Lecture 09 · Particles

Fourier Analysis, Particle Mechanics & Wave Packets

The Fourier transform as the bridge between position and momentum, the free particle Schrödinger equation, wave packet dynamics, dispersion relations, group versus phase velocity, and Ehrenfest's theorem connecting quantum to classical mechanics.

Part I — From Spins to Particles: A New Physical Setting

Throughout Lectures 1–7, we worked exclusively with finite-dimensional Hilbert spaces: the qubit ($\mathbb{C}^2$), multi-spin systems ($\mathbb{C}^2 \otimes \mathbb{C}^2 \otimes \cdots$). We now make a fundamental leap to an infinite-dimensional Hilbert space: a particle moving on a line.

The new setting introduces two key continuous observables:

ObservableOperatorSpectrumEigenstates
Position$\hat{x}$$x \in \mathbb{R}$ (continuous)$|x\rangle$: delta-localized
Momentum$\hat{p}$$p \in \mathbb{R}$ (continuous)$|p\rangle$: plane waves

The state is represented by a wave function $\psi(x) = \langle x|\psi\rangle$ — a complex-valued function on $\mathbb{R}$. The physical content: $|\psi(x)|^2$ is a probability density, not a probability. The probability of finding the particle in the interval $[a,b]$ is:

$$P(a < x < b) = \int_a^b |\psi(x)|^2\,dx$$

Part II — The Fourier Transform

The Fourier transform is the mathematical bridge between the position and momentum representations. It converts a function of $x$ into a function of $k$ (wavenumber, with $p = \hbar k$) and vice versa. Using the symmetric convention with $\hbar = 1$:

Fourier Transform Pair
$$\tilde{\psi}(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\psi(x)\,e^{-ikx}\,dx \qquad \psi(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\tilde{\psi}(k)\,e^{ikx}\,dk$$

Parseval's theorem states that the Fourier transform preserves norms:

$$\int_{-\infty}^\infty|\psi(x)|^2\,dx = \int_{-\infty}^\infty|\tilde{\psi}(k)|^2\,dk$$

This is essential: it means $|\tilde{\psi}(k)|^2$ is a valid probability density in momentum space. The physical meaning of $\tilde{\psi}(k)$: it is the amplitude to have momentum $p = \hbar k$. The key duality: if $\psi(x)$ is sharply peaked (definite position), then $\tilde{\psi}(k)$ is spread out (indefinite momentum) — the Heisenberg uncertainty principle expressed as Fourier duality.

The canonical Fourier pair: Gaussian ↔ Gaussian. If $\psi(x) \sim e^{-x^2/(4\sigma^2)}$ has width $\sigma$ in position, then $\tilde{\psi}(k) \sim e^{-k^2\sigma^2}$ has width $1/(2\sigma)$ in momentum, with the product $\sigma \cdot \frac{1}{2\sigma} = \frac{1}{2}$ — saturating the Heisenberg bound.

Part III — The Momentum Operator

In position space, the action of the momentum operator on the wave function is derived using completeness:

$$\langle x|\hat{p}|\psi\rangle = \int\langle x|\hat{p}|p\rangle\langle p|\psi\rangle\,dp = \int p\cdot\frac{e^{ipx}}{\sqrt{2\pi}}\tilde{\psi}(p)\,dp$$

Since $p\,e^{ipx} = -i\,\dfrac{d}{dx}\,e^{ipx}$, we can pull the derivative outside the integral:

$$\langle x|\hat{p}|\psi\rangle = -i\frac{d}{dx}\left(\frac{1}{\sqrt{2\pi}}\int\tilde{\psi}(p)\,e^{ipx}\,dp\right) = -i\frac{d\psi}{dx}$$
Momentum Operator (position space, $\hbar=1$)
$$\hat{p} = -i\hbar\frac{d}{dx}$$

Part IV — Free Particle Dynamics

The Hamiltonian for a free particle (no potential) is purely kinetic:

$$H = \frac{\hat{p}^2}{2m} = \frac{1}{2m}\left(-\hbar^2\frac{d^2}{dx^2}\right) = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}$$

The time-dependent Schrödinger equation $i\hbar\,\partial\psi/\partial t = H\psi$ becomes:

$$i\hbar\frac{\partial\psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}$$

The plane wave $\psi_k(x,t) = e^{i(kx - \omega t)}$ is a solution, provided the dispersion relation is satisfied:

Free Particle Dispersion Relation
$$\omega(k) = \frac{\hbar k^2}{2m}$$

This is a quadratic dispersion — different wavenumber components travel at different speeds. This is what causes wave packets to spread over time. A plane wave is not normalizable (it extends to $\pm\infty$), so physical states must be superpositions of plane waves.

