3.2.3 Free Particle Propagator

3.2.3 Free Particle Propagator#

Prompts

  • What is a plane wave solution to the free Schrödinger equation? How does it relate to momentum eigenstates?

  • How can we build a propagator by superposing plane waves over all momenta?

  • Evaluate the Gaussian integral over \(k\) to find the closed-form free-particle propagator. What is the normalization factor?

  • Identify the classical action for a free particle in the exponent of the propagator. Does this match the Phase = Action theme from §3.1.3?

  • How does the free-particle propagator close the circle: from the path integral proposal ψ ∼ exp(iS/ℏ) in §3.1, through formal path integral quantization in §3.2.1–2, to exact realization in §3.2.3?

Lecture Notes#

Overview#

This section solves the Schrödinger equation for a free particle (\(V = 0\)) and builds the propagator from plane wave solutions. By superposing plane waves over all momenta and evaluating a Gaussian integral in momentum space, we recover the exact propagator:

\[K_\text{free}(x, t; x_0, 0) = \left(\frac{m}{2\pi \mathrm{i}\hbar t}\right)^{1/2} \exp\left(\frac{\mathrm{i}m}{2\hbar t}(x - x_0)^2\right)\]

The exponent contains the classical action for motion from \(x_0\) to \(x\) in time \(t\). This confirms the central theme: Phase = Action is not metaphor—it is exact for the free particle.

Plane Wave Solution#

The free Schrödinger equation (with \(V = 0\)) is:

\[\mathrm{i}\hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \psi\]

A natural ansatz is a plane wave:

\[\psi(x, t) = \mathrm{e}^{\mathrm{i}(kx - \omega t)}\]

Substitute into the Schrödinger equation

\[\mathrm{i}\hbar (-\mathrm{i}\omega) \mathrm{e}^{\mathrm{i}(kx - \omega t)} = -\frac{\hbar^2}{2m} (-k^2) \mathrm{e}^{\mathrm{i}(kx - \omega t)}\]
\[\hbar\omega = \frac{\hbar^2 k^2}{2m}\]

Plane Wave Dispersion Relation

\[\omega(k) = \frac{\hbar k^2}{2m}\]

A plane wave with wave vector \(k\) oscillates as \(\exp(\mathrm{i}(kx - \omega t))\) and carries energy \(E = \hbar\omega = \frac{\hbar^2 k^2}{2m} = \frac{p^2}{2m}\) (momentum eigenstate with \(p = \hbar k\)).

Building the Propagator from Plane Waves#

The key insight: the propagator is a superposition of all plane wave solutions. A particle can take any momentum, so we sum over all \(k\):

\[K_\text{free}(x, t; x_0, 0) = \int \frac{\mathrm{d}k}{(2\pi)^d} \, \mathrm{e}^{\mathrm{i}(k(x - x_0) - \omega(k) t)}\]
\[= \int \frac{\mathrm{d}k}{(2\pi)^d} \exp\left(\mathrm{i}k(x - x_0) - \mathrm{i}\frac{\hbar k^2}{2m} t\right)\]

This is exactly the form anticipated by the path integral: the phase is \(S/\hbar\) where \(S\) is the action, here summed over all possible momenta.

Note: Plane Waves as Momentum Eigenstates

Each plane wave with momentum \(p = \hbar k\) is a momentum eigenstate \(\vert p \rangle = (2\pi\hbar)^{-d/2} \mathrm{e}^{\mathrm{i}px/\hbar}\). The propagator is the momentum-space integral of the projector \(\vert p \rangle \langle p \vert\) evolved by the time operator \(\mathrm{e}^{-\mathrm{i}\hat{H}t/\hbar}\) with \(\hat{H} = \hat{p}^2/(2m)\). This connection to spectral decomposition will be formalized later.

Evaluating the Gaussian Integral in Momentum Space#

The integral over \(k\) is Gaussian; completing the square and shifting variables yields \(K_\text{free}\) in closed form.

The Free-Particle Propagator

\[K_\text{free}(x, t; x_0, 0) = \left(\frac{m}{2\pi \mathrm{i}\hbar t}\right)^{1/2} \exp\left(\mathrm{i}\frac{m}{2\hbar t}(x - x_0)^2\right)\]

This agrees with the one-step free kernel (65) from §3.2.2: set the elapsed time to \(\delta t\) (so \(t \to \delta t\)) and identify endpoints as in time-slicing (\(x_0 \leftrightarrow x'\)). The Gaussian prefactor and phase then match exactly—self-consistency between the short-time path-integral factor and the exact momentum superposition.

