University Physics IV · The Schrödinger Equation · 6.5

Operators & the Measurement Postulate

Quantum mechanics splits a measurement in two. Solve one eigenvalue problem and you have every result the meter can ever show. Expand your state on those eigenfunctions and you have how often each one shows. This lesson builds both halves, and marks the line the theory refuses to cross.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Classical mechanics keeps an observable and its value in the same object: a particle has a position, and measuring it copies that number out. Quantum mechanics breaks the pair apart and gives each half its own mathematical home. The observable becomes a Hermitian operator — position multiplies by x, momentum differentiates as −iħ ∂/∂x, kinetic energy is −(ħ²/2m) ∂²/∂x², and the Hamiltonian is their sum with V(x) — while the value belongs to the state, if it exists at all.

Solve Q̂φₙ = qₙφₙ once and the eigenvalues qₙ are the complete menu of readings that observable can ever produce; anything else is not unlikely, it is unavailable. The state Ψ then supplies only odds: expand it as Ψ = Σ cₙφₙ and P(qₙ) = |cₙ|². And the reading rewrites the state — once qₙ comes back, what is left is φₙ, which is why an immediate repeat returns qₙ again.

That last clause is where the model charges you. The projection is not a consequence of the Schrödinger equation; it is a second rule, abrupt, random and irreversible, bolted alongside a smooth deterministic one, kept because laboratories behave that way. What it delivers is the statistics of an ensemble of identically prepared runs, and it says nothing whatever about what the next single run will show.

Simple definition: An observable is represented by a Hermitian operator, the possible results of measuring it are exactly that operator's eigenvalues, and the probability of each is the squared modulus of the state's coefficient on the matching eigenfunction.
Example: An electron in a 0.50 nm infinite well prepared as Ψ = (ψ₁ + 2ψ₂)/√5 can read only 1.504 eV or 6.017 eV, with probabilities 1/5 = 0.20 and 4/5 = 0.80 — never the in-between mean of 5.11 eV.

The quantisation dictionaryx̂ψ = xψ, p̂ψ = −iħ ∂ψ/∂x, T̂ψ = −(ħ²/2m) ∂²ψ/∂x²

Write the classical expression, then substitute. Every operator in this course is assembled from these two.

ħ = 1.055×10⁻³⁴ J s, so ħ ∂/∂x carries kg m s⁻¹ and T̂ψ carries joules × ψ

The HamiltonianĤ = T̂ + V(x) = −(ħ²/2m) d²/dx² + V(x)

One operator does two jobs: its eigenvalues are the energies you can read, its eigenfunctions the stationary states.

V(x) in joules, multiplying ψ; Ĥ is the total-energy observable and also the generator of the evolution

Eigenvalue problem: the whole dialQ̂φₙ = qₙφₙ, qₙ real

Solved once, before any state is named. A number outside (qₙ) is not improbable — it cannot be read.

φₙ is the eigenfunction, n labels the spectrum; the boundary conditions decide whether n is discrete or continuous

Expansion and Born's ruleΨ = Σₙ cₙφₙ, P(qₙ) = |cₙ|², Σₙ |cₙ|² = 1

Converts a wavefunction into a probability table over the dial. If the squares miss 1, a term or a normaliser was dropped.

cₙ = ∫ φₙ* Ψ dx, complex, and dimensionless when Ψ and φₙ are both normalised

State update on reading qₙΨ → φₙdegenerate case Ψ → P̂ₙΨ / ‖P̂ₙΨ‖

The only reason a repeated measurement reproduces itself — and the one step no Schrödinger evolution supplies.

