Wave Packets, Uncertainty & the Momentum Operator

A plane wave has perfect momentum and no location; a real particle needs both, approximately. The compromise is the wave packet — and Fourier analysis charges a fee for it: sharpen the position, and the momentum spread grows. That fee, priced in units of $\hbar$, is the Heisenberg uncertainty principle. Along the way we extract $\hat p = -i\hbar\,\partial_x$ from the Schrödinger equation itself, and find that a cat curled into a smaller box necessarily fidgets more.

Learning Objectives

After this lesson you will be able to:

1. Explain why free particles are described by wave packets, and derive the group velocity $v_g = d\omega/dk$ by the stationary-phase argument.
2. Compute $\Delta x$ and $\Delta k$ for a Gaussian packet, show it saturates the Fourier reciprocity bound, and derive the spreading law $\Delta x(t)$.
3. State and interpret the Heisenberg uncertainty principle $\Delta x\,\Delta p \ge \hbar/2$ as Fourier analysis plus $p = \hbar k$.
4. Use the momentum-space wavefunction $\Phi(p,t)$ and the Born rule in momentum space.
5. Derive the momentum operator $\hat p = -i\hbar\,\partial_x$ from $d\langle x\rangle/dt$, compute the canonical commutator $[\hat x,\hat p] = i\hbar$, and derive Ehrenfest's theorem.

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Intuition

The free-particle stationary states are plane waves $e^{i(kx-\omega t)}$ — but P.4.1 already ruled them out as physical states: $|e^{ikx}|^2 = 1$ everywhere is not normalizable. Nature's free particles are instead wave packets: superpositions of plane waves with neighboring wavenumbers, interfering constructively in a small region and destructively everywhere else. A packet is localized because it contains a spread of momenta — that trade-off is not a technical nuisance but the deepest structural fact in this lesson.

Two questions drive everything. How fast does the lump move? Answer: at the group velocity, which turns out to be exactly the classical particle velocity $p/m$ — de Broglie's picture survives. What is momentum, really, in wave mechanics? Answer: a derivative. Watching how $\langle x\rangle$ moves under the Schrödinger equation forces $\hat p = -i\hbar\,\partial/\partial x$ on us — and with it the commutator $[\hat x, \hat p] = i\hbar$, the quantum fingerprint of the Poisson bracket from P.1.3.

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Theory

Wave packets: superposing plane waves

For $V = 0$ every $e^{ikx}$ solves the TISE with $E = \hbar^2k^2/2m$, so the general free solution is a continuous superposition (the $k$-integral analogue of $\sum_n c_n\psi_n$ from P.4.2):

$$\Psi(x,t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\phi(k)\,e^{i(kx - \omega(k)t)}\,dk, \qquad \omega(k) = \frac{\hbar k^2}{2m},$$

with $\phi(k)$ fixed by the initial condition via the inverse Fourier transform ($\phi(k) = \frac{1}{\sqrt{2\pi}}\int\Psi(x,0)e^{-ikx}dx$, 0.3.2). If $\phi(k)$ is concentrated near $k_0$, the integrand's phases agree near one point in space and cancel elsewhere: a wave packet.

Group velocity: the stationary-phase derivation

Where is the packet at time $t$? The integrand's phase is $\theta(k) = kx - \omega(k)t$. For most $x$, $\theta$ varies rapidly across the window where $\phi(k)$ lives, and the contributions average to zero. The packet sits where the phase is stationary at the center of the window:

$$\left.\frac{d\theta}{dk}\right|_{k_0} = 0 \quad\Longrightarrow\quad x - \omega'(k_0)\,t = 0 \quad\Longrightarrow\quad x_{\text{packet}}(t) = \frac{d\omega}{dk}\bigg|_{k_0}\,t .$$

More explicitly: expand $\omega(k) \approx \omega_0 + \omega_0'(k - k_0)$ near $k_0$ (with $\omega_0' \equiv \omega'(k_0)$) and substitute, writing $s = k - k_0$:

$$\Psi(x,t) \approx e^{i(k_0x - \omega_0 t)}\cdot \frac{1}{\sqrt{2\pi}}\int \phi(k_0+s)\,e^{is\,(x - \omega_0' t)}\,ds = e^{i(k_0x - \omega_0 t)}\; F\!\big(x - \omega_0' t\big).$$

