Operators and expectation values

The state is a whole function ψ(x), not a list of values, so a measurable quantity cannot be read off it the way a speed is read off a trajectory. What it can be is a rule for turning one state into another. Two requirements then narrow the rules almost completely. The rule must be linear, Â(αψ₁ + βψ₂) = αÂψ₁ + βÂψ₂, because superposition is the entire content of the previous topics and a nonlinear rule would wreck it. And the numbers the rule produces — the averages — must be real, since no laboratory ever reports 3 + 2i metres. Linearity makes  an operator; reality makes it a Hermitian one. In the position representation the first assignment is almost trivial: x̂ acts by multiplication, x̂ψ = xψ, which is why ⟨x̂⟩ = ∫x|ψ|²dx looks like an ordinary weighted average with |ψ|² as the weight. Nothing else on the list looks like that, and expecting it to is where most of the trouble starts.

Momentum is not defined by decree. A free particle of definite momentum is a plane wave ψ = e^(ikx) carrying p = ħk, so whatever operator represents momentum must return ħk times that wave. Differentiate: d/dx e^(ikx) = ik e^(ikx), so multiplying by −iħ gives exactly ħk e^(ikx). Hence p̂ = −iħ d/dx, with the i sitting there precisely to cancel the i the derivative produces and leave a real eigenvalue behind. Everything else is then built by substitution. Kinetic energy p²/2m becomes −(ħ²/2m) d²/dx², total energy adds V(x) as a multiplication, and angular momentum r × p becomes −iħ r × ∇. The one place substitution turns ambiguous is a classical product of x and p, because x̂p̂ ≠ p̂x̂ while the classical xp and px are the same number. That ordering problem has no classical counterpart and has to be settled by hand.

The recipe is ⟨Â⟩ = ∫ψ*(x) [Âψ(x)] dx, and the brackets matter: the operator acts on ψ first, and only then is the result multiplied by ψ*. Take the infinite-well ground state ψ₁ = √(2/L) sin(πx/L). Position is easy, because x̂ merely multiplies: |ψ₁|² is symmetric about L/2, so ⟨x̂⟩ = L/2. Momentum is not. Here p̂ψ₁ = −iħ√(2/L)(π/L)cos(πx/L), so ⟨p̂⟩ ∝ ∫₀^L sin(πx/L)cos(πx/L)dx, which is zero. That is a general result for any real normalisable ψ: ⟨p̂⟩ = −iħ∫ψψ′dx = −iħ[ψ²/2] = 0. It does not mean the particle is at rest. Apply p̂ a second time and p̂²ψ₁ = (πħ/L)²ψ₁, so ⟨p̂²⟩ = (πħ/L)² — decidedly nonzero, and for an electron in a well of width L = 0.50 nm that is a kinetic energy of 2.41 × 10⁻¹⁹ J, or 1.50 eV. A standing wave is an equal mixture of +ħk and −ħk: the mean cancels, the mean square does not.

Write the test out for p̂ on the line. Integrating by parts, ∫φ*(−iħψ′)dx = −iħ[φ*ψ] + ∫(−iħφ′)*ψ dx, so ∫φ*(p̂ψ)dx equals ∫(p̂φ)*ψ dx exactly when the boundary term [φ*ψ] vanishes at both limits. For square-integrable states on the whole line it does, and p̂ is Hermitian. Change the domain and the conclusion changes with it. On the half-line 0 ≤ x < ∞ the term at x = 0 cannot be made to vanish for every pair of states, and p̂ admits no self-adjoint extension there — radial momentum in three dimensions inherits exactly this problem. Inside a hard-walled box, ψ(0) = ψ(L) = 0 does kill the boundary term, so ⟨p̂⟩ comes out real; but no plane wave e^(ikx) vanishes at two points, so p̂ has no eigenfunctions in that domain and momentum is not a good observable in a box, even though p̂² is. An operator is its formula plus the functions it is allowed to act on.

Two consequences follow from the Hermitian condition in about three lines each, and the measurement postulate needs both. Let Âψ = aψ with ψ normalised. Then ∫ψ*(Âψ)dx = a, while ∫(Âψ)*ψ dx = a*. Hermiticity equates them, so a = a*: every eigenvalue is real, which is why the possible outcomes of a measurement are real numbers. Now take two eigenfunctions, Âψₘ = aₘψₘ and Âψₙ = aₙψₙ with aₘ ≠ aₙ. The same equality gives aₙ∫ψₘ*ψₙdx = aₘ*∫ψₘ*ψₙdx, and aₘ* = aₘ, so (aₙ − aₘ)∫ψₘ*ψₙdx = 0 and the overlap vanishes. Distinct outcomes therefore live in orthogonal states. That orthogonality is what pins the coefficients in ψ = Σcₙψₙ down uniquely and lets Born's rule read |cₙ|² = |∫ψₙ*ψ dx|² as the probability of outcome aₙ. Drop Hermiticity and the outcomes need not be real and the coefficients need not be unique, so the probability reading of |cₙ|² goes with them.

The spread of outcomes is σ_A² = ⟨²⟩ − ⟨Â⟩², and ⟨²⟩ means applying the operator twice inside one integral, ∫ψ*Â(Âψ)dx. It is not ⟨Â⟩². The difference is the whole physics: for the well ground state ⟨p̂⟩² = 0 while ⟨p̂²⟩ = (πħ/L)². Two facts make the variance worth the integral. The spread collapses to zero precisely for an eigenfunction, since Âψ = aψ makes ⟨²⟩ = a² = ⟨Â⟩²; away from one, the width of the outcome distribution is a fact about the state rather than about the instrument. And for p̂ one integration by parts turns ⟨p̂²⟩ = −ħ²∫ψ*ψ″dx into ħ²∫|ψ′|²dx, manifestly non-negative. Put both spreads together for the well ground state: σₓ = 0.181L and σₚ = πħ/L, so σₓσₚ = 0.568ħ, comfortably above the ħ/2 floor that the commutator topic later derives.