Fourier decomposition: position space and momentum space
One state in two bases: position and momentum.
Position space is where ψ was first written down; momentum space is where every question about p turns into arithmetic. This lesson builds the dictionary between them: computing φ(k), checking it with Plancherel, carrying shifts, phase ramps and squeezes across, and seeing why a state squeezed in x must spread in k.
Build the model
One state, written once as a function of x and once as a function of k.
1The big idea
A wavefunction ψ(x) answers position questions directly, but a momentum question asks how much of each plane wave e(ikx) the state contains, and that amount is the Fourier transform φ(k) = (1/√(2π)) ∫ ψ(x) e(−ikx) dx. Because the plane waves e(ikx)/√(2π) are normalised to a Dirac delta rather than to one, the map runs back with the same kernel and the opposite sign, and it preserves every overlap, so ∫|ψ|² dx = ∫|φ|² dk. φ is therefore not a statistic extracted from ψ but the same state in another basis, and |φ(k)|² dk is the probability of finding the momentum within ħ dk of ħk. The change of basis pays twice. It turns the derivative in p̂ = −iħ d/dx into multiplication by ħk, so every momentum average becomes a weighted integral with no derivatives in it. And it lays the structure bare: a shift in x is a phase ramp in k and a phase ramp in x is a shift in k, a squeeze in one space is a stretch in the other, smoothness in x is fast decay in k, and the product of the two widths never falls below ½. The machinery has two limits. It needs ψ square-integrable: a plane wave, or anything else of infinite norm, has a delta or nothing for a transform. And it simplifies only operators built from p̂: a potential V(x), a multiplication in x, becomes an integral operator in k, so the space to work in is the one where your operator is a multiplication.
2In one sentence
The momentum-space wavefunction φ(k) is the Fourier transform of ψ(x): the amplitude with which each plane wave e(ikx) appears in the state, so that |φ(k)|² dk is the probability that a momentum measurement returns a value between ħk and ħ(k + dk).
Example. For ψ(x) = e(−|x|/a)/√a with a = 0.50 nm, φ(k) = √(2a/π)/(1 + a²k²). The density |φ|² peaks at 2a/π = 0.32 nm at k = 0, falls to a quarter of that at k = 1/a = 2.0 nm⁻¹, and integrates to exactly 1, as ∫|ψ|² dx does.
3One state, two sets of coordinates
Any square-integrable ψ(x) is a continuous sum of plane waves, ψ(x) = (1/√(2π)) ∫ φ(k) e(ikx) dk, and the weights come back from the analysis integral φ(k) = (1/√(2π)) ∫ ψ(x) e(−ikx) dx. The two formulas differ only in the sign of the exponent. Splitting the 2π symmetrically, √(2π) on each side rather than 2π on one, is the convention in which the map keeps norms with no stray factors, and it is the one this course uses. Nothing is lost in either direction: φ is ψ written in the plane-wave basis, as a vector can be given by its components along a different set of axes. The units agree. ψ is in m⁻¹⁄² and φ in m¹⁄², so |ψ|² dx and |φ|² dk are both pure numbers, and |φ(k)|² dk is the probability that a momentum measurement returns a value between ħk and ħ(k + dk). One quick check fits every transform: φ(0) is the area under ψ divided by √(2π).
4Computing one transform by hand
Take ψ(x) = e(−|x|/a)/√a, normalised because ∫ e(−2|x|/a) dx = a. The modulus splits the analysis integral in two. For x ≥ 0 the integrand is e(−x(1/a + ik)), which integrates to 1/(1/a + ik); for x ≤ 0 the same steps give 1/(1/a − ik). Their sum is (2/a)/(1/a² + k²) = 2a/(1 + a²k²), so φ(k) = √(2a/π)/(1 + a²k²). Three checks: √a carries m¹⁄², as φ must; φ(0) = √(2a/π) is the area 2√a under ψ divided by √(2π); and φ is real and even because ψ is. The shape has changed on the way. A cusped exponential became a smooth Lorentzian, and the cusp has moved into the tails, which fall only as k⁻², so |φ|² falls as k⁻⁴. With a = 0.50 nm, |φ(0)|² = 2a/π = 0.32 nm, and at k = 1/a = 2.0 nm⁻¹ the density has dropped to a quarter of that.
5Plancherel keeps the norm, and needs it finite
Substitute the synthesis integral for ψ and for ψ* in ∫|ψ|² dx. The x integral left over is ∫ e(i(k′ − k)x) dx = 2π δ(k′ − k), which collapses one k integral and cancels both factors of 1/√(2π), leaving ∫|ψ(x)|² dx = ∫|φ(k)|² dk. The same steps with two different states give ∫ψ₁*ψ₂ dx = ∫φ₁*φ₂ dk, so states orthogonal in x are orthogonal in k. On the exponential, (2a/π) ∫ dk/(1 + a²k²)² = (2a/π)(π/(2a)) = 1, as it must be. The theorem needs ψ square-integrable, and that is not a technicality. A plane wave e(ik₀x) has infinite norm, and its transform is √(2π) δ(k − k₀), which has no finite square to integrate. A state such as √(b/π) sin(x/b)/x is square-integrable even though ∫|ψ| dx diverges: its transform is the limit of the integral over |x| ≤ X, Plancherel guarantees that limit in the mean-square sense, and it is a flat box on |k| ≤ 1/b.