The packet factorizes into a fast carrier wave moving at the phase velocity $v_{\text{ph}} = \omega_0/k_0$, and a slow envelope $F$ that translates rigidly at the group velocity

$$\boxed{\;v_g = \frac{d\omega}{dk}\bigg|_{k_0}\;}$$

For matter waves, $\omega = \hbar k^2/2m$ gives

$$v_g = \frac{\hbar k_0}{m} = \frac{p_0}{m} = v_{\text{classical}}, \qquad v_{\text{ph}} = \frac{\omega_0}{k_0} = \frac{\hbar k_0}{2m} = \frac{v_g}{2}.$$

The apparent paradox — "the wave moves at half the particle's speed!" — dissolves once you ask what carries the physics. The phase velocity is the speed of individual crests of the unobservable carrier; crests are features of $\Psi$, not of $|\Psi|^2$, and indeed they continually slide backward through the envelope, being born at the rear and dying at the front. The probability density — the thing the Born rule ties to experiment — rides the envelope, at $v_g = p/m$. (Phase velocity even depends on the arbitrary zero of energy: shift $V \to V + V_0$ and $v_{\text{ph}}$ changes while $v_g$ and all physics do not.)

The Gaussian packet and Fourier reciprocity

Take the standard Gaussian packet at $t = 0$:

$$\Psi(x,0) = \big(2\pi a^2\big)^{-1/4} e^{-x^2/4a^2}\,e^{ik_0x} \quad\Longleftrightarrow\quad \phi(k) = \left(\frac{2a^2}{\pi}\right)^{1/4} e^{-a^2(k-k_0)^2},$$

(a Gaussian transforms to a Gaussian, 0.3.2). The densities $|\Psi|^2 \propto e^{-x^2/2a^2}$ and $|\phi|^2 \propto e^{-2a^2(k-k_0)^2}$ are normal distributions with standard deviations

$$\Delta x = a, \qquad \Delta k = \frac{1}{2a} \qquad\Longrightarrow\qquad \Delta x\,\Delta k = \frac12 .$$

This is no accident of the Gaussian: for any Fourier pair, $\Delta x\,\Delta k \ge 1/2$, with equality iff the function is Gaussian — pure Fourier reciprocity, proved in 0.3.2. Narrow in $x$ means broad in $k$, unavoidably.

The Heisenberg uncertainty principle

Now inject one piece of physics: de Broglie, $p = \hbar k$, so $\Delta p = \hbar\,\Delta k$. Fourier reciprocity becomes

$$\boxed{\;\Delta x\,\Delta p \ \ge\ \frac{\hbar}{2}\;}$$

— the Heisenberg uncertainty principle, saturated exactly by Gaussian packets. Nothing about measurement apparatus entered this derivation: the bound is a property of the state $\Psi$, i.e. of which functions exist. In Term 1 the same inequality reappears as the position–momentum case of the general Robertson relation $\sigma_A\sigma_B \ge \frac12|\langle[\hat A,\hat B]\rangle|$ (1.3.2 Expectation & Uncertainty), with $[\hat x,\hat p] = i\hbar$ (derived below) supplying the right-hand side.

Caution. Uncertainty is a property of the state — pure Fourier analysis dressed in $\hbar$ — not a disturbance inflicted by a clumsy measurement. Heisenberg's original "microscope" story (photon kicks electron) is a heuristic, and a misleading one: even before anyone measures anything, a particle simply does not possess a sharp $x$ and a sharp $p$ simultaneously, because no square-integrable function is narrow in both $x$ and $k$. The statistics of $x$-measurements on one ensemble and $p$-measurements on another, both prepared in $\Psi$, already obey the bound.

Spreading of the Gaussian packet

Evolve the Gaussian exactly: multiply $\phi(k)$ by $e^{-i\omega(k)t} = e^{-i\hbar k^2t/2m}$ and transform back. The exponent stays quadratic in $k$, so the $k$-integral is a Gaussian integral (complete the square; all steps are the standard $\int e^{-\alpha k^2 + \beta k}dk = \sqrt{\pi/\alpha},e^{\beta^2/4\alpha}$with complex$\alpha$). The result for the density is again normal:

$$|\Psi(x,t)|^2 = \frac{1}{\sqrt{2\pi}\,\Delta x(t)}\, \exp\!\left[-\frac{(x - v_g t)^2}{2\,\Delta x(t)^2}\right],$$

centered on the classical trajectory $x = v_gt$, with width

$$\boxed{\;\Delta x(t) = \Delta x_0\sqrt{1 + \left(\frac{\hbar t}{2m\,\Delta x_0^2}\right)^2}\;} \qquad (\Delta x_0 = a).$$

