University Quantum Mechanics II · Semiclassical Methods: WKB and the Path Integral · 11.1

The propagator, and the free particle's in closed form

Every way you have evolved a state so far began by expanding it. The propagator does that work once: a single function of two points and a time that carries any initial wavefunction forward. Here you build it, derive the free particle's exactly, and find classical mechanics already written into its phase.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

For a time-independent Hamiltonian the evolution operator U(t) = e^(−iHt/ħ) is known formally and useless until it is put in a basis. The propagator puts it in the position basis: K(x, t; x′, 0) = ⟨x|U(t)|x′⟩ is the amplitude to be found at x a time t after release exactly at x′, and inserting position completeness turns ψ(x, t) into one integral of K against ψ(x′, 0), so K is found once and serves every initial state. Each of its properties is a property of U rewritten: it obeys the Schrödinger equation in x and starts as δ(x − x′), it composes by an integral over an intermediate point because U(t₁ + t₂) = U(t₂)U(t₁), and it equals Σ ψₙ(x)ψₙ*(x′)e^{(−iEₙt/ħ)} because U is diagonal in energy.

For a free particle U is diagonal in momentum instead, and momentum completeness leaves one Gaussian integral, which converges once t is given a small negative imaginary part. The result, √(m/(2πiħt)) e^{(im(x − x′)²/(2ħt))}, is exact: its modulus is the same at every x, and its phase is the classical action of the straight path over ħ, whose slope is the classical momentum. The exactness is also the limit.

Only p²/2m enters, so K₀ reaches every x at once, with local momenta above mc beyond |x − x′| = ct; and beyond quadratic potentials a closed form is the exception, which is why the rest of this unit builds approximations.

Simple definition: The propagator K(x, t; x′, 0) = ⟨x|e^(−iHt/ħ)|x′⟩ is the amplitude for a particle released exactly at x′ to be found at x a time t later; integrating it against any initial wavefunction gives the wavefunction at t.
Example: For an electron 1.0 fs after release, the free propagator has |K₀|² = m/(2πħt) = 1.37 nm⁻² at every x, while its phase m x²/(2ħt) is 4.3 rad at 1 nm from the source and 17 rad at 2 nm: all the position dependence is in the phase.

Evolution by one integralψ(x, t) = ∫ K(x, tx′, 0) ψ(x′, 0) dx′

Derived by inserting ∫|x′⟩⟨x′| dx′ = 1. Find K once and every initial state is carried forward by the same integral.

K = ⟨x|e^(−iHt/ħ)|x′⟩ in m⁻¹ in one dimension; ψ in m^(−1/2); x, x′ in m; t in s

The problem that defines Kiħ ∂K/∂t = −(ħ²/2m) ∂²K/∂x² + V(x) K, with K → δ(x − x′) as t → 0⁺

The δ start makes K the response to a perfectly localised release, and the pair fixes K uniquely: anything passing both tests is K.

From iħ ∂U/∂t = HU and U(0) = 1; x′ is a fixed label in m, not a variable of the equation

Composition over an intermediate pointK(x, t₁ + t₂x′, 0) = ∫ K(x, t₂x″, 0) K(x″, t₁x′, 0) dx″

It is U(t₁ + t₂) = U(t₂)U(t₁) with completeness inserted between. Sliced N times, it becomes the next topic's path integral.

Exact for time-independent H; x″ in m runs over the whole line; t₁, t₂ in s

Spectral expansionK(x, tx′, 0) = Σₙ ψₙ(x) ψₙ*(x′) e^{(−iEₙt/ħ)}

Any solved spectrum hands you K at once. At t = 0 the sum is the completeness relation, which is the δ start again.

ψₙ normalised eigenfunctions of H in m^(−1/2), Eₙ in J; a continuous spectrum turns the sum into an integral

Free propagator, exactK₀(x, tx′, 0) = √(m/(2πiħt)) exp[i m(x − x′)²/(2ħt)]

One Gaussian integral over p. The exponent is iS_cl/ħ with S_cl = m(x − x′)²/(2t), the straight path's action, and |K₀| is the same at every x.

m in kg, ħ in J s, t in s and positive; √i = e^(iπ/4), the branch fixed by t → t − iε

Gaussian spreading, from K₀σ(t) = σ₀ √(1 + (ħt/(2mσ₀²))²)

One more Gaussian integral of K₀ against Quantum Mechanics I's packet. At fixed t the narrowest final width, √(ħt/m), needs σ₀ = √(ħt/(2m)).

