Position and momentum representations

In a finite-dimensional space you expand a vector on a basis and keep the list of coefficients; change the basis and the list changes while the vector does not. Quantum states work the same way. Project |ψ⟩ on the position eigenstates and you recover ψ(x) = ⟨x|ψ⟩, the function you have been drawing all along. Project the same ket on the eigenstates of p̂ = −iħ d/dx, which are uₚ(x) = e^(ipx/ħ)/√(2πħ) with p̂ uₚ = p uₚ, and you get Φ(p) = ⟨uₚ|ψ⟩ = ∫ uₚ*(x) ψ(x) dx = (2πħ)^(−1/2) ∫ ψ(x) e^(−ipx/ħ) dx. The integral happens to be a Fourier transform, but it did not arrive as a technique borrowed from signal processing: it is an overlap, computed the way every overlap is. Because the inverse integral rebuilds ψ from Φ, neither list outranks the other. Which one you write down is a question about the problem in front of you, not about the electron: a well with walls is easiest in x, a beam through a monochromator is easiest in p.

Fourier inversion fixes only the product of the two prefactors: substitute one transform into the other and ψ comes back only if the forward and inverse constants multiply to 1/2πħ. The symmetric convention splits that factor evenly, (2πħ)^(−1/2) on each side, so the two transforms differ only in the sign of the exponent, and it is the one split that makes the transform unitary, as the next section shows. The same factor is what δ-normalises the plane waves themselves, ⟨uₚ′|uₚ⟩ = δ(p − p′); that continuous-spectrum bookkeeping is worked in Continuous spectra and the momentum-space wavefunction (UPIV-06.07), and nothing here needs more of it. Now count units. In one dimension [ψ] = m^(−1/2), so ∫ψ dx carries m¹⁄²; and [ħ] = J s = (kg m s⁻¹)⋅m, so dividing by √(2πħ) leaves [Φ] = (kg m s⁻¹)^(−1/2). |Φ(p)|² is therefore a probability per unit momentum while |ψ(x)|² is one per unit length: the two numbers are never comparable, and a peak height in one picture tells you nothing about a peak height in the other. Asymmetric conventions exist, with all of the 2πħ on one side; they are legal, but silently mixing two of them breaks Parseval, so declare yours once and keep it.

Substitute the transform into ∫|Φ(p)|² dp and it becomes (2πħ)⁽⁻¹⁾ ∫dp ∫dx ∫dx′ ψ*(x′) ψ(x) e^{(ip(x′ − x)/ħ)}. Do the p integral first: it returns 2πħ δ(x − x′), which cancels the prefactor and collapses the double integral to ∫|ψ(x)|² dx. So a state normalised in position is already normalised in momentum, and there is no second constant to hunt for. Run the same argument on two different states and you get ∫φ*(x) ψ(x) dx = ∫Φ_φ*(p) Φ_ψ(p) dp: the transform preserves every inner product, not merely norms. That is what unitary means, and it is the formal content of the claim that the two representations are equally valid rather than one being derived from the other. It also hands you a free check on any transform you compute by hand — if the two integrals of the squared modulus disagree, a factor of 2πħ has gone astray.

Three theorems handle most problems without a single new integral. Translate the packet, ψ(x) → ψ(x − a), and Φ(p) → Φ(p) e^(−ipa/ħ): the momentum density is untouched, because moving a packet does not change what it is doing. Multiply by a position-dependent phase, ψ(x) → ψ(x) e^(ip₀x/ħ), and Φ(p) → Φ(p − p₀): now the position density is untouched and the whole momentum distribution slides onto p₀. Rescale, ψ(x) → λ^(−1/2) ψ(x/λ), and Φ(p) → λ¹⁄² Φ(λp), so widening in x narrows in p by exactly the same factor. Behind all three sits one fact worth memorising: writing ψ = |ψ| e^(iθ), the mean momentum is ⟨p⟩ = ħ ∫ |ψ|² (dθ/dx) dx. Momentum lives in the gradient of the phase — precisely the part of ψ that squaring the modulus throws away. A real wavefunction, however elaborate its bumps, always has ⟨p⟩ = 0.

Take ψ(x) = (2πσ²)^(−1/4) e^{(−x²/4σ²)}, so |ψ|² is a normalised Gaussian of standard deviation σ. In the transform integral, complete the square: −x²/4σ² − ipx/ħ = −(1/4σ²)(x + 2iσ²p/ħ)² − σ²p²/ħ². The bracket integrates to 2σ√π, and the prefactor tidies to give Φ(p) = (2σ²/πħ²)¹⁄⁴ e^{(−σ²p²/ħ²)}. Square it: |Φ|² is Gaussian with σₚ = ħ/2σ, so σₓ σₚ = ħ/2 exactly. Numbers make the reciprocity concrete. For an electron with σ = 1.0 nm, σₚ = ħ/2σ = 5.3 × 10⁻²⁶ kg m s⁻¹, and the kinetic energy that spread alone implies is σₚ²/2mₑ = 1.5 × 10⁻²¹ J = 9.5 meV. Squeeze the same electron to σ = 0.10 nm and σₚ grows tenfold while that energy grows a hundredfold, to 0.95 eV. That factor is why atomic energies come out in electron-volts, and why nothing confines an electron for free.

Push σ → ∞ in the Gaussian pair and |Φ|² collapses towards δ(p) while ψ flattens into a constant, the plane wave of zero momentum; boost the packet first and the same limit delivers δ(p − p₀) instead. Push σ → 0 and the roles swap. Neither endpoint is a state, because neither is square-integrable: ∫|e^(ipx/ħ)|² dx diverges, so the plane wave has no norm and no Born probability. It is a basis element, exactly like the δ(x − x₀) it is dual to. How a basis of non-normalisable functions is handled, by delta normalisation or a large periodic box, is the business of Continuous spectra and the momentum-space wavefunction (UPIV-06.07); here only the limit matters, and no preparation reaches it. Real beams settle the argument anyway. Trim a 100 eV electron beam to Δp/p = 10⁻⁴; since pc = √(2mₑc²E) = √(2 × 511 keV × 100 eV) = 10.1 keV, the spread is Δp c ≈ 1.0 eV, and the coherence length is ħ/Δp = ħc/(Δp c) = 197 eV nm / 1.0 eV ≈ 0.2 μm. Long compared with a lattice spacing, which is why diffraction works — and finite, which is why the transform pair is the honest description.