Operators and the measurement postulate

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.

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. Writing sin(πx/L) = (e^(iπx/L) − e^(−iπx/L))/2i tempts you to predict ±ħπ/L half the time each, but the sine stops at the walls, so those truncated waves are not momentum eigenfunctions, and a momentum measurement returns a continuous spread of values, symmetric about zero, not two spikes. What the state does fix are the moments: ⟨p̂⟩ = 0 because ψ₁ is real, and ⟨p̂²⟩ = ħ²∫|dψ₁/dx|² dx = (ħπ/L)², so √⟨p̂²⟩ = ħπ/L = 6.63×10⁻²⁵ kg m s⁻¹ — and p²/2m for that magnitude is 1.504 eV, exactly E₁, as it must be where V = 0.

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, 6.016, 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 6.016, 24.07 and 54.15 eV.

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(6.016 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.

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 6.016 eV, and the ψ₁ term is gone: the state is ψ₂, and a second energy measurement made straight away returns 6.016 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. Whether that step is a physical event or only an update of our description, the postulate does not say; that question is where the interpretations of quantum mechanics part company.

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 × 6.016 = 5.11 eV and σ_E = |E₂ − E₁|√(P₁P₂) = 4.512 × 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 6.016 eV. Every prediction it makes is a claim about many identically prepared runs.