# 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).
For an electron, $\lambda = h/m_ev$. Drag its speed and watch the wavelength fall as $1/v$:
```qm-demo de-broglie-speed
```
### 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.
Wrap the wave yourself, and push $n$ off a whole number to see the orbit fail:
```qm-demo bohr-standing-wave
```
### 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.
### 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](../../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).
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).
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.
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).
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
$(1 + 50/1022)^{-1/2} = 0.976$.
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.