Matter as Waves

Light, a certified wave, turned out to arrive in particles (P.3.1). In 1924 de Broglie asked the symmetric question: do electrons, certified particles, travel as waves? Three years later two accidental experiments said yes, and the double slit — run one electron at a time — became the cleanest window into what quantum mechanics actually is: amplitudes that superpose, and detections that are always whole.

Learning Objectives

After this lesson you will be able to:

1. Derive the two-slit interference pattern $I(\theta) = 4I_0\cos^2(\pi d\sin\theta/\lambda)$ two ways — by far-field path difference and by the paraxial expansion — and invert the fringe spacing $\lambda L/d$ into a measurement of $\lambda$.
2. State the de Broglie hypothesis $\lambda = h/p$, $\nu = E/h$ and compute matter wavelengths from electrons to baseballs.
3. Derive Bohr's quantization $L = n\hbar$ from a standing matter wave on a circular orbit.
4. Verify the Davisson–Germer data quantitatively against $\lambda = h/p$.
5. Explain single-electron interference build-up and why which-way information destroys the fringes.
6. Map the two-path interferometer onto the qubit superposition of the main program.

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Intuition

Nature showed no interest in keeping "wave" and "particle" as separate castes for light; de Broglie bet it would not do so for matter either. If a photon of momentum $p$ has wavelength $\lambda = h/p$, assign the same wavelength to an electron of momentum $p$ and see what follows. Two things follow immediately. First, electrons fired at a crystal should diffract like X-rays — testable. Second, an electron bound in an atom is a wave on a closed loop, and a wave on a loop only fits if a whole number of wavelengths fits — exactly like the discrete notes of the clamped string in P.1.1. Discreteness of atomic orbits stops being a decree and becomes what waves always do in confinement.

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Theory

Young's double slit for light

Monochromatic light of wavelength $\lambda$ falls on two narrow slits a distance $d$ apart; a screen sits at $L \gg d$. Each slit acts as a coherent point source. At screen angle $\theta$ the path difference is $\delta = r_2 - r_1 = d\sin\theta$ (far field: the rays are effectively parallel). The fields at the screen are $E_1 = E_0\cos(kr_1 - \omega t)$ and $E_2 = E_0\cos(kr_2 - \omega t)$, $k = 2\pi/\lambda$; the sum-to-product identity gives

$$E = E_1 + E_2 = 2E_0\,\cos\!\Big(\frac{k\,d\sin\theta}{2}\Big)\,\cos\!\Big(k\,\frac{r_1 + r_2}{2} - \omega t\Big).$$

A detector responds to the time-averaged squared field, $\langle\cos^2(\cdot-\omega t)\rangle = \tfrac12$, so with $I_0$ the intensity from one slit alone,

$$I(\theta) = 4I_0\cos^2\!\Big(\frac{\pi d\sin\theta}{\lambda}\Big).$$

Maxima where whole wavelengths fit the path difference, $d\sin\theta = m\lambda$ ($m = 0, \pm1, \pm2, \dots$); zeros at half-integers. For small angles $\sin\theta \approx y/L$, bright fringes sit at $y_m = m\lambda L/d$, with uniform spacing $\Delta y = \lambda L/d$. Two checks: averaged over fringes, $\langle 4I_0\cos^2\rangle = 2I_0 = I_1 + I_2$ — interference redistributes energy, it does not create it; and real slits of finite width $a$ multiply the pattern by the single-slit envelope $[\sin\beta/\beta]^2$, $\beta = \pi a\sin\theta/\lambda$, included in the Hands-on — the diffraction envelope of P.1.2, with the two-slit fringes riding inside it.

The same result without angles: the paraxial derivation

The $d\sin\theta$ argument quietly assumed the two rays are parallel. Doing the geometry exactly is worth the five lines, because it shows precisely what is approximated and it hands back the fringe spacing in the coordinates an experimenter actually measures. Put the slits at $\pm d/2$ on the mask, the screen at distance $L$, and ask for the field at height $y$. The two path lengths are exact Pythagoras:

$$r_\mp = \sqrt{\big(y \mp d/2\big)^2 + L^2} = L\sqrt{1 + \big(y \mp d/2\big)^2/L^2} \;\simeq\; L + \frac{\big(y \mp d/2\big)^2}{2L} = \underbrace{L + \frac{y^2}{2L} + \frac{d^2}{8L}}_{\textstyle \equiv\, \alpha/k} \;\mp\; \frac{dy}{2L},$$

