Continuous spectra and the momentum-space wavefunction

Momentum in one dimension is the operator −iħ d/dx, so the eigenvalue problem −iħ du/dx = pu integrates at once to uₚ(x) = A e^(ipx/ħ). On the whole line there is no wall to impose a boundary condition. The only demand is that u stay bounded at both infinities, which forces p to be real, since a complex p makes |u| grow exponentially in one direction, and leaves every real value allowed. So the spectrum is the entire real axis, a continuum rather than a list. The trouble arrives at the next step. |uₚ(x)|² = |A|² is the same number at every point, so ∫|uₚ|²dx over the line diverges for every A ≠ 0. Read physically, the state claims the particle is exactly as likely to be found in a 1 nm window here as in a 1 nm window a light-year away, and no rescaling turns that into a total of one. This is not a defect peculiar to momentum. The same happens to the energy eigenfunctions of any unbound problem — a free particle, or a scattering state above a barrier. A continuous spectrum simply does not come with normalisable eigenfunctions.

Keep the eigenfunctions and weaken the condition they must satisfy. Take the overlap of two of them: ∫uₚ* uₚ′ dx = |A|²∫e^{(i(p′−p)x/ħ)}dx = |A|² 2πħ δ(p′ − p), using the Fourier representation of the Dirac δ. Choosing |A|² = 1/2πħ makes that overlap exactly δ(p′ − p), the continuum's replacement for δₘₙ: zero for different labels, and for the same label a spike whose integral is one. The delta is a distribution, meaningful only inside an integral, and it carries the unit of 1/p, which is what forces uₚ to carry (J s)^(−1/2). If distributions grate, put the system in a box of length L with periodic ends instead: momenta are then discrete, pₙ = 2πnħ/L, spaced by 2πħ/L, every eigenfunction is normalisable, and sums pass to integrals through Σₙ → (L/2πħ)∫dp. Every physical answer must come out independent of L, and letting L → ∞ returns the delta convention.

The box shows exactly how Σₙ turns into ∫dp. Its normalised states are ψₙ = e^{(ipₙ x/ħ)}/√L, which is √(2πħ/L) times the δ-normalised uₚ taken at p = pₙ. For a state that fits well inside the box, the projection cₙ = ∫ψₙ*ψ dx is therefore √(2πħ/L) φ(pₙ), where φ(p) = ∫uₚ*ψ dx is the same projection taken on the δ-normalised basis. The discrete probabilities are |cₙ|² = (2πħ/L)|φ(pₙ)|²: the density times the spacing. Sum them over any window and you have a Riemann sum for ∫|φ|²dp, one bin per allowed momentum. Let L grow and every |cₙ|² falls as 1/L while the number of states in the window rises as L, so the sum converges on the integral and L drops out. The expansion follows the same path, ψ = Σₙ cₙ ψₙ → ∫φ(p) uₚ(x) dp, and it works for every square-integrable ψ because the plane waves are complete: ∫uₚ(x) uₚ*(x′) dp = δ(x − x′), the continuum's version of Σₙ ψₙ(x) ψₙ*(x′) = δ(x − x′). Written out, φ(p) = (2πħ)^(−1/2)∫ψ(x)e^(−ipx/ħ)dx is the forward half of the transform pair derived in the position and momentum representations lesson (UPIV-05.04), and its Parseval theorem, ∫|φ|²dp = ∫|ψ|²dx = 1, is the statement that the |cₙ|² summed to one all along.

Because ∫|φ|²dp = 1 and p carries a unit, |φ(p)|² must carry the unit of 1/p, inverse kg m s⁻¹. It is a probability per unit momentum, and it becomes a probability only after an integral: P(p₁ ≤ p ≤ p₂) = ∫|φ|²dp. That the density can exceed one is no paradox but a consequence of that unit, exactly as |ψ(0)|² = κ reaches 1.0 × 10¹⁰ m⁻¹ for the exponential state of the third example, whose decay length is 0.10 nm. Take the 1 nm electron packet: σₚ = ħ/2σₓ = 5.3 × 10⁻²⁶ kg m s⁻¹, so the peak density is 1/(σₚ√(2π)) = 7.6 × 10²⁴ (kg m s⁻¹)⁻¹. Ask for the probability of a momentum inside a window of width 1.0 × 10⁻²⁷ kg m s⁻¹ at that peak and the answer is 7.6 × 10²⁴ × 1.0 × 10⁻²⁷ = 7.6 × 10⁻³. Ask for the probability of exactly p = 0 and the answer is zero, for every state — which is why a continuum is described by a density and never by a list of |cₙ|².

The measurement postulate says a reading is an eigenvalue of the observable. For momentum every real p is one, so any value where φ(p) ≠ 0 can come out; what changes is how probability attaches to it. Since no single value carries any, a real reading is always a bin: a spectrometer of resolution δp reports that p lies within δp of some p₀, with probability ∫|φ|²dp over that bin, close to |φ(p₀)|²δp when the bin is narrow. Repeated on many identical preparations, the fraction of readings in each bin, divided by δp, reproduces |φ(p)|² averaged over that bin. The spread is a property of the prepared state, not damage done by the apparatus. What a reading cannot do is leave the system in a momentum eigenstate. In the textbook projection rule — an idealised apparatus, and one interpretation of what the update means — the state afterwards is φ cut down to the reported bin and renormalised: a packet of momentum spread about δp, and so, through the transform, spread over roughly h/δp in space, 0.66 μm for δp = 1.0 × 10⁻²⁷ kg m s⁻¹. An immediate repeat returns a value inside the same bin, not the same number to every digit as a discrete level would. Sharpening the resolution towards zero sends the probability of the reading to zero and the extent of the state left behind to infinity, which is why the plane wave is a limit no measurement reaches.

Momentum on the line is purely continuous, but most Hamiltonians are mixed. A finite square well, or hydrogen, has a discrete set of bound levels below the threshold for escape and a continuum of unbound states above it, and neither part is complete alone. The closure relation needs both, Σₙ ψₙ(x) ψₙ*(x′) + ∫ψ_E(x) ψ_E*(x′) dE = δ(x − x′), with the continuum states δ-normalised in E and the integral also summing the two directions of travel at each E in one dimension. A normalised state then splits its probability as Σₙ |cₙ|² + ∫|c(E)|²dE = 1: an energy measurement returns a bound level with a finite probability, or finds the particle unbound with a probability that only a window of energies can carry. The same bookkeeping shows that a density belongs to its label as well as to the state. A free particle's energy E = p²/2m is reached from both p and −p, so the density per unit energy is (m/|p|)(|φ(p)|² + |φ(−p)|²) at |p| = √(2mE), which diverges at E = 0 whenever φ(0) ≠ 0 yet stays integrable. Quote momenta as k = p/ħ and the density per unit k is ħ|φ(p)|², a different number for the same state and the same outcomes. Only the probability of a window is label-free.