3.2.2 Schrödinger Equation#

Prompts

Why does the slice integral \(\psi(x,t+\delta t) = \int K_{\delta t}\,\psi(x',t)\,\mathrm{d}x'\) produce only an infinitesimal change in \(\psi\) as \(\delta t\to 0\)?
Compute the Gaussian moments \(M_0\), \(M_1\), \(M_2\) of the slice kinetic kernel. Why do only these three contribute at order \(\delta t\)?
The slice ansatz carries an undetermined amplitude \(A(\delta t)\). What self-consistency condition fixes it, and what value does it take?
Derive the free Schrödinger equation from the small-\(\delta t\) expansion. In what sense is this a derivation rather than a postulate?
How does adding a potential \(V(x)\) modify the slice phase and produce the full Schrödinger equation \(\mathrm{i}\hbar\,\partial_t\psi = \hat{H}\psi\)?

Lecture Notes#

Overview#

In §3.1 we met the path-integral idea as a global principle: quantum amplitudes are organized by phase = action and summed over histories. In §3.2.2 we now turn that abstract principle into practical machinery. We analyze one infinitesimal time slice, then connect the slices in a controlled limit, and each step sharpens the same physical picture: local wave evolution emerges from interference of nearby paths. The purpose is singular and concrete: derive the Schrödinger equation — the working equation we use to solve quantum-mechanical problems — directly from the path-integral formulation rather than introducing it as an independent postulate.

Setting Up One Slice#

Combining the wave-evolution rule (63) with the slice ansatz (68), the wavefunction one slice later is

(71)#\[ \psi(x,t+\delta t) = A(\delta t)\int\exp\!\left[\frac{\mathrm{i}\,m\,(x-x')^{2}}{2\hbar\,\delta t}\right]\psi(x',t)\,\mathrm{d}x'. \]

Physical picture. The amplitude at the observation point \((x,t+\delta t)\) is built by collecting contributions from every source point \(x'\) on the previous wavefront \(\psi(\cdot,t)\), propagated to \(x\) by the slice propagator. Different sources interfere through their relative phases. To make the geometry transparent, parameterize a source by its displacement from \(x\),

\[ x' = x - \delta x, \]

so \(\delta x\) is the (signed) distance the wave travels in time \(\delta t\). The slice integral becomes

(72)#\[ \psi(x,t+\delta t) = A(\delta t)\int\mathrm{e}^{\mathrm{i}\alpha(\delta x)^{2}}\,\psi(x-\delta x,t)\,\mathrm{d}(\delta x), \]

with the kinetic-phase coefficient

(73)#\[ \alpha = \frac{m}{2\hbar\,\delta t}. \]

For small \(\delta t\), \(\alpha\) is large, so \(\mathrm{e}^{\mathrm{i}\alpha(\delta x)^{2}}\) oscillates rapidly except where the displacement is small — only sources within roughly \(\vert\delta x\vert\lesssim 1/\sqrt{\alpha}\sim\sqrt{\hbar\,\delta t/m}\) contribute coherently. The slowly varying \(\psi\) may therefore be Taylor-expanded around \(x\).

Step 1 — Taylor Expand the Source Wave#

To second order (we will see only the \(n\le 2\) moments survive in the limit),

(74)#\[ \psi(x-\delta x,t) = \psi(x,t) - \delta x\,\partial_{x}\psi(x,t) + \frac{1}{2}(\delta x)^{2}\,\partial_{x}^{2}\psi(x,t) + O((\delta x)^{3}). \]

Substitute into (72). The integral splits term by term into Gaussian moments,

(75)#\[ \psi(x,t+\delta t) = A(\delta t)\Bigl[\,M_{0}\,\psi - M_{1}\,\partial_{x}\psi + \tfrac{1}{2}M_{2}\,\partial_{x}^{2}\psi + O(\delta t^{\,2})\Bigr], \]

with

(76)#\[ M_{n} \;\equiv\; \int_{-\infty}^{\infty}(\delta x)^{n}\,\mathrm{e}^{\mathrm{i}\alpha(\delta x)^{2}}\,\mathrm{d}(\delta x). \]

Step 2 — Compute the Gaussian Moments#

Three short calculations give all we need.

