University Physics IV · Foundations of Quantum Mechanics · 5.5

Operators & Expectation Values

Once the state is a whole function, a measurable quantity can no longer be a number stored inside it. It has to be something you do to it. This lesson fixes which operation goes with which observable, turns that operation into an average and a spread, and shows that the reality of every average hangs on a boundary term.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Born's rule hands you a probability density, but almost nothing you measure is the position. The repair is to stop hunting for observables inside ψ and start representing them as linear operators acting on it: position multiplies, momentum differentiates as p̂ = −iħ ∂/∂x, and every classical quantity built from x and p is assembled from those two. Momentum is not guessed — a de Broglie plane wave e^(ikx) carries momentum ħk, and −iħ ∂/∂x returns exactly ħk times it, so the operator is read off the relation rather than decreed.

Numbers come back by sandwiching: ⟨Â⟩ = ∫ψ*Âψ dx is the mean of many measurements on identically prepared systems, and σ_A² = ⟨Â²⟩ − ⟨Â⟩² is their spread. Demanding that every such mean be real, for every state, is precisely what forces the operator to be Hermitian — and Hermiticity pays for itself twice over: real eigenvalues, so outcomes are real numbers, and orthogonal eigenfunctions, so the expansion that Born's rule needs actually exists. The cost is that Hermiticity belongs to the formula and its domain together.

It survives only while the boundary terms thrown up by integration by parts vanish, so p̂ is Hermitian on the whole line and hopeless on a half-line; and the product of two Hermitian operators need not be Hermitian, so a classical product like xp has to be symmetrised before it means anything at all.

Simple definition: An observable is represented by a Hermitian linear operator Â acting on the wave function, and the mean of many measurements of it on identically prepared systems is the integral ⟨Â⟩ = ∫ψ*(x) Âψ(x) dx.
Example: For the ground state of a well of width L, ⟨x̂⟩ = L/2 by symmetry, but ⟨x̂²⟩ = L²(1/3 − 1/(2π²)) = 0.2827L², so σₓ = 0.181L. The particle is centred on the middle, not sitting at it.

Expectation value⟨Â⟩ = ∫ ψ*(x) [Â ψ(x)] dx

The mean over many measurements on identically prepared systems — not a value one system holds.

ψ normalised; ⟨Â⟩ carries the unit of A. The operator acts on ψ before ψ* multiplies.

Position and momentum operatorsx̂ψ = xψ · p̂ψ = −iħ ∂ψ/∂x

p̂ e^(ikx) = ħk e^(ikx): the operator is read straight off de Broglie's p = ħk, not postulated.

Position representation. ħ = 1.055 × 10⁻³⁴ J s, so p̂ returns kg m s⁻¹.

The Hermiticity condition∫ φ*(Âψ) dx = ∫ (Âφ)*ψ dx for all φ, ψ in the domain

Guarantees every ⟨Â⟩ is real; it constrains the domain as much as it constrains the formula.

For p̂ it holds only if the boundary term −iħ[φ*ψ] vanishes at both limits.

Variance and standard deviationσ_A² = ⟨Â²⟩ − ⟨Â⟩² = ⟨(Â − ⟨Â⟩)²⟩

σ_A = 0 exactly when ψ is an eigenfunction of Â, so a spread reports superposition, not a poor apparatus.

σ_A shares the unit of A. ⟨Â²⟩ means applying Â twice, never squaring the number ⟨Â⟩.

Kinetic energy as curvature⟨p̂²⟩ = −ħ² ∫ ψ* ψ″ dx = ħ² ∫ |ψ′|² dx

Makes ⟨p̂²⟩ ≥ 0 obvious: kinetic energy is paid for in the curvature of ψ, not its value.

Second form after one integration by parts, valid whenever ψ → 0 at the limits.

What Hermiticity buysÂψₙ = aₙψₙ ⟹ aₙ real, and ∫ ψₘ* ψₙ dx = 0 when aₘ ≠ aₙ

Supplies the real outcomes and the orthonormal basis that Born's rule expands the state in.

The same vanishing boundary terms are required for both halves of the proof.

Why an observable has to become an operator

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 Â 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.

Reading p̂ = −iħ d/dx off the de Broglie relation

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.

Sandwiching: the expectation-value integral

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 L = 0.50 nm it gives 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.

Hermiticity is a condition on the boundary too

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.

Real eigenvalues and orthogonal eigenfunctions come free

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 makes the expansion ψ = Σcₙψₙ unique and lets Born's rule read |cₙ|² = |∫ψₙ*ψ dx|² as the probability of outcome aₙ. Remove Hermiticity and there is no basis, no probabilities, and nothing to normalise.

