University Physics IV · Foundations of Quantum Mechanics · 5.4

Position & Momentum Representations

Position is not where quantum mechanics lives; it is one basis among several. Learn to resolve the same state on momentum eigenstates, read probabilities straight off the transform, and see why nothing ever has to be normalised twice.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

A quantum state is a vector, and ψ(x) is only its component list on one particular basis — the eigenstates of position. The momentum operator has its own complete set, the plane waves uₚ(x) = e^(ipx/ħ)/√(2πħ), and resolving the same vector on those gives a second component list Φ(p) = ⟨uₚ|ψ⟩, which works out to be the Fourier transform of ψ with ħ carrying the units. Nothing is lost in the change: the inverse integral rebuilds ψ from Φ, Parseval's theorem shows the two lists share a norm, and Born's rule applies to each in its own variable — |ψ|² per unit length, |Φ|² per unit momentum.

What the change buys is that momentum questions become ordinary averages, ⟨p²⟩ = ∫p²|Φ|² dp instead of a derivative sandwiched inside an integral, and that the reciprocity of the two widths becomes a theorem about transforms rather than an extra postulate. What it costs is the pretence that sharp momentum is available: the momentum eigenfunctions are not square-integrable, so they are basis elements and not states. Every real preparation is a packet whose Φ has finite width, and the two pictures constrain each other precisely because they are two readings of one object, not two independent things a particle separately has.

Simple definition: The momentum representation of a state is the list of amplitudes Φ(p) got by projecting ψ onto the momentum eigenfunctions — the Fourier transform of ψ(x) with ħ setting the scale — whose squared modulus is the probability density per unit momentum.
Example: For an electron with ψ(x) = (2πσ²)^(−1/4) e^{(−x²/4σ²)} and σ = 1.0 nm, Φ(p) = (2σ²/πħ²)¹⁄⁴ e^{(−σ²p²/ħ²)}: a Gaussian of width σₚ = ħ/2σ = 5.3 × 10⁻²⁶ kg m s⁻¹, or about 99 eV/c.

Momentum eigenfunctions, δ-normaliseduₚ(x) = e^(ipx/ħ)/√(2πħ), ⟨uₚ′|uₚ⟩ = δ(p − p′)

The 1/√(2πħ) is not decoration — it is what makes the pair below symmetric and Parseval exact.

[uₚ] = (J s)^(−1/2); the delta carries units of inverse momentum

The transform pairΦ(p) = (2πħ)^(−1/2) ∫ψ e^(−ipx/ħ) dx ⇄ ψ(x) = (2πħ)^(−1/2) ∫Φ e^(+ipx/ħ) dp

Φ(p) = ⟨uₚ|ψ⟩ — an overlap integral, a change of basis, not a change of variable.

Opposite signs in the exponents; both integrals run over the whole line

Born's rule in the momentum pictureP(p₁ ≤ p ≤ p₂) = ∫ |Φ(p)|² dp over [p₁, p₂]

Momentum statistics come from Φ alone; the modulus of ψ cannot supply them.

[Φ] = (kg m s⁻¹)^(−1/2), so |Φ|² is a probability per unit momentum

Parseval–Plancherel∫|ψ(x)|² dx = ∫|Φ(p)|² dp = 1

Normalise once, in whichever picture is easier, and the other is normalised too.

True for any square-integrable ψ, with no extra constant anywhere

Operators swap roles between the picturesp-picture: p̂ → p ×, x̂ → +iħ ∂/∂p x-picture: x̂ → x ×, p̂ → −iħ ∂/∂x

⟨pⁿ⟩ = ∫pⁿ|Φ|² dp is an ordinary weighted average — no derivatives to integrate by parts.

Note the sign flip on the derivative when you change picture

The Gaussian pair and reciprocal widthsψ = (2πσ²)^(−1/4) e^{(−x²/4σ²)} ⇄ Φ = (2σ²/πħ²)¹⁄⁴ e^{(−σ²p²/ħ²)}

The one shape whose transform is the same shape, and the only one sitting on the ħ/2 floor.

σₓ = σ and σₚ = ħ/2σ; σ = 1.0 nm gives σₚ = 5.3 × 10⁻²⁶ kg m s⁻¹

ψ and Φ are two component lists for one vector

Write the state as a vector |ψ⟩. The position eigenstates give one set of components, ψ(x) = ⟨x|ψ⟩, and that is the function you have been drawing all along. The momentum operator −iħ d/dx has its own complete set, uₚ(x) = e^(ipx/ħ)/√(2πħ) with p̂ uₚ = p uₚ, and resolving the same vector on those gives a second set of components: Φ(p) = ⟨uₚ|ψ⟩ = ∫ uₚ*(x) ψ(x) dx = (2πħ)^(−1/2) ∫ ψ(x) e^(−ipx/ħ) dx. That integral is a Fourier transform, but it did not arrive as a technique borrowed from signal processing — it is an overlap integral, the same construction as any other change of basis. The inverse transform rebuilds ψ from Φ, so neither list is the more fundamental one. Which 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.

