3.2.2 Schrödinger Equation#

Prompts

How does taking the limit \(\delta t \to 0\) in the path integral formula transform it into a differential equation?
What role do Gaussian integrals play in deriving the Laplacian term? Why does the kinetic energy lead to \(\frac{\partial^2}{\partial x^2}\)?
In what sense is the Schrödinger equation derived rather than postulated? What assumptions underlie the small-\(\delta t\) expansion?
How does the propagator \(K(x, t; x', 0)\) encode the same physics as the wavefunction evolution \(\psi(x, t) = \int K \, \psi_0(x') \, \mathrm{d}x'\)?

Lecture Notes#

Overview#

The Schrödinger equation is traditionally postulated as a fundamental axiom. Here we derive it from the path integral by taking the limit \(\delta t \to 0\) in the time-slice propagator formula. This reveals that the two formulations of quantum mechanics—the path integral (global sum over histories) and the Schrödinger equation (local differential equation)—are mathematically equivalent. The propagator emerges as the Green’s function of the Schrödinger equation.

The Time-Slice Propagator Revisited#

Recall from §3.2.1 that for a small time interval \(\delta t\), the propagator factorizes as

\[ \psi(x, t + \delta t) = \int K_{\delta t}^{\text{free}}(x, x') \, \mathrm{e}^{-\mathrm{i}V(x') \delta t / \hbar} \, \psi(x', t) \, \mathrm{d}x' \]

where the free-particle (kinetic) kernel is

(65)#\[ K_{\delta t}^{\text{free}}(x, x') = \left(\frac{m}{2\pi \mathrm{i}\hbar \delta t}\right)^{1/2} \exp\left(\frac{\mathrm{i}m(x - x')^2}{2\hbar \delta t}\right) \]

To derive the Schrödinger equation, we expand both sides in powers of \(\delta t\) and identify the \(O(\delta t)\) term.

Step 1: Gaussian Integral and Taylor Expansion#

Substitute \(\delta x = x - x'\). The kinetic phase

\[ \exp\left(\frac{\mathrm{i} m(\delta x)^2}{2\hbar \delta t}\right) \]

oscillates rapidly unless \(|\delta x| \sim \sqrt{\hbar \delta t / m}\), a width that shrinks to zero as \(\delta t \to 0\). This allows us to Taylor-expand \(\psi(x', t)\) around \(x'\):

\[ \psi(x' + \delta x, t) = \psi(x', t) + \delta x \cdot \frac{\partial}{\partial x} \psi(x', t) + \frac{1}{2}(\delta x)^2 \frac{\partial^2 \psi}{\partial x^2} + \ldots \]

Changing variables to \(x' = x - \delta x\):

\[ \psi(x, t + \delta t) = \int \mathrm{d}(\delta x) \, K_{\delta t}^{\text{free}}(\delta x) \left[ \psi + \delta x \cdot \frac{\partial}{\partial x} \psi + \frac{1}{2}(\delta x)^2 \partial_x^2 \psi \right] \mathrm{e}^{-\mathrm{i}V(x) \delta t / \hbar} \]

where \(K_{\delta t}^{\text{free}}(\delta x) = (m / 2\pi \mathrm{i}\hbar \delta t)^{1/2} \exp(\mathrm{i}m(\delta x)^2 / 2\hbar \delta t)\).

Step 2: Gaussian Moments#

Key Gaussian Integrals

For the normalized Gaussian kernel \(K_{\delta t}^{\text{free}}(\delta x) = \left(\frac{m}{2\pi\mathrm{i}\hbar\,\delta t}\right)^{1/2} \exp\!\left(\frac{\mathrm{i}m(\delta x)^2}{2\hbar\,\delta t}\right)\):

\[ \int K_{\delta t}^{\text{free}}(\delta x) \, \mathrm{d}(\delta x) = 1 \]

\[ \int K_{\delta t}^{\text{free}}(\delta x) \, \delta x \, \mathrm{d}(\delta x) = 0 \quad \text{(odd integrand)} \]

\[ \int K_{\delta t}^{\text{free}}(\delta x) \, (\delta x)^2 \, \mathrm{d}(\delta x) = \frac{\mathrm{i}\hbar\,\delta t}{m} \]

Derivation: Gaussian Moments

Write \(\alpha = \mathrm{i}m/(2\hbar\,\delta t)\) so that \(K_{\delta t}^{\text{free}}(\delta x) = (\alpha/\pi)^{1/2}\,\mathrm{e}^{\alpha(\delta x)^2}\).