Variance: applying twice, not squaring the answer

The spread of outcomes is σ_A² = ⟨Â²⟩ − ⟨Â⟩², and ⟨Â²⟩ 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. It vanishes exactly when ψ is an eigenfunction of Â, since Âψ = aψ makes ⟨Â²⟩ = a² = ⟨Â⟩²; a nonzero σ_A is therefore a statement that the state is a superposition of outcomes, not that the instrument is poor. 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.

Change one variable at a time

Make the relationship visible.

Interactive model

Packet centre ⟨x̂⟩1.20 nm

Packet width σₓ0.24 nm

Central wavenumber k₀8 nm⁻¹

Pull k₀ to zero: the dashed imaginary part flattens and ⟨p̂⟩ goes with it, which is why any real ψ has zero mean momentum. Now sweep k₀ to −16 and watch the lower panel not move at all — |ψ|² is blind to the phase, so ∫(−iħ d/dx)|ψ|²dx can only ever return zero.

⟨x̂⟩1.20 nm

σₓ0.24 nm

⟨p̂⟩ = ħk₀8.44 ×10⁻²⁵ kg m/s

σₚ = ħ / 2σₓ2.20 ×10⁻²⁵ kg m/s

Live interpretation⟨x̂⟩: 1.20 nm. σₓ: 0.24 nm. ⟨p̂⟩ = ħk₀: 8.44 ×10⁻²⁵ kg m/s. σₚ = ħ / 2σₓ: 2.20 ×10⁻²⁵ kg m/s

Catch the common trap

Explain before calculating.

Misconception check

The commonest error is to slide the operator out of the sandwich and write ⟨Â⟩ = ∫Â|ψ|²dx. For position that happens to be right, since x̂ only multiplies and ∫x|ψ|²dx is the correct weighted average — which is exactly why the habit survives. Try it on momentum and it collapses: ∫(−iħ d/dx)|ψ|²dx = −iħ[|ψ|²] evaluated at ±∞ = 0, for every normalisable state whatever, including a packet visibly travelling at 10⁶ m s⁻¹. The operator must act on ψ before ψ* multiplies it, because differentiating |ψ|² throws the phase away — and for momentum the phase is the whole answer: ψ = f(x)e^(ik₀x) with f real has ⟨p̂⟩ = ħk₀, while f and f e^(ik₀x) have identical probability densities. A second version of the same slip is writing ⟨p̂²⟩ = ⟨p̂⟩², which sets the kinetic energy of every bound state to zero.

A normalised Gaussian packet on the line is ψ(x) = (2a/π)¹⁄⁴ e^(−ax²) e^(ik₀x), with a > 0 and k₀ > 0. Which statement about its momentum expectation value and spread is correct?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAn electron sits in the ground state of an infinite well of width L = 0.50 nm, ψ₁(x) = √(2/L) sin(πx/L) on 0 ≤ x ≤ L. Find ⟨x̂⟩, ⟨p̂⟩ and ⟨p̂²⟩, then convert ⟨p̂²⟩ into a kinetic energy in eV. Take ħ = 1.055 × 10⁻³⁴ J s and mₑ = 9.11 × 10⁻³¹ kg.

x̂ only multiplies, so ⟨x̂⟩ = ∫₀^L x|ψ₁|²dx. The density (2/L)sin²(πx/L) is symmetric about x = L/2, so the integral is L/2 = 0.25 nm with no work at all.
For momentum the operator must act first: p̂ψ₁ = −iħ√(2/L)(π/L)cos(πx/L). Then ⟨p̂⟩ = −iħ(2/L)(π/L)∫₀^L sin(πx/L)cos(πx/L)dx, and that integral is ½∫₀^L sin(2πx/L)dx = (L/4π)[−cos(2πx/L)]₀^L = 0. So ⟨p̂⟩ = 0.
Apply p̂ again: p̂²ψ₁ = −ħ²d²ψ₁/dx² = ħ²(π/L)²ψ₁, so ψ₁ is an eigenfunction of p̂² and ⟨p̂²⟩ = (πħ/L)² with no integral needed. Numerically πħ/L = π(1.055 × 10⁻³⁴)/(0.50 × 10⁻⁹) = 6.63 × 10⁻²⁵ kg m s⁻¹.
K = ⟨p̂²⟩/2mₑ = (6.63 × 10⁻²⁵)²/(2 × 9.11 × 10⁻³¹) = 4.39 × 10⁻⁴⁹/1.822 × 10⁻³⁰ = 2.41 × 10⁻¹⁹ J, and dividing by 1.602 × 10⁻¹⁹ J eV⁻¹ gives 1.50 eV.

