University Physics IV · Foundations of Quantum Mechanics · 5.2

The Wave Function as the State

Before you can solve anything you have to know what you are solving for. This lesson fixes the object quantum mechanics computes with: a complex, normalisable function whose overall phase you may throw away and whose internal phases you may not.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Classically the state of a particle is a point (x, p) — six numbers, and everything else is computed from them. Quantum mechanics throws that away and puts in its place a single complex-valued function ψ(x, t) spread over all space. Two experimental facts force the shape of the replacement. Electron diffraction says alternatives must be combined by adding quantities that can cancel, so the state space has to be linear and its scalars complex; Born's rule says |ψ|² is a probability density, so ψ must be square-integrable and normalisable to one.

Together they name the space exactly — L², the complex square-integrable functions — and hand it an inner product ⟨φ|ψ⟩ = ∫φ*ψ dx that measures how much of one state is contained in another. Everything later in this unit is built from that pair: probabilities, expectation values, expansions on a basis, the uncertainty bound. The model charges for it in two currencies. Redundancy first: multiplying ψ by a constant e^(iα) changes nothing measurable, so the physical state is a whole ray of functions, and only phases taken relative to one another inside a superposition mean anything.

Then scope: this ψ describes one non-relativistic spinless particle. Give it spin and the state gains a component index; add a second particle and ψ lives on a six-dimensional configuration space rather than in the room.

Simple definition: The wave function is a complex, square-integrable function of position and time that holds everything predictable about one spinless non-relativistic particle — with its overall phase holding nothing at all.
Example: ψ(x) = A e^{(−x²/2a²)} with a = 1.0 nm normalises when |A|² a√π = 1, so |A| = 0.75 nm^(−1/2). Replacing A by −A, or by 0.75i nm^(−1/2), changes no probability anywhere.

The state, normalised⟨ψ|ψ⟩ = ∫ |ψ(x, t)|² dx = 1

Membership of L² is what makes ψ a state at all; the integral fixes |A| and fixes the units.

x in m ⇒ |ψ|² in m⁻¹ and ψ in m^(−1/2); in three dimensions ψ carries m^(−3/2).

Superpositionψ = c₁ψ₁ + c₂ψ₂ for any c₁, c₂ ∈ ℂ

Linearity is the postulate interference forces on you, not a small-amplitude approximation you may later drop.

Normalised only when |c₁|² + |c₂|² + 2Re(c₁*c₂⟨ψ₁|ψ₂⟩) = 1.

Inner product and overlap⟨φ|ψ⟩ = ∫ φ*(x) ψ(x) dx = ⟨ψ|φ⟩*

|⟨φ|ψ⟩|² is the probability that a system prepared as ψ passes a test for φ.

Complex; dimensionless for normalised states; linear in the ket, antilinear in the bra.

Expansion on an orthonormal basisψ = Σₙ cₙuₙ, cₙ = ⟨uₙ|ψ⟩, Σₙ |cₙ|² = 1

Turns a function into a column of numbers — and the coefficients are amplitudes, so square the modulus before calling one a probability.

Needs ⟨uₘ|uₙ⟩ = δₘₙ. Non-orthogonal components leave overlap terms behind.

Global phase: the state is a rayψ ≃ e^(iα)ψ, α real and constant

Nothing measures α, so physical states are counted as rays — whole families of functions — and not as functions.

The same α at every x and every t; so ψ and −ψ are one and the same physical state.

Relative phase: the interference term|ψ₁ + e^(iθ)ψ₂|² = |ψ₁|² + |ψ₂|² + 2Re(e^(iθ) ψ₁*ψ₂)

The only route by which phase reaches an experiment, and the whole difference between a superposition and a mixture.

θ is one term's phase against the other's; it moves probability about but never creates any.