using $\sqrt{1+u} \simeq 1 + u/2$ for $|y|, d \ll L$ — the paraxial (Fresnel) approximation. The three collected terms are common to both paths; only the last differs, and it differs by a sign. Superposing $e^{ikr_-} + e^{ikr_+}$ therefore factorizes cleanly:

$$\psi(y) \;\propto\; e^{i\alpha}\Big[e^{\,ik\,dy/2L} + e^{-ik\,dy/2L}\Big] = 2e^{i\alpha}\cos\!\Big(\frac{k\,dy}{2L}\Big) = 2e^{i\alpha}\cos\!\Big(\frac{\pi dy}{\lambda L}\Big),$$

and the common factor $e^{i\alpha}$ — carrying the whole messy $L + y^2/2L + d^2/8L$ — has modulus 1, so it vanishes from the intensity:

$$\lvert\psi(y)\rvert^2 \;\propto\; \cos^2\!\Big(\frac{\pi dy}{\lambda L}\Big) = \tfrac12\Big[1 + \cos\Big(\frac{2\pi dy}{\lambda L}\Big)\Big].$$

Adjacent maxima are one period of that cosine apart, giving the fringe spacing

$$\boxed{\ \Delta y = \frac{\lambda L}{d}\ } \qquad\Longleftrightarrow\qquad \lambda = \frac{\Delta y\, d}{L},$$

identical to the far-field answer, as it must be. The second form is the one that matters: $\Delta y$, $d$, and $L$ are all measurable with a ruler, so two slits turn an unmeasurably small wavelength into a millimetre-scale one. The apparatus magnifies by $L/d$. That is how $\lambda$ was ever measured for light, and — with the electron gun of Example 2 — how it is measured for matter, which is the entire point of the next section.

Caution. Only relative phase survives. The discarded factor $e^{i\alpha}$ is enormous — at $L = 1$ m and $\lambda = 550$ nm the phase $kL$ is $\sim 10^7$ radians — yet it is invisible, because both paths carry it. Physical observables never depend on a common phase, only on the difference $k\,dy/L$ between paths. This is the classical rehearsal for the global-vs-relative phase distinction that runs through all of quantum computing (1.1.1): a global phase is unobservable; a relative phase is the signal.

The de Broglie hypothesis

For photons, P.3.1 established $E = h\nu$ and $p = h/\lambda$. De Broglie (1924) postulated that the same two relations attach a wave to every particle:

$$\lambda = \frac{h}{p}, \qquad \nu = \frac{E}{h}.$$

Pure symmetry — no derivation, no mechanism; a doctoral thesis so bold the committee mailed it to Einstein for a verdict ("he has lifted a corner of the great veil"). One subtlety: for a nonrelativistic free particle $E = p^2/2m$, so the phase velocity is $v_p = \nu\lambda = E/p = v/2$ — half the particle's speed. No contradiction: the particle rides the group velocity of a wave packet, which comes out exactly $v$ (P.4.3).

For an electron, $\lambda = h/m_ev$. Drag its speed and watch the wavelength fall as $1/v$:

Standing matter waves make Bohr's rule natural

Wrap the de Broglie wave around a circular orbit of radius $r$. The wave must return in phase with itself after one lap — otherwise successive turns interfere destructively and the wave annihilates itself. Constructive self-interference requires a whole number of wavelengths on the circumference, and substituting $\lambda = h/p$ turns that into Bohr's rule:

$$n\lambda = 2\pi r_n \quad\xrightarrow{\;\lambda = h/p\;}\quad n\,\frac{h}{p} = 2\pi r_n \quad\Longrightarrow\quad L = p\,r_n = n\,\frac{h}{2\pi} = n\hbar, \qquad n = 1, 2, 3, \dots$$

Bohr's quantization postulate (P.2.2) is no longer a postulate: atomic orbits are the standing-wave modes of the electron wave, the circular cousins of the clamped-string modes of P.1.1. Confinement plus waves equals discreteness, always.

Wrap the wave yourself, and push $n$ off a whole number to see the orbit fail:

Orders of magnitude: who gets to diffract

For a nonrelativistic particle $p = \sqrt{2mE}$, so $\lambda = h/\sqrt{2mE}$; for electrons this packages neatly as $\lambda \approx 1.226\,\mathrm{nm}/\sqrt{E/\mathrm{eV}}$.