Gaussian moments

(77)#\[\begin{split} \begin{split} M_{0} &= \sqrt{\frac{\mathrm{i}\pi}{\alpha}} = \sqrt{\frac{2\pi\mathrm{i}\hbar\,\delta t}{m}},\\ M_{1} &= 0,\\ M_{2} &= \frac{\mathrm{i}\hbar\,\delta t}{m}\,M_{0}. \end{split} \end{split}\]

Derivation: \(M_{0}\) from the 2D polar trick

Start from the real Gaussian \(I = \int_{-\infty}^{\infty}\mathrm{e}^{-x^{2}}\,\mathrm{d}x\). Square it and pass to polar coordinates:

\[\begin{split} \begin{split} I^{2} &= \int\!\!\int\mathrm{e}^{-(x^{2}+y^{2})}\,\mathrm{d}x\,\mathrm{d}y\\ &= \int_{0}^{2\pi}\mathrm{d}\theta\int_{0}^{\infty}\mathrm{e}^{-r^{2}}\,r\,\mathrm{d}r = 2\pi\cdot\tfrac{1}{2} = \pi, \end{split} \end{split}\]

so \(I = \sqrt{\pi}\). By scaling, \(\int\mathrm{e}^{-ax^{2}}\,\mathrm{d}x = \sqrt{\pi/a}\). Analytically continuing to imaginary \(a = -\mathrm{i}\alpha\) (rotating the contour by \(\pi/4\) keeps the integral convergent for \(\alpha>0\)),

\[ M_{0} = \sqrt{\frac{\pi}{-\mathrm{i}\alpha}} = \sqrt{\frac{\mathrm{i}\pi}{\alpha}}. \]

Derivation: \(M_{1}\) by parity

The integrand \(\delta x\,\mathrm{e}^{\mathrm{i}\alpha(\delta x)^{2}}\) is odd under \(\delta x\mapsto-\delta x\), while the integration domain \((-\infty,\infty)\) is symmetric. Hence \(M_{1} = 0\). Every odd moment vanishes for the same reason.

Derivation: \(M_{2}\) by the derivative trick

Differentiate \(M_{0}(\alpha)\) with respect to \(\alpha\) to lower an extra \((\delta x)^{2}\) from the exponent:

\[ \frac{\partial M_{0}}{\partial\alpha} = \mathrm{i}\int(\delta x)^{2}\,\mathrm{e}^{\mathrm{i}\alpha(\delta x)^{2}}\,\mathrm{d}(\delta x) = \mathrm{i}\,M_{2}. \]

From \(M_{0} = \sqrt{\mathrm{i}\pi/\alpha}\), \(\partial M_{0}/\partial\alpha = -M_{0}/(2\alpha)\). Therefore

\[ M_{2} = \frac{1}{\mathrm{i}}\,\frac{\partial M_{0}}{\partial\alpha} = \frac{\mathrm{i}\,M_{0}}{2\alpha} = \frac{\mathrm{i}\hbar\,\delta t}{m}\,M_{0}. \]

Step 3 — Self-Consistency Fixes \(A(\delta t)\)#

Substitute the moments (77) into (75):

(78)#\[ \psi(x,t+\delta t) = A(\delta t)\,M_{0}\,\Bigl[\,\psi(x,t) + \frac{\mathrm{i}\hbar\,\delta t}{2m}\,\partial_{x}^{2}\psi(x,t) + O(\delta t^{\,2})\,\Bigr]. \]

Take \(\delta t\to 0\). The right-hand side must reduce to \(\psi(x,t)\), which forces \(A(\delta t)\,M_{0} = 1\):

Slice normalization (free particle)

(79)#\[ A(\delta t) = \frac{1}{M_{0}} = \sqrt{\frac{m}{2\pi\mathrm{i}\hbar\,\delta t}}. \]

The normalization is not postulated. It is fixed by the requirement that a slice of zero duration leave the wavefunction unchanged.

The free slice propagator is therefore the closed-form expression

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

The Free Schrödinger Equation#

With \(A\,M_{0} = 1\), the small-\(\delta t\) expansion (78) simplifies to

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

Divide by \(\delta t\), take \(\delta t\to 0\), and multiply through by \(\mathrm{i}\hbar\):

Free Schrödinger equation

(81)#\[ \mathrm{i}\hbar\,\partial_{t}\psi(x,t) = -\frac{\hbar^{2}}{2m}\,\partial_{x}^{2}\psi(x,t). \]

Historically this equation entered physics as an oracle from the sky. Here it has been derived from the same principle phase = action that organized §3.1.3. Wave mechanics is not an extra postulate; it is the local content of the global phase = action.