Part V — Wave Packets

A physical particle is described by a wave packet — a superposition of plane waves localized in both position and momentum:

Wave Packet (General)
$$\psi(x,t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \tilde{\psi}(k)\,e^{i(kx - \omega(k)t)}\,dk$$

At $t = 0$, a Gaussian wave packet centered at $x = 0$ with width $\sigma$ and mean momentum $p_0 = \hbar k_0$:

$$\psi(x,0) = (2\pi\sigma^2)^{-1/4}\,e^{ik_0 x}\,e^{-x^2/(4\sigma^2)}$$

The dynamics are governed by two effects:

  • Translation: the center moves at the group velocity $v_g = \hbar k_0/m = p_0/m$
  • Spreading: the width grows as $\sigma(t)^2 = \sigma^2 + \left(\dfrac{\hbar t}{2m\sigma}\right)^2$

Heavy particles spread slowly; light particles spread rapidly. This spreading is purely quantum — it has no classical analogue.

Part VI — Group Velocity vs Phase Velocity

A plane wave $e^{i(kx-\omega t)}$ has two velocities:

VelocityDefinitionFormula ($\omega = \hbar k^2/2m$)Physical meaning
Phase velocity$v_\mathrm{ph} = \omega/k$$\hbar k/(2m) = p/(2m)$Speed of phase fronts — not physical
Group velocity$v_g = d\omega/dk$$\hbar k/m = p/m$Speed of wave packet envelope — physical

For the free non-relativistic particle: $v_g = 2v_\mathrm{ph}$. The particle moves at the group velocity — this is what corresponds to the classical particle velocity $p/m$. The phase velocity of $p/(2m)$ has no direct physical interpretation.

The expectation value of position moves at the group velocity: $\langle x\rangle(t) = \langle x\rangle(0) + v_g\,t$, recovering Newton's first law for the quantum average.

Part VII — The Hamiltonian $H = (\omega/2)\sigma_z$ as Linear Dispersion

Our familiar spin Hamiltonian $H = (\omega/2)\sigma_z$ has exactly two energy levels $\pm\omega/2$ — analogous to a "two-site dispersion relation." This connects to a broader principle about dispersion. For a linear dispersion $\omega = vk$ (like a relativistic photon or massless Dirac fermion):

  • Group velocity = phase velocity = $v$ for all $k$
  • The Schrödinger equation becomes $i\hbar\partial\psi/\partial t = -i\hbar v\,\partial\psi/\partial x$
  • Exact solution: $\psi(x,t) = f(x - vt)$ — any initial packet translates rigidly at speed $v$, no spreading

The spreading of the non-relativistic wave packet comes from the quadratic dispersion $\omega = \hbar k^2/2m$: different wavenumber components travel at different group velocities $\hbar k/m$, so the packet stretches over time. Linear dispersion is dispersionless — all components travel at the same speed.

Part VIII — Connection to Classical Mechanics (Ehrenfest)

Quantum mechanics must recover classical mechanics for macroscopic systems. Ehrenfest's theorem makes this precise. For a free particle:

$$\frac{d\langle x\rangle}{dt} = \frac{\langle p\rangle}{m}, \qquad \frac{d\langle p\rangle}{dt} = 0$$

Together: $\langle x\rangle(t) = \langle x\rangle(0) + \dfrac{\langle p\rangle(0)}{m}\,t$ — Newton's first law for expectation values.

Adding a potential $V(x)$:

$$\frac{d\langle p\rangle}{dt} = -\left\langle V'(x)\right\rangle \approx -V'\!\left(\langle x\rangle\right) \quad\text{(when the packet is narrow)}$$

In the narrow-packet limit, the quantum averages satisfy exactly the classical equations of motion. Classical mechanics is the limit of quantum mechanics when wave packets are so narrow that $V'(x)$ is approximately constant across the packet width.

