University Physics V · The Quantum Wavefunction · 3.1
The State Vector & Its Position Representation
Before any dynamics, get the ontology straight: the particle's state is a single vector in L²(ℝ), the wavefunction is that vector's list of position components, and every measurable number the theory will ever produce comes out of one integral — the inner product.
Build the model
Connect the measurement to the mechanism.
Quantum mechanics does not begin with a function of x; it begins with a vector. The state of one spinless non-relativistic particle is a ray in the Hilbert space L²(ℝ) — the complex vector space of square-integrable functions, complete in the norm its inner product induces — and the familiar wavefunction is nothing but that vector's components along the position basis, ψ(x) = ⟨x|ψ⟩, the continuous analogue of vᵢ = ⟨eᵢ|v⟩. This one move buys the whole formalism: the inner product ⟨φ|ψ⟩ = ∫φ*ψ dx supplies every norm, overlap and — one topic from now — every probability; linearity makes superposition automatic; and the same vector can later be reread as momentum components without anything physical changing.
The cost is paid at the basis itself. A continuous label forces delta normalisation, ⟨x|x′⟩ = δ(x − x′), so the basis "vectors" |x⟩ have infinite norm and live outside the very space they coordinate. And because predictions reach experiment only through |⟨φ|ψ⟩|², the state is a ray rather than a vector: ψ and e(iα)ψ are one state, and only relative phases inside a superposition are ever observable.
- Simple definition
- The state of the particle is a unit vector |ψ⟩ in the Hilbert space L²(ℝ), defined only up to a global phase; its position representation ψ(x) = ⟨x|ψ⟩ lists its components along the δ-normalised position basis.
- Example
- ψ(x) = (1/√a)e(−|x|/a) with a = 2.0 nm is a legal state: ∫|ψ|² dx = 1 exactly, and its component along |x = a⟩ is ψ(a) = e⁻¹/√a ≈ 0.26 nm(−1/2) — one complex number per position, carrying units, and not itself a probability.
The wavefunction is the coordinate list of |ψ⟩ along the position basis, exactly as vᵢ = ⟨eᵢ|v⟩ lists a finite vector.
ψ(x): one complex number per x, units L(−1/2) in one dimension; the ket |ψ⟩ itself is unitless
Every norm, every overlap and, from the next topic on, every probability in this course is this single integral.
conjugate on the first (bra) slot — the physics convention; ⟨φ|ψ⟩ = ⟨ψ|φ⟩*
Any finite norm rescales to 1. e(ikx) has no finite norm at all — sharp momentum is not a state.
the integral runs over all x; finiteness of ∫|ψ|² is the entry condition for L²(ℝ)
Insert it between a bra and a ket and the abstract bracket becomes the wavefunction integral ∫φ*ψ dx.
the continuous analogue of Σₙ |n⟩⟨n| = I over an orthonormal basis
Orthogonality for a continuous label — and the proof that position kets are instruments, not states.
δ(x − x′) carries units L⁻¹, so ⟨x|x⟩ = δ(0) diverges: |x⟩ is not in L²(ℝ)
No |⟨φ|ψ⟩|² can see α. A relative phase between components survives and interferes.
α real and applied to the whole ket; a phase on one branch of a superposition is not global
The vector is the state; the wavefunction is its components
Start in two dimensions, where nothing is hidden. A ket |v⟩ in C² is one arrow; pick an orthonormal basis (|e₁⟩, |e₂⟩) and it acquires components vᵢ = ⟨eᵢ|v⟩ — two complex numbers that change the moment you rotate the basis, while the arrow does not. Position representation is the same move with a continuous label: the basis vectors are |x⟩, one for each real number, the components are ψ(x) = ⟨x|ψ⟩, and the sum over an index becomes an integral over x. So "the wavefunction" is a coordinate list — an uncountable column vector — and writing |ψ⟩ instead of ψ(x) is not pedantry: it is the reminder that the same state will later be reread as momentum components φ(p) = ⟨p|ψ⟩ without anything physical changing. The notation splits the labour cleanly: kets carry the physics, brackets extract numbers from it.
Square-integrability is the membership test
The space is L²(ℝ): complex-valued functions of x with ∫|ψ|²dx finite. Run candidates through the test. A Gaussian e(−x²/2) passes. 1/(1+x²) passes: its squared modulus falls as x⁻⁴. The plane wave e(ikx) fails — |ψ|² = 1 everywhere integrates to infinity — so a state of perfectly sharp momentum is not in the space, however often the phrase gets used. Boundedness is not the point either: ψ = |x|(−1/4) on [−1, 1] blows up at the origin yet ∫|x|(−1/2)dx = 4 is finite, so it qualifies, while the gentler-looking |x|(−1/2) fails on the same interval because ∫|x|⁻¹dx diverges logarithmically. And "Hilbert" adds one clause beyond the inner product: the space is complete, so Cauchy sequences of functions converge to members. That clause is what will later let an infinite eigenfunction expansion actually equal the state it represents.