Verification without the full integral: the momentum distribution $|\phi(k)|^2$ never changes (free evolution only multiplies $\phi$ by a phase), so the packet permanently contains velocities spread by $\Delta v = \hbar\Delta k/m = \hbar/2ma$. Ballistically, positions then spread as $\Delta x(t)^2 \approx \Delta x_0^2 + (\Delta v\,t)^2 = \Delta x_0^2\big[1 + (\hbar t/2m\Delta x_0^2)^2\big]$ — matching the exact law. Note the revenge of uncertainty: the smaller you squeeze $\Delta x_0$, the faster the packet explodes ($\Delta x(t) \approx \hbar t/2m\Delta x_0$ at late times). Meanwhile $\Delta x(t)\Delta p = \frac{\hbar}{2}\sqrt{1 + (\hbar t/2m\Delta x_0^2)^2} \ge \hbar/2$: saturation holds only at $t = 0$.

Momentum space and the Born rule there

Define the momentum-space wavefunction as the ($p$-scaled) Fourier transform of $\Psi$:

$$\Phi(p,t) = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}\Psi(x,t)\,e^{-ipx/\hbar}\,dx, \qquad \Psi(x,t) = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty}\Phi(p,t)\,e^{ipx/\hbar}\,dp .$$

Parseval's theorem gives $\int|\Phi|^2dp = \int|\Psi|^2dx = 1$, and the Born rule extends: $|\Phi(p,t)|^2\,dp$ is the probability of finding momentum in $[p, p+dp]$. Position space and momentum space are two representations of one state — in Term 1 language, $\Psi(x,t) = \langle x|\Psi(t)\rangle$and$\Phi(p,t) = \langle p|\Psi(t)\rangle$, two bases for the same ket (P.4.1).

The momentum operator (Griffiths route)

How do we compute $\langle p\rangle$ without leaving position space? Follow the motion. Start from $\langle x\rangle = \int x\,|\Psi|^2dx$ and differentiate, using the continuity equation $\partial_t|\Psi|^2 = -\partial_x j$ with $j = \frac{i\hbar}{2m}(\Psi,\partial_x\Psi^* - \Psi^*\partial_x\Psi)$ from P.4.1:

$$\frac{d\langle x\rangle}{dt} = \int x\,\frac{\partial|\Psi|^2}{\partial t}\,dx = -\int x\,\frac{\partial j}{\partial x}\,dx = \int j\,dx = -\frac{i\hbar}{2m}\int\left(\Psi^*\frac{\partial\Psi}{\partial x} - \Psi\frac{\partial\Psi^*}{\partial x}\right)dx,$$

where the first integration by parts dropped the boundary term $[xj]_{-\infty}^{\infty} = 0$. Now integrate the second term by parts once more, $\int\Psi,\partial_x\Psi^dx = [,|\Psi|^2,]_{-\infty}^{\infty} - \int\Psi^\partial_x\Psi,dx = -\int\Psi^*\partial_x\Psi,dx$:

$$\frac{d\langle x\rangle}{dt} = -\frac{i\hbar}{m}\int\Psi^*\frac{\partial\Psi}{\partial x}\,dx .$$

Identify momentum as mass times the velocity of the mean: $\langle p\rangle \equiv m\,\frac{d\langle x\rangle}{dt}$, giving

$$\langle p\rangle = \int\Psi^*\left(-i\hbar\frac{\partial}{\partial x}\right)\Psi\,dx \qquad\Longrightarrow\qquad \boxed{\;\hat p = -i\hbar\frac{\partial}{\partial x}\;}$$

Momentum acts in position space as a derivative — exactly as de Broglie hinted: $\hat p\,e^{ikx} = \hbar k\,e^{ikx}$, plane waves are momentum eigenfunctions. Consistency check in momentum space: $\langle p\rangle = \int\Phi^*\,p\,\Phi\,dp$ (there, $\hat p$ is just multiplication — each representation diagonalizes its own operator).

