Wave Mechanics

Chapter 5's bridge asked the question; here is the answer, and it is smaller than you fear. For the qubit, a complete question had two answers, so a state was two amplitudes. For a particle on a line, the complete question is “where are you?” — and it has a continuum of answers. So the state is a continuum of amplitudes: one complex number for each point,

The wavefunction is not a new kind of object. It is the same list of amplitudes you have squared and summed since Chapter 2 — just indexed by a real number instead of ±. Sums become integrals; the bookkeeping survives untouched.

What is momentum in this language? Take the state whose behavior under motion is simplest — the one that looks the same everywhere except for a steadily turning phase, ${\psi }_k(x)=e^{ikx}$: a corkscrew of pitch $\lambda =2\pi /k$. Chapter 1 already told us what turns that crank: the phase of a path advanced by $kr$, and interference patterns measured $p=h/\lambda =\hslash k$ directly — the matter-wave relation Louis de Broglie dared to write down in 1924. So definite momentum is definite wavelength, and the operator that reads the local crank rate is a derivative:

Check that commutator in one line — apply $\hat{x}\hat{p}-\hat{p}\hat{x}$ to any function and watch the product rule leave $i\hslash$ behind. It is a constant, not an operator that could vanish in some clever state — which is why, by Robertson (4.5), no state of anything anywhere escapes $\Delta x\Delta p\ge \hslash /2$. Position and momentum are the same information read in two bases — a function and its Fourier transform (the same shape rewritten as a recipe of pure sine and cosine waves — which wavelengths, in what amounts — i.e. its momentum content) — and a function cannot be narrow together with its own transform. That is the uncertainty principle's deepest costume: a theorem about waves, older than quantum mechanics itself.

Now feed the energy operator $\hat{H}={\hat{p}}^2/2m+V(\hat{x})$ into Chapter 5's law of motion. In the position basis it becomes a partial differential equation:

For a free particle each corkscrew $e^{ikx}$ is a stationary state with $E={\hslash }^2{k}^2/2m$, so it turns at $\omega (k)=\hslash k^2/2m$ — and there is the strangeness: quadratic dispersion. Light has $\omega =ck$, all colors ride together, and a light pulse holds its shape. Matter waves of different wavelength travel at different speeds, so any lump built from them must come apart. Build the lump — a packet of corkscrews centered on $k_0$ — and stationary phase (the rule that a sum of many waves cancels itself everywhere except where their phases momentarily stop changing with $k$, so the lump lives where its component clocks briefly agree) gives its two speeds:

The envelope moves at the classical velocity $p/m$ — the correspondence principle riding in the group velocity — while the ripples inside crawl at half that speed, and the mismatch between neighboring speeds spreads the envelope as

Enough derivation — fly the thing. This widget is a real Schrödinger solver running in your browser (the same split-operator algorithm as the Rust lab). Start with Free flight and watch both facts of (6.4) at once: the green ripples lagging inside the blue envelope, and the envelope quietly fattening:

One more connection before the walls arrive: the Gaussian packet is exactly the minimum-uncertainty state that saturated Robertson's bound in Chapter 4 — and equation (6.5) says even perfection cannot last. The moment $t>0$, the momentum spread that localization demanded starts cashing itself out as position spread. For an electron localized to an atom's width, the doubling time is a few times 10⁻¹⁶ seconds; for a thrown baseball, longer than the age of the universe by a factor so large it has no name. That single formula is why your desk does not visibly obey this chapter.

Free flight was only the overture. The rest of this chapter throws the wave at a small family of potential landscapes — the standard obstacle course of one-dimensional quantum mechanics, collected in Figure 6.1. A step of height $V_0$ where the wavelength changes abruptly; a rectangular barrier of width $a$ and height $V_0$ that a wave of energy $E<V_0$ has no classical right to cross; a finite square well of depth $V_0$ deep enough to trap a few standing waves; and the infinite box of width $L$, walls so high nothing leaks out. One wave equation (6.3) throughout — only the landscape $V(x)$ changes.