Exercises — Lecture 9

Nine exercises covering Fourier transforms, wave packet dynamics, dispersion relations, and the uncertainty principle. Click any exercise to reveal the full worked solution.

1
Fourier Transform of Delta Function
EasyFourier Transform

Compute the Fourier transform $\tilde{\psi}(k) = \dfrac{1}{\sqrt{2\pi}}\displaystyle\int \delta(x - x_0)\,e^{-ikx}\,dx$.

What does this say about the momentum distribution of a perfectly localized particle?

Complete Solution
compute

Apply the defining property of the delta function with $f(x) = e^{-ikx}$ and $x_0$ fixed:

$$\tilde{\psi}(k) = \frac{1}{\sqrt{2\pi}}\int\delta(x - x_0)\,e^{-ikx}\,dx = \frac{1}{\sqrt{2\pi}}\,e^{-ikx_0}$$
interpret

The probability density in momentum space:

$$|\tilde{\psi}(k)|^2 = \frac{1}{2\pi}\,|e^{-ikx_0}|^2 = \frac{1}{2\pi}$$
Physical Interpretation
The momentum distribution is completely uniform — all momenta $k$ are equally probable with density $1/(2\pi)$. A perfectly localized particle ($\Delta x = 0$) has completely undefined momentum ($\Delta p = \infty$). This is the extreme limit of the Heisenberg uncertainty principle: $\Delta x \cdot \Delta p = 0 \cdot \infty$ (formally). Perfect position localization demands infinite momentum spread.
2
Group and Phase Velocity
EasyDispersion

For a free non-relativistic particle with dispersion $\omega = \hbar k^2/2m$:

(a) Find the phase velocity $v_\mathrm{ph} = \omega/k$.

(b) Find the group velocity $v_g = d\omega/dk$.

(c) Show $v_g = 2v_\mathrm{ph}$. What is the physical speed of a particle with momentum $p = \hbar k$?

Complete Solution
a
$$v_\mathrm{ph} = \frac{\omega}{k} = \frac{\hbar k^2/(2m)}{k} = \frac{\hbar k}{2m} = \frac{p}{2m}$$
b
$$v_g = \frac{d\omega}{dk} = \frac{d}{dk}\left(\frac{\hbar k^2}{2m}\right) = \frac{\hbar k}{m} = \frac{p}{m}$$
c
$$v_g = \frac{\hbar k}{m} = 2\cdot\frac{\hbar k}{2m} = 2\,v_\mathrm{ph} \checkmark$$
Physical Answer
The physical velocity of the particle is the group velocity $v_g = p/m$ — the classical result from Newton's second law. The phase velocity $p/(2m)$ is half this and has no direct physical meaning. Never use phase velocity to describe particle motion.
3
Parseval's Theorem
MediumFourier Transform

Verify Parseval's theorem for the Gaussian $\psi(x) = \pi^{-1/4}\,e^{-x^2/2}$ (which is its own Fourier transform: $\tilde{\psi}(k) = \pi^{-1/4}\,e^{-k^2/2}$).

Show that $\displaystyle\int|\psi(x)|^2\,dx = \int|\tilde{\psi}(k)|^2\,dk = 1$.

Complete Solution
position space
$$\int_{-\infty}^\infty|\psi(x)|^2\,dx = \pi^{-1/2}\int e^{-x^2}\,dx = \pi^{-1/2}\cdot\sqrt{\pi} = 1 \checkmark$$

Using the standard Gaussian integral $\int e^{-x^2}\,dx = \sqrt{\pi}$.

momentum space
$$\int_{-\infty}^\infty|\tilde{\psi}(k)|^2\,dk = \pi^{-1/2}\int e^{-k^2}\,dk = \pi^{-1/2}\cdot\sqrt{\pi} = 1 \checkmark$$
Key Point
Both integrals equal 1, confirming Parseval's theorem. The Fourier transform is a unitary operation — it preserves the norm of the state. This is essential because it means $|\tilde{\psi}(k)|^2$ has the correct total probability for the momentum distribution. The Fourier transform maps one valid probability density to another.
4
Plane Wave Action of Momentum Operator
MediumOperators

Apply $\hat{p} = -i\,d/dx$ ($\hbar = 1$) to $\psi(x) = A\,e^{ik_1 x} + B\,e^{ik_2 x}$ with $|A|^2 + |B|^2 = 1$.