General observables. Any classical $Q(x,p)$ is promoted by the substitution rule $\hat Q = Q(\hat x, \hat p)$, with $\hat x = x$ (multiplication) and $\hat p = -i\hbar\partial_x$:

$$\langle Q\rangle = \int\Psi^*\,\hat Q\!\left(x,\,-i\hbar\tfrac{\partial}{\partial x}\right)\Psi\,dx .$$

E.g. kinetic energy $T = p^2/2m \to -\frac{\hbar^2}{2m}\partial_x^2$ — which is exactly the kinetic term of the Schrödinger equation: $i\hbar\,\partial_t\Psi = \hat H\Psi$ with $\hat H = \hat p^2/2m + V(\hat x)$. The construction of P.4.1 was quantization in disguise.

The canonical commutator

Operators need not commute. Apply $\hat x\hat p - \hat p\hat x$ to a test function $f(x)$:

$$[\hat x, \hat p]f = x\left(-i\hbar\frac{df}{dx}\right) - \left(-i\hbar\frac{d}{dx}\right)(xf) = -i\hbar\,x f' + i\hbar\,(f + xf') = i\hbar f .$$

Since $f$ was arbitrary,

$$\boxed{\;[\hat x, \hat p] = i\hbar\;}$$

— the canonical commutation relation, the irreducible quantum of non-commutativity from which the uncertainty principle, and much of Term 1, flows.

Ehrenfest's theorem

Finally, how does $\langle p\rangle$ itself move? Differentiate $\langle p\rangle = -i\hbar\int\Psi^*\partial_x\Psi\,dx$ and substitute $\partial_t\Psi = \frac{1}{i\hbar}\hat H\Psi$, $\partial_t\Psi^* = -\frac{1}{i\hbar}(\hat H\Psi)^*$:

$$\frac{d\langle p\rangle}{dt} = -i\hbar\int\left(\partial_t\Psi^*\,\partial_x\Psi + \Psi^*\,\partial_x\partial_t\Psi\right)dx = \int\left[(\hat H\Psi)^*\,\partial_x\Psi - \Psi^*\,\partial_x(\hat H\Psi)\right]dx .$$

The kinetic pieces cancel after two integrations by parts (same boundary arguments as always): $\int\big(\!-\tfrac{\hbar^2}{2m}\big)\big[\Psi''^*\Psi' - \Psi^*\Psi'''\big]dx = 0$. The potential pieces give

$$\int V\Psi^*\,\partial_x\Psi\,dx - \int\Psi^*\,\partial_x(V\Psi)\,dx = -\int\Psi^*\,\frac{\partial V}{\partial x}\,\Psi\,dx,$$

by the product rule $\partial_x(V\Psi) = V'\Psi + V\Psi'$. Hence Ehrenfest's theorem:

$$\boxed{\;\frac{d\langle p\rangle}{dt} = -\left\langle\frac{\partial V}{\partial x}\right\rangle\;} \qquad\text{together with}\qquad \frac{d\langle x\rangle}{dt} = \frac{\langle p\rangle}{m}.$$

Expectation values obey (almost) Newton's second law — "almost," because $\langle\partial_xV(x)\rangle \neq \partial_xV(\langle x\rangle)$ in general; they coincide for potentials up to quadratic, which is why Gaussian packets in a harmonic trap follow classical trajectories exactly (the coherent states of P.5.3 and, much later, Term 4.4).

Connections: from Poisson brackets to commutators

• Canonical quantization. P.1.3 ended with the fundamental Poisson bracket $\{x, p\} = 1$. Dirac's rule $\{\cdot,\cdot\} \to \frac{1}{i\hbar}[\cdot,\cdot]$ maps it to $[\hat x,\hat p] = i\hbar$ — which we have now computed concretely with $\hat x = x$, $\hat p = -i\hbar\partial_x$. The classical bracket is the $\hbar \to 0$ shadow of the commutator; this lesson is canonical quantization made flesh.
• Robertson relation. $\Delta x\,\Delta p \ge \hbar/2$ is the special case $\hat A = \hat x$, $\hat B = \hat p$ of $\sigma_A\sigma_B \ge \frac12|\langle[\hat A,\hat B]\rangle| = \hbar/2$ (1.3.2): here we got it from Fourier analysis, Term 1 gets it from operator algebra — same theorem, two proofs.
• Correspondence. Ehrenfest's theorem is the correspondence principle of P.2.3 fulfilled: where packets stay narrow compared to the scale on which $V'$ varies, quantum averages trace classical orbits.

