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Derivation

Expectation Values and the Operator Dictionary

Statement

For a normalised state ψ(x), the average value of an observable A over repeated identical measurements is the inner product ⟨A⟩ = ⟨ψ|Â|ψ⟩ = ∫ ψ*(x) Â ψ(x) dx. Imposing the Born rule for position fixes x̂ = x (multiplication), and imposing the de Broglie relation for momentum eigenstates fixes p̂ = −iℏ ∂/∂x. These two representations are the seed of the entire operator dictionary.

Why it matters

The wavefunction is not itself measurable; only statistical averages of observables are. The expectation-value integral is the bridge from the abstract state to numbers a laboratory reports, and it is the object that appears in Ehrenfest's theorem, the variational principle, and every uncertainty relation.

Fixing and as concrete differential/multiplicative operators is what turns quantum mechanics from an abstract postulate set into a computational engine: once these two are known, every classical observable built from x and p — energy, angular momentum, current — is obtained by substitution.

Assumptions
The state is normalisable and normalised, ∫|ψ|² dx = 1.Without normalisation the integrals define relative weights only; ⟨A⟩ must be divided by ⟨ψ|ψ⟩, and unnormalisable states (e.g. pure plane waves) have no ordinary expectation value. The Born rule holds: |ψ(x)|² dx is the probability of finding the particle in [x, x+dx].If |ψ|² were not the probability density, the identification ⟨x⟩ = ∫ x|ψ|² dx would fail and x̂ = x would lose its justification. The de Broglie relation p = ℏk assigns momentum ℏk to the plane wave eikx.Drop it and the momentum operator is undetermined; any operator with the same eigenvalues on non-plane-wave bases would compete, and the Fourier duality between x and p collapses. Observables are represented by Hermitian operators (prior result), so wavefunctions and their derivatives vanish at ±∞ fast enough for boundary terms to drop.If  is not Hermitian, ⟨A⟩ can be complex and boundary terms in the integration by parts survive, so would not be self-adjoint and ⟨p⟩ would not be real.
Derivation
1
⟨x⟩ = ∫ x P(x) dx = ∫ x |ψ(x)|² dx
The classical mean of a random variable weighted by its probability density; P(x)=|ψ|² is the Born rule. A
2
⟨x⟩ = ∫ ψ*(x) [ x ψ(x) ] dx
Write |ψ|² = ψ*ψ and slot the real scalar x between them; this is the inner product of ψ with , which defines the action of as multiplication by x. A
3
⟨f(x)⟩ = ∫ ψ* f(x) ψ dx ⟹ f(x̂)ψ = f(x)ψ
Any function of position averages the same way, so every position-only observable is a multiplicative operator. A
4
ψp(x) = (2πℏ)−1/2 eipx/ℏ , (−iℏ ∂/∂x) eipx/ℏ = p eipx/ℏ
de Broglie assigns momentum p = ℏk to eikx; the operator −iℏ∂/∂x is the unique first-order operator returning that eigenvalue on every plane wave. This constructs . B
5
ψ(x) = (2πℏ)−1/2 ∫ φ(p) eipx/ℏ dp
Any normalisable state is a superposition of momentum eigenstates (Fourier / completeness); |φ(p)|² is the momentum probability density by the Born rule in the momentum basis. C
6
⟨p⟩ = ∫ p |φ(p)|² dp = ∫∫∫ p φ*(p′)φ(p) (2πℏ)−1 ei(p−p′)x/ℏ dx dp dp′
Start from the momentum-space mean, then insert the closure ∫ei(p−p′)x/ℏdx/2πℏ = δ(p−p′) written explicitly to move back toward position space. C
7
p eipx/ℏ = −iℏ ∂/∂x ( eipx/ℏ )
Replace the scalar factor p by the derivative acting on the exponential — legal because they are equal on each plane wave. C
8
⟨p⟩ = ∫ [ (2πℏ)−1/2∫φ*(p′)e−ip′x/ℏdp′ ] (−iℏ ∂/∂x) [ (2πℏ)−1/2∫φ(p)eipx/ℏdp ] dx
Reassemble the two Fourier integrals into ψ*(x) and ψ(x); the derivative passes through the p-integral since it acts only on x. C
9
⟨p⟩ = ∫ ψ*(x) (−iℏ ∂/∂x) ψ(x) dx = ⟨ψ|p̂|ψ⟩
The same inner-product form as position, now with p̂ = −iℏ∂/∂x. Hermiticity (prior result) guarantees the result is real. B
Result
⟨A⟩ = ∫ ψ*(x) Â ψ(x) dx , x̂ = x , p̂ = −iℏ ∂/∂x

