# 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**.
6. State the **energy–time relation** in its precise (Mandelstam–Tamm) form, say what its $\Delta t$ means and does not mean, and apply it to lifetimes and natural linewidths.
---
## Intuition
The free-particle stationary states are plane waves $e^{i(kx-\omega t)}$ — but
[P.4.1](01-schrodinger-equation-born-rule.md) 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.1.4](../c01-theoretical-mechanics/04-hamiltonian-mechanics.md).
---
## 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](02-stationary-states-superposition.md)):
$$
\Psi(x,t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\phi(k)\,e^{i(kx - \omega(k)t)}\,dk,
\qquad
\omega(k) = \frac{\hbar k^2}{2m},
$$
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](../../term-0-foundations/c03-fourier-methods/02-fourier-series-transforms.md)). 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](../../term-0-foundations/c03-fourier-methods/02-fourier-series-transforms.md)). 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](../../term-0-foundations/c03-fourier-methods/02-fourier-series-transforms.md). 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](../../term-1-quantum-mechanics/c03-measurement/02-expectation-uncertainty.md)),
with $[\hat x,\hat p] = i\hbar$ (derived below) supplying the right-hand side.
Heisenberg's own 1927 argument was a thought experiment, worth seeing once before it is put in its
place. To locate an electron, scatter light off it into a microscope whose objective accepts rays
within a half-angle $\alpha$. Diffraction limits the resolution to $\Delta x \approx \lambda/\sin\alpha$
([P.1.2](../c01-theoretical-mechanics/02-wave-propagation-diffraction.md)). The scattered photon, of
momentum $h/\lambda$, may enter the lens anywhere within the cone, so the recoil it gives the electron
along $x$ is uncertain by $\Delta p_x \approx (h/\lambda)\sin\alpha$. A shorter wavelength or a wider
lens sharpens $\Delta x$ and blurs $\Delta p_x$ in exact proportion, and
$\Delta x\,\Delta p_x \approx h$ whatever $\lambda$ and $\alpha$ are. The scale comes out right; the
story behind it does not, as the caution explains.
> **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.
One consequence is easy to lose: a particle has no size of its own. Its spatial extent is the
$\Delta x$ of whatever state it was prepared in:
```qm-demo wave-packet-width
```
### 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](01-schrodinger-equation-born-rule.md)).
### 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](01-schrodinger-equation-born-rule.md):
$$
\frac{d\langle x\rangle}{dt}
= \int x\,\frac{\partial|\Psi|^2}{\partial t}\,dx
= -\int x\,\frac{\partial j}{\partial x}\,dx
= \int j\,dx
= -\frac{i\hbar}{2m}\int\left(\Psi^*\frac{\partial\Psi}{\partial x} - \Psi\frac{\partial\Psi^*}{\partial x}\right)dx,
$$
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](01-schrodinger-equation-born-rule.md) 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).
### The energy–time relation
Position and momentum are both observables, measured at one instant on one ensemble. Energy and time
are not a pair of that kind: $t$ is the parameter that labels the evolution, read off a laboratory
clock, and there is no time operator conjugate to $\hat H$ (Pauli showed none can exist when the energy
is bounded below). A relation of the form $\Delta E\,\Delta t \gtrsim \hbar$ still holds, provided
$\Delta t$ is read correctly.
**The wave-packet estimate.** A packet of length $\Delta x$ moving at group velocity $v_g$ takes
$\Delta t \approx \Delta x/v_g$ to pass a fixed detector, and its energy spread is
$\Delta E \approx (dE/dp)\,\Delta p = v_g\,\Delta p$. Multiplying,
$\Delta E\,\Delta t \approx \Delta x\,\Delta p \gtrsim \hbar/2$: a pulse that passes quickly must
contain a broad range of energies. This is Fourier reciprocity again, now between $t$ and
$\omega = E/\hbar$.
**The precise statement.** Let $\hat Q$ be any observable without explicit time dependence. The
Robertson relation $\sigma_H\sigma_Q \ge \tfrac12\lvert\langle[\hat H,\hat Q]\rangle\rvert$
([1.3.2](../../term-1-quantum-mechanics/c03-measurement/02-expectation-uncertainty.md)) combined with
the generalized Ehrenfest relation $d\langle Q\rangle/dt = \tfrac{i}{\hbar}\langle[\hat H,\hat Q]\rangle$
(E5) gives $\sigma_H\sigma_Q \ge \tfrac{\hbar}{2}\lvert d\langle Q\rangle/dt\rvert$. Define
$\Delta t_Q \equiv \sigma_Q/\lvert d\langle Q\rangle/dt\rvert$, the time it takes $\langle Q\rangle$ to
move by one standard deviation of $Q$. Then
$$
\boxed{\;\sigma_H\,\Delta t_Q \ \ge\ \frac{\hbar}{2}\;}
$$
the **Mandelstam–Tamm relation** (1945). Here $\Delta E = \sigma_H$ is a genuine spread of energy in the
state, and $\Delta t$ is not an uncertainty in any time measurement: it is the time the system needs to
change appreciably. Three consequences:
- **Stationary states never change.** $\sigma_H = 0$ forces $\Delta t_Q = \infty$ for every $Q$, which
is the frozen dynamics of [P.4.2](02-stationary-states-superposition.md) seen from a new angle. Only
a state with an energy spread can evolve, and the faster it evolves the larger the spread must be.