One integral supplies every number
The inner product ⟨φ|ψ⟩ = ∫φ*(x)ψ(x)dx — conjugate on the first slot, in the physics convention — is the only measuring device the space owns. It gives lengths: ‖ψ‖² = ⟨ψ|ψ⟩, real and non-negative, scaled to 1 by normalisation. It gives geometry: swapping the slots conjugates the number, ⟨φ|ψ⟩ = ⟨ψ|φ⟩*, and Cauchy–Schwarz bounds |⟨φ|ψ⟩| by ‖φ‖‖ψ‖, so for unit vectors the squared overlap sits in [0, 1] — exactly the interval a probability needs, which is why Born's rule (next topic) can be stated at all. And it gives distinguishability a shape: two unit-norm Gaussians of width σ whose centres sit a distance d apart overlap by e(−d²/4σ²). At d = 2σ that is e⁻¹ ≈ 0.37; at d = 6σ it is e⁻⁹ ≈ 1.2 × 10⁻⁴, and the states are orthogonal for every practical purpose. Overlap, not distance in x, is how the space measures how different two states are.
The position basis and the delta it costs
A basis labelled by a continuum cannot be normalised to 1. Demand that the expansion ψ(x) = ⟨x|ψ⟩ and the completeness relation I = ∫|x⟩⟨x|dx hold together, and orthonormality is forced into the distributional form ⟨x|x′⟩ = δ(x − x′): insert the identity into ⟨φ|ψ⟩ and the abstract bracket becomes ∫φ*(x)ψ(x)dx, which is the whole point of the basis. The price appears at x′ = x: ⟨x|x⟩ = δ(0) is infinite, so |x⟩ fails the very membership test it administers — it is an improper vector, a distribution that only means something under an integral. Physically this is no defect. No instrument resolves position exactly; a real position measurement of width ε leaves the particle in a normalisable packet of that width, and |x⟩ is the idealisation those packets approach in the bookkeeping but never in the space.
Global phase is invisible; relative phase is not
Multiply the state by e(iα): every component becomes e(iα)ψ(x), every inner product picks up the same phase, and every measurable — each |⟨φ|ψ⟩|², each |ψ(x)|² — is untouched. So the physical state is not a unit vector but a ray, the set (e(iα)|ψ⟩), and "normalise ψ" fixes only the modulus of the constant. The immunity stops inside a superposition. Take |ψ⟩ = (|a⟩ + e(iθ)|b⟩)/√2 with ⟨a|b⟩ = 0, and test it against |c⟩ = (|a⟩ + |b⟩)/√2, for which ⟨c|a⟩ = ⟨c|b⟩ = 1/√2: the amplitude is (1 + e(iθ))/2, so the probability is cos²(θ/2) — 1 at θ = 0, 1/2 at θ = π/2, 0 at θ = π. That θ lives between two components of one vector, and no global rotation can remove it. One rule carries the whole distinction: a phase you can factor out of the entire ket is gauge; a phase you cannot is physics.
On a grid, the vector talk becomes literal
Discretise the line: N points xⱼ spaced Δx apart, and ψ collapses to the column vector ψⱼ = ψ(xⱼ) in CN. The inner product becomes a conjugate dot product carrying the measure, ⟨φ|ψ⟩ ≈ Δx Σ φⱼ* ψⱼ — in NumPy, dx * np.vdot(φ, ψ), and vdot rather than dot because vdot conjugates its first argument exactly as the bra does. Try it: a unit-norm Gaussian of σ = 1 sampled on [−8, 8] with N = 256 (Δx = 0.0625) returns dx * np.vdot(ψ, ψ) = 1 to machine precision, because the tails the box cuts off are already below 10⁻¹⁴. Two habits transfer from the mathematics: check the norm before trusting any other number, and remember what truncation assumes — amplitude outside the box, or structure finer than Δx, is simply not representable, and any probability that belongs there is silently lost. Every numerical method later in the course — eigh, the FFT, split-step propagation — is built on this identification.
Change one variable at a time
Make the relationship visible.
With σ₁ = σ₂ = 1, drag d from 0 to 4 and watch the dashed product — not the packets — collapse: the overlap falls as e(−d²/4), through 0.37 at d = 2. Then return d to 0 and mismatch the widths; identical centres still lose overlap, because the inner product compares whole functions.
OVERLAP ⟨φ|ψ⟩0.527
PROBABILITY |⟨φ|ψ⟩|²0.278
ANGLE BETWEEN RAYS58.2 °
Live interpretationOVERLAP ⟨φ|ψ⟩: 0.527. PROBABILITY |⟨φ|ψ⟩|²: 0.278. ANGLE BETWEEN RAYS: 58.2 °
Catch the common trap
Explain before calculating.
Two normalised states of a single particle satisfy ⟨φ|ψ⟩ = (1 + i)/2. What does this number tell you?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyNormalise ψ(x) = N e(−|x|/a) on the whole line for a = 2.0 nm, choose the conventional phase, and evaluate the position component ⟨x|ψ⟩ at x = a. What units does it carry?