Where 2πħ comes from, and what it does to the units

The plane waves are not normalisable, so their scale is fixed by a delta convention instead of by ∫|u|² = 1. One integral settles it: ∫ e^{(i(p − p′)}x/ħ) dx = 2πħ δ(p − p′) over the whole line. Dividing each wave by √(2πħ) makes ⟨uₚ′|uₚ⟩ = δ(p − p′) exactly, and shares that constant symmetrically between the forward and inverse transforms — which is why both carry (2πħ)^(−1/2) and only the sign in the exponent tells them apart. 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.

Parseval: normalise once, and both pictures are normalised

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.

The dictionary: a shift, a boost, and a squeeze

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.

Work the Gaussian, then put numbers on it

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.

Sharp momentum is a limit the pair never occupies

Push σ → ∞ in the Gaussian pair and |Φ|² collapses towards δ(p − p₀) while ψ spreads into a plane wave; 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. Two repairs are standard — a box of length L, which discretises momentum to pₙ = 2πnħ/L and restores normalisability, or the continuum with δ normalisation as above. 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.

Change one variable at a time

Make the relationship visible.

Interactive model

Packet width σ in x1.0 nm

Centre shift a in x0.0 nm

Boost phase p₀0.0 ħ/nm

Move the shift slider: only the left curve travels. Move the boost slider: only the right one does. The width slider is the one that touches both — and it drives the two ±σ bars in opposite directions, because there is one state and one ħ/2 to share.

σx POSITION SPREAD1.00 nm

σp MOMENTUM SPREAD0.50 ħ/nm

σx σp (floor is ħ/2)0.50 ħ

σp IN BEAM UNITS99 eV/c

Live interpretationσx POSITION SPREAD: 1.00 nm. σp MOMENTUM SPREAD: 0.50 ħ/nm. σx σp (floor is ħ/2): 0.50 ħ. σp IN BEAM UNITS: 99 eV/c

Catch the common trap

Explain before calculating.

An electron is prepared twice: first in ψ₁(x) = (2πσ²)^(−1/4) e^{(−x²/4σ²)}, then in ψ₂(x) = ψ₁(x) · e^(ip₀x/ħ) with p₀ > 0. Which statement about the two preparations is correct?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAn electron is prepared in the Gaussian state ψ(x) = (2πσ²)^(−1/4) e^{(−x²/4σ²)} with σ = 2.0 nm. Write down Φ(p), give the standard deviation σₚ of the momentum density, and find the kinetic energy σₚ²/2mₑ that this spread alone implies. Take ħ = 1.055 × 10⁻³⁴ J s and mₑ = 9.11 × 10⁻³¹ kg.

The transform of this Gaussian is Φ(p) = (2σ²/πħ²)¹⁄⁴ e^{(−σ²p²/ħ²)}, so |Φ(p)|² ∝ e^{(−2σ²p²/ħ²)}. Match that to e^{(−p²/2σₚ²)} and read the width off the exponent, never off the amplitude: σₚ = ħ/2σ.
σₚ = ħ/(2σ) = (1.055 × 10⁻³⁴ J s)/(2 × 2.0 × 10⁻⁹ m) = 2.6 × 10⁻²⁶ kg m s⁻¹.
Check the product: σₓ σₚ = (2.0 × 10⁻⁹ m)(2.64 × 10⁻²⁶ kg m s⁻¹) = 5.3 × 10⁻³⁵ J s, which is ħ/2. The Gaussian sits exactly on the floor, as it must.
σₚ²/2mₑ = (2.64 × 10⁻²⁶)² / (2 × 9.11 × 10⁻³¹) = (6.95 × 10⁻⁵²)/(1.82 × 10⁻³⁰) = 3.8 × 10⁻²² J = 2.4 meV.
Beam units are quicker: with ħc = 197 eV nm, σₚ c = ħc/2σ = 197/(2 × 2.0) = 49 eV, so σₚ ≈ 49 eV/c — the same number, no powers of ten to lose.

AnswerΦ(p) = (2σ²/πħ²)¹⁄⁴ e^{(−σ²p²/ħ²)}; σₚ = 2.6 × 10⁻²⁶ kg m s⁻¹ (≈ 49 eV/c); σₚ²/2mₑ = 3.8 × 10⁻²² J = 2.4 meV.