• Electron, $E = 100\,\mathrm{eV}$: $\lambda = 1.226/\sqrt{100} = 0.123\,\mathrm{nm} \approx 1.2$ Å — the scale of atomic spacings. Crystals are free diffraction gratings for electrons.
• Baseball, $m = 0.15\,\mathrm{kg}$, $v = 40\,\mathrm{m/s}$: $p = 6.0\,\mathrm{kg\,m/s}$, so $\lambda = 6.626\times10^{-34}/6.0 = 1.1\times10^{-34}\,\mathrm{m}$ — nineteen orders of magnitude below the size of a nucleus.

Macroscopic diffraction is not forbidden; it is unobservable, because nothing offers slits or lattice structure at $10^{-34}\,\mathrm{m}$. The wave is always there; $h$ is just very small on kitchen scales.

Davisson–Germer and G. P. Thomson

Davisson–Germer (1927). Firing $54\,\mathrm{eV}$ electrons at nickel, they found a strong scattered peak at $\varphi = 50^\circ$ from the incident beam — after an accidental vacuum break forced them to anneal the target, which crystallized it. The surface atom rows, spacing $d = 0.215\,\mathrm{nm}$, act as a plane grating, so the grating equation of P.1.2, $d\sin\varphi = n\lambda$, applies unchanged — and read backwards, the peak measures a wavelength: $\lambda_{exp} = 0.215\sin 50^\circ = 0.165\,\mathrm{nm}$. De Broglie predicts $\lambda = 1.226/\sqrt{54} = 0.167\,\mathrm{nm}$ — agreement to about 1% (the residual is understood: the crystal's inner potential slightly refracts the electron wave). A particle's momentum had fixed a wavelength, exactly as prescribed.

G. P. Thomson (1927). Independently, Thomson passed keV electrons through thin polycrystalline foils and photographed concentric diffraction rings, geometrically identical to X-ray Debye–Scherrer rings from the same foils; he shared the 1937 Nobel Prize with Davisson. The family irony is standard-issue: J. J. Thomson won a Nobel for showing the electron is a particle; his son won one for showing it is a wave. Both were right.

Bullets, waves and electrons

Feynman sharpened the double slit into a three-way comparison ([FLP] Vol. III, Ch. 1): fire three kinds of thing at the same two-slit wall and record what arrives on the screen behind it. Write $P_1(x)$ for the arrival distribution with only slit 1 open and $P_2(x)$ with only slit 2.

• Classical particles (bullets, marbles). Each arrives whole, at one point, having passed through one slit or the other. Opening the second slit cannot change what happens to particles using the first, so with both open the distribution is $P_{12} = P_1 + P_2$: a broad hump with no fringes. Probabilities add.
• Waves (water ripples, light). Nothing arrives in lumps; the detector reads a continuous intensity. The disturbances from the two slits add, $h = h_1 + h_2$, and the intensity is $I_{12} = \lvert h_1 + h_2\rvert^2 = I_1 + I_2 + 2\sqrt{I_1I_2}\cos\delta$, where the cross term draws the fringes. Amplitudes add.
• Electrons. They arrive like bullets, whole, one at a time and at one point each, yet they distribute like waves: $P_{12} = \lvert\psi_1 + \psi_2\rvert^2 \ne P_1 + P_2$. At the center of a dark fringe, opening the second slit reduces the number of electrons arriving there, which no account of particles each traveling through one definite slit can explain.

Lumps on arrival, waves in the distribution: that is wave–particle duality in one sentence. The next two subsections push it to its hardest form, one electron at a time and then with the path watched.

One electron at a time

The deepest version: a double slit (in practice an electron biprism) with the source turned so far down that electrons traverse the apparatus one at a time — in Tonomura's 1989 Hitachi experiment successive electrons were separated by kilometers of flight path, so there is nothing for an electron to interfere with except itself. The detector records: each arrival as a single localized dot — one whole electron, never a fraction; no structure at all after the first $\sim 100$ electrons; and the clean $\cos^2$ two-slit fringes, with $\lambda = h/p$, once $\sim 10^4$–$10^5$ dots accumulate. The interference pattern is therefore a single-particle probability distribution: each electron's amplitude passes through both slits and interferes, and each detection delivers one whole electron at one point with density $\propto |\psi_1 + \psi_2|^2$. Wave propagation, particle detection — the photon's duality, now for matter.