Two demands, and the space they pick out

Start from what has to be reproduced. Electrons fired one at a time at a double slit build up a fringe pattern, so the two alternatives must be combined by adding quantities that can cancel — probabilities, being non-negative, cannot do that. Complex numbers can, and smoothly, so the state space is a complex vector space and any superposition c₁ψ₁ + c₂ψ₂ of allowed states is itself allowed. Now add Born's rule: |ψ(x)|² is a probability density for position. Total probability is one, so ∫|ψ|² dx must converge and be scalable to unity. Those two demands name the space exactly — the complex square-integrable functions, written L² — and they come with an inner product ⟨φ|ψ⟩ = ∫φ*ψ dx for free. The wave function is not a picture of the particle. It is a vector in that space, and it is the entire state: give me ψ and I can compute every prediction the theory makes about this system.

Normalisation fixes the constant, the units, and nothing else

Any equation you solve for ψ leaves an overall constant free, and normalisation spends it. Take ψ(x) = A e^{(−x²/2a²)} with a = 1.0 nm. Then ∫|ψ|² dx = |A|² ∫e^(−x²/a²) dx = |A|² a√π, so |A| = (a√π)^(−1/2) = 0.75 nm^(−1/2). Two consequences. The units of ψ come from this integral and not from the dynamics: in one dimension |ψ|² is a probability per unit length, so ψ carries m^(−1/2); in three dimensions it carries m^(−3/2), and a two-particle state carries m⁽⁻³⁾. And the condition pins only the modulus |A|. The phase of A is left completely free — the first appearance of the redundancy that runs through the rest of this lesson. You normalise once and not again, because evolution under a real potential preserves the norm; that is the content of the continuity equation you meet in the next topic.

The inner product does all the work

⟨φ|ψ⟩ = ∫φ*(x)ψ(x) dx is a complex number: linear in ψ, antilinear in φ, and equal to ⟨ψ|φ⟩*. Put φ = ψ and it returns the squared norm, so normalisation reads ⟨ψ|ψ⟩ = 1. Between two normalised states the Cauchy–Schwarz inequality bounds it, 0 ≤ |⟨φ|ψ⟩| ≤ 1, and the two ends are the extreme cases: modulus 1 means the same state up to a phase, and 0 means orthogonal — a system prepared as ψ never passes a test for φ. In between, |⟨φ|ψ⟩|² is the probability that it does. Two normalised Gaussians of width σ whose centres are d apart overlap as e^{(−d²/8σ²)}; at d = 2σ that is e^(−0.5) = 0.607, so 0.368 of the time one is mistaken for the other. Choose an orthonormal set uₙ and the same integral extracts coordinates, cₙ = ⟨uₙ|ψ⟩ with Σ|cₙ|² = 1: the state as a column of numbers.

Global phase is invisible; relative phase is not

Replace ψ by e^(iα)ψ, with α a real constant — the same at every point and every time. Then |ψ|² is untouched, and every overlap picks up that one factor, so every |⟨φ|ψ⟩|² is untouched too. Nothing measures α. The physical state is therefore not the function but the ray, the whole family (e^(iα)ψ), and ψ and −ψ are the same state. Now put a phase on one term of a superposition instead. For ψ = ψ₁ + e^(iθ)ψ₂ the density is |ψ₁|² + |ψ₂|² + 2Re(e^(iθ)ψ₁*ψ₂), and that last term swings with θ: it adds at θ = 0, vanishes at θ = π/2, subtracts at θ = π. Throughout, the weights |c₁|² and |c₂|² never move — probability is being pushed from one place to another, not made or destroyed. That term is interference, and it is why a coherent superposition and a statistical mixture of the same two states are different physical objects.

Dirac notation strips out the representation

Write the state as a ket |ψ⟩, an abstract vector, and the wave function becomes one reading of it: ψ(x) = ⟨x|ψ⟩ is the coefficient of |ψ⟩ on the position basis, and φ(p) = ⟨p|ψ⟩ the coefficient on the momentum basis. The bra ⟨φ| is the object that eats a ket and returns the overlap. Insert the completeness relation ∫|x⟩⟨x| dx = 1 and the integral comes back: ⟨φ|ψ⟩ = ∫⟨φ|x⟩⟨x|ψ⟩ dx = ∫φ*(x)ψ(x) dx. Nothing new has been asserted. The point of the notation is that ⟨φ|ψ⟩, and so every probability built from it, is a property of the vectors and is indifferent to the basis you expanded them on. One and the same |ψ⟩ can look like a narrow spike in x and a broad smear in p, and that is a fact about two bases, not about two states.