(a) Is $\psi$ an eigenfunction of $\hat{p}$?

(b) Compute $\langle p\rangle$ and the uncertainty $\Delta p$.

Complete Solution
a
$$\hat{p}\,\psi = -i\frac{d}{dx}(Ae^{ik_1 x} + Be^{ik_2 x}) = k_1 Ae^{ik_1 x} + k_2 Be^{ik_2 x}$$

This equals $\lambda\psi$ only if $k_1 = k_2$ (a single plane wave). No — it is not an eigenfunction when $k_1 \neq k_2$; the two components have different momenta.

b

Since the state is a superposition of momentum eigenstates $|k_1\rangle$ and $|k_2\rangle$ with weights $A$ and $B$:

$$\langle p\rangle = |A|^2\,k_1 + |B|^2\,k_2$$
$$\langle p^2\rangle = |A|^2\,k_1^2 + |B|^2\,k_2^2$$
$$\Delta p^2 = \langle p^2\rangle - \langle p\rangle^2 = |A|^2|B|^2(k_1 - k_2)^2$$
Answer
$\Delta p = |A||B||k_1 - k_2|$. The uncertainty is zero only if $k_1 = k_2$ (single momentum) or if one of $|A|, |B|$ is zero (also a single momentum). The two-component superposition has momentum uncertainty proportional to the separation $|k_1 - k_2|$.
5
Wave Packet Spreading
MediumDynamics

A Gaussian wave packet has $\sigma(t)^2 = \sigma_0^2 + \left(\dfrac{\hbar t}{2m\sigma_0}\right)^2$.

(a) Find the "doubling time" $t_\mathrm{spread}$ when $\sigma(t) = 2\sigma_0$.

(b) For an electron ($m \approx 9\times10^{-31}$ kg) with $\sigma_0 = 1$ nm and $\hbar = 10^{-34}$ J·s, estimate $t_\mathrm{spread}$.

(c) For a baseball ($m = 0.1$ kg, $\sigma_0 = 10^{-15}$ m), estimate $t_\mathrm{spread}$.

Complete Solution
a

Set $\sigma(t) = 2\sigma_0$, so $\sigma(t)^2 = 4\sigma_0^2$:

$$4\sigma_0^2 = \sigma_0^2 + \left(\frac{\hbar\,t_\mathrm{spread}}{2m\sigma_0}\right)^2 \implies \left(\frac{\hbar\,t_\mathrm{spread}}{2m\sigma_0}\right)^2 = 3\sigma_0^2$$
$$t_\mathrm{spread} = \frac{2\sqrt{3}\,m\sigma_0^2}{\hbar}$$
b

Electron: $m = 9\times10^{-31}$ kg, $\sigma_0 = 10^{-9}$ m, $\hbar = 10^{-34}$ J·s:

$$t_\mathrm{spread} = \frac{2\sqrt{3}\cdot 9\times10^{-31}\cdot(10^{-9})^2}{10^{-34}} = \frac{2\cdot 1.73\cdot 9\times10^{-49}}{10^{-34}} \approx 3.1\times10^{-14}\,\mathrm{s}$$

About 30 femtoseconds — an electron wave packet spreads extremely rapidly!

c

Baseball: $m = 0.1$ kg, $\sigma_0 = 10^{-15}$ m, $\hbar = 10^{-34}$ J·s:

$$t_\mathrm{spread} = \frac{2\sqrt{3}\cdot 0.1\cdot(10^{-15})^2}{10^{-34}} = \frac{3.46\cdot0.1\cdot10^{-30}}{10^{-34}} = 3.46\times10^3\,\mathrm{s} \approx 1\,\mathrm{hour}$$
Insight
Even for a baseball with nuclear-scale initial localization, the spreading time is hours. For any macroscopic object with realistic initial uncertainty, the spreading time is astronomically large — quantum spreading is negligible for macroscopic objects. This is why we observe classical behavior in everyday life.
6
Schrödinger Equation for Wave Packet
MediumDynamics

A free particle wave packet: $\psi(x,t) = \dfrac{1}{\sqrt{2\pi}}\displaystyle\int\tilde{\psi}(k)\,e^{i(kx - \omega_k t)}\,dk$ with $\omega_k = k^2/2m$ ($\hbar = 1$).