Reading. The measurable average of any observable is the overlap of the state with the operator acting on the state. Position acts by plain multiplication because |ψ|² already lives in position space; momentum acts by differentiation because a spatial derivative is what reads off the local wavenumber k = p/ℏ of the wave. The two operators together generate the whole dictionary A(x,p) → Â(x̂, p̂).

Units check. ⟨x⟩: ψ has units m−1/2 in 1D, so ∫ψ*·x·ψ dx = (m−1/2)(m)(m−1/2)(m) = m. ✓ ⟨p⟩: −iℏ∂xψ has units (J·s)(m−1)(m−1/2) = kg·m1/2·s−1; multiply by ψ*dx = (m−1/2)(m) to get kg·m·s−1. ✓

Limiting cases
  • Real ψ (bound stationary state): ⟨p⟩ = ∫ψ(−iℏψ′)dx is i times a real number, but must be real, so ⟨p⟩ = 0 — no net current, as expected for a standing wave.
  • Momentum eigenstate limit: as φ(p) → δ(p−p₀), ⟨p⟩ → p₀ and the variance ⟨p²⟩−⟨p⟩² → 0, recovering a sharp de Broglie wave.
  • Classical / Ehrenfest limit: the operator averages obey d⟨x⟩/dt = ⟨p⟩/m and d⟨p⟩/dt = −⟨∂xV⟩, so ⟨x⟩, ⟨p⟩ track Newton's equations for slowly varying potentials.
  • Position-diagonal observable: if  = f(x̂) only, the integral collapses to the classical average ∫f(x)|ψ|²dx.
Breaks when
  • Non-normalisable states. A plane wave or a free particle in infinite space has ∫|ψ|² dx = ∞; ⟨x⟩ and ⟨p⟩ are ill-defined and must be replaced by wavepacket averages or box-normalised densities.
  • Heavy-tailed wavefunctions. If |ψ|² decays only as 1/x² (Lorentzian-like), the integral ∫x|ψ|²dx diverges: ⟨x⟩ does not exist even though ψ is normalisable. The moment integral must converge, not just the norm.
  • Boundaries and singular domains. On a half-line or a box, p̂ = −iℏ∂x is Hermitian only with correct boundary conditions; on [0,∞) it has no self-adjoint extension, so ⟨p⟩ is not guaranteed real and the naive formula fails.
  • Operator ordering ambiguity. For observables mixing x and p (e.g. xp), the classical product has no unique quantum image; x̂p̂ ≠ p̂x̂ and one must symmetrise, so the dictionary is not a plain substitution.
Failure modes
  • Dropping the conjugate: writing ⟨A⟩ = ∫ ψ Â ψ dx instead of ∫ ψ* Â ψ dx — gives wrong (often complex) answers for any state with a complex phase.
  • Sign/factor slip in p̂: using +iℏ∂x or −iℏ/2π · ∂x; the sign is fixed by e+ipx/ℏ having momentum +p, and there is no once (not h) is used.
  • Letting p̂ act on ψ* : differentiating the wrong factor, or integrating by parts and forgetting the sign flip — ⟨p⟩ should come out real; a residual i signals this error.
  • Confusing ⟨x²⟩ with ⟨x⟩² : the variance ⟨x²⟩ − ⟨x⟩² is not zero in general; students plug ⟨x⟩ into .
  • Treating x̂ and p̂ as commuting: evaluating ⟨xp⟩ by multiplying ⟨x⟩⟨p⟩, ignoring [x̂,p̂]=iℏ.
Discussion