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Worked Examples

Example 1 — How fast does an electron packet spread?

An electron ($m = 9.11\times10^{-31}\,$kg) is localized to $\Delta x_0 = 1\,\text{Å} = 10^{-10}\,$m (atomic size). Find the doubling time of its width, and compare with a dust grain ($m = 10^{-9}\,$kg, $\Delta x_0 = 1\,\mu$m).

The width doubles when $\hbar t/2m\Delta x_0^2 = \sqrt{3}$, i.e. $t_2 = 2\sqrt3\,m\,\Delta x_0^2/\hbar$. For the electron:

$$t_2 = \frac{2\sqrt3\,(9.11\times10^{-31})(10^{-10})^2}{1.055\times10^{-34}} \approx 3.0\times10^{-16}\,\text{s}.$$

A free electron confined to atomic size delocalizes essentially instantly — which is why atoms need a binding potential to hold their shape. For the dust grain:

$$t_2 = \frac{2\sqrt3\,(10^{-9})(10^{-6})^2}{1.055\times10^{-34}} \approx 3.3\times10^{13}\,\text{s} \approx 10^{6}\,\text{years}.$$

Quantum spreading is real for electrons and utterly negligible for anything you can see — the correspondence principle in numbers.

Example 2 — Momentum statistics of a "particle in a box" ground state

At $t=0$ the well is removed and we ask for the momentum distribution of what was the infinite-well ground state, $\psi(x) = \sqrt{2/L}\,\sin(\pi x/L)$ on $[0,L]$ (zero outside). Compute $\Phi(p)$.

Write $\sin(\pi x/L) = \frac{1}{2i}\big(e^{i\pi x/L} - e^{-i\pi x/L}\big)$ and integrate:

$$\Phi(p) = \frac{1}{\sqrt{2\pi\hbar}}\sqrt{\frac{2}{L}}\int_0^L \sin\frac{\pi x}{L}\,e^{-ipx/\hbar}dx = \frac{1}{\sqrt{\pi\hbar L}}\,\frac{\pi/L}{(\pi/L)^2 - (p/\hbar)^2}\, \Big(1 + e^{-ipL/\hbar}\Big),$$

after elementary exponential integrals. Then

$$|\Phi(p)|^2 = \frac{4\pi L}{\hbar}\, \frac{\cos^2\!\left(\dfrac{pL}{2\hbar}\right)}{\left[\pi^2 - \left(\dfrac{pL}{\hbar}\right)^2\right]^2},$$

using $|1 + e^{-ipL/\hbar}|^2 = 4\cos^2(pL/2\hbar)$.

The distribution peaks near $p = \pm\pi\hbar/L$ (the limit is finite there — l'Hôpital), i.e. near the de Broglie momenta $\pm\hbar k_1$ of the standing wave, but with tails: a bound state has no sharp momentum. A standing wave is an equal-weight superposition of left- and right-movers, so $\langle p\rangle = 0$ while $\langle p^2\rangle = \pi^2\hbar^2/L^2 = 2mE_1 \neq 0$ — the Born rule in momentum space, doing real work.

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Hands-on (Python)

import numpy as np
import matplotlib.pyplot as plt

# --- Free Gaussian packet: EXACT evolution in k-space via FFT ----------------
# Numerical values hbar = m = 1 (physics keeps ħ explicit; arrays don't care).
hbar, m = 1.0, 1.0
N, Lbox = 4096, 400.0
x = np.linspace(-Lbox/2, Lbox/2, N, endpoint=False)
dx = x[1] - x[0]
k = 2*np.pi*np.fft.fftfreq(N, d=dx)          # FFT wavenumber grid

a, k0 = 1.0, 5.0                              # width Δx0 = a, mean momentum ħk0
psi0 = (2*np.pi*a**2)**(-0.25) * np.exp(-x**2/(4*a**2)) * np.exp(1j*k0*x)
print(np.trapz(np.abs(psi0)**2, x))           # 1.0 (normalized)

def evolve(psi, t):
    """Free evolution: FFT -> multiply by e^{-i ħ k² t / 2m} -> inverse FFT (exact)."""
    return np.fft.ifft(np.fft.fft(psi) * np.exp(-1j*hbar*k**2*t/(2*m)))

for t in [0, 4, 8, 12]:
    plt.plot(x, np.abs(evolve(psi0, t)), label=f"t = {t}")
plt.xlim(-15, 80); plt.xlabel("x"); plt.ylabel(r"$|\Psi|$"); plt.legend(); plt.show()
# Expected: |Ψ| translates at v_g = ħk0/m = 5 (center at x = 5t) while widening.