Adding the Potential#

For a particle in a potential the Lagrangian is \(L = \tfrac{1}{2}m\dot x^{2} - V(x)\), so the slice action picks up an additional \(-V(x)\,\delta t\) (HW 3.2.1.4). The slice propagator therefore acquires an additional phase factor:

(82)#\[ K_{\delta t}(x,x') = K_{\delta t}^{\text{free}}(x,x')\;\exp\!\left[-\frac{\mathrm{i}\,V(x)\,\delta t}{\hbar}\right]. \]

Because the kinetic Gaussian peaks at \(\delta x = 0\), the choice of where to evaluate \(V\) within the slice (endpoint, midpoint, or starting point) only changes the slice action by \(O(\delta t^{\,2})\) and does not affect the limit. Denote the kinetic-only result (78) (now with \(A\) fixed) by

(83)#\[ \psi_{\text{kin}}(x,t+\delta t) \;\equiv\; \psi(x,t) + \frac{\mathrm{i}\hbar\,\delta t}{2m}\,\partial_{x}^{2}\psi(x,t). \]

The full slice evolution is then \(\psi_{\text{kin}}\) multiplied by the potential phase and expanded to first order:

\[ \psi(x,t+\delta t) = \Bigl[1 - \frac{\mathrm{i}\,V(x)\,\delta t}{\hbar} + O(\delta t^{\,2})\Bigr]\,\psi_{\text{kin}}(x,t+\delta t). \]

Multiplying out and keeping \(O(\delta t)\),

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

Dividing by \(\delta t\), taking \(\delta t\to 0\), and multiplying by \(\mathrm{i}\hbar\) gives the result:

Schrödinger equation

(84)#\[ \mathrm{i}\hbar\,\partial_{t}\psi(x,t) = \hat{H}\,\psi(x,t),\qquad \hat{H} = -\frac{\hbar^{2}}{2m}\,\partial_{x}^{2} + V(x). \]

The kinetic energy emerges from the second Gaussian moment of the slice propagator; the potential energy emerges from the extra slice phase \(-V(x)\,\delta t/\hbar\). The Hamiltonian operator \(\hat{H}\) is the small-\(\delta t\) generator of the propagator.

Poll: what fixes the slice normalization?

In the derivation above the slice normalization \(A(\delta t) = \sqrt{m/(2\pi\mathrm{i}\hbar\,\delta t)}\) was determined by demanding a specific consistency condition. Which condition is it?

(A) Probability conservation: \(\int\vert\psi(x,t+\delta t)\vert^{2}\,\mathrm{d}x = 1\) at every step.

(B) Composition of two slices reproduces a slice of double duration: \(K_{2\delta t} = K_{\delta t}\circ K_{\delta t}\).

(D) The classical limit \(\hbar\to 0\) should reproduce Newton’s equation \(m\ddot x = -V'(x)\).

Summary#

One slice → PDE. Taylor-expanding the source wavefunction inside the slice integral reduces the time evolution at order \(\delta t\) to a sum of Gaussian moments.
Three moments suffice. Only \(M_{0}\), \(M_{1} = 0\), and \(M_{2}\) contribute at order \(\delta t\) ().
Self-consistency fixes \(A\). Demanding \(\psi(x,0) = \psi(x,0)\) at \(\delta t = 0\) forces \(A(\delta t) = 1/M_{0}\) (); the slice propagator is then complete.
Free Schrödinger follows from the second moment.
Adding \(V(x)\) appends a slice phase \(-V(x)\,\delta t/\hbar\); the full Schrödinger equation follows, and the Hamiltonian is read off as the infinitesimal generator.

Homework#

1. Zeroth Gaussian moment. Use the polar-coordinate trick to derive \(\int\mathrm{e}^{-r^{2}}\mathrm{d}^{2}r = \pi\) in the plane.

(a) From this, derive \(\int\mathrm{e}^{-bx^{2}}\,\mathrm{d}x = \sqrt{\pi/b}\) for \(\operatorname{Re}b>0\) by separating the 2D integral and rescaling.