The expectation-value postulate is where the linear, unitary machinery of Hilbert space makes contact with irreversible measurement. The number ⟨A⟩ is not the outcome of any single experiment — measurements yield eigenvalues of  — but the mean of a long run of identically prepared measurements. Because  is Hermitian (prior result), its spectrum is real and this mean is real, so the formalism never predicts an imaginary voltage.

The asymmetry between (multiplicative) and (differential) is a choice of representation, not a physical fact: in the momentum representation the roles reverse, with p̂ = p and x̂ = +iℏ∂/∂p. Position and momentum are Fourier conjugates, and it is exactly the impossibility of a function being sharply peaked in both x and its Fourier variable that becomes the Heisenberg uncertainty relation once you compute the two variances from these operators.

The canonical commutator [x̂, p̂] = x(−iℏ∂x) − (−iℏ∂x)x = iℏ follows immediately from these representations and is the true content of the dictionary: it is the quantum echo of the classical Poisson bracket {x,p}=1, and every later structure — ladder operators, angular-momentum algebra, canonical quantisation of fields — is built on it.

At the deepest level the "operator dictionary" is Dirac's canonical quantisation prescription {A,B}PB → (1/iℏ)[Â,B̂], and its ambiguities are not defects but signposts. The ordering problem for xp, the Groenewold–van Hove no-go theorem forbidding a fully consistent quantisation of all classical observables, and the need for symmetrisation all show that x̂ = x, p̂ = −iℏ∂x is a minimal, physically anchored choice rather than a derivation of quantum mechanics from classical mechanics. The Born rule and de Broglie relation are inputs; what is derived is their self-consistent operational form.

Common misconceptions. "The expectation value is the most likely outcome" — false; it need not even be an allowed eigenvalue (e.g. ⟨x⟩ = L/2 in a box where the particle is never most likely at the exact centre for excited states). "Operators are matrices only" — in a continuous basis they are differential/multiplicative; the matrix picture is the discrete-basis shadow of the same object.

Worked examples
1
Ground state of an infinite square well, width L: find ⟨x⟩ and ⟨p⟩.
Setup: ψ₁(x) = √(2/L) sin(πx/L) on [0,L], real and normalised. A
2
⟨x⟩ = ∫₀L x (2/L) sin²(πx/L) dx = (2/L)·(L²/4) = L/2
Symmetry of sin² about x=L/2; the standard integral ∫₀Lx sin²(πx/L)dx = L²/4. A
3
⟨p⟩ = ∫₀L ψ₁ (−iℏ∂x) ψ₁ dx = −iℏ(2/L)(π/L)∫₀L sin(πx/L)cos(πx/L) dx = 0
ψ₁ real ⟹ integrand is (real, odd about centre) ⟹ integral vanishes; also required by reality of ⟨p⟩. B
4
Numbers: L = 1.00 nm = 1.00×10−9 m ⟹ ⟨x⟩ = 5.00×10−10 m, ⟨p⟩ = 0
Direct substitution. A
⟨x⟩ = L/2 = 0.500 nm , ⟨p⟩ = 0 kg·m·s−1

Reading. The particle sits, on average, at the centre with no net drift — a standing wave carries no mean momentum, though ⟨p²⟩ ≠ 0.