# --- Δx(t) vs the analytic spreading law; uncertainty product over time ------
def moments(psi):
    """Return (Δx, Δp) — Δx from |Ψ(x)|², Δp = ħΔk from the FFT spectrum |Φ(k)|²."""
    rho = np.abs(psi)**2
    xm  = np.trapz(x*rho, x)
    dx_ = np.sqrt(np.trapz((x - xm)**2 * rho, x))
    phi2 = np.abs(np.fft.fft(psi))**2         # ∝ |Φ(k)|² on the FFT grid
    km  = np.sum(k*phi2) / np.sum(phi2)
    dk_ = np.sqrt(np.sum((k - km)**2 * phi2) / np.sum(phi2))
    return dx_, hbar*dk_

ts = np.linspace(0, 12, 25)
dxs, dps = np.array([moments(evolve(psi0, t)) for t in ts]).T
dx_theory = a*np.sqrt(1 + (hbar*ts/(2*m*a**2))**2)   # Δx(t) = Δx0 √(1+(ħt/2mΔx0²)²)

plt.plot(ts, dxs, "o", label=r"numerical $\Delta x(t)$")
plt.plot(ts, dx_theory, "k-", label="analytic law")
plt.xlabel("t"); plt.legend(); plt.show()
# Expected: dots on the curve to ~4 decimal places.

print(dxs[0]*dps[0] / hbar)                   # ≈ 0.5000 — saturation ΔxΔp = ħ/2 at t = 0
print((dxs*dps/hbar).min())                   # ≥ 0.5 at ALL times (grows after t = 0):
                                              # Δp stays constant (free), Δx grows.

# --- Fourier reciprocity: narrow in x <-> broad in p --------------------------
fig, ax = plt.subplots(2, 2, figsize=(9, 5))
for col, aa in enumerate([0.5, 4.0]):         # a narrow and a wide packet
    psi = (2*np.pi*aa**2)**(-0.25)*np.exp(-x**2/(4*aa**2))*np.exp(1j*k0*x)
    phi = np.fft.fftshift(np.fft.fft(psi))
    ks  = np.fft.fftshift(k)
    ax[0, col].plot(x, np.abs(psi)**2); ax[0, col].set_xlim(-15, 15)
    ax[0, col].set_title(f"$\\Delta x_0$ = {aa}")
    ax[1, col].plot(ks, np.abs(phi)**2); ax[1, col].set_xlim(k0-4, k0+4)
    ax[1, col].set_xlabel("k")
plt.tight_layout(); plt.show()
# Expected: left column narrow-in-x / broad-in-k; right column the reverse.
# The product of widths is the same (= 1/2): you can trade, never win.

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Exercises

E1 (easy). Deep-water gravity waves obey $\omega = \sqrt{gk}$. Compute $v_{\text{ph}}$ and $v_g$ and show $v_g = v_{\text{ph}}/2$. Compare with matter waves: which velocity does a surfer care about, and which does a wave packet of storm swell arrive at?

Solution

$v_{\text{ph}} = \omega/k = \sqrt{g/k}$; $v_g = d\omega/dk = \frac12\sqrt{g/k} = v_{\text{ph}}/2$. For water the group is slower than the crests (crests run forward through the packet and die at the front); for matter waves it is the opposite ($v_g = 2v_{\text{ph}}$). A surfer rides one crest ($v_{\text{ph}}$); the swell energy from a distant storm arrives at $v_g$ — in every wave theory, the transportable stuff moves at the group velocity.

E2 (easy). Verify $\hat p\,e^{ik_0x} = \hbar k_0\,e^{ik_0x}$, and compute $[\hat x, \hat p^2]$ by applying it to a test function. (Answer: $2i\hbar\,\hat p$.)