Derivation: Gaussian Evaluation

The exponent is quadratic in \(k\):

\[\mathrm{i}k(x - x_0) - \mathrm{i}\frac{\hbar k^2}{2m} t = -\mathrm{i}\frac{\hbar t}{2m}\left(k^2 - \frac{2m}{\hbar t}k(x - x_0)\right)\]

Complete the square in \(k\):

\[k^2 - \frac{2m}{\hbar t}k(x - x_0) = \left(k - \frac{m}{\hbar t}(x - x_0)\right)^2 - \frac{m^2}{(\hbar t)^2}(x - x_0)^2\]

The \(k\) integral becomes

\[K_\text{free} = \int \frac{\mathrm{d}k}{(2\pi)^d} \exp\left(-\mathrm{i}\frac{\hbar t}{2m}\left[\left(k - \frac{m}{\hbar t}(x - x_0)\right)^2 - \frac{m^2}{(\hbar t)^2}(x - x_0)^2\right]\right)\]

The integral over the shifted quadratic term is a standard Gaussian:

\[\int \mathrm{d}k \exp\left(-\mathrm{i}\frac{\hbar t}{2m} k'^2\right) = \left(\frac{\pi m}{-\mathrm{i}\hbar t/2}\right)^{1/2} = \left(\frac{2\pi m}{\mathrm{i}\hbar t}\right)^{1/2}\]

(where \(k' = k - m(x - x_0)/(\hbar t)\) and we use \(\sqrt{-\mathrm{i}} = (1-\mathrm{i})/\sqrt{2}\)).

The remaining exponent is

\[\exp\left(\mathrm{i}\frac{\hbar t}{2m} \cdot \frac{m^2}{(\hbar t)^2}(x - x_0)^2\right) = \exp\left(\mathrm{i}\frac{m}{2\hbar t}(x - x_0)^2\right)\]

Combining,

\[K_\text{free} = \frac{1}{(2\pi)^d} \left(\frac{2\pi m}{\mathrm{i}\hbar t}\right)^{1/2} \exp\left(\mathrm{i}\frac{m}{2\hbar t}(x - x_0)^2\right) = \left(\frac{m}{2\pi \mathrm{i}\hbar t}\right)^{1/2} \exp\left(\mathrm{i}\frac{m}{2\hbar t}(x - x_0)^2\right)\]

Phase = Classical Action#

The exponent can be rewritten by recognizing the classical action. A particle traveling from \(x_0\) to \(x\) in time \(t\) with constant velocity has action:

\[S_\text{cl} = \int_0^t \frac{m}{2}\dot{x}^2 \, \mathrm{d}t' = \frac{m}{2} v^2 t = \frac{m}{2t}(x - x_0)^2\]

Thus the exponent in the propagator is precisely \(\mathrm{i}S_\text{cl}/\hbar\):

\[K_\text{free} = \left(\frac{m}{2\pi \mathrm{i}\hbar t}\right)^{1/2} \exp\left(\mathrm{i}\frac{S_\text{cl}}{\hbar}\right)\]

Phase = Action Realized

The free-particle propagator embodies the central principle: the quantum amplitude is the exponential of the classical action divided by \(\hbar\).

The normalization prefactor \((m/2\pi \mathrm{i}\hbar t)^{1/2}\) is the “quantum correction” to the amplitude. The exponent \(\mathrm{i}S_\text{cl}/\hbar\) is pure classical action. Together they form the complete quantum propagator.

This is the exact realization of the ansatz from §3.1.3: \(\psi \sim \exp(\mathrm{i}S/\hbar)\) is not metaphorical—for the free particle it is mathematically exact.

Closing the Circle#

Recall the structure of unit 3.2:

  • §3.1: Proposed ψ ∼ exp(iS/ℏ) as the guiding principle

  • §3.2.1: Formalized the path integral as ∫ 𝒟x exp(iS/ℏ)

  • §3.2.2: Derived the Schrödinger equation from the path integral (initial value problem)

  • §3.2.3 (now): Solved the Schrödinger equation for the free particle and found K = (normalization) × exp(iS_cl/ℏ)

The propagator closes the loop: the path integral (which sums all classical actions) produces a result whose exponent is exactly the classical action. No approximation. No stationary-phase argument yet. This is exact.

In §3.3 we will use the stationary-phase approximation to extend this insight to general potentials: the dominant classical path contributions to the integral will emerge.

Discussion: Why Is the Factor i Important?

The exponent is \(+\mathrm{i}S_\text{cl}/\hbar\), not \(-\mathrm{i}S_\text{cl}/\hbar\) (as in the path integral formulation itself, which has \(\mathcal{S}/\hbar\) with \(\mathcal{S}\) defined with a +i sign in the path integral definition). This is crucial:

  • A particle with momentum \(p = m v\) travels distance \(\Delta x = v t = (p/m) t\), so \(S_\text{cl} = p \Delta x / 2 = m (\Delta x)^2 / (2t) > 0\).