Answer⟨x̂⟩ = 0.25 nm, ⟨p̂⟩ = 0, ⟨p̂²⟩ = (πħ/L)² = 4.39 × 10⁻⁴⁹ kg² m² s⁻², so σₚ = 6.63 × 10⁻²⁵ kg m s⁻¹ and K = 1.50 eV. Zero mean with a nonzero mean square: the standing wave carries +ħk and −ħk equally.

MediumFor the same ground state, evaluate ⟨x̂²⟩ and hence σₓ, then form the product σₓσₚ using σₚ = πħ/L from the previous problem. Compare the result with ħ/2, and say whether the answer depends on L. Use L = 0.50 nm.

⟨x̂²⟩ = (2/L)∫₀^L x² sin²(πx/L)dx. Write sin²θ = (1 − cos2θ)/2, giving ⟨x̂²⟩ = (1/L)∫₀^L x²dx − (1/L)∫₀^L x²cos(2πx/L)dx.
The first integral is L³/3, so that term is L²/3. For the second, with b = 2π/L, ∫₀^L x²cos(bx)dx = [x²sin(bx)/b + 2x cos(bx)/b² − 2sin(bx)/b³]₀^L = 2L/b² = L³/(2π²), so that term is L²/(2π²).
⟨x̂²⟩ = L²(1/3 − 1/(2π²)) = L²(0.33333 − 0.05066) = 0.28267L². Subtract ⟨x̂⟩² = (L/2)² = 0.25L²: σₓ² = 0.03267L², so σₓ = 0.1808L = 0.0904 nm.
Form the product: σₓσₚ = (0.1808L)(πħ/L) = 0.1808π ħ = 0.568ħ. The L cancels, so every well width gives the same number — the state is squeezed in x and stretched in p by exactly compensating factors.
Numerically 0.0904 × 10⁻⁹ × 6.63 × 10⁻²⁵ = 5.99 × 10⁻³⁵ J s against ħ/2 = 5.28 × 10⁻³⁵ J s, a ratio of 1.14.

Answer⟨x̂²⟩ = 0.2827L², so σₓ = 0.1808L = 0.0904 nm, and σₓσₚ = 0.568ħ = 5.99 × 10⁻³⁵ J s — a factor 1.14 above ħ/2. The L cancels, so every well width gives the same product; only a Gaussian reaches the floor.

HardShow that the product x̂p̂ is not Hermitian even though x̂ and p̂ separately are, evaluate ⟨x̂p̂⟩ for an arbitrary real normalised ψ(x) on the whole line, and build from x̂ and p̂ an operator whose expectation value is guaranteed real.

Adjoints reverse order: (x̂p̂)† = p̂†x̂† = p̂x̂. So x̂p̂ equals its own adjoint only if x̂ and p̂ commute — and they do not, since [x̂, p̂] = iħ gives p̂x̂ = x̂p̂ − iħ ≠ x̂p̂. The product of two Hermitian operators is in general not Hermitian.
Evaluate it directly for real ψ: ⟨x̂p̂⟩ = ∫ψ x(−iħ dψ/dx)dx = −iħ∫x ψψ′dx, and ψψ′ = ½ d(ψ²)/dx.
Integrate by parts: ∫₋∞^∞ x d(ψ²)/2 = [xψ²/2]₋∞^∞ − ½∫ψ²dx = 0 − ½. The boundary term vanishes for any state with finite ⟨x̂²⟩, and the remaining integral is 1 by normalisation.
So ⟨x̂p̂⟩ = −iħ(−½) = iħ/2 — purely imaginary, so no apparatus can be reporting it. The failure of Hermiticity has shown up as an unmeasurable number, not as an algebraic curiosity.
Repair it by symmetrising: Â = ½(x̂p̂ + p̂x̂). Since p̂x̂ = x̂p̂ − iħ, ⟨p̂x̂⟩ = iħ/2 − iħ = −iħ/2, so ⟨Â⟩ = ½(iħ/2 − iħ/2) = 0, real as required.
Check the adjoint algebraically: Â = x̂p̂ − iħ/2, so Â† = p̂x̂ + iħ/2 = (x̂p̂ − iħ) + iħ/2 = x̂p̂ − iħ/2 = Â. Symmetrising is the standard cure for every ordering ambiguity inherited from classical mechanics.

Answer(x̂p̂)† = p̂x̂ = x̂p̂ − iħ, so x̂p̂ is not Hermitian. For any real normalised ψ, ⟨x̂p̂⟩ = iħ/2 — imaginary, hence unmeasurable. The symmetrised ½(x̂p̂ + p̂x̂) = x̂p̂ − iħ/2 is Hermitian, with expectation 0.