σ₀, σ(t) the standard deviations of |ψ|² in m; τ = 2mσ₀²/ħ in s is when spreading takes over

From the evolution operator to a kernel

Quantum Mechanics I gave |ψ(t)⟩ = U(t)|ψ(0)⟩ with U(t) = e^(−iHt/ħ). Project onto ⟨x| and insert ∫|x′⟩⟨x′| dx′ = 1 in front of the initial state: ψ(x, t) = ∫⟨x|U(t)|x′⟩ ψ(x′, 0) dx′. That matrix element is the propagator K(x, t; x′, 0). Read the integral as superposition: each point x′ of the initial wavefunction sends out the wavelet K, weighted by ψ(x′, 0), and the wavelets add. Units check it. Because ⟨x|x′⟩ = δ(x − x′), K carries m⁻¹ in one dimension, and ψ(x, t) comes out in m^(−1/2) as it must. K is not itself a state: it grows from a δ-function and, like one, is not square-integrable. Two exact identities follow from unitarity at the cost of one line each: K(x, t; x′, 0)* = K(x′, −t; x, 0), and ∫K(x, t; x″, 0)* K(x, t; x′, 0) dx = δ(x″ − x′), which is U†U = 1 written in the position basis.

Four properties, each a property of U

Everything K does is something U does, written in position space. Since iħ ∂U/∂t = HU, and ⟨x|H acts as −(ħ²/2m)∂²/∂x² + V(x) on whatever follows it, K obeys the Schrödinger equation in x at fixed x′. Since U(0) = 1, K → ⟨x|x′⟩ = δ(x − x′) as t → 0⁺, and that equation with that start fixes K uniquely. Since U(t₁ + t₂) = U(t₂)U(t₁), inserting completeness at the intermediate time gives composition: an integral over every intermediate point x″, not a product at one of them. Since U is diagonal in the energy basis, inserting Σ|n⟩⟨n| gives K = Σ ψₙ(x)ψₙ*(x′)e^{(−iEₙt/ħ)}, which at t = 0 is the completeness of the ψₙ and so the δ start again. All four are exact, derived for any time-independent H; a time-dependent H keeps composition but needs a time-ordered U. For the free particle the spectrum is a continuum of momentum states, and the sum becomes the integral of the next step.

The free propagator is one Gaussian integral

With V = 0 the Hamiltonian p̂²/2m is diagonal in momentum, so insert ∫|p⟩⟨p| dp with ⟨x|p⟩ = e^(ipx/ħ)/√(2πħ): K₀ = (1/(2πħ)) ∫ exp[−itp²/(2mħ) + ip(x − x′)/ħ] dp. This is ∫e^{(−ap² + bp)} dp = √(π/a) e^(b²/(4a)) with a = it/(2mħ) and b = i(x − x′)/ħ, but that formula needs a positive real part, and here a is purely imaginary: the integrand has modulus 1 and the integral does not converge absolutely. Replace t by t − iε, ε → 0⁺. Then Re a = ε/(2mħ) is positive, a factor e^{(−εp²/(2mħ))} damps large momenta, and the limit exists. Completing the square gives b²/(4a) = im(x − x′)²/(2ħt) and √(π/a)/(2πħ) = √(m/(2πiħt)), with √i = e^(iπ/4) chosen by the same ε. Checks: √(m/(ħt)) has units m⁻¹; in iħ ∂K₀/∂t = −(ħ²/2m)∂²K₀/∂x² the 1/t and (x − x′)²/t² terms balance separately; ∫K₀ dx = 1 at every t; and as t → 0⁺ it is a normalised Gaussian of complex width √(iħt/m), which tends to δ(x − x′). The result is exact, derived, not approximate.