The zeroth moment is normalized by construction: \(\int K_{\delta t}^{\text{free}}\,\mathrm{d}(\delta x) = 1\).

The first moment vanishes by symmetry (odd integrand).

For the second moment, use the standard Gaussian results \(\int \mathrm{e}^{\alpha x^2}\,\mathrm{d}x = \sqrt{\pi/(-\alpha)}\) and \(\int x^2\,\mathrm{e}^{\alpha x^2}\,\mathrm{d}x = \tfrac{1}{2}(-\alpha)^{-3/2}\sqrt{\pi}\):

\[ \int K_{\delta t}^{\text{free}}(\delta x)\,(\delta x)^2\,\mathrm{d}(\delta x) = \frac{1}{-2\alpha} = \frac{\mathrm{i}\hbar\,\delta t}{m} \]

Step 3: Collecting O(δt) Terms#

Using the moments from Step 2:

The zeroth-order term: \(\int K_{\delta t}^{\text{free}} \psi \, \mathrm{e}^{-\mathrm{i}V \delta t/\hbar} \approx \psi(1 - \mathrm{i}V \delta t/\hbar) = \psi - (\mathrm{i}V \delta t/\hbar) \psi\)
The first-moment term vanishes: \(\int K_{\delta t}^{\text{free}} \, \delta x \cdot \frac{\partial}{\partial x} \psi \, \ldots = 0\)
The second-moment term: \(\int K_{\delta t}^{\text{free}} \, (\delta x)^2 \, \partial_i \partial_j \psi \, \ldots = -\frac{\mathrm{i}\hbar \delta t}{m} \frac{\partial^2}{\partial x^2} \psi\)

Therefore:

\[ \psi(x, t + \delta t) - \psi(x, t) = -\frac{\mathrm{i}\hbar \delta t}{m} \frac{\partial^2}{\partial x^2} \psi - \frac{\mathrm{i}}{\hbar} V(x) \delta t \, \psi + O((\delta t)^2) \]

Dividing by \(\delta t\) and taking \(\delta t \to 0\):

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

Multiplying by \(\mathrm{i}\hbar\):

The Schrödinger Equation

\[ \mathrm{i}\hbar \frac{\partial \psi}{\partial t} = \left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x)\right) \psi = \hat{H} \psi \]

The Schrödinger equation is derived from the path integral, not postulated. The kinetic term produces the Laplacian \(\frac{\partial^2}{\partial x^2}\), and the potential appears as a multiplicative factor.

Two Equivalent Formulations#

The derivation reveals a deep equivalence:

Path Integral ↔ Schrödinger Equation

Quantum mechanics admits two mathematically equivalent formulations:

Aspect	Path Integral	Schrödinger Equation
Perspective	Global: sum over all histories	Local: differential equation
Central Object	Propagator \(K = \sum \mathrm{e}^{\mathrm{i}S/\hbar}\)	Wavefunction \(\psi(x, t)\)
Equation Satisfied	Schrödinger equation with delta-function initial condition	Schrödinger equation with arbitrary initial state

The propagator \(K(x, t_f; x', t_i)\) is the Green’s function of the Schrödinger equation:

\[ \mathrm{i}\hbar \frac{\partial K}{\partial t_f} = \hat{H}_{x_f} K(x_f, t_f; x', t_i) \]

with the initial condition \(K(x_f, t_i; x', t_i) = \delta(x_f - x')\).

The wavefunction at any time is obtained by convolving the propagator with the initial state:

\[ \psi(x, t) = \int K(x, t; x', 0) \, \psi(x', 0) \, \mathrm{d}x' \]

This shows that solving the Schrödinger equation is equivalent to evaluating the path integral.