- **Beats.** The superposition $(\psi_1 + \psi_2)/\sqrt2$ has $\sigma_H = (E_2 - E_1)/2$ and its
observables oscillate with period $2\pi\hbar/(E_2 - E_1)$, so $\sigma_H$ times the period is $\pi\hbar$:
comfortably above the bound (E6).
- **Lifetimes and linewidths.** An excited state that decays with lifetime $\tau$ has no sharp energy.
Exponential decay gives its emission line a Lorentzian profile of full width at half maximum
$\Gamma = \hbar/\tau$, the **natural linewidth**. Hydrogen's 2p state lives $\tau = 1.6$ ns, so
$\Gamma = 4.1\times10^{-7}$ eV, a frequency width $\Gamma/h = 1/2\pi\tau \approx 100$ MHz on the
$2.5\times10^{15}$ Hz Lyman-α line (E6). Short-lived states have broad lines; the qubits of Term 4 obey
the same trade between how long a state survives and how sharply its frequency is defined.
> **Caution.** The energy–time relation does not let a system "borrow" energy $\Delta E$ for a time
> $\Delta t$, and it does not limit how precisely an isolated system's energy is conserved: $\langle H\rangle$
> and every moment of $H$ are exactly constant under the Schrödinger equation. It constrains how fast a
> state with a given energy spread can change, nothing more.
### Connections: from Poisson brackets to commutators
- **Canonical quantization.** [P.1.4](../c01-theoretical-mechanics/04-hamiltonian-mechanics.md)
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](../../term-1-quantum-mechanics/c03-measurement/02-expectation-uncertainty.md)): 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](../c02-radiation-old-quantum/03-correspondence-and-limits.md) fulfilled: where packets stay
narrow compared to the scale on which $V'$ varies, quantum averages trace classical orbits.
---
## 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.
---
## Hands-on (Python)
```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.
```
```python
# --- 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.
```
```python
# --- 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.
```
---
## 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.
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
$[\hat x,\hat p] = i\hbar$.)
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.
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
$\Delta x = \Delta p = 0$.
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.1.4](../c01-theoretical-mechanics/04-hamiltonian-mechanics.md).
Solution
(a) $\Gamma = \hbar/\tau = 6.58\times10^{-16}\ \mathrm{eV\,s}/1.6\times10^{-9}\ \mathrm s = 4.1\times10^{-7}$
eV, i.e. $\Gamma/h = 1/2\pi\tau = 99$ MHz. The line itself sits at
$10.2\ \mathrm{eV}/h = 2.47\times10^{15}$ Hz, so the fractional width is $4\times10^{-8}$: a very sharp
line, because the state lives for some $10^{6}$ optical periods. (b) Energy outcomes $E_1$ and $E_2$
with probability $\tfrac12$ each give $\sigma_H = (E_2 - E_1)/2 = 5.1$ eV. Observables oscillate at the
Bohr frequency, period
$T = h/(E_2 - E_1) = 4.14\times10^{-15}\ \mathrm{eV\,s}/10.2\ \mathrm{eV} = 0.41$ fs. Then $\sigma_HT = \tfrac{E_2 - E_1}{2}\cdot\tfrac{2\pi\hbar}{E_2 - E_1} = \pi\hbar \ge \hbar/2$ ✓:
the faster the beat, the wider the energy spread it requires.
Solution
$\langle p\rangle_n = \frac{m}{i\hbar}\langle\psi_n\lvert[\hat x,\hat H]\rvert\psi_n\rangle = \frac{m}{i\hbar}\big(\langle\psi_n\lvert\hat x\hat H\rvert\psi_n\rangle - \langle\psi_n\lvert\hat H\hat x\rvert\psi_n\rangle\big)$.
In the first term $\hat H$ acts on $\psi_n$ to the right, giving $E_n\langle x\rangle_n$; in the
second, Hermiticity lets it act to the left, $\langle\psi_n\rvert\hat H = E_n\langle\psi_n\rvert$, giving
$E_n\langle x\rangle_n$ again. The difference vanishes, so $\langle p\rangle_n = 0$. (Equivalently,
$\langle x\rangle$ is constant in a stationary state, and $\langle p\rangle = m\,d\langle x\rangle/dt$.)
The argument needs $\langle x\rangle_n$ to exist and the integrations by parts behind Hermiticity to
have vanishing boundary terms. Both fail for $e^{ikx}$, which is not normalizable: it is not a state in
$L^2$ at all, and it carries a steady current to infinity. Bound states stay put; only unnormalizable
idealizations can "move forever" with sharp energy.
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).
6. $\sigma_H\,\Delta t_Q \ge \hbar/2$ with $\Delta t_Q = \sigma_Q/\lvert d\langle Q\rangle/dt\rvert$, from
Robertson plus generalized Ehrenfest. $\Delta t$ is the time for an observable's mean to move by one
standard deviation, a rate of change, not a time-measurement error. A stationary state has
$\sigma_H = 0$, so every $\Delta t_Q$ is infinite: nothing changes.