  • With exponent \(\mathrm{i}S_\text{cl}/\hbar > 0\), the phase oscillates: \(\exp(\mathrm{i}\theta)\) with \(\theta = m(\Delta x)^2 / (2\hbar t) \gg 1\) for macroscopic systems. This oscillation suppresses interference from distant paths (destructive interference).

  • If the sign were reversed, far paths would constructively interfere—wrong physics.

The +i is the quantum signature of stationary-phase dominance.

Summary#

  • Plane wave: \(\psi = \exp(\mathrm{i}(kx - \omega t))\) with \(\omega = \hbar k^2/(2m)\) solves the free Schrödinger equation; each plane wave is a momentum eigenstate.

  • Propagator superposition: Sum all plane waves over momentum to build the propagator. This is exactly the path integral idea: paths have different momenta.

  • Gaussian evaluation: Complete the square in momentum space; the integral yields a Gaussian in position with normalization \((m/2\pi \mathrm{i}\hbar t)^{1/2}\).

  • Classical action in exponent: The exponent equals \(\mathrm{i}S_\text{cl}/\hbar\) where \(S_\text{cl}\) is the classical action for constant-velocity motion. Phase = Action is exact.

  • Closing the circle: From the ansatz ψ ∼ exp(iS/ℏ) in §3.1, through path integral quantization in §3.2.1–2, to exact realization in §3.2.3.

See Also

Homework#

1. Plane Wave Verification

Verify directly that the plane wave \(\psi(x, t) = \exp(\mathrm{i}(kx - \omega t))\) satisfies the free Schrödinger equation \(\mathrm{i}\hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} \psi\) if and only if \(\omega = \hbar k^2 / (2m)\). What is the energy and momentum of this eigenstate?

2. Momentum Superposition

A general solution to the free Schrödinger equation is a superposition of plane waves:

\[\psi(x, t) = \int \frac{\mathrm{d}k}{(2\pi)^d} \phi(k) \exp\left(\mathrm{i}(kx - \omega(k) t)\right)\]

What is \(\phi(k)\) for the free-particle propagator \(K(x, t; x_0, 0)\) acting on a delta-function initial condition \(\psi(x, 0) = \delta(x - x_0)\)?

3. Completing the Square in Momentum

Show that:

\[k(x - x_0) - \frac{\hbar k^2}{2m} t = -\frac{\hbar t}{2m}\left(k - \frac{m}{\hbar t}(x - x_0)\right)^2 + \frac{m}{2\hbar t}(x - x_0)^2\]

What do the shifted and constant terms represent physically?

4. Gaussian Integral in d Dimensions

The standard result for a multi-dimensional Gaussian is:

\[\int \mathrm{d}k \exp\left(-\mathrm{i}\frac{\hbar t}{2m} k^2\right) = \left(\frac{2\pi m}{\mathrm{i}\hbar t}\right)^{1/2}\]

Use this to derive the prefactor \((m/2\pi \mathrm{i}\hbar t)^{1/2}\) in the free-particle propagator. (Hint: extract the \(1/(2\pi)^d\) from the momentum integration measure.)

5. Normalization and Spreading

The probability density magnitude of the propagator is \(\vert K \vert^2 \propto (m/\hbar t)^d\). In 1D, show that \(\vert K \vert^2 \propto t^{-1}\). What does this scaling tell you about how the amplitude spreads as the wavepacket evolves?

6. Classical Action for Free Particle

For a particle moving with constant velocity from \(x_0\) to \(x\) in time \(t\), the classical action is \(S_\text{cl} = \frac{m}{2t}(x - x_0)^2\). Verify that this equals \(pq\) where \(p = m(x - x_0)/t\) is the momentum and \(q = x - x_0\) is the displacement.

7. Exponent and Quantum Oscillations

For a macroscopic object (\(m = 1\) kg) traveling 1 meter in 1 second, compute the exponent \(S_\text{cl}/\hbar\) in the propagator. For an electron (\(m \approx 10^{-30}\) kg) traveling 1 nanometer in 1 picosecond, compute the same quantity. Which case shows rapid quantum oscillations? Why?

8. Delta-Function Initial Condition

The propagator \(K(x, t; x_0, 0)\) should satisfy the initial condition:

\[\lim_{t \to 0^+} K(x, t; x_0, 0) = \delta(x - x_0)\]

The exponent becomes \(\mathrm{i}m(x - x_0)^2 / (2\hbar t) \to \infty\) as \(t \to 0^+\). Explain why this rapid oscillation produces the delta function. (Hint: Consider how the oscillations cancel except near \(x = x_0\).)