University Physics V · Mathematical Foundations of Quantum Physics · 1.1
Complex Amplitudes & the Phase That Survives
Quantum mechanics computes with numbers no detector ever reads out. This topic teaches how to write and multiply them, then drills the one reflex the whole course leans on: before discarding any phase, ask whether it multiplies the entire state or only one branch of a superposition.
Build the model
Connect the measurement to the mechanism.
Quantum mechanics superposes states linearly but pays out probabilities quadratically, and that pairing is why the amplitudes are complex: the theory needs numbers that can cancel at any angle, not merely reinforce or oppose. Euler's formula supplies their working form, z = r e(iφ), in which multiplication is geometry — moduli multiply, arguments add — and e(ikθ) emerges as the eigenfunction of the derivative operator, (d/dθ) e(ikθ) = ik e(ikθ), which is why exponentials rather than sines will diagonalise momentum all course long. The cost is that no measurement reads an amplitude out: the only exit to experiment is |z|² = z*z, and that squaring erases exactly one phase — a factor e(iα) common to the whole state, which cancels in every probability and every expectation value, so a physical state is a ray, not a vector.
Everything else survives: each relative phase inside a superposition sits in the cross term 2r₁r₂ cos(φ₂ − φ₁) whenever two amplitudes for the same outcome add before the square. And unlike the phasor of AC circuits, the exponential here is not scaffolding to be stripped by Re at the end: the i is load-bearing, and removing it undoes the eigenfunction property the amplitude was built to carry.
- Simple definition
- A complex amplitude z = r e(iφ) carries a modulus, which fixes the probability through |z|² = z*z = r², and a phase, which is measurable only as a difference — when two amplitudes for the same outcome are added before the modulus is squared.
- Example
- Add z₁ = 0.5 and z₂ = 0.5 e(iδ): |z₁ + z₂|² = 0.25 + 0.25 + 0.5 cos δ = 0.5(1 + cos δ) — equal to 1 at δ = 0, 0.5 at δ = π/2, and 0 at δ = π, though each modulus stayed 0.5 throughout.
Multiplication turns into geometry: z₁z₂ = r₁r₂ e(i(φ₁+φ₂)) — moduli multiply, phases add.
r ≥ 0 is the modulus; φ the argument, in radians, defined only modulo 2π — use atan2, not arctan, to keep the quadrant
Every unit-modulus number is a pure phase; multiplying an amplitude by one rotates it in the plane without touching any probability built from it alone.
φ real, in radians: e(iπ/2) = i, e(iπ) = −1, e(2πi) = 1; |e(iφ)| = 1 always
cos and sin are not eigenfunctions — d/dθ swaps them into each other. This is why quantum mechanics expands on exponentials, and it previews p̂ = −iħ d/dx.
on 2π-periodic functions single-valuedness forces k ∈ ℤ; then −i d/dθ is Hermitian with real eigenvalues k
Every probability passes through this product, and z*z is blind to the argument of z — a lone amplitude's phase never reaches a detector.
z* = x − iy = r e(−iφ) is the conjugate; z*z is real and ≥ 0, as a probability must be
Add amplitudes first, square once. The cross term is where a phase becomes measurable: the whole two-slit pattern is this one identity.
r₁, r₂ the moduli, φ₁, φ₂ the arguments; only the difference φ₂ − φ₁ enters the cross term, which runs from −2r₁r₂ to +2r₁r₂
e(iα) factors out of every amplitude and dies at the squaring. Rephase one branch of a superposition instead and no factor pulls out — the cross term moves.
expectation values match too: ⟨ψ'|A|ψ'⟩ = ⟨ψ|A|ψ⟩ for every observable A — two kets on the same ray are one physical state
Amplitudes live in the plane, not on the line
A real amplitude added to another can only reinforce it or oppose it. A complex amplitude z = x + iy is a point in a plane, addition is head-to-tail vector addition there, and two amplitudes can therefore partially cancel at any angle — exactly the freedom interference patterns demand. Cartesian form suits addition; polar form z = r e(iφ) suits multiplication. Converting between them has one trap: the argument is φ = atan2(y, x), not arctan(y/x). The point z = −1 − i sits at φ = −3π/4, while blind arctan of y/x = 1 reports +π/4; that sign error later flips a cross term from constructive to destructive. Conjugation z* = x − iy = r e(−iφ) reflects across the real axis, and the product z*z = r² is real and non-negative — unlike z², which for z = e(iπ/4) equals i: still complex, still unit modulus. Quantum mechanics squares moduli, never the numbers themselves.
Euler's formula is the multiplication table
Define f(θ) = e(iθ) as the solution of f′ = i f with f(0) = 1. Then (f*f)′ = −i f*f + i f*f = 0, so |f| stays 1: the solution runs around the unit circle at unit speed, starting at 1 and turning anticlockwise — which is the point (cos θ, sin θ). That is Euler's formula, and it makes phase arithmetic trivial: e(iφ₁) e(iφ₂) = e(i(φ₁+φ₂)) turns composing rotations into adding angles, and de Moivre's theorem (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ is one line. Read backwards it dissolves trigonometry: cos θ = (e(iθ) + e(−iθ))/2 and sin θ = (e(iθ) − e(−iθ))/(2i), which is how every trig integral in this course becomes an exponential integral. Numerically e(iπ/3) = 0.500 + 0.866i, e(iπ) = −1, e(2πi) = 1: the phase is periodic, which is why arguments are only ever defined modulo 2π.