(b) Analytically continue \(b\to-\mathrm{i}\alpha\) with \(\alpha>0\) (rotate the contour by \(\pi/4\) and check convergence) to obtain \(M_{0} = \sqrt{\mathrm{i}\pi/\alpha}\).

2. Second Gaussian moment. Starting from \(M_{0}(\alpha) = \sqrt{\mathrm{i}\pi/\alpha}\), use the derivative trick \(\partial_{\alpha}M_{0} = \mathrm{i}M_{2}\) to compute \(M_{2}\) and verify the result \(M_{2} = (\mathrm{i}\hbar\,\delta t/m)\,M_{0}\) stated in (77).

3. Verifying the slice normalization. For the closed-form free slice propagator (80), evaluate \(\int K_{\delta t}^{\text{free}}(x,x')\,\mathrm{d}x'\) by Gaussian integration. Show that the result equals \(1\), independent of \(x\) and \(\delta t\). What does this say about how a constant initial wavefunction evolves over one slice?

4. Composition of two slices. Compute

\[ K_{2\delta t}^{\text{free}}(x,x'') \;=\; \int K_{\delta t}^{\text{free}}(x,x')\,K_{\delta t}^{\text{free}}(x',x'')\,\mathrm{d}x' \]

by completing the square in \(x'\).

(a) Show that the resulting kernel has the same functional form as \(K_{\delta t}^{\text{free}}\) with \(\delta t\) replaced by \(2\delta t\).

(b) Comment on what this verifies about the slice propagator (cf. HW 3.2.1.6).

5. Probability conservation. Starting from (84) with real \(V(x)\), derive the continuity equation \(\partial_{t}\rho + \partial_{x}j = 0\) with \(\rho = \vert\psi\vert^{2}\) and \(j = (\hbar/m)\operatorname{Im}(\psi^{*}\partial_{x}\psi)\).

(a) Compute \(\partial_{t}\vert\psi\vert^{2}\) using the Schrödinger equation and its complex conjugate.

(b) Identify the probability current \(j\) and verify \(\partial_{t}\rho + \partial_{x}j = 0\).

(c) Integrate over all \(x\), assuming \(\psi\to 0\) at infinity, to conclude that \(\frac{\mathrm{d}}{\mathrm{d}t}\int\vert\psi\vert^{2}\,\mathrm{d}x = 0\).

6. Higher moments do not matter. The Taylor expansion (74) produces moments \(M_{n}\) for \(n = 0, 1, 2, 3, 4, \ldots\).

(a) Use the derivative trick repeatedly to show that \(M_{2k} = c_{k}\,M_{0}\,\alpha^{-k}\) for some dimensionless constants \(c_{k}\).

(b) Substituting \(\alpha = m/(2\hbar\,\delta t)\), conclude that \(A\,M_{2k} = O(\delta t^{\,k})\). Why does the small-\(\delta t\) limit therefore truncate exactly at the second moment?

7. Wavepacket spreading time scale. A free particle is initially localized to a position uncertainty \(\sigma\).

(a) Use the de Broglie / Heisenberg estimate \(\sigma_{p}\sim\hbar/\sigma\) to argue that the typical velocity uncertainty is \(v\sim\hbar/(m\sigma)\).

(b) Estimate the time \(\tau\) at which the wavepacket has spread to roughly twice its initial width.

(c) Evaluate \(\tau\) for an electron with \(\sigma = 1\) Å and for a baseball with \(\sigma = 1\) mm. Comment on why classical baseballs behave classically.

8. Potential placement freedom. Repeat the derivation of the full Schrödinger equation, but evaluate the slice potential at the source point \(x' = x - \delta x\) instead of the endpoint \(x\).

(a) Show that the additional slice phase becomes \(\exp[-\mathrm{i}V(x-\delta x)\,\delta t/\hbar]\).

(b) Expand \(V(x-\delta x) = V(x) - \delta x\,V'(x) + O((\delta x)^{2})\) and use the Gaussian moments to show that the difference relative to evaluating \(V\) at \(x\) contributes only at \(O(\delta t^{\,2})\).

(c) Conclude that the Schrödinger equation is independent of where \(V\) is sampled within the slice, and explain in one sentence why this freedom matches the analogous freedom for the slice action (HW 3.2.1.4).