MediumAn electron is confined to a hard-edged region of length L: ψ(x) = 1/√L for |x| ≤ L/2 and zero outside. Find Φ(p), locate the first zeros of |Φ(p)|², and evaluate them for L = 1.0 nm. Then explain why quoting an rms momentum spread for this state is a mistake.

Normalisation first: ∫|ψ|² dx = (1/L)⋅L = 1, so no constant is missing.
Φ(p) = (2πħ)^(−1/2) L^(−1/2) ∫ from −L/2 to L/2 of e^(−ipx/ħ) dx. That integral is 2ħ sin(pL/2ħ)/p, so Φ(p) = √(L/2πħ) · sin(pL/2ħ)/(pL/2ħ) — a sinc of argument pL/2ħ, with Φ(0) = √(L/2πħ).
Units check: [√(L/ħ)] = (m/(J s))¹⁄² = (kg m s⁻¹)^(−1/2), so |Φ|² is indeed a density per unit momentum.
Zeros where sin(pL/2ħ) = 0 with p ≠ 0: pL/2ħ = ±π, i.e. p = ±2πħ/L = ±h/L. With hc = 1240 eV nm and L = 1.0 nm, pc = 1240 eV, so p = ±6.6 × 10⁻²⁵ kg m s⁻¹. About 90% of the probability lies inside that central lobe.
Now the trap: |Φ|² falls only as p⁻² in the tails, so ∫p²|Φ|² dp diverges. The rms spread — and with it ⟨p²⟩/2mₑ — is infinite, because ψ′ jumps at the two walls and an infinitely sharp kink demands arbitrarily large momenta. Quote the lobe half-width h/L instead, or round the corners of ψ first.

AnswerΦ(p) = √(L/2πħ) · sin(pL/2ħ)/(pL/2ħ); first zeros at p = ±h/L, which for L = 1.0 nm is ±6.6 × 10⁻²⁵ kg m s⁻¹ (±1240 eV/c). The rms width is infinite: the hard edges leave |Φ|² ~ p⁻² tails.

HardAn electron is prepared in ψ(x) = N[g(x − a) + g(x + a)], with g(x) = (2πσ²)^(−1/4) e^{(−x²/4σ²)}, σ = 1.0 nm and a = 5.0 nm. Find Φ(p) and the momentum density, give the fringe spacing in p and the width of the envelope, and say what a 50:50 incoherent mixture of the same two lumps would give instead.

Normalise. The overlap is ∫g(x − a) g(x + a) dx = e^{(−a²/2σ²)}; with a/σ = 5 that is e^(−12.5) = 3.7 × 10⁻⁶, so N = [2(1 + 3.7 × 10⁻⁶)]^(−1/2) ≈ 1/√2.
Apply the shift theorem to each lump: g(x ∓ a) ⇄ G(p) e^(∓ipa/ħ), with G(p) = (2σ²/πħ²)¹⁄⁴ e^{(−σ²p²/ħ²)}. Adding them, Φ(p) = N G(p)[e^(−ipa/ħ) + e^(+ipa/ħ)] = 2N G(p) cos(pa/ħ) ≈ √2 G(p) cos(pa/ħ).
So |Φ(p)|² = 2|G(p)|² cos²(pa/ħ): a Gaussian envelope of width σₚ = ħ/2σ = 5.3 × 10⁻²⁶ kg m s⁻¹ (99 eV/c), ruled by interference fringes. Two lumps in x have become fringes in p — the double slit, read in the momentum picture.
Fringe maxima at cos² = 1, i.e. pa/ħ = nπ, so pₙ = nh/2a and the spacing is Δp = h/2a. With 2a = 10 nm, Δp c = hc/2a = 1240/10 = 124 eV, i.e. Δp = 6.6 × 10⁻²⁶ kg m s⁻¹.
Count what survives: Δp/σₚ = (h/2a)/(ħ/2σ) = 2πσ/a = 2π/5 = 1.26, so only two or three fringes fit under the envelope. Separating the lumps further packs in finer fringes; narrowing each lump widens the envelope and reveals more of them.
The mixture has no cross term, so its momentum density is just |G(p)|² — the same envelope, no fringes at all — while its position density matches the superposition's to within 4 × 10⁻⁶. The position picture cannot tell the two preparations apart; the momentum picture does it at a glance.

AnswerΦ(p) = √2 G(p) cos(pa/ħ), G(p) = (2σ²/πħ²)¹⁄⁴ e^{(−σ²p²/ħ²)}; fringes spaced Δp = h/2a = 6.6 × 10⁻²⁶ kg m s⁻¹ (124 eV/c) under an envelope of width ħ/2σ = 5.3 × 10⁻²⁶. The mixture gives the same |ψ|² but a smooth |G(p)|², with no fringes.