(a) Show this solves the Schrödinger equation $i\,\partial_t\psi = -\dfrac{1}{2m}\,\partial_x^2\psi$.

(b) For $\tilde{\psi}(k)$ peaked at $k = k_0$, describe qualitatively how $\langle x\rangle(t)$ evolves.

Complete Solution
a

Compute the left-hand side by differentiating under the integral:

$$i\frac{\partial\psi}{\partial t} = \frac{1}{\sqrt{2\pi}}\int\tilde{\psi}(k)\cdot i(-i\omega_k)\,e^{i(kx-\omega_k t)}\,dk = \frac{1}{\sqrt{2\pi}}\int\tilde{\psi}(k)\cdot\omega_k\,e^{i(kx-\omega_k t)}\,dk$$

For the right-hand side:

$$-\frac{1}{2m}\frac{\partial^2\psi}{\partial x^2} = \frac{1}{\sqrt{2\pi}}\int\tilde{\psi}(k)\cdot\frac{k^2}{2m}\,e^{i(kx-\omega_k t)}\,dk$$

These are equal since $\omega_k = k^2/(2m)$. ✓

b
Group Velocity
The Fourier transform $\tilde{\psi}(k)$ peaks at $k = k_0$, so the dominant contribution to the packet comes from wavevectors near $k_0$. These travel at group velocity $v_g = d\omega/dk|_{k_0} = k_0/m$. By Ehrenfest's theorem: $\langle x\rangle(t) = \langle x\rangle(0) + k_0 t/m = p_0 t/m$. The center of the packet moves at the classical velocity $p_0/m$ — Newton's first law.
7
Linear Dispersion: No Spreading
HardDispersion

For linear dispersion $\omega = vk$ (massless particle, e.g., photon):

(a) Show $v_g = v_\mathrm{ph} = v$ (no dispersion).

(b) Show that any initial wave packet satisfies $\psi(x,t) = \psi(x-vt, 0)$ — exact translation, no spreading.

(c) Contrast with $\omega = k^2/2m$: why does the quadratic case spread?

Complete Solution
a
$$v_\mathrm{ph} = \frac{\omega}{k} = \frac{vk}{k} = v, \qquad v_g = \frac{d\omega}{dk} = \frac{d(vk)}{dk} = v$$

Equal velocities — all wavenumber components travel at the same speed. ✓

b
$$\psi(x,t) = \frac{1}{\sqrt{2\pi}}\int\tilde{\psi}(k)\,e^{i(kx - vkt)}\,dk = \frac{1}{\sqrt{2\pi}}\int\tilde{\psi}(k)\,e^{ik(x-vt)}\,dk = \psi(x-vt,0)$$

The factor $e^{-ivkt}$ does not depend on $k$ non-linearly — it simply shifts the argument of the inverse Fourier transform from $x$ to $x - vt$. ✓

c
Why Quadratic Disperses
For $\omega = k^2/2m$, the group velocity is $v_g(k) = k/m$ — different for each component. High-$k$ (high-momentum) components travel faster than low-$k$ components. The packet stretches over time as its Fourier components separate. With linear dispersion, all components travel at the same speed, so no stretching occurs. Dispersion = velocity spread across wavenumber components.
8
Position-Momentum Uncertainty from Gaussian
HardUncertainty Principle

For $\psi(x) = (2\pi\sigma^2)^{-1/4}\,e^{-x^2/(4\sigma^2)}$ ($\hbar = 1$):

(a) Compute $\langle x\rangle$, $\langle x^2\rangle$, and $\Delta x$. (Use $\int x^2 e^{-x^2/(2\sigma^2)}\,dx = \sigma^3\sqrt{2\pi}$.)

(b) Compute $\langle p\rangle$ and $\langle p^2\rangle = \int\psi^*(-d^2\psi/dx^2)\,dx$, hence $\Delta p$.

(c) Verify $\Delta x\,\Delta p = 1/2$ — the minimum uncertainty state.