Name the operator: d/dθ and its eigenfunctions
Work on the circle: square-integrable functions with the boundary condition f(θ + 2π) = f(θ). Apply d/dθ to cos θ and you get −sin θ — a different function, so cos θ is no eigenfunction; the derivative keeps knocking sines and cosines into each other. Apply it to e(ikθ) and the function returns as ik e(ikθ): an eigenfunction with eigenvalue ik, and the boundary condition admits only integer k. Integrating by parts around the closed circle kills the boundary terms and shows d/dθ is anti-Hermitian, so its eigenvalues are pure imaginary; multiply by −i and L̂ = −i d/dθ is Hermitian, with real eigenvalues k ∈ ℤ and the orthonormal eigenbasis uₖ(θ) = e(ikθ)/√(2π), ⟨uⱼ|uₖ⟩ = δⱼₖ. This is the template the whole course stamps out: name the space, the boundary condition and the operator, then read the eigenvalue equation as the list of measurable values. Attach ħ and it is already physics: L̂z = −iħ d/dθ on the ring, p̂ = −iħ d/dx on the line.
Squaring erases one phase — exactly one
The exit to experiment is |z|² = z*z, and it is worth being exact about what that step forgets. Multiply an entire state by e(iα): every amplitude ⟨φ|ψ⟩ acquires the same factor, and |e(iα)⟨φ|ψ⟩|² = |⟨φ|ψ⟩|² for every |φ⟩; every expectation value ⟨ψ|A|ψ⟩ is equally untouched, since e(−iα) e(iα) = 1. No experiment, in any basis, distinguishes e(iα)|ψ⟩ from |ψ⟩ — the physical state is the whole ray of kets e(iα)|ψ⟩, and normalisation ⟨ψ|ψ⟩ = 1 fixes the overall modulus while leaving the overall phase forever free. But that is the only phase the squaring erases. Rephase one branch of a superposition, c₂ → e(iδ)c₂, and no common factor pulls out: (|0⟩ + |1⟩)/√2 and (|0⟩ − |1⟩)/√2 have identical moduli in this basis — probabilities ½, ½ for both — yet their inner product is zero. They are orthogonal: as different as two states can be, on the strength of one relative sign.
Where the relative phase shows: a rotated basis
A relative phase hides in the basis that carries it and appears in any rotated one. Prepare |ψδ⟩ = (|0⟩ + e(iδ)|1⟩)/√2 and measure in |±⟩ = (|0⟩ ± |1⟩)/√2. The amplitude for + is ⟨+|ψδ⟩ = (1 + e(iδ))/2, so P(+) = (1 + cos δ)/2 and P(−) = (1 − cos δ)/2: certainty of + at δ = 0, an even split at δ = π/2, certainty of − at δ = π. The pattern generalises through one identity: whenever two amplitudes reach the same outcome they add before the square, and |z₁ + z₂|² = r₁² + r₂² + 2r₁r₂ cos(φ₂ − φ₁). The first two terms are the incoherent sum a classical mixture would give; the cross term, bounded by ±2r₁r₂, is interference, and only the phase difference enters it. Interference is not a wave phenomenon bolted onto the formalism — it is complex addition followed by a single squaring.
This is not a phasor: no real part at the end
You have used complex exponentials before, in AC circuits, and the habit learned there must be unlearned here. A phasor computes with V₀ e(iωt), but the physical voltage is Re[V₀ e(iωt)]: the imaginary part is scaffolding, removable at the price of clumsier trigonometry. The quantum amplitude is not scaffolding. Take the real part of the momentum eigenfunction e(ikx) and you get cos kx = (e(ikx) + e(−ikx))/2 — no longer an eigenfunction of −iħ d/dx but an equal superposition of momenta +ħk and −ħk: Re destroys the very property the exponential was chosen for. The time factor tells the same story. A stationary state evolves by e(−iEt/ħ), whose unit modulus keeps |Ψ|² frozen; a real cos(Et/ħ) in its place would make every atom's charge density pulse at angular frequency 2E/ħ, which stable atoms do not do. The i in iħ ∂Ψ/∂t = ĤΨ is load-bearing. No measurement ever returns Re z, Im z, or φ itself; what is measurable is |z|², and phase differences through the cross terms they feed.
Change one variable at a time
Make the relationship visible.
Sweep δ: the resultant arrow folds up and the bar swings either side of the dashed line, reaching zero at δ = 180° when r₂ = 1. Then sweep α: the whole diagram rotates, arg(z₁+z₂) tracks it, and the bar never moves — that is the phase no measurement can see.
|z₁+z₂|²2.44
CROSS TERM 2r₂cos δ0.80
VISIBILITY0.98
ARG(z₁+z₂)26 °
Live interpretation|z₁+z₂|²: 2.44. CROSS TERM 2r₂cos δ: 0.80. VISIBILITY: 0.98. ARG(z₁+z₂): 26 °
Catch the common trap
Explain before calculating.