P̂ₙ projects onto the whole eigenspace of qₙ; the update is instantaneous, non-unitary and irreversible

Ensemble mean and spread⟨Q̂⟩ = ∫ Ψ* Q̂Ψ dx = Σₙ qₙ|cₙ|²

The average of many runs. It need not be an eigenvalue, so it is often a number no single run can ever show.

carries the unit of q; the variance is Σₙ qₙ²|cₙ|² − ⟨Q̂⟩², and σ is its square root

Build the operator from the classical expression

Start from the plane wave ψ = A e^{(i(kx − ωt)}), the state de Broglie gave momentum p = ħk. Differentiate it: −iħ ∂ψ/∂x = −iħ(ik)ψ = ħk ψ = p ψ. The derivative pulled the momentum out as a factor, so the operator standing for momentum is p̂ = −iħ ∂/∂x — and the i is not decoration, since without it the eigenvalues come out imaginary. Position needs no derivative at all: x̂ψ = xψ, multiplication by the coordinate. Everything else is assembled by writing the classical expression and substituting. Kinetic energy p²/2m becomes T̂ = −(ħ²/2m) ∂²/∂x², total energy becomes Ĥ = T̂ + V(x). Check units as you go: ħ = 1.055×10⁻³⁴ J s, so ħ ∂/∂x carries J s m⁻¹ = kg m s⁻¹, a momentum. The recipe has one soft spot. Classically xp and px are the same number, but x̂p̂ and p̂x̂ are different operators, so a classical product of position and momentum leaves the ordering undecided and the physics, not the dictionary, has to choose.

Test whether the state has a value at all

There is exactly one test for whether an observable has a definite value in a given state: act with the operator and see whether what comes back is that same function times a constant. Take an electron in an infinite well of width L = 0.50 nm, in its ground state ψ₁ = √(2/L) sin(πx/L). Apply Ĥ and out comes E₁ψ₁ with E₁ = h²/8mL² = 2.410×10⁻¹⁹ J = 1.504 eV: energy is sharp, and every run of an energy measurement reads 1.504 eV. Apply p̂ and you get −iħ√(2/L)(π/L) cos(πx/L) — a cosine, not a multiple of the sine. Momentum has no value in this state. To see what it does have, write sin(πx/L) = (e^(iπx/L) − e^(−iπx/L))/2i: an equal-weight superposition of the two momentum eigenfunctions ±ħπ/L = ±6.63×10⁻²⁵ kg m s⁻¹. Each comes up half the time, so ⟨p̂⟩ = 0 while √⟨p̂²⟩ = 6.63×10⁻²⁵ kg m s⁻¹ — and p²/2m for that magnitude is 1.504 eV, exactly E₁, as consistency demands.

The spectrum is the entire menu of readings

Solving Q̂φₙ = qₙφₙ is a one-off cost, paid before any particular state is considered, and what it returns is the whole dial. For the electron in the 0.50 nm well the energies are Eₙ = n²h²/8mL² = 1.504, 6.017, 13.54 and 24.07 eV, …, so an energy measurement on any state of that electron returns one of those numbers. A reading of 3.0 eV is not unlikely; it is not on the dial. Two ingredients fix the list, and students routinely credit only the first. The operator supplies the differential equation; the boundary conditions supply the rest. The same −(ħ²/2m) d²/dx² gives a discrete ladder inside a box and an unbroken continuum for a free electron, where every real p is an eigenvalue of p̂ and the sum over states becomes an integral. It is also why the dial moves when the apparatus does: halve the well to 0.25 nm and every level rises by a factor of four, to 6.017, 24.07 and 54.15 eV.

Expand the state to get the odds

With the eigenfunctions in hand, any admissible state is a sum over them: Ψ = Σ cₙφₙ. Orthonormality extracts the coefficients — multiply by φₘ* and integrate, and every term but one dies, leaving cₘ = ∫ φₘ* Ψ dx. Born's rule reads the probability straight off: P(qₘ) = |cₘ|². Take Ψ = N(ψ₁ + 2ψ₂) in the well. Normalising gives |N|²(1 + 4) = 1, so N = 1/√5, c₁ = 1/√5 and c₂ = 2/√5, and the table is P(1.504 eV) = 1/5 = 0.20 and P(6.017 eV) = 4/5 = 0.80. Notice that the coefficient 2 became a probability of 0.80, not 2/3: it is the squares that are shared out, not the amplitudes. Notice too what the modulus throws away. Replace c₂ by 2e^(iθ)/√5 for any θ and the energy table does not move — but |Ψ(x)|² does, because its cross term 2Re(c₁*c₂ψ₁ψ₂) carries the relative phase. Phase is invisible to one observable's statistics and plainly visible in the density.