Its phase is the classical action, and its slope the momentum

A free particle that leaves x′ at time 0 and reaches x at time t moves at v = (x − x′)/t, so its action is S_cl = ∫½mv² dt = m(x − x′)²/(2t). The exponent of K₀ is exactly iS_cl/ħ, so all of the propagator's x-dependence is the straight path's action in units of ħ, while |K₀|² = m/(2πħt) is flat. The slope of the phase, (1/ħ)∂S_cl/∂x = m(x − x′)/(ħt), is p_cl/ħ, so the local wavelength at x is h/p_cl, the de Broglie wavelength of the one classical trajectory making that trip in time t: fast components arrive far away, slow ones stay near x′. For an electron after 2.0 fs, at 2.0 nm S_cl/ħ = 8.64 rad and the local wavelength is 0.727 nm, which is h/(mv) at 1.0 × 10⁶ m s⁻¹. That reading needs S_cl ≫ ħ, several radians of phase, which is the semiclassical condition of this unit. For a beam crossing a distance L at speed v along a second axis, t = L/v turns the phase into k(x − x′)²/(2L) with k = mv/ħ: the paraxial Fresnel kernel of optics. That the prefactor follows from S_cl too, as √(−∂²S_cl/∂x∂x′/(2πiħ)), is quoted here and taken up in the stationary-phase topic.

One more Gaussian integral recovers the spreading law

Feed in the Gaussian ψ(x′, 0) = (2πσ₀²)^(−1/4) e^{(−x′²/(4σ₀²))}. The integrand's exponent, −x′²/(4σ₀²) + im(x − x′)²/(2ħt), has an x′² coefficient with a negative real part of its own, so no ε is needed now. Completing the square, done in full in the Hard example, leaves |ψ(x, t)|² Gaussian with σ(t)² = σ₀² + (ħt/(2mσ₀))², Quantum Mechanics I's law σ₀√(1 + (t/τ)²) with τ = 2mσ₀²/ħ, reached this time without Fourier transforming the state. The two terms name their sources: σ₀ is the initial size, and ħ/(2σ₀) is the momentum spread, which adds a width σₚt/m in quadrature. For t ≫ τ that second term wins and the packet's shape becomes its momentum distribution with x read as pt/m, the far-field limit of K₀ and the reason time-of-flight images of released atoms measure momenta. An electron packet with σ₀ = 0.50 nm has τ = 4.32 fs and reaches 1.26 nm after 10 fs; a 1 μg grain with σ₀ = 1 μm has τ = 1.9 × 10¹³ s, about 6 × 10⁵ years.

Checks, and where the closed form stops

Dimensions: √(m/(ħt)) is m⁻¹ and m(x − x′)²/(ħt) is a pure number. Limits: as t → 0⁺, K₀ → δ(x − x′); as m → ∞ at fixed t the phase varies ever faster away from x′ and nothing spreads, the classical limit. No coupling constant appears, so there is no order in α to check: the corrections to K₀ are relativistic, of relative size (p/mc)², not radiative. K₀ is non-zero at every x for every positive t, and its local momentum m(x − x′)/t passes mc once |x − x′| exceeds ct, so p²/2m is being used where it is wrong. In the figure's window, 3 nm after 1 fs, v = 3.0 × 10⁶ m s⁻¹ and (v/c)² = 1.0 × 10⁻⁴, harmless. With √(p²c² + m²c⁴) the amplitude outside the light cone is exponentially small over a Compton wavelength, a result quoted from relativistic quantum mechanics. Measured check: the free phase ħk²t/(2m) that the spectral form attaches to each momentum is what recoil atom interferometers read; Parker and colleagues (2018) turned it into h/m for caesium and, with the Rydberg constant and the caesium-to-electron mass ratio, α⁻¹ = 137.035999046(27).

Change one variable at a time

Make the relationship visible.

Interactive model

Time since release t2.0 fs

Marked point x = d2.0 nm

Width of a launched packet σ₀0.50 nm

Put σ₀ at 0.20 nm and drag t from 1 to 10 fs: the flat envelope sinks as 1/√t, the chirp relaxes, and the narrow packet ends at 2.90 nm, wider than a 1.00 nm one (1.16 nm). Then slide d outward: the local wavelength shrinks as ht/(md).

|K₀|² AT EVERY x0.687 nm⁻²

PHASE S_cl/ħ AT x = d8.64 rad

LOCAL λ AT x = d0.727 nm

PACKET WIDTH σ(t)0.55 nm

Live interpretation|K₀|² AT EVERY x: 0.687 nm⁻². PHASE S_cl/ħ AT x = d: 8.64 rad. LOCAL λ AT x = d: 0.727 nm. PACKET WIDTH σ(t): 0.55 nm

Catch the common trap

Explain before calculating.