Start with the simplest. Switch the widget to Step and send the packet at a cliff of height $V_0$ lower than its energy. A classical marble rolls up, slows, and continues — always. The wave does something a marble never does: part of it comes back. Match $\psi$ and ${\psi }^{\prime }$ at the cliff edge (the wavefunction cannot kink, or ${\hat{p}}^2\psi$ would blow up) and for a plane wave the arithmetic is four lines:

which an optician will recognize instantly: it is the Fresnel reflection formula, the same law that makes window glass throw back your face. Any abrupt change of wavelength reflects waves — light at glass, matter at cliffs; the medium does not matter, the waveness does. And with the cliff higher than the energy, the wave does not stop at the wall like a marble: it seeps in, an evanescent tail $e^{-\kappa x}$ with $\kappa =\sqrt{2m(V_0-E)}/\hslash$ — exponentially dying, but not zero. Hold that tail. It is about to become a superpower.

Make the wall thin. The evanescent tail seeps in from the left face — and if the wall ends before the tail dies, the remnant walks out the far side as a traveling wave again, feebler but free. Switch the widget to Barrier, set the height above the packet's energy, and watch: most of the packet bounces, and a ghost of it steps through six feet of classically solid wall. For a rectangular barrier of width $a$, matching wavefunctions at both faces gives

Read the punchline exponent: every unit of thickness costs a factor $e^{-2\kappa }$. Nature taxes tunneling exponentially, and that exponential sensitivity is not a nuisance — it is a product line. Alpha decay: the same nucleus, give or take a few percent of barrier, decays in microseconds or in billions of years — Gamow explained a 20-order-of-magnitude spread of lifetimes with equation (6.7)'s exponent. The scanning tunneling microscope: park a needle an atom's breadth above a surface and the tunneling current changes by a factor of ten per ångström of gap — a distance meter so absurdly sensitive it draws individual atoms. And the flash memory holding this page: electrons tunneled through an oxide wall to get in, and the same wall's exponential opacity keeps them there for decades, unpowered.

One scene remains: trap the wave. Between two walls a wave can only survive by interfering constructively with its own reflections — the round trip must advance the phase by a whole number of turns. For hard walls a distance $L$ apart, that means fitting half-wavelengths exactly, and the allowed energies snap to a ladder:

There it is — the answer to the oldest question in this book. Why are atomic energies discrete? Not because anyone postulated orbits: because bound waves are standing waves, and standing waves come in whole numbers, for exactly the reason a guitar string has a fundamental and overtones and nothing in between. Confinement quantizes. Note also what (6.8) says at $n=1$: the caged wave cannot have zero energy. A confined particle must bend its wavefunction inside the box, and curvature is kinetic energy — the zero-point energy that keeps helium liquid at absolute zero, enforced by $\Delta x\Delta p\ge \hslash /2$.

Real wells have soft walls and finite depth, and the leaky edges do exactly what Section 6.3 taught: the standing wave spills its evanescent tails into the forbidden region, the effective box widens, the energies slide down, and only a finite handful of states fit. The lab found this well's entire population — two:

But how did the solver find those two states without ever touching a transcendental equation? With a trick worth savoring on its own, because here the method is the lesson. Rotate time itself: replace $t$ with $-i\tau$, and every stationary state's phase factor $e^{-iEt/\hslash }$ — a clock hand that merely spins — turns into a real decay $e^{-E\tau /\hslash }$. Start from any random lump, let it evolve in this imaginary time, and every component fades — but the higher its energy, the faster it dies. Renormalize after each step so the total probability stays 1, and what survives the cooling is the slowest-decaying piece of all: the ground state, distilled straight out of noise. Want the first excited state? Project the ground state out at every step (the Gram–Schmidt move of Chapter 3) so it can never regrow, and the next-slowest survivor rises to take its place; repeat to climb the ladder one rung at a time.

This is not a classroom toy. Imaginary-time evolution, in exactly this form, is how modern quantum chemistry hunts molecular ground states, how lattice QCD reads the lightest particle off the long-time tail of a correlator, and how diffusion Monte Carlo cools many-electron systems that no transcendental equation could ever corner. The finite well's two roots were a warm-up; the same descent scales to problems with no closed form at all — which is why you will meet it again in Chapter 16.