Caution. The electron does not split in two, half through each slit — no experiment has ever caught half an electron. What superposes is the amplitude; the detection density is $|\psi_1 + \psi_2|^2$, and every detection is a whole electron. Add probabilities instead, $|\psi_1|^2 + |\psi_2|^2$, and the cross term — with every fringe — is erased. Amplitudes first, squared modulus second: that ordering is the entire content of the mystery.

Which-way information and complementarity

Add a monitor that records which slit each electron used. Every such scheme — light probe, spin tag, anything — leaves the monitor in a different state for the two paths, and the fringes vanish: the screen shows the structureless $|\psi_1|^2 + |\psi_2|^2$. Qualitatively, gaining path information is a measurement, and measurement disturbs the superposition it interrogates; the precise statement is Term 1's 1.3.1 Projective Measurement. Bohr packaged this as complementarity: wave behavior (fringes) and particle behavior (a definite path) are complementary aspects, and one experimental arrangement can fully exhibit only one of them. Modern interferometry makes the trade-off quantitative, but the 1927 slogan already contains the point.

Connections ahead: the interferometer is a qubit

Look at the two-path experiment with Term 1 eyes. Between slits and screen the electron has exactly two available paths, so its state lives in a two-dimensional complex vector space with basis $\{\lvert\text{slit }1\rangle, \lvert\text{slit }2\rangle\}$ — the primal qubit. A balanced superposition with relative phase $\delta$,

$$\lvert\psi\rangle = \tfrac{1}{\sqrt2}\big(\lvert\text{slit }1\rangle + e^{i\delta}\,\lvert\text{slit }2\rangle\big),$$

is precisely the superposition of 1.1.1 The State Postulate, and sweeping $\delta$ across the screen traces the fringes — relative phase made visible, the equator of 1.2.1 The Qubit. This is not an analogy but an identity, and it is the working capital of the whole program: quantum algorithms choreograph exactly such amplitude interference so that wrong answers cancel and right answers reinforce (Term 3).

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Worked Examples

Example 1 — Fringes with light

Green light, $\lambda = 550\,\mathrm{nm}$; slits $d = 0.25\,\mathrm{mm}$ apart; screen at $L = 1.5\,\mathrm{m}$. Fringe spacing: $\Delta y = \lambda L/d = 550\times10^{-9}\times1.5/(2.5\times10^{-4}) = 3.3\,\mathrm{mm}$ — comfortably visible by eye. First maximum at $\sin\theta = \lambda/d = 2.2\times10^{-3}$: the small-angle approximation is excellent.

Example 2 — Fringes with electrons

Electrons accelerated through $1.00\,\mathrm{kV}$: $\lambda = 1.226/\sqrt{1000} = 0.0388\,\mathrm{nm}$ ($3.88\times10^{-11}\,\mathrm{m}$; nonrelativistic is fine at 1 keV — see E5). Slits $d = 1.0\,\mathrm{\mu m}$ apart, screen at $L = 1.0\,\mathrm{m}$: $\Delta y = \lambda L/d = 3.88\times10^{-11}/10^{-6} = 3.9\times10^{-5}\,\mathrm{m} \approx 39\,\mathrm{\mu m}$. Small but resolvable — essentially Jönsson's 1961 geometry, later repeated one electron at a time. Note the trade: $\lambda$ shrank by four orders of magnitude versus light, so $d$ had to shrink almost as much to keep the fringes visible.

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Hands-on (Python)

Two-slit intensity for the electron parameters of Example 2, including the single-slit envelope.

import numpy as np
import matplotlib.pyplot as plt

h, me, eV = 6.62607015e-34, 9.1093837015e-31, 1.602176634e-19
E   = 1000 * eV                          # 1 keV electrons
lam = h / np.sqrt(2 * me * E)            # de Broglie wavelength = 3.88e-11 m
d, a, L = 1.0e-6, 0.3e-6, 1.0            # slit separation, slit width, screen distance (m)

y     = np.linspace(-150e-6, 150e-6, 4001)   # screen coordinate, m
sin_t = y / L                                 # small-angle sin(theta)
beta  = np.pi * a * sin_t / lam               # single-slit phase
delta = np.pi * d * sin_t / lam               # two-slit phase
env   = np.sinc(beta / np.pi) ** 2            # np.sinc(x) = sin(pi x)/(pi x)
I     = 4 * env * np.cos(delta) ** 2          # intensity in units of one-slit peak I0

plt.plot(y * 1e6, I, lw=0.8, label="two-slit pattern")
plt.plot(y * 1e6, 4 * env, "--", label="single-slit envelope")
plt.xlabel("y (μm)"); plt.ylabel("I / I0"); plt.legend(); plt.show()
# Fringes spaced lam * L / d = 38.8 μm under a sinc^2 envelope
# whose first zero sits at y = lam * L / a ≈ 129 μm.