Complete Solution
a

$\langle x\rangle = 0$ since $|\psi|^2$ is an even function and $x|\psi|^2$ is odd. For $\langle x^2\rangle$:

$$\langle x^2\rangle = (2\pi\sigma^2)^{-1/2}\int x^2\,e^{-x^2/(2\sigma^2)}\,dx = (2\pi\sigma^2)^{-1/2}\cdot\sigma^3\sqrt{2\pi} = \sigma^2$$
$$\Delta x = \sqrt{\langle x^2\rangle - \langle x\rangle^2} = \sigma$$
b

$\langle p\rangle = 0$: the state is real, so $\psi^*\psi' = \psi\psi' = d(\psi^2/2)/dx$, which integrates to zero. For $\langle p^2\rangle$: compute $\psi'' = \left[\dfrac{x^2}{4\sigma^4} - \dfrac{1}{2\sigma^2}\right]\psi$, so:

$$\langle p^2\rangle = \int\psi\!\left(-\psi''\right)\,dx = \int|\psi|^2\!\left(\frac{1}{2\sigma^2} - \frac{x^2}{4\sigma^4}\right)\,dx = \frac{1}{2\sigma^2} - \frac{\langle x^2\rangle}{4\sigma^4} = \frac{1}{2\sigma^2} - \frac{\sigma^2}{4\sigma^4} = \frac{1}{4\sigma^2}$$
$$\Delta p = \frac{1}{2\sigma}$$
c
Minimum Uncertainty
$$\Delta x\,\Delta p = \sigma\cdot\frac{1}{2\sigma} = \frac{1}{2} = \frac{\hbar}{2} \checkmark$$ The Gaussian wave packet saturates the Heisenberg inequality $\Delta x\,\Delta p \geq \hbar/2$ with equality. It is the unique state of minimum uncertainty — often called a "coherent state" in the ground state of a harmonic oscillator. Any other wave function satisfies $\Delta x\,\Delta p > \hbar/2$.
9
Expectation Value of Position Under Free Evolution
HardEhrenfest

For the Gaussian wave packet $\psi(x,0) = (2\pi\sigma^2)^{-1/4}\,e^{ik_0 x}\,e^{-x^2/(4\sigma^2)}$:

(a) Write the Fourier transform $\tilde{\psi}(k)$ and identify where it peaks.

(b) Using Ehrenfest's theorem, show $\langle x\rangle(t) = k_0\,t/m$ ($\hbar = 1$).

(c) Show $\langle p\rangle(t) = k_0$ (constant). Conclude Newton's first law for expectation values.

Complete Solution
a

The factor $e^{ik_0 x}$ shifts the Fourier transform by $k_0$. The Fourier transform of $e^{ik_0 x}f(x)$ is $\tilde{f}(k - k_0)$:

$$\tilde{\psi}(k) = (2\sigma^2/\pi)^{1/4}\cdot (2\sigma^2)^{1/2}\,e^{-(k-k_0)^2\sigma^2}$$

(The exact prefactor follows from the Gaussian Fourier transform computed in Exercise 7.) This is a Gaussian centered at $k = k_0$ with width $1/(2\sigma)$.

b

Ehrenfest's theorem for a free particle ($V = 0$):

$$\frac{d\langle p\rangle}{dt} = -\left\langle\frac{\partial V}{\partial x}\right\rangle = 0 \qquad\Rightarrow\qquad \langle p\rangle(t) = \langle p\rangle(0) = \text{const}$$
$$\frac{d\langle x\rangle}{dt} = \frac{\langle p\rangle}{m} = \frac{k_0}{m} \qquad\Rightarrow\qquad \langle x\rangle(t) = \langle x\rangle(0) + \frac{k_0}{m}t = \frac{k_0\,t}{m}$$

using $\langle x\rangle(0) = 0$ (symmetric Gaussian centered at origin).

c
Newton's First Law
$\langle p\rangle(t) = k_0 = \mathrm{const}$ (no force, no change in momentum). $\langle x\rangle(t) = \langle x\rangle(0) + \dfrac{\langle p\rangle(0)}{m}\,t$ — uniform motion at velocity $p_0/m$. This is Newton's first law for quantum expectation values. Quantum mechanics does not simply "reduce to" classical mechanics — but for the expectation values of well-localized wave packets, the classical equations of motion hold exactly.