A qubit is prepared in |ψ⟩ = (|0⟩ + |1⟩)/√2. Which ket describes a physically different state — one that some measurement could distinguish from |ψ⟩?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyOne path of an interferometer contributes the amplitude z = (1 − i)/2. Write z in polar form, state the probability it alone would deliver, and show that multiplying by the global factor i changes neither the modulus nor that probability.
- Modulus first: r = √(x² + y²) = √((1/2)² + (−1/2)²) = √(1/2) = 1/√2 ≈ 0.707.
- Argument with the quadrant kept: φ = atan2(−1/2, 1/2) = −π/4, so z = (1/√2) e(−iπ/4). Naive arctan agrees here only because x > 0 — for x < 0 it would be off by π.
- Probability: P = z*z = ((1 + i)/2)((1 − i)/2) = (1 − i²)/4 = 2/4 = 1/2. The phase −π/4 has dropped out, as it must for a lone amplitude.
- Global factor i = e(iπ/2): iz = (i − i²)/2 = (1 + i)/2 = (1/√2) e(+iπ/4). The argument moved by π/2; the modulus, and with it P = 1/2, did not.
Answerz = (1/√2) e(−iπ/4) with P = 1/2; the global i rotates the phase to +π/4 and leaves P = 1/2 untouched.
MediumTwo paths reach the same detector with amplitudes z₁ = 0.30 and z₂ = 0.40 e(iδ), the remaining probability going elsewhere on the screen. Find the detection probability as a function of δ, its extreme values, the fringe visibility, and the phases at which interference contributes nothing.
- Add before squaring: P(δ) = |z₁ + z₂|² = 0.30² + 0.40² + 2(0.30)(0.40) cos δ = 0.25 + 0.24 cos δ.
- Extremes: δ = 0 gives P = 0.49 (constructive); δ = π gives P = 0.01 (destructive) — a 49 : 1 contrast from amplitudes whose moduli never changed.
- Visibility: V = (Pₘₐₓ − Pₘᵢₙ)/(Pₘₐₓ + Pₘᵢₙ) = 0.48/0.50 = 0.96, which is the general 2r₁r₂/(r₁² + r₂²). Only equal moduli reach V = 1.
- The cross term 2r₁r₂ cos δ vanishes when cos δ = 0, at δ = π/2 and 3π/2. There P = r₁² + r₂² = 0.25, exactly the incoherent sum — the phases are in quadrature, not absent.
AnswerP = 0.25 + 0.24 cos δ: maximum 0.49, minimum 0.01, visibility 0.96; at δ = π/2 and 3π/2 the detector sees the incoherent 0.25.
HardOn the ring 0 ≤ θ < 2π with periodic boundary condition, L̂z = −iħ d/dθ has orthonormal eigenfunctions uₘ(θ) = e(imθ)/√(2π), m ∈ ℤ. A particle is prepared in ψ(θ) = [3u₁(θ) + 4i u₋₂(θ)]/5. Find the possible L̂z outcomes with their probabilities, ⟨L̂z⟩, and the probability density at θ = 0 — then identify which of these the factor i actually affects.
- Check the state: c₁ = 3/5 and c₋₂ = 4i/5, so ⟨ψ|ψ⟩ = |3/5|² + |4i/5|² = 9/25 + 16/25 = 1 — normalised, because |4i/5|² = 16/25: the i has unit modulus.
- Outcomes are eigenvalues: L̂z uₘ = mħ uₘ, so only +ħ and −2ħ can appear, with P(+ħ) = |c₁|² = 0.36 and P(−2ħ) = |c₋₂|² = 0.64. Single-valuedness ψ(θ + 2π) = ψ(θ) is what restricted m to integers in the first place.
- ⟨L̂z⟩ = Σ mħ |cₘ|² = (0.36)(+ħ) + (0.64)(−2ħ) = (0.36 − 1.28)ħ = −0.92ħ.
- Density at θ = 0: ψ(0) = (3 + 4i)/(5√(2π)), so |ψ(0)|² = |3 + 4i|²/(25 · 2π) = 25/(50π) = 1/(2π) ≈ 0.159. The cross term 2 Re(3 · (4i)*)/(50π) vanished because the two amplitudes sit 90° apart.
- Replace 4i by 4 (relative phase 0): every |cₘ|², and hence the L̂z distribution and ⟨L̂z⟩, is unchanged — but |ψ(0)|² = |3 + 4|²/(50π) = 49/(50π) ≈ 0.312, nearly double. The relative phase is invisible in the basis that carries it and measurable in the conjugate one.
AnswerP(+ħ) = 0.36, P(−2ħ) = 0.64, ⟨L̂z⟩ = −0.92ħ; |ψ(0)|² = 1/(2π) ≈ 0.159, rising to 49/(50π) ≈ 0.312 if the i is replaced by 1 — the phase moves the density, not the L̂z statistics.