What ψ is not, and where this model stops

ψ is not a classical field. It is complex, so no instrument reads it off directly; only bilinear quantities such as |ψ|² and ⟨φ|ψ⟩ reach an experiment. Nor is it a smeared-out particle: a detector fires at one point, and what is spread out is the probability, not the electron. The scope here is narrow and worth saying aloud. One particle, non-relativistic, spinless — so ψ(x, t), one complex number at each point. Give the particle spin and the state acquires a discrete index, two components for spin-½. Add a second particle and the state is ψ(x₁, x₂, t) on a six-dimensional configuration space, not two functions living in the room; that is where entanglement comes from. And L² membership rules out two of the most-used functions in the subject: a plane wave e^(ikx) and a Dirac delta are not normalisable, so they are limits and conventions rather than states.

Change one variable at a time

Make the relationship visible.

Interactive model

Mixing angle φ45 °

Relative phase θ0 °

Packet separation d2.0 nm

Hold φ at 45° and swing θ from 0° to 180°: the solid curve drains the midpoint to zero while the dashed mixture does not stir. Then slide the packets apart — as the overlap falls towards zero the two curves close on each other and phase has nothing left to do.

OVERLAP ⟨u₁|u₂⟩0.249

NORM FACTOR N0.895

|ψ(0)|² SUPERPOSITION0.265 nm⁻¹

|ψ(0)|² MIXTURE0.166 nm⁻¹

Live interpretationOVERLAP ⟨u₁|u₂⟩: 0.249. NORM FACTOR N: 0.895. |ψ(0)|² SUPERPOSITION: 0.265 nm⁻¹. |ψ(0)|² MIXTURE: 0.166 nm⁻¹

Catch the common trap

Explain before calculating.

Misconception check

The standard slip is to conclude from Born's rule that phase never matters: |ψ|² is all you measure, the argument runs, so multiply ψ by whatever phase you like. Half of that is right. A global factor e^(iα) — one constant, the same at every x and t — cancels out of |ψ|² and out of every |⟨φ|ψ⟩|, which is exactly why the physical state is the ray and not the function, and why ψ and −ψ describe the same system. The other half is false. Inside a superposition, the phase of one term relative to another survives the squaring: |c₁u₁ + c₂u₂|² carries the cross term 2Re(c₁*c₂u₁*u₂), and flipping the sign of c₂ alone turns (u₁ + u₂)/√2 into (u₁ − u₂)/√2 — two states with identical |cₙ|², yet orthogonal, so a single well-chosen measurement separates them with certainty. Global phase is a redundancy in the notation; relative phase is the entire content of interference.

u₁ and u₂ are real, orthonormal, and overlap over the same region of space. Compare the states ψ_A = (u₁ + u₂)/√2 and ψ_B = (u₁ − u₂)/√2. Which statement is correct?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA particle in one dimension has ψ(x) = A(1 − |x|/L) for |x| ≤ L and ψ = 0 outside, with L = 2.0 nm. Find |A| with its units, and the probability of finding the particle within L/2 of the origin.

Normalisation is ∫|ψ|² dx = 1. The function is even, so ∫₋L^L |A|²(1 − |x|/L)² dx = 2|A|² ∫₀^L (1 − x/L)² dx.
Substituting u = 1 − x/L gives ∫₀^L (1 − x/L)² dx = L ∫₀¹ u² du = L/3. So 2|A|²L/3 = 1 and |A| = √(3/2L).
With L = 2.0 nm: |A| = √(3/4.0 nm) = √(0.75 nm⁻¹) = 0.87 nm^(−1/2). Note where the units come from — |ψ|² is a probability per unit length, so ψ must carry nm^(−1/2). Normalisation sets that, not the dynamics.
The same substitution gives ∫₀^(L/2) (1 − x/L)² dx = L(1 − (1/2)³)/3 = 7L/24, so P(|x| < L/2) = 2|A|²(7L/24) = (2|A|²L/3)(7/8) = 7/8 = 0.875.
Normalisation fixed only the modulus of A. The values A = 0.87, A = −0.87 and A = 0.87i nm^(−1/2) all satisfy it, and all three describe the same physical state.