Solution

$-i\hbar\,\partial_x e^{ik_0x} = -i\hbar(ik_0)e^{ik_0x} = \hbar k_0 e^{ik_0x}$: plane waves are $\hat p$-eigenfunctions with eigenvalue $\hbar k_0$. For the commutator: $[\hat x,\hat p^2]f = -\hbar^2\big(x f'' - (xf)''\big) = -\hbar^2\big(xf'' - 2f' - xf''\big) = 2\hbar^2 f' = 2i\hbar,(-i\hbar f') = 2i\hbar,\hat p f$. (Or use the identity $[\hat A,\hat B\hat C] = [\hat A,\hat B]\hat C + \hat B[\hat A,\hat C]$ with $[\hat x,\hat p] = i\hbar$.)

E3 (medium). For the normalized Gaussian $\Psi(x,0) = (2\pi a^2)^{-1/4}e^{-x^2/4a^2}e^{ik_0x}$, compute $\langle p\rangle$ and $\langle p^2\rangle$ directly in position space with $\hat p = -i\hbar\partial_x$, and confirm $\Delta p = \hbar/2a$.

Solution

$\partial_x\Psi = \big(ik_0 - \frac{x}{2a^2}\big)\Psi$, so $\langle p\rangle = -i\hbar\int|\Psi|^2\big(ik_0 - \frac{x}{2a^2}\big)dx = \hbar k_0$ (the $x$ term integrates to zero by symmetry). For the second moment, $\hat p^2\Psi = -\hbar^2\big[\big(ik_0 - \frac{x}{2a^2}\big)^2 - \frac{1}{2a^2}\big]\Psi$, so $\langle p^2\rangle = -\hbar^2\big[-k_0^2 + \frac{\langle x^2\rangle}{4a^4} - \frac{ik_0}{a^2} \langle x\rangle - \frac{1}{2a^2}\big] = \hbar^2\big[k_0^2 - \frac{a^2}{4a^4} + \frac{1}{2a^2}\big] = \hbar^2k_0^2 + \frac{\hbar^2}{4a^2}$, using$\langle x\rangle = 0$,$\langle x^2\rangle = a^2$. Hence $\Delta p^2 = \hbar^2/4a^2$, i.e. $\Delta p = \hbar/2a$, and with $\Delta x = a$: $\Delta x\Delta p = \hbar/2$ — saturation, as claimed.

E4 (medium). Use the uncertainty principle to estimate the ground-state energy of the harmonic oscillator, $V = \frac12m\omega^2x^2$: write $E \approx \frac{\Delta p^2}{2m} + \frac12m\omega^2\Delta x^2$with$\Delta p = \hbar/2\Delta x$, minimize over$\Delta x$, and compare with the exact $E_0 = \hbar\omega/2$ (P.5.3).

Solution

$E(\Delta x) = \frac{\hbar^2}{8m\Delta x^2} + \frac12m\omega^2\Delta x^2$. Setting $dE/d\Delta x = 0$: $-\frac{\hbar^2}{4m\Delta x^3} + m\omega^2\Delta x = 0 \Rightarrow \Delta x^2 = \frac{\hbar}{2m\omega}$. Then $E_{\min} = \frac{\hbar\omega}{4} + \frac{\hbar\omega}{4} = \frac{\hbar\omega}{2}$ — exactly right (because the true ground state is Gaussian, saturating the bound). Moral: zero-point energy is mandatory — a particle cannot sit at the bottom of the well with $x = p = 0$, since that would mean $\Delta x = \Delta p = 0$.

E5 (hard). Derive the generalized Ehrenfest relation $\frac{d\langle Q\rangle}{dt} = \frac{i}{\hbar}\langle[\hat H, \hat Q]\rangle$ for any time-independent $\hat Q$ (assume you may integrate by parts freely, i.e. $\hat H$ is Hermitian). Recover from it both $\frac{d\langle x\rangle}{dt} = \langle p\rangle/m$ and $\frac{d\langle p\rangle}{dt} = -\langle V'\rangle$ using commutators only.