Discussion

The derivation assumed that the \(\delta t \to 0\) limit can be taken order-by-order, with Gaussian integrals evaluated exactly at each step. Several questions arise:

Is this limiting procedure always justified mathematically, or are there boundary cases where it breaks down?
Starting from the Schrödinger equation, can we uniquely reconstruct the path integral?
For time-dependent Hamiltonians \(\hat{H}(t)\), how does the path integral formulation change?
Does the choice of splitting (kinetic vs potential factor ordering) affect the \(\delta t \to 0\) limit?

Summary#

The Schrödinger equation emerges from the path integral via small-time-step expansion: each \(\delta t\) slice contributes a kinetic Gaussian and a potential phase.
The three-step derivation shows: (1) Taylor expand the wavefunction, (2) evaluate Gaussian moments, (3) collect \(O(\delta t)\) terms.
The kinetic energy produces the Laplacian \(\frac{\partial^2}{\partial x^2}\); the potential acts as a multiplicative factor.
The path integral (global, sum over histories) and the Schrödinger equation (local, differential) are mathematically equivalent.
The propagator is the Green’s function of the Schrödinger equation and solves it with a delta-function source.

Homework#

1. Explain why the kinetic phase \(\exp(\mathrm{i}m(x - x')^2 / 2\hbar \delta t)\) in the propagator oscillates rapidly as \(|x - x'|\) increases. For what range of displacements \(|x - x'|\) is the integrand significant? How does this range shrink as \(\delta t \to 0\)?

2. Verify the Gaussian integrals:

\(\int_{-\infty}^{\infty} \mathrm{e}^{\mathrm{i}ax^2} \mathrm{d}x = \sqrt{\pi / (-\mathrm{i}a)}\) for \(a > 0\)
Use this to compute \(\int K_{\delta t}^{\text{free}}(\delta x) \, (\delta x)^2 \, \mathrm{d}(\delta x)\) where \(K_{\delta t}^{\text{free}}\) is the kinetic propagator. Show that the result is \(-\mathrm{i}\hbar \delta t \, 1 / m\).

3. Starting from \(\psi(x, t + \delta t) = \int K_{\delta t}^{\text{free}} \, \mathrm{e}^{-\mathrm{i}V \delta t/\hbar} \psi(x', t) \mathrm{d}x'\), derive the Schrödinger equation by expanding to first order in \(\delta t\). Identify which Gaussian moment (zeroth, first, second) contributes to each term (kinetic, potential).

4. Suppose instead of \(V(x')\) in the action, we used \(V(x)\) (the endpoint) or \(V((x + x')/2)\) (the midpoint). Does the first-order Schrödinger equation change? Argue whether the choice matters in the \(\delta t \to 0\) limit.

5. Show that the Schrödinger equation \(\mathrm{i}\hbar \partial_t \psi = \hat{H} \psi\) with Hermitian \(\hat{H}\) conserves probability. Derive the continuity equation \(\partial_t \rho + \frac{\partial}{\partial x} \cdot j = 0\) for \(\rho = |\psi|^2\) and identify the probability current \(j(x, t)\).

6. The propagator \(K(x, t_f; x', t_i)\) satisfies \(\mathrm{i}\hbar \partial_{t_f} K = \hat{H}_{x_f} K\) with \(K(x, t_i; x', t_i) = \delta(x - x')\). Verify that this is consistent with \(\psi(x, t) = \int K(x, t; x', 0) \psi(x', 0) \mathrm{d}x'\). Why does this justify calling \(K\) the Green’s function?

7. In the path integral derivation, the kinetic Gaussian has width \(\sim \sqrt{\hbar \delta t / m}\). What is the physical meaning of this length scale? How does it relate to the de Broglie wavelength \(\lambda = h/p\)?

8. The path integral sums over all trajectories (global picture), while the Schrödinger equation describes evolution at each point (local picture). Explain how these perspectives are reconciled by the derivation. Can you recover the path integral starting from the Schrödinger equation?