Expectation values, variances and moments

The bracket ⟨ψ|A|ψ⟩ is an instruction with an order: act with A on |ψ⟩, then take the inner product of the result with ⟨ψ|. In the position representation that is ∫ψ*(Aψ)dx, which equals ∫A|ψ|²dx only when A is multiplication by a function of x. Position qualifies, momentum does not. Take the box ground state ψ₁ = √(2/L) sin(πx/L), which is real. Then ⟨p⟩ = ∫ψ₁(−iℏψ₁′)dx = −iℏ[ψ₁²/2] evaluated at the walls, which is zero because Dirichlet conditions kill ψ₁ there. The expression ∫p|ψ|²dx is not merely wrong, it is not well formed: p is not a function of x, so there is nothing to put in the integrand. And the second moment is emphatically not the square of the first — ⟨p²⟩ = (πℏ/L)² while ⟨p⟩² = 0, and it is that gap that gives a confined electron its zero-point energy.

Insert the resolution of the identity Σ|n⟩⟨n| = 1 on both sides of A and use A|n⟩ = aₙ|n⟩. Everything collapses to ⟨A⟩ = Σ aₙ|cₙ|² with cₙ = ⟨n|ψ⟩. Three facts follow at once: ⟨A⟩ is real because the aₙ are; it lies between the smallest and largest aₙ; and it is generally not an eigenvalue, so it is not a possible reading. Numbers make the point. An electron in a well of width L = 0.50 nm has E₁ = h²/(8mL²) = 1.50 eV, so the levels are 1.50 eV, 6.02 eV, 13.5 eV. Prepare |ψ⟩ = (|1⟩ + i|2⟩)/√2 and ⟨H⟩ = ½(1.50) + ½(6.02) = 3.76 eV, with σ_H = 1.5E₁ = 2.26 eV. No apparatus ever displays 3.76 eV; each run returns 1.50 eV or 6.02 eV, half the time each. Notice too that the factor i vanished: H is diagonal in this basis, so only |cₙ|² enters and a relative phase cannot touch any energy statistic.

Subtracting the conjugate gives ⟨A⟩ − ⟨A⟩* = ⟨ψ|(A − A†)|ψ⟩, so a real expectation value is a statement about the adjoint, and the adjoint is a statement about the domain. Test it on p = −iℏ d/dx. On the whole line, one integration by parts gives ⟨p⟩ − ⟨p⟩* = −iℏ[|ψ|²] between ±∞, which vanishes because a ψ in the domain of p, with ψ and ψ′ both square-integrable, dies at infinity. Restrict the same operator to the half-line [0, ∞) and the boundary term at the origin survives: ⟨p⟩ − ⟨p⟩* = iℏ|ψ(0)|², so Im⟨p⟩ = ℏ|ψ(0)|²/2. For a state with |ψ(0)|² = 1.0 nm⁻¹ that is 5.3 × 10⁻²⁶ kg m s⁻¹ of imaginary momentum. The symbol −iℏ d/dx did not change; the domain did. So an imaginary part well above machine precision in a numerical expectation value is a diagnostic, not noise: go and find the boundary term you discarded.

Define |δ⟩ = (A − ⟨A⟩)|ψ⟩. Because A is Hermitian and ⟨A⟩ is a real number, ⟨δ|δ⟩ = ⟨ψ|(A − ⟨A⟩)²|ψ⟩ = ⟨A²⟩ − ⟨A⟩², so the variance is a squared norm. Two consequences are free: it can never be negative, and it is zero only if |δ⟩ is the zero vector, that is if A|ψ⟩ = ⟨A⟩|ψ⟩ — an eigenvector. A spread of zero is therefore not a well-run experiment but a statement about which eigenbasis the state sits in. For the ground state of a 1.00 nm box, ⟨x⟩ = 0.500 nm and ⟨x²⟩ = L²(1/3 − 1/(2π²)) = 0.2827 nm², so σₓ = √(0.2827 − 0.2500) = 0.181 nm, and with σₚ = πℏ/L the product is 0.568ℏ: above ℏ/2, not at it. One numerical warning follows from the same algebra. ⟨A²⟩ − ⟨A⟩² subtracts two nearly equal large numbers, so on a grid accumulate Σ(x − ⟨x⟩)²|ψ|²Δx instead and keep your digits.

The Born rule delivers frequencies, so ⟨A⟩ and σ_A describe N systems identically prepared in |ψ⟩ and each measured once. They are not properties of a single run, and one system cannot be re-measured to sample ψ afresh: under the projection postulate (the Lüders rule) an ideal measurement leaves the system in the normalised projection of |ψ⟩ onto the eigenspace of the outcome, so an immediate second measurement of A simply repeats the first result. Estimating ⟨A⟩ from N runs carries a standard error σ_A/√N, and this is the only quantity in sight that improves with data. To pin ⟨x⟩ for the 1.00 nm box ground state to ±0.010 nm you need N = (0.181/0.010)² ≈ 3.3 × 10² runs. Meanwhile σₓ itself stays at 0.181 nm however many runs you take and however good the detector is, because it is a property of ψ. Keep the two apart in your notation: σ_A is the width of the distribution, σ_A/√N the width of your ignorance about its centre.

⟨ψ|ψ⟩ = 1 is one convergent integral and it certifies nothing about the others. Take ψ(x) = √(a/π)/√(x² + a²). It is normalised, since ∫dx/(x² + a²) = π/a, but its density is a Lorentzian falling only as x⁻², so x²|ψ|² tends to the constant a/π and ⟨x²⟩ diverges linearly. Even ⟨x⟩ is only conditionally zero: the integrand is odd, but ∫|x||ψ|²dx diverges logarithmically, so the cancellation is a convention rather than a value. Momentum is untroubled — ⟨p²⟩ = ℏ²∫|ψ′|²dx = ℏ²/(8a²) is finite, so ψ sits in the domain of p but not in the domain of x. The numerical trap is that a grid never reports infinity. On [−X, X] the truncated moment is ⟨x²⟩ = (2a/π)(X − a arctan(X/a)), which for a = 1.0 nm gives 11.8 nm² at X = 20 nm and 24.5 nm² at X = 40 nm. A number that doubles when the box doubles has not converged.