- ⟨ψ|ψ⟩ = |N|² ∫ e(−2|x|/a) dx. The integrand is even, so the integral is 2 ∫₀^∞ e(−2x/a) dx = 2 · (a/2) = a, giving ⟨ψ|ψ⟩ = |N|² a.
- Set |N|² a = 1: N = e(iα)/√a for any real α — normalisation fixes only the modulus, because the state is a ray. Take α = 0, so N = 1/√a = 1/√(2.0 nm) = 0.71 nm(−1/2).
- The component along |x = a⟩ is ψ(a) = ⟨a|ψ⟩ = (1/√a) e(−a/a) = e⁻¹/√a = 0.368 × 0.707 nm(−1/2) = 0.26 nm(−1/2).
- Units check: ⟨ψ|ψ⟩ = ∫|ψ|² dx must be a pure number, so [ψ] = L(−1/2). A wavefunction is not dimensionless, and its value at a point is an amplitude, never a probability.
AnswerN = 1/√a = 0.71 nm(−1/2) up to an arbitrary phase; ψ(a) = 0.26 nm(−1/2), with units of length(−1/2).
MediumTwo normalised Gaussian states of equal width σ = 0.50 nm sit a distance d apart: ψd(x) = (πσ²)(−1/4) e(−(x−d)²/2σ²). Show ⟨ψ₀|ψd⟩ = e(−d²/4σ²), evaluate it at d = 1.0 nm, and find the separation at which the overlap probability drops to 1%.
- Write the integrand: ψ₀*ψd = (πσ²)(−1/2) e(−[x² + (x−d)²]/2σ²). Complete the square: x² + (x−d)² = 2(x − d/2)² + d²/2, so the exponent splits into a shifted Gaussian times the constant factor e(−d²/4σ²).
- Integrate: ∫ e(−(x−d/2)²/σ²) dx = √π σ, so ⟨ψ₀|ψd⟩ = (πσ²)(−1/2) · √π σ · e(−d²/4σ²) = e(−d²/4σ²). The prefactors cancel exactly because both states were normalised, and the result is real, as ⟨ψd|ψ₀⟩ = ⟨ψ₀|ψd⟩* requires here.
- At d = 1.0 nm: d²/4σ² = 1.0/(4 × 0.25) = 1.0, so the overlap is e⁻¹ = 0.368 and the overlap probability is e⁻² = 0.135.
- For 1%: |⟨ψ₀|ψd⟩|² = e(−d²/2σ²) = 0.01 → d² = 2σ² ln 100 = 2(0.25)(4.605) = 2.30 nm² → d = 1.52 nm.
Answer⟨ψ₀|ψd⟩ = e(−d²/4σ²): 0.37 at d = 1.0 nm, and the overlap probability reaches 1% at d ≈ 1.5 nm — barely three widths of separation makes two states nearly orthogonal.
HardModel a would-be position eigenstate at x₀ = 0 by the normalised top hat ψε(x) = 1/√ε on [−ε/2, ε/2], zero elsewhere. Against the fixed state φ(x) = π(−1/4) e(−x²/2), compute ⟨ψε|φ⟩ for ε = 0.10 and ε = 0.01, and show why the ε → 0 limit exists as a δ-normalised object but not as a state.
- Norm first: ⟨ψε|ψε⟩ = ∫ (1/ε) dx over a width ε = 1 for every ε — each ψε is a genuine member of L², however narrow.
- Overlap: ⟨ψε|φ⟩ = (1/√ε) ∫ from −ε/2 to ε/2 of φ dx. For ε ≪ 1 the integrand is flat, ≈ φ(0) ε, so ⟨ψε|φ⟩ ≈ √ε φ(0), with φ(0) = π(−1/4) = 0.751.
- Numbers: ε = 0.10 → √0.10 × 0.751 = 0.238. ε = 0.01 → 0.10 × 0.751 = 0.075. The overlap with every fixed state dies as √ε.
- So a limit vector would need norm 1 yet zero inner product with every state in the space — but a vector orthogonal to everything is the zero vector, which has norm 0. Contradiction: the unit-norm sequence ψε has no limit in L².
- The delta convention rescales instead: |0⟩ε = ψε/√ε gives ⟨0ε|φ⟩ → φ(0), the sifting property — but ⟨0ε|0ε⟩ = 1/ε, which is 10 at ε = 0.10 and 100 at ε = 0.01, diverging as δ(0). |x⟩ delivers components ψ(x) = ⟨x|ψ⟩ and the identity ∫|x⟩⟨x|dx, and can never be occupied.
Answer⟨ψε|φ⟩ = 0.238 (ε = 0.10) and 0.075 (ε = 0.01), vanishing as √ε; the rescaled kets have norms 10 and 100 → ∞. Position kets are δ-normalised instruments, not states.