1
Travelling Gaussian packet: find ⟨p⟩ for ψ(x) = (πσ²)−1/4 eik₀x e−x²/2σ².
Setup: normalised Gaussian modulated by a plane-wave phase of wavenumber k₀. B
2
−iℏ∂xψ = −iℏ( ik₀ − x/σ² )ψ = ( ℏk₀ + iℏx/σ² )ψ
Differentiate the product; the phase gives ik₀, the envelope gives −x/σ². B
3
⟨p⟩ = ∫ |ψ|²( ℏk₀ + iℏx/σ² ) dx = ℏk₀ + (iℏ/σ²)∫ x|ψ|² dx = ℏk₀
|ψ|² = (πσ²)−1/2e−x²/σ² is even, so ∫x|ψ|²dx = 0; the imaginary piece cancels, as Hermiticity demands. B
4
Numbers: k₀ = 1.00×1010 m−1, ℏ = 1.055×10−34 J·s ⟹ ⟨p⟩ = ℏk₀
Substitute. A
⟨p⟩ = ℏk₀ = 1.06×10−24 kg·m·s−1

Reading. The mean momentum is set entirely by the phase gradient k₀, independent of the packet width σ — the envelope contributes spread (via ⟨p²⟩) but no mean.

Problems
  1. (A) A particle has uniform probability density on [0, L], i.e. |ψ|² = 1/L. Find ⟨x⟩ and ⟨x²⟩.
    Solution⟨x⟩ = ∫₀L x (1/L) dx = (1/L)(L²/2) = L/2. ⟨x²⟩ = ∫₀L x²(1/L)dx = (1/L)(L³/3) = L²/3. Variance = L²/3 − L²/4 = L²/12, so Δx = L/√12 ≈ 0.289 L.
  2. (B) Show that for any real normalisable wavefunction ψ(x), ⟨p⟩ = 0.
    Solution⟨p⟩ = ∫ ψ(−iℏψ′) dx = −iℏ∫ ψψ′ dx = −iℏ·½∫ (ψ²)′ dx = −iℏ·½[ψ²]−∞ = 0, since ψ→0 at infinity. A real wavefunction is an equal superposition of +p and −p components, so the mean momentum cancels.
  3. (B) For the SHO ground state ψ₀(x) = (mω/πℏ)1/4 e−mωx²/2ℏ, compute ⟨x⟩ and ⟨x²⟩.
    Solution⟨x⟩ = 0 by even symmetry of |ψ₀|². With a = mω/ℏ, |ψ₀|² = √(a/π) e−ax² and ∫x²e−ax²dx = ½√(π/a³), so ⟨x²⟩ = √(a/π)·½√(π/a³) = 1/(2a) = ℏ/(2mω). Hence Δx = √(ℏ/2mω), the ground-state spread.
  4. (C) Prove ⟨p⟩ is real for any normalisable ψ by showing ⟨p⟩ = ⟨p⟩*.
    Solution⟨p⟩ = ∫ ψ*(−iℏψ′) dx. Take the complex conjugate: ⟨p⟩* = ∫ ψ(+iℏψ*′) dx. Integrate this by parts: = iℏ[ψψ*]−∞ − iℏ∫ ψ′ψ* dx = 0 − iℏ∫ ψ*ψ′ dx = ∫ ψ*(−iℏψ′)dx = ⟨p⟩. The boundary term vanishes because ψ→0. Thus ⟨p⟩ = ⟨p⟩* is real — this is exactly the Hermiticity of in action.
  5. (C) For the infinite-well ground state ψ₁ = √(2/L) sin(πx/L), compute ⟨p²⟩ and confirm E₁ = ⟨p²⟩/2m.
    Solutionp̂²ψ₁ = −ℏ²∂x²ψ₁ = −ℏ²·(−(π/L)²)ψ₁ = ℏ²(π/L)²ψ₁. Then ⟨p²⟩ = ℏ²(π/L)²∫|ψ₁|²dx = (ℏπ/L)². Energy: E₁ = ⟨p²⟩/2m = ℏ²π²/(2mL²), the known ground-state energy, and since V=0 inside, all energy is kinetic. Note Δp = ℏπ/L while ⟨p⟩=0.