What the reading leaves behind

The postulate has a second clause. When the meter shows qₙ, the state immediately afterwards is φₙ, renormalised. Measure the energy of Ψ = (ψ₁ + 2ψ₂)/√5, read 6.017 eV, and the ψ₁ term is gone: the state is ψ₂, and a second energy measurement made straight away returns 6.017 eV with probability 1. Reproducibility is not a bonus feature; it is what the projection clause exists to deliver. Two refinements matter. If qₙ is degenerate the state does not collapse onto a single eigenfunction but onto its projection into the whole eigenspace, P̂ₙΨ/‖P̂ₙΨ‖, so a measurement can perfectly well leave a superposition standing. And if the second measurement is of an incompatible observable, the projection it performs scrambles the first table again. Set the two kinds of change side by side: Schrödinger evolution is continuous, deterministic and reversible; projection is abrupt, random and irreversible. Nothing in the equation produces the second, which is precisely why it is a separate postulate.

Statistics of an ensemble, not a forecast for one run

Because no single outcome is fixed, the content of the postulate is a distribution, and the two numbers usually quoted are its mean and its spread. For the 20/80 state, ⟨Ĥ⟩ = 0.20 × 1.504 + 0.80 × 6.017 = 5.11 eV and σ_E = |E₂ − E₁|√(P₁P₂) = 4.513 × 0.400 = 1.8 eV. Read that mean carefully: it is the average of a long run of measurements, and it is a number no single measurement can ever display, because it is not in the spectrum. The distribution is also not ignorance. A coherent superposition and a statistical mixture of 20% ψ₁ with 80% ψ₂ give identical energy tables, so no quantity of energy data will separate them; what separates them is the interference term in |Ψ(x, t)|², present in one and absent in the other. Finally, be blunt about the gap. The postulate offers no rule, hidden or otherwise, for whether the next electron will show 1.504 eV or 6.017 eV. Every prediction it makes is a claim about many identically prepared runs.

Change one variable at a time

Make the relationship visible.

Interactive model

Coefficient c₁ on ψ₁1.0

Coefficient c₂ on ψ₂2.0

Coefficient c₃ on ψ₃0.0

Drive the c₂ and c₃ sliders to zero: one bar goes to 1.000 and the dashed mean snaps exactly onto the 1.504 eV rung, because only an eigenstate has a definite energy. Any other setting strands the mean between rungs, at a value no single measurement can ever read.

MEAN ⟨E⟩5.11 eV

P(E₁) = |c₁|²0.200

P(E₂) = |c₂|²0.800

P(E₃) = |c₃|²0.000

Live interpretationMEAN ⟨E⟩: 5.11 eV. P(E₁) = |c₁|²: 0.200. P(E₂) = |c₂|²: 0.800. P(E₃) = |c₃|²: 0.000

Catch the common trap

Explain before calculating.

An electron in a 0.50 nm infinite well is prepared in Ψ = (ψ₁ + 2ψ₂)/√5, where ψ₁ and ψ₂ are energy eigenfunctions with E₁ = 1.504 eV and E₂ = 6.017 eV. One energy measurement is made on it. What can that single measurement give?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA free electron is in the state ψ(x) = A e^{ikx} with k = 1.20×10¹⁰ m⁻¹. Show that momentum has a definite value in this state while position does not, and find the momentum and the kinetic energy.

Apply the momentum operator: p̂ψ = −iħ dψ/dx = −iħ(ik)A e^{ikx} = ħk ψ. What comes back is ψ times a constant, so ψ is an eigenfunction of p̂ and the eigenvalue is p = ħk.
p = 1.055×10⁻³⁴ J s × 1.20×10¹⁰ m⁻¹ = 1.266×10⁻²⁴ kg m s⁻¹. Every run of a momentum measurement on this state returns that one number, so the spread is zero.
Kinetic energy is built from the same operator, T̂ = p̂²/2m, so ψ is an eigenfunction of it too: T = p²/2m = (1.266×10⁻²⁴)²/(2 × 9.11×10⁻³¹) = 8.80×10⁻¹⁹ J = 5.49 eV.
Now the position operator: x̂ψ = x A e^{ikx}. The result is ψ multiplied by the variable x, not by a constant, so ψ is not an eigenfunction of x̂ and position has no value here. Consistently, |ψ|² = |A|² is flat — every position is equally likely and none of them is the value.