Now the Tonomura experiment in silico: treat the normalized intensity as $|\psi|^2$ and draw whole electrons from it, one at a time.

rng = np.random.default_rng(42)
p = I / I.sum()                          # normalized |psi|^2 -> pmf on the screen grid

fig, axes = plt.subplots(3, 1, figsize=(6, 7), sharex=True)
for ax, N in zip(axes, (100, 1_000, 100_000)):
    hits = rng.choice(y, size=N, p=p)    # each hit = ONE whole electron at ONE point
    ax.hist(hits * 1e6, bins=150)
    ax.set_ylabel(f"N = {N}")
axes[-1].set_xlabel("y (μm)")
plt.tight_layout(); plt.show()
# N = 100: random-looking specks.  N = 1,000: bands suggest themselves.
# N = 100,000: crisp cos^2 fringes — the pattern assembles one cat-step at a
# time, yet every single event was a whole electron at a single point.

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Exercises

E1 (easy). A neutron ($m_n = 1.675\times10^{-27}\,\mathrm{kg}$) is in thermal equilibrium at $T = 300\,\mathrm{K}$, with kinetic energy $E = \tfrac32 k_BT$. Find its de Broglie wavelength. Why are "thermal neutrons" a standard crystallography probe?

Solution

$E = 1.5\times(8.617\times10^{-5}\,\mathrm{eV/K})\times300 = 0.0388\,\mathrm{eV} = 6.21\times10^{-21}\,\mathrm{J}$, so $p = \sqrt{2m_nE} = 4.56\times10^{-24}\,\mathrm{kg\,m/s}$ and $\lambda = h/p = 1.45\times10^{-10}\,\mathrm{m} = 0.145\,\mathrm{nm}$ — right at typical lattice spacings, so crystals diffract thermal neutrons strongly (and, being uncharged, neutrons probe the bulk).

E2 (easy). A helium–neon laser ($\lambda = 632.8\,\mathrm{nm}$) illuminates slits with $d = 0.40\,\mathrm{mm}$, screen at $L = 2.0\,\mathrm{m}$. Find the fringe spacing and the position of the third-order bright fringe.

Solution

$\Delta y = \lambda L/d = 632.8\times10^{-9}\times2.0/(4.0\times10^{-4}) = 3.2\,\mathrm{mm}$; third order at $y_3 = 3\Delta y = 9.5\,\mathrm{mm}$ (small-angle check: $\sin\theta_3 = 3\lambda/d = 4.7\times10^{-3}$, amply small).

E3 (medium). For hydrogen's ground state ($n = 1$, $r_1 = a_0 = 0.0529\,\mathrm{nm}$), find the electron's de Broglie wavelength from the standing-wave condition and verify it equals the orbit's circumference. What is $\lambda$ for the $n = 3$ orbit ($r_n = n^2a_0$)?

Solution

Standing-wave condition $n\lambda_n = 2\pi r_n$. For $n = 1$: $\lambda_1 = 2\pi a_0 = 0.332\,\mathrm{nm}$ — exactly one wavelength around the circumference, by construction. For $n = 3$: $\lambda_3 = 2\pi(9a_0)/3 = 6\pi a_0 = 0.997\,\mathrm{nm}$ — higher orbits carry slower electrons (smaller $p$), hence longer wavelengths, with exactly three fitting the $n = 3$ loop.

E4 (medium). At a bright fringe both slits contribute equal amplitudes $\psi_0$ in phase. Compare the detection density predicted by (a) adding amplitudes and (b) adding probabilities, and compute the fringe visibility $V = (I_{max} - I_{min})/(I_{max} + I_{min})$ for each.

Solution

(a) Amplitudes: $|\psi_0 + \psi_0|^2 = 4|\psi_0|^2$ at maxima, $|\psi_0 - \psi_0|^2 = 0$ at minima: $V = (4-0)/(4+0) = 1$. (b) Probabilities: $|\psi_0|^2 + |\psi_0|^2 = 2|\psi_0|^2$ everywhere: $V = 0$, no fringes. The interference term $2\,\mathrm{Re}(\psi_1^*\psi_2)$ is exactly what which-way information deletes — and note that (a) gives twice the classical density at maxima while conserving total counts (the deficit sits in the zeros).