Misconception check

The commonest wrong picture treats the free propagator as a bump that starts at x′ and spreads like heat from a point source, so that a particle released at x′ is most likely still near x′. The free Schrödinger equation does look like the diffusion equation ∂u/∂t = D ∂²u/∂x², but with D = iħ/(2m) imaginary, and that i turns the heat kernel's decaying e^{(−(x − x′)²/(4Dt))} into a pure phase. |K₀|² = m/(2πħt) is the same at every x: a δ-function contains every momentum with equal weight, so after any positive t every distance has been reached with the same probability density. The spreading bump belongs to a normalisable initial state, such as a Gaussian of width σ₀ whose momentum spread is a finite ħ/(2σ₀); K itself is not a state, and ∫|K₀|² dx diverges.

An electron is released from x′ = 0 in the idealised δ-function state. According to the free propagator, what is the local wavelength of its oscillations at x = 3.0 nm, a time t = 2.0 fs after release?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyTest K₀ on a state whose evolution you already know. Propagate the plane wave ψ(x′, 0) = e^(ikx′) with the free propagator, and evaluate the phase its time factor turns through in 1.00 fs for an electron with k = 10.0 nm⁻¹.

Write ψ(x, t) = √(m/(2πiħt)) ∫ exp[im(x − x′)²/(2ħt) + ikx′] dx′ and shift to u = x′ − x, which pulls out e^(ikx) and leaves ∫ exp[−au² + iku] du with a = −im/(2ħt).
With t → t − iε, Re a is positive and the Gaussian formula applies: √(π/a) = √(2πiħt/m), which cancels the prefactor exactly.
The remaining factor is e^{((ik)²/(4a))} = e^{(−iħk²t/(2m))}, so ψ(x, t) = e^{(i(kx − ωt))} with ω = ħk²/(2m): K₀ reproduces the free dispersion relation, as it must.
Numbers: ħ/(2m) = 0.05788 nm² fs⁻¹ for an electron, so ω = 0.05788 nm² fs⁻¹ × 100 nm⁻² = 5.79 fs⁻¹ = 5.79 × 10¹⁵ rad s⁻¹, an energy ħω = 3.81 eV.
Check: the phase velocity ω/k = 5.79 × 10⁵ m s⁻¹ is half the group velocity ħk/m = 1.16 × 10⁶ m s⁻¹, as quadratic dispersion requires; ħk/(mc) = 0.004, so p²/2m is safe.

Answerψ(x, t) = e^{(i(kx − ωt))} with ω = ħk²/(2m) = 5.79 × 10¹⁵ rad s⁻¹; after 1.00 fs the phase factor e^(−iωt) has turned through 5.79 rad.

MediumA ⁴⁰Ca⁺ ion (m = 6.636 × 10⁻²⁶ kg) moves along the axis of a Paul trap, a harmonic well with ω = 2π × 1.00 MHz. Use the spectral expansion of K to show that, for any initial state, a quarter period later the position wavefunction is the initial momentum wavefunction with x read as p/(mω). The ion starts in the ground state, and at t = 0 a short laser pulse gives it a velocity v₀ = 0.200 m s⁻¹: when is that quarter period, where is the ion centred then, and how wide is it?