Answerp = ħk = 1.266×10⁻²⁴ kg m s⁻¹ and T = 5.49 eV, both sharp; position has no value at all, since x̂ψ is not a multiple of ψ.

MediumThe electron is now trapped in an infinite well of width L = 0.50 nm and prepared as Ψ = N(ψ₁ + 2ψ₂), with ψₙ the normalised energy eigenfunctions. Normalise the state, write the table of possible energy readings with their probabilities, and find the mean and standard deviation of the energy.

Normalise using orthonormality: ∫|Ψ|² dx = |N|²(1 + 4) = 1, because the cross terms ∫ψ₁*ψ₂ dx vanish. So N = 1/√5, giving c₁ = 1/√5 and c₂ = 2/√5.
Born's rule: P(E₁) = |c₁|² = 1/5 = 0.20 and P(E₂) = |c₂|² = 4/5 = 0.80. They sum to 1, as any complete table must.
The eigenvalues come from the well, not from the state: E₁ = h²/8mL² = (6.626×10⁻³⁴)²/(8 × 9.11×10⁻³¹ × (5.0×10⁻¹⁰)²) = 2.410×10⁻¹⁹ J = 1.504 eV, and E₂ = 4E₁ = 6.017 eV.
Mean: ⟨Ĥ⟩ = Σ Eₙ|cₙ|² = 0.20 × 1.504 + 0.80 × 6.017 = 0.301 + 4.814 = 5.11 eV.
Spread, from the two-outcome shortcut: σ_E = |E₂ − E₁|√(P₁P₂) = 4.513 × √0.16 = 4.513 × 0.400 = 1.8 eV. Check it the long way: Σ Eₙ²|cₙ|² − ⟨Ĥ⟩² = 29.42 − 26.16 = 3.26 eV², whose root is 1.8 eV.

AnswerN = 1/√5; the readings are 1.504 eV with probability 0.20 and 6.017 eV with probability 0.80; ⟨Ĥ⟩ = 5.11 eV with σ_E = 1.8 eV, and 5.11 eV is itself never a reading.

HardThat same state is left alone for 1.0 ps, then its energy is measured and reads 6.017 eV, and the energy is measured again immediately afterwards. (a) Do the outcome probabilities drift during the 1.0 ps? (b) What does the second reading give, and with what probability? (c) What are ⟨Ĥ⟩ and σ_E after it?

Each eigenstate picks up its own phase, cₙ → cₙ e^{(−iEₙt/ħ)}. That factor has unit modulus, so |cₙ|² is untouched: the table is still 0.20 / 0.80 at t = 1.0 ps and at every other time. So the answer to (a) is no.
What does move is the density. Its cross term oscillates at ω₂₁ = (E₂ − E₁)/ħ, and E₂ − E₁ = 4.5124 eV = 7.230×10⁻¹⁹ J, so ω₂₁ = 7.230×10⁻¹⁹ / 1.0546×10⁻³⁴ = 6.86×10¹⁵ rad s⁻¹ — a period of 0.92 fs, about 1100 sloshes in 1.0 ps while the energy statistics stood perfectly still.
The first reading returns 6.017 eV, an outcome of probability 0.80, and the projection clause then replaces the state by the matching eigenfunction: Ψ → ψ₂, up to a global phase. The ψ₁ amplitude is deleted, not merely made small.
Expand that new state on the energy eigenbasis: c₂ = 1 and every other coefficient is 0. The second reading therefore gives 6.017 eV with probability |1|² = 1, which answers (b).
For (c), a single term leaves ⟨Ĥ⟩ = 6.017 eV and σ_E = 0. The mean jumped from 5.11 eV to 6.017 eV at the instant of the first reading — a change no solution of the Schrödinger equation produces, which is exactly why projection has to be posted as a separate rule.

Answer(a) No: the phases have unit modulus, so the table stays 0.20 / 0.80. (b) 6.017 eV, with probability 1. (c) ⟨Ĥ⟩ = 6.017 eV and σ_E = 0.