E5 (hard). For electrons accelerated through a potential $V$, derive the relativistically correct de Broglie wavelength $\lambda = \frac{h}{\sqrt{2m_eeV}}\big(1 + \frac{eV}{2m_ec^2}\big)^{-1/2}$, and evaluate it for Tonomura's $V = 50\,\mathrm{kV}$ electrons. How large is the error of the nonrelativistic formula?

Solution

With $K = eV$ and total energy $E = K + m_ec^2$, the relation $E^2 = (pc)^2 + (m_ec^2)^2$ from P.3.1 gives $(pc)^2 = (K + m_ec^2)^2 - (m_ec^2)^2 = K^2 + 2Km_ec^2 = 2m_ec^2K\big(1 + \frac{K}{2m_ec^2}\big)$, so $\lambda = h/p = \frac{h}{\sqrt{2m_eK}}(1 + K/2m_ec^2)^{-1/2}$. Numbers: $K = 50\,\mathrm{keV}$, $m_ec^2 = 511\,\mathrm{keV}$, so $pc = \sqrt{50^2 + 2\times50\times511} = 231.5\,\mathrm{keV}$ and $\lambda = hc/pc = 1240\,\mathrm{eV\,nm}/2.315\times10^5\,\mathrm{eV} = 5.36\,\mathrm{pm}$. The nonrelativistic formula gives $1.226/\sqrt{5\times10^4} = 5.48\,\mathrm{pm}$ — about 2.4% high; the correction factor is $(1 + 50/1022)^{-1/2} = 0.976$.

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Checkpoint

1. Derive the two-slit intensity $I(\theta) = 4I_0\cos^2(\pi d\sin\theta/\lambda)$ and the fringe spacing on a distant screen.
2. State the de Broglie relations and explain what motivated them.
3. Show how a standing matter wave on a circular orbit yields $L = n\hbar$.
4. In the single-electron double slit, what exactly is wave-like and what is particle-like?
5. Why does recording which slit the electron used destroy the fringes, and what does the two-path experiment have to do with a qubit?
6. Compare the two-slit distributions for bullets, water waves and electrons. Which observation rules out "each electron goes through one definite slit"?

Answers

1. Superpose two equal-amplitude coherent fields with path difference $d\sin\theta$; the sum-to-product identity gives amplitude $2E_0\cos(\pi d\sin\theta/\lambda)$, hence the $\cos^2$ intensity; small angles put maxima at $y_m = m\lambda L/d$, spacing $\lambda L/d$.
2. $\lambda = h/p$ and $\nu = E/h$ — the photon relations of P.3.1, postulated to hold for all particles on grounds of symmetry between light and matter.
3. Single-valuedness demands $n\lambda = 2\pi r_n$; substituting $\lambda = h/p$ gives $L = pr_n = nh/2\pi = n\hbar$ — Bohr's rule as a standing-wave condition.
4. Wave-like: the propagation of the amplitude through both slits, producing the $\cos^2$ distribution. Particle-like: every detection is one whole electron at one point. The distribution is $|\psi_1 + \psi_2|^2$, built up dot by dot.
5. Which-way monitoring is a measurement: it correlates the paths with distinguishable monitor states, killing the cross term, so probabilities add and $V = 0$ (formally: 1.3.1). The two paths span a two-dimensional state space; a superposition with a relative phase across them is a qubit state, and the fringes are that phase made visible.
6. Bullets: $P_{12} = P_1 + P_2$, no fringes. Waves: $I_{12} = \lvert h_1 + h_2\rvert^2$, with a cross term that makes fringes. Electrons: detected as whole lumps like bullets, but distributed as $\lvert\psi_1 + \psi_2\rvert^2$ like waves. At a dark fringe, opening the second slit lowers the count; if every electron used one definite slit, opening another could only add arrivals.

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Further Reading

• [ER] Eisberg & Resnick, §3-1–3-2 — de Broglie's postulate and the Davisson–Germer and Thomson experiments; the primary treatment.
• [FLP] Feynman, Leighton & Sands, Vol. III, Ch. 1 — bullets, water waves and electrons at the double slit; the classic statement of "the only mystery".
• [Sha] Shankar, Ch. 3 — the double-slit autopsy: why amplitudes, not probabilities, and what must replace classical mechanics.
• [Gri] Griffiths & Schroeter, §1.2 — the statistical interpretation the fringes force on us.

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