Answer|A| = 0.87 nm^(−1/2), the unit forced by |ψ|² being a probability per unit length; P(|x| < L/2) = 7/8 = 0.875. The phase of A stays free.

Mediumu₁ and u₂ are orthonormal. A system is prepared as ψ = N(3u₁ + 4i u₂). (a) Find N and the probability of each basis state. (b) A detector projects onto φ = (u₁ + u₂)/√2; find |⟨φ|ψ⟩|². (c) Repeat (b) for ψ′ = N(3u₁ + 4u₂), and for e^(iπ/3)ψ.

(a) ⟨ψ|ψ⟩ = |N|²(|3|² + |4i|²) = 25|N|², since ⟨u₁|u₂⟩ = 0 kills the cross terms and |4i|² = 16. So N = 1/5, and the probabilities are P₁ = 9/25 = 0.36 and P₂ = 16/25 = 0.64.
(b) On an orthonormal basis the overlap is the sum of conjugated-coefficient products: ⟨φ|ψ⟩ = (1/√2)(3/5) + (1/√2)(4i/5) = (3 + 4i)/(5√2).
|3 + 4i| = 5, so |⟨φ|ψ⟩| = 5/(5√2) = 1/√2 and |⟨φ|ψ⟩|² = 0.50.
(c) ψ′ has coefficients of the same moduli, so P₁ and P₂ are still 0.36 and 0.64. But ⟨φ|ψ′⟩ = 7/(5√2) = 0.9899, giving |⟨φ|ψ′⟩|² = 49/50 = 0.98. Changing only the relative phase, from i to 1, moved the detection probability from 0.50 to 0.98.
Multiplying ψ by e^(iπ/3) multiplies ⟨φ|ψ⟩ by that same factor and leaves its modulus alone, so the probability is 0.50 again. A global phase is invisible; a phase on one term is not.

AnswerN = 1/5, P₁ = 0.36, P₂ = 0.64. |⟨φ|ψ⟩|² = 0.50; |⟨φ|ψ′⟩|² = 0.98; for e^(iπ/3)ψ it is 0.50, unchanged.

Hardψ_L and ψ_R are normalised real Gaussians of standard deviation σ = 0.80 nm centred at x = −0.80 nm and +0.80 nm, so d = 1.6 nm and ⟨ψ_L|ψ_R⟩ = S = exp(−d²/8σ²). Normalise ψ_± = N_±(ψ_L ± ψ_R), then compare the two densities at the midpoint x = 0.

S = exp(−d²/8σ²) = exp(−2.56/5.12) = e^(−0.5) = 0.6065. The two packets are nowhere near orthogonal, so their coefficients are not probabilities and the cross term cannot be dropped.
⟨ψ_±|ψ_±⟩ = N_±²(1 + 1 ± 2S) = 1, so N₊ = 1/√(2(1 + S)) = 1/√3.213 = 0.5579 and N₋ = 1/√(2(1 − S)) = 1/√0.787 = 1.1272. The relative phase has changed the normalisation itself.
Each packet at the midpoint: ψ(0) = (2πσ²)^(−1/4) exp(−(d/2)²/4σ²) = (4.021)^(−1/4) e^(−0.25) = 0.7062 × 0.7788 = 0.5500 nm^(−1/2), the same for both by symmetry.
|ψ₊(0)|² = (2 × 0.5579 × 0.5500)² = 0.6136² = 0.377 nm⁻¹, while |ψ₋(0)|² = 0 exactly: the two amplitudes are equal there, so a relative phase of π cancels them.
Both states weight the two packets equally and differ only in the sign of one term, yet ⟨ψ₊|ψ₋⟩ = N₊N₋(1 − S + S − 1) = 0 — orthogonal, so one suitable measurement tells them apart with certainty.

AnswerS = 0.607; N₊ = 0.558 and N₋ = 1.127; |ψ₊(0)|² = 0.377 nm⁻¹ against |ψ₋(0)|² = 0. The two are orthogonal despite identical packet weights.