Solution

$\frac{d\langle Q\rangle}{dt} = \int(\partial_t\Psi^)\hat Q\Psi,dx + \int\Psi^\hat Q(\partial_t\Psi)dx = -\frac{1}{i\hbar}\int(\hat H\Psi)^\hat Q\Psi,dx + \frac{1}{i\hbar}\int\Psi^\hat Q\hat H\Psi,dx$. Hermiticity moves $\hat H$ off $\Psi^*$: $\int(\hat H\Psi)^*\hat Q\Psi\,dx = \int\Psi^*\hat H\hat Q\Psi\,dx$. Hence $\frac{d\langle Q\rangle}{dt} = \frac{1}{i\hbar}\langle\hat Q\hat H - \hat H\hat Q\rangle = \frac{i}{\hbar}\langle[\hat H,\hat Q]\rangle$. Now with$\hat H = \frac{\hat p^2}{2m} + V(\hat x)$: $[\hat H,\hat x] = \frac{1}{2m}[\hat p^2,\hat x] = -\frac{2i\hbar\hat p}{2m} = -\frac{i\hbar}{m}\hat p$ (E2), so $\frac{d\langle x\rangle}{dt} = \frac{i}{\hbar}\big(-\frac{i\hbar}{m}\big)\langle\hat p\rangle = \frac{\langle p\rangle}{m}$. And$[\hat H,\hat p] = [V,\hat p]$; on a test function, $[V,\hat p]f = V(-i\hbar f') + i\hbar(Vf)' = i\hbar V'f$, so $[\hat H,\hat p] = i\hbar V'$ and $\frac{d\langle p\rangle}{dt} = \frac{i}{\hbar}\langle i\hbar V'\rangle = -\langle V'\rangle$. ✓ Both Ehrenfest relations drop out of one commutator identity — the identity that becomes the Heisenberg equation of motion in Term 1, and whose classical twin is $\dot Q = \{Q, H\}$ from P.1.3.

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Checkpoint

1. Why must a physical free particle be a wave packet, and what does the stationary-phase argument say about where the packet is?
2. For matter waves, what are $v_g$ and $v_{\text{ph}}$, and why is $v_{\text{ph}} = v_g/2$ not a contradiction?
3. State the uncertainty principle and identify precisely which two ingredients produce it.
4. How is $\hat p = -i\hbar\,\partial_x$ derived rather than postulated in this lesson?
5. State Ehrenfest's theorem. In what sense, and under what condition, do quantum expectation values follow classical trajectories?

Answers

1. Plane waves are not square-integrable, hence not states; a normalizable free state is a continuous superposition $\int\phi(k)e^{i(kx-\omega t)}dk$. Stationary phase: the packet sits where $d\theta/dk|_{k_0} = 0$, i.e. at $x = \omega'(k_0)t$ — it moves at the group velocity.
2. $v_g = d\omega/dk = \hbar k_0/m = p/m$ (the classical velocity); $v_{\text{ph}} = \omega/k = \hbar k_0/2m = v_g/2$. No contradiction: crests of the unobservable carrier move at $v_{\text{ph}}$, but $|\Psi|^2$ — the measurable density — rides the envelope at $v_g$.
3. $\Delta x\,\Delta p \ge \hbar/2$. Ingredients: (i) Fourier reciprocity $\Delta x\,\Delta k \ge 1/2$ (pure mathematics of wave packets), (ii) de Broglie $p = \hbar k$ (physics). It is a property of states, not of measurements.
4. From dynamics: differentiate $\langle x\rangle$ using the continuity equation, integrate by parts twice, and define $\langle p\rangle = m\,d\langle x\rangle/dt$; the integrand that emerges is $\Psi^*(-i\hbar\partial_x)\Psi$, identifying $\hat p = -i\hbar\partial_x$.
5. $d\langle x\rangle/dt = \langle p\rangle/m$ and $d\langle p\rangle/dt = -\langle\partial_xV\rangle$. Expectation values obey Newton-like equations; they follow genuinely classical trajectories when $\langle V'(x)\rangle \approx V'(\langle x\rangle)$ — packets narrow on the scale over which the force varies (exact for potentials up to quadratic).

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Further Reading

• [Gri] Griffiths & Schroeter, §1.5–1.6 & §2.4 — momentum, the uncertainty principle, and the free-particle wave packet.
• [Gri] Griffiths & Schroeter, §3.5 — the general uncertainty principle (Robertson), for the Term 1 view.
• [ER] Eisberg & Resnick, Ch. 3 — de Broglie waves, group velocity, and the uncertainty principle historically.
• [Gold] Goldstein, Poole & Safko, §9.5–9.6 — Poisson brackets and canonical transformations, the classical side of $[\hat x,\hat p] = i\hbar$.
• [Sha] Shankar, Ch. 6 & §9.3 — the classical limit and Ehrenfest's theorem done carefully.

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