# Matter as Waves > Light, a certified wave, turned out to arrive in particles ([P.3.1](01-light-as-particles.md)). 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. --- ## 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](../c01-theoretical-mechanics/01-waves-and-the-wave-equation.md). Discreteness of atomic orbits stops being a decree and becomes what waves always do in confinement. --- ## 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](../c01-theoretical-mechanics/02-wave-propagation-diffraction.md), 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](../../term-1-quantum-mechanics/c01-postulates/01-state-postulate.md)): a global phase is > unobservable; a relative phase is the signal. ### The de Broglie hypothesis For photons, [P.3.1](01-light-as-particles.md) 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). ### 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](../c01-theoretical-mechanics/01-waves-and-the-wave-equation.md). Confinement plus waves equals discreteness, always. ### 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](../c01-theoretical-mechanics/02-wave-propagation-diffraction.md), $d\sin\varphi = n\lambda$, 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. ### 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](../../term-1-quantum-mechanics/c03-measurement/01-projective-measurement.md). 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](../../term-1-quantum-mechanics/c01-postulates/01-state-postulate.md), and sweeping $\delta$ across the screen traces the fringes — relative phase made visible, the equator of [1.2.1 The Qubit](../../term-1-quantum-mechanics/c02-qubits-bloch/01-the-qubit.md). 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). --- ## 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. --- ## Hands-on (Python) Two-slit intensity for the electron parameters of Example 2, including the single-slit envelope. ```python 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. ```python 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. ``` --- ## 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](01-light-as-particles.md) 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 $(1 + 50/1022)^{-1/2} = 0.976$.

--- ## 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?

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.

--- ## Further Reading - **[ER]** Eisberg & Resnick, §3-1–3-2 — de Broglie's postulate and the Davisson–Germer and Thomson experiments; the primary treatment. - **[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. --- ← Prev: [Light as Particles](01-light-as-particles.md) · Up: [Pre-Term](../README.md) · Next: [The Schrödinger Equation & the Born Rule](../c04-schrodinger-equation/01-schrodinger-equation-born-rule.md) →