Quote the oscillator from Quantum Mechanics I: real eigenfunctions ψₙ(x) with Eₙ = (n + ½)ħω, and, because a† = (mωx − ip)/√(2mħω) keeps its form in the momentum representation up to a factor −i, momentum eigenfunctions φₙ(p) = (−i)ⁿ ψₙ(p/(mω))/√(mω). The spectral form is K = Σ ψₙ(x)ψₙ(x′) e^{(−i(n + ½)ωt)}.
At t = π/(2ω) every phase is e^(−iπ/4)(−i)ⁿ, and (−i)ⁿψₙ(x) = √(mω) φₙ(mωx). So K = e^(−iπ/4)√(mω) Σ φₙ(mωx)ψₙ(x′), and completeness, Σ⟨p|n⟩⟨n|x′⟩ = ⟨p|x′⟩ = e^{(−ipx′/ħ)}/√(2πħ), sums it: K(x, π/(2ω); x′, 0) = √(mω/(2πħ)) e^(−iπ/4) e^{(−imωxx′/ħ)}.
Then ψ(x, π/(2ω)) = ∫ K ψ(x′, 0) dx′ = e^(−iπ/4)√(mω) φ(mωx, 0), exact for every state: position now displays the initial momentum, scaled by 1/(mω). Units: √(mω/ħ) is m⁻¹ for K, and √(mω) φ is m^(−1/2) for ψ.
Numbers: the quarter period is π/(2ω) = 1/(4 × 1.00 MHz) = 250 ns. The kick multiplies the ground state by e^(imv₀x/ħ), so φ(p, 0) = φ₀(p − mv₀), and the ion is centred at x = mv₀/(mω) = v₀/ω = 0.200 m s⁻¹ ÷ 6.283 × 10⁶ s⁻¹ = 31.8 nm.
Its width is σₚ/(mω), with σₚ = ħ/(2σ₀) and σ₀ = √(ħ/(2mω)) = 11.2 nm; since ħ = 2mωσ₀², that is σ₀ again, 11.2 nm. Check: the classical orbit x(t) = (v₀/ω) sin ωt reaches v₀/ω at the quarter period, as Ehrenfest's theorem requires, and the kicked ground state keeps its shape.
Further on, at t = π/ω the phases are −i(−1)ⁿ and ψₙ(−x) = (−1)ⁿψₙ(x), so ψ(x, π/ω) = −iψ(−x, 0), the mirror image; at 2π/ω every phase is −1. Both are exact only because the levels are exactly evenly spaced.

Answerψ(x, π/(2ω)) = e^(−iπ/4)√(mω) φ(mωx, 0), exactly. For the ion the quarter period is 250 ns; it is then centred at v₀/ω = 31.8 nm and still σ₀ = 11.2 nm wide.

HardA ⁸⁷Rb atom (m = 1.443 × 10⁻²⁵ kg) is cooled to the ground state of an optical tweezer with trap frequency ω = 2π × 20.0 kHz along a horizontal x, then released. Propagate the state with K₀ to find σ(t), show the spreading time is 1/ω, find σ after 100 μs of flight, and find the trap frequency that would give the narrowest cloud after that flight.

Quote the oscillator ground state: a Gaussian with σ₀² = ħ/(2mω), so σ₀ = √(1.0546 × 10⁻³⁴ ÷ (2 × 1.443 × 10⁻²⁵ × 1.257 × 10⁵)) m = 53.9 nm.
ψ(x, t) ∝ ∫ exp[−Ax′² + iB(x − x′)²] dx′ with A = 1/(4σ₀²) and B = m/(2ħt). The x′² coefficient A − iB has a positive real part, so the integral converges without ε, and completing the square leaves the exponent iABx²/(A − iB).
Its real part is −AB²x²/(A² + B²), so σ(t)² = 1/(4A) + A/(4B²) = σ₀² + (ħt/(2mσ₀))². The spreading time is τ = 2mσ₀²/ħ = 1/ω = 7.96 μs, so σ(t) = σ₀√(1 + ω²t²).
At t = 100 μs, ωt = 12.57 and σ = 53.9 nm × √158.9 = 680 nm. That is essentially the velocity spread times t: ħ/(2mσ₀) = 6.78 mm s⁻¹, times 100 μs, is 678 nm.
At fixed t, dσ²/dσ₀ = 0 gives σ₀² = ħt/(2m), that is ω = 1/t = 1.00 × 10⁴ s⁻¹, or 1.59 kHz, and a minimum width √(ħt/m) = 0.270 μm.
So the 20 kHz trap, tighter by a factor 12.6, ends 2.5 times wider. Limits check: t → 0 returns σ₀, and ħ → 0 removes the spreading.

Answerσ(t) = σ₀√(1 + ω²t²) with σ₀ = 53.9 nm and τ = 1/ω = 7.96 μs; after 100 μs, σ = 0.680 μm. A 1.59 kHz trap (ω = 1/t) would give the minimum, √(ħt/m) = 0.270 μm.