 as. Slices per period: 304 as ÷ 2.42 as = 126.
- Limiting case: bind a heavier particle in the same Coulomb field and a₀ shrinks as 1/m, so ma₀²/ħ falls as 1/m and the slices must be thinner still. Near r = 0 the Coulomb V is unbounded, so these are estimates for where the 1s electron mostly is, not bounds.
Answerε ≤ 2.4 as from each condition, and the two coincide because |V(a₀)| = ħ²/(ma₀²). The Fresnel step is then 16.7 pm, a third of a₀, and one 304 as period of the 1s phase takes about 126 slices.
HardModel H₂'s vibration as an oscillator whose quantum matches the measured fundamental, 4161 cm⁻¹, and evolve it with the plain split Uε = e(−iTε/ħ) e(−iVε/ħ), where V = ½μω²x² with μ the reduced mass. Derive the level spacing the sliced evolution produces, evaluate it for ε = 1.00 fs, find the largest ε for which it has bound levels at all, and the ε that brings the fundamental within 1 cm⁻¹.
- Each factor moves x and p exactly in the Heisenberg picture, because its exponent is quadratic and the commutator series stops after one term: e(iVε/ħ) p e(−iVε/ħ) = p − μω²εx and e(iTε/ħ) x e(−iTε/ħ) = x + εp/μ. One slice therefore maps (x, p) to ((1 − ω²ε²)x + εp/μ, p − μω²εx), a matrix of determinant 1 and trace 2 − ω²ε².
- Its eigenvalues are e(±iθ) with cos θ = 1 − ω²ε²/2, that is sin(θ/2) = ωε/2. The matching combination b = αx + βp obeys Uε†bUε = e(−iθ)b, so Uε commutes with b†b and each quantum adds θ to the phase per slice: the levels are spaced by ħω̃ with ω̃ = θ/ε = (2/ε) arcsin(ωε/2). The mass has dropped out.
- Checks: ω̃ has units s⁻¹; as ε → 0, ω̃ → ω; and for small ε, ω̃ ≈ ω(1 + ω²ε²/24), an error of order ε² in the spacing even though the plain split's total error in U is of order ε.
- Numbers: ω = 2πc × 4161 cm⁻¹ = 7.838 × 10¹⁴ s⁻¹, a period of 8.02 fs. At ε = 1.00 fs, ωε = 0.7838, θ = 2 arcsin(0.3919) = 0.8054 and ω̃/ω = 1.0275, so the sliced fundamental is 4276 cm⁻¹, 115 cm⁻¹ above the measured 4161 cm⁻¹. The series gives 1.0256, already off in the third decimal place.
- Stability: arcsin needs ωε ≤ 2. Beyond ε = 2/ω = 2.55 fs the trace falls below −2, the eigenvalues are real, and the sliced oscillator has no bound levels at all: ⟨x⟩ grows exponentially from slice to slice. The split is then not inaccurate but a different system.
- Accuracy: ω²ε²/24 ≤ 1/4161 needs ωε ≤ 0.0759, so ε ≤ 0.0759 ÷ 7.838 × 10¹⁴ s⁻¹ = 9.69 × 10⁻¹⁷ s, about 97 as or 83 slices per vibrational period; the exact formula at that ε gives an error of 1.00 cm⁻¹.
AnswerSpacing ħω̃ with ω̃ = (2/ε) arcsin(ωε/2): at ε = 1.00 fs the fundamental comes out at 4276 cm⁻¹, 2.8% above the measured 4161 cm⁻¹; there are no bound levels for ε above 2.55 fs; 1 cm⁻¹ accuracy needs ε ≤ 97 as.