The lab Rust-QP/ch06-wavemech earns this chapter's pictures. The engine is the split-operator method, and it is built on a piece of luck worth savoring: $e^{-i(T+V)dt}$ is hard because $\hat{T}$ and $\hat{V}$ do not commute (Chapter 3's lesson, now charging rent) — but each factor alone is trivial: $V$ is diagonal in position, $T$ is diagonal in momentum, and the FFT teleports between the two in $N\log N$ operations. Trotter's splitting stitches them together with error $O(d{t}^3)$ per step, and because every factor is a pure phase, the evolution is exactly unitary — the norm cannot drift except by float roundoff:

1/// One Strang step: half-V, full-T (in k-space), half-V.
2/// `dt` may be complex: real dt = real time, -i*tau = imaginary time.
3fn step(&self, psi: &mut [C], dt: C) {
4    let n = self.grid.n as f64;
5    for (p, &v) in psi.iter_mut().zip(&self.v) {
6        *p *= (-c(0.0, 1.0) * v * dt * 0.5).exp();
7    }
8    self.fft.process(psi);
9    for (p, &k) in psi.iter_mut().zip(&self.grid.k) {
10        *p *= (-c(0.0, 1.0) * (k * k / 2.0) * dt).exp();
11    }
12    self.ifft.process(psi);
13    for p in psi.iter_mut() {
14        *p /= n; // rustfft is unnormalized
15    }
16    for (p, &v) in psi.iter_mut().zip(&self.v) {
17        *p *= (-c(0.0, 1.0) * v * dt * 0.5).exp();
18    }
19}

Now the war story, because this lab bit its author twice. The first tunneling run compared packet counts against the rectangle formula (6.7) — and failed its own assertion by 1.6%. The culprit: a sharp-edged barrier sampled on a grid has no definite width at the scale of one cell, and (6.7) is exponentially sensitive to width. The fix was not to loosen the tolerance — it was to get honest: make the barrier smooth (real barriers are), and check it against a second, grid-free method — direct integration of the stationary equation:

1/// Numerically exact transmission for ANY smooth barrier: integrate the
2/// stationary Schrödinger equation u'' = 2(V - E)u from the transmitted
3/// side (pure e^{ikx}) leftward through the barrier with RK4, then read
4/// off the incident amplitude A from u = A e^{ikx} + B e^{-ikx}.
5/// T = 1/|A|^2. No grid, no edges — a second, independent method.
6fn t_numeric(e: f64, v: &dyn Fn(f64) -> f64, x_l: f64, x_r: f64) -> f64 {
7    // …
8    let a_inc = (-ik * x).exp() * (u + du / ik) * 0.5;
9    1.0 / a_inc.norm_sqr()
10}
11
12// …
13
14// referee the referee: t_numeric on a TRUE rectangle must reproduce
15// the analytic formula (integrate with edges just inside the range)
16let rect = move |x: f64| -> f64 { if x.abs() < a / 2.0 { v0 } else { 0.0 } };
17let mut method_err = 0.0f64;
18for &e in &[0.3, 0.6, 0.9, 1.1, 1.6] {
19    let tn = t_numeric(e, &rect, -8.0, 8.0);
20    method_err = method_err.max((tn - t_exact(e, v0, a)).abs());
21}
22
23// …
24
25/// One referee: what it achieved, the tolerance it had to beat, and the verdict.
26/// `pass` demands finiteness FIRST — a NaN is not a small number.
27#[derive(Serialize)]
28struct Check {
29    value: f64,
30    tol: f64,
31    pass: bool,
32}
33
34fn check(value: f64, tol: f64) -> Check {
35    Check { value, tol, pass: value.is_finite() && value < tol }
36}
37
38// …
39
40// THE REFEREES. Tolerances sit at the achieved accuracy — tighten these
41// when the numerics improve, never loosen them to make a run pass.
42let referee = RefereeReport {
43    // Roundoff accumulates with step count: the barrier runs take up to
44    // 26,667 steps on 8192 points, the free run 5,000 on 2048.
45    norm_drift_free: check(norm_drift_free, 3e-12),
46    norm_drift_barrier: check(norm_drift_barrier, 2e-11),
47    spread_err: check(spread_err, 1e-13),
48    method_err: check(method_err, 2e-7),
49    tun_err: check(tun_err, 8e-6),
50    bound_err: check(bound_err, 2e-6),
51};

Every number in the table below is read out of the JSON this lab writes, next to the tolerance it had to beat — so the scoreboard cannot go stale behind the prose, and you can see exactly how much margin each claim has left. Two of the rows deserve a word of warning about what they do not prove. The free-flight norm drift is measured where $V\equiv 0$, so both potential half-steps are $e^0=1$: it tests the FFT round trip and nothing else, and it is named for that. Unitarity with a potential acting is a separate row, measured during the tunneling runs where $V$ is nonzero on every one of 8192 points. A referee that can only pass is not a referee; a referee whose name overstates it is worse.

The bound-state row came from §6.5's imaginary-time descent — running time imaginary, $t\to -i\tau$, so that $e^{-\hat{H}\tau }$ starves every state but the lowest — cooling as arithmetic. It needed the same edge-alignment care as the barrier: the well's walls are placed exactly midway between grid samples, so the sampled well and the ideal rectangle are the same object. The sharp edge bit twice.

1. (F) Derive the spreading law (6.5): write the packet in momentum space, evolve each mode by $e^{-i\hslash k^{2}t/2m}$, and compute the position variance of the result. Where exactly does the spreading come from in your algebra?
2. (F) Derive the step formulas (6.6) by matching $\psi$ and ${\psi }^{\prime }$ at $x=0$, and check the strange corner: at $E\to V_0^{+}$, what fraction reflects? (All of it — a wave barely cresting a cliff almost surely comes back.)
3. (C) In the widget's barrier mode set V₀ = 1.0. The widget's wall is now the lab's wall to the digit — same width, same softness, same crest — so it is fair to read the chart's red curve (the smooth barrier; the gold one is the sharp rectangle) against it. Find the k₀ where roughly half the packet gets through, convert to $E=k_0^2/2$, and compare. Expect a few percent, not more: as the widget's fine print says, its packet is narrower in x than the lab's and therefore averages T over a wider band of k. Then lower k₀ until the transmitted ghost is barely visible and estimate T by eye from the left/right counters — how many e-foldings of suppression is that? (And why must you watch the absorbed fraction before you trust the ratio?)
4. (F, hard) Ehrenfest with teeth: prove $\frac{d}{dt}⟨\hat{x}⟩=⟨\hat{p}⟩/m$ and $\frac{d}{dt}⟨\hat{p}⟩=-⟨V^{\prime }(\hat{x})⟩$ from (5.4). Then find the precise crime: $⟨V^{\prime }(\hat{x})⟩\ne V^{\prime }(⟨\hat{x}⟩)$ in general. Expand $V^{\prime }$ about $⟨x⟩$ to show the error is driven by ${\sigma }^2{V}^{\prime \prime \prime }$ — so packets follow Newton exactly in uniform fields and harmonic wells (Chapter 7!), and betray him wherever the force curves across the packet's width.
5. (F, hard) Derive the tunneling formula (6.7) by the transfer-matrix method: write plane waves in the three regions, match at both faces to get a 2×2 matrix per interface, multiply, and extract T. Then prove the resonance result your chart shows above the barrier: T = 1 exactly when $k_{2}a=n\pi$ — the wall turns invisible when it holds a whole number of half-wavelengths (the Ramsauer–Townsend effect, seen in electron–atom scattering in 1921, before anyone could explain it).
6. (P) Add a “double barrier” mode to the lab (or the widget): two barriers separated by a gap. Sweep energy finely and find the sharp transmission resonances where T → 1 through two nearly opaque walls — you are simulating a Fabry–Pérot cavity for electrons, the working principle of the resonant-tunneling diode.
7. (P, hard) Quantum revivals. Put a Gaussian in a hard box of width L and evolve for a long time. The packet disperses into apparent chaos — but because all box energies are multiples of $E_1$, the state is exactly periodic with $T_{\text{rev}}=4m{L}^2/\pi \hslash$. Verify the full revival numerically via the autocorrelation $|⟨\psi (0)|\psi (t)⟩{|}^2$, then look for the fractional revivals at $T_{\text{rev}}/2$ and $T_{\text{rev}}/4$ — the state resurrects as two and four ghost copies of itself. (Plot |ψ|² at those instants; the structure is startling.)