6Differentiation in x is multiplication in k
Differentiate the synthesis integral under the integral sign: dψ/dx = (1/√(2π)) ∫ ik φ(k) e(ikx) dk. The transform of dψ/dx is therefore ik φ(k), and p̂ = −iħ d/dx acts on φ as plain multiplication by ħk. That is why momentum questions belong in k-space: ⟨p⟩ = ∫ ħk |φ|² dk and ⟨p²⟩ = ∫ ħ²k² |φ|² dk are weighted averages with no derivatives anywhere. Plancherel applied to dψ/dx gives the x-space form ⟨p²⟩ = ħ² ∫ |dψ/dx|² dx, and the two must agree. For the exponential with a = 0.50 nm, ⟨k²⟩ = 1/a² = 4.0 nm⁻², so an electron in that state has mean kinetic energy ħ²⟨k²⟩/2m = (0.0381 eV nm²)(4.0 nm⁻²) = 0.152 eV. The dictionary runs both ways: multiplying ψ by x becomes i dφ/dk. A potential V(x), a multiplication in x, becomes an integral over k in momentum space, so choose the space in which your operator is a multiplication.
7Where position and momentum are stored
Two rules follow from one change of variable in the analysis integral. A translation, ψ(x − x₀), multiplies the transform by e(−ikx₀): |φ|² is untouched, so moving a particle changes no momentum probability, and the displacement is stored in a phase that turns at the rate −x₀ as k increases. A phase ramp, e(ik₀x) ψ(x), shifts the transform to φ(k − k₀): |ψ|² is untouched, yet every momentum rises by ħk₀. The rules are mirror images, and together they say that neither |ψ|² nor |φ|² is the whole state: the phase in each space holds what the other space displays. A real ψ is the extreme case. Then φ(−k) = φ*(k), |φ|² is even and ⟨p⟩ = 0 exactly, which is why a real stationary state has no net drift. For an electron, a ramp with k₀ = 4.0 nm⁻¹ adds ħk₀ = 4.2 × 10⁻²⁵ kg m s⁻¹ to ⟨p⟩, a classical speed ħk₀/m = 4.6 × 10⁵ m s⁻¹, without moving |ψ|² at all.
8Narrow in x forces broad in k, by theorem
Squeeze a state by a factor s while keeping it normalised, ψₛ(x) = ψ(x/s)/√s. Changing variable in the analysis integral gives φₛ(k) = √s φ(sk), the original transform stretched by 1/s. So Δx scales by s, Δk by 1/s, and the product Δx Δk depends on the shape alone: for the exponential it is (a/√2)(1/a) = 1/√2 = 0.71 for every a. Across all shapes the product has a floor, Δx Δk ≥ ½, reached only by a Gaussian, while a box, whose jumps give |φ|² ∝ k⁻², has infinite Δk. Multiplied by ħ this is Δx Δp ≥ ħ/2, obtained with no measurement anywhere in the argument. It has consequences. Confine an electron to Δx = 0.10 nm and Δk must be at least 5.0 nm⁻¹, so its mean kinetic energy is at least ħ²Δk²/2m = 0.95 eV even when ⟨p⟩ = 0. In the opposite limit, stretching the exponential with a → ∞, φ collapses toward a spike at k = 0: the plane-wave limit, approached but never reached.
Key equations
What each relation says, and the units it is written in.
Synthesis: ψ from its plane waves
ψ(x) = (1/√(2π)) ∫ φ(k) e(ikx) dk
Every normalisable state is a continuous sum of plane waves, each carrying momentum ħk, with φ(k) as its amplitude. Nothing about the state is lost in writing it this way.
x in m, k in m⁻¹; ψ in m⁻¹⁄² and φ in m¹⁄²; the integral runs over every real k
Analysis: φ from ψ
φ(k) = (1/√(2π)) ∫ ψ(x) e(−ikx) dx
Reads off how much of each plane wave the state contains. At k = 0 it is the area under ψ divided by √(2π), a one-line check on any transform you compute.
the same 1/√(2π) as the synthesis and the opposite sign in the exponent; |φ(k)|² dk is a pure number
Plancherel's theorem
∫ |ψ(x)|² dx = ∫ |φ(k)|² dk
Normalise in whichever space the integral is easier and the other description is normalised for free. The same steps keep overlaps, so states orthogonal in x stay orthogonal in k.
both sides are dimensionless; it holds for every square-integrable ψ and has no meaning for a plane wave
Momentum averages in k-space
⟨pⁿ⟩ = ∫ (ħk)ⁿ |φ(k)|² dk
A derivative d/dx on ψ becomes a factor ik on φ, so p̂ = −iħ d/dx is plain multiplication by ħk: a momentum average needs no derivatives, only a weighted integral.
ħ = 1.055 × 10⁻³⁴ J s, k in m⁻¹; finite only if |φ|² falls faster than k(−n−1) at large k
A phase ramp shifts the momentum
e(ik₀x) ψ(x) ↔ φ(k − k₀)
Every momentum rises by ħk₀ while the position density stays put, so ⟨p⟩ gains ħk₀ and the spread Δp is unchanged. Momentum lives in how the phase of ψ varies with x.
k₀ in m⁻¹; ↔ pairs a function with its transform; the ramp leaves |ψ(x)|² unchanged
Bandwidth theorem
Δx Δk ≥ ½
Squeezing ψ by s stretches φ by 1/s, so the product depends on shape alone: 1/√2 for e(−|x|/a). Times ħ it is Δx Δp ≥ ħ/2, with no measurement in the argument.
Δx in m and Δk in m⁻¹ are the standard deviations of |ψ|² and |φ|²; equality holds only for a Gaussian
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Drag a from 0.25 to 1.00 nm: the x envelope spreads fourfold, the k envelope narrows fourfold, and the product readout never leaves 0.71. Slide x₀ to 0: |φ| does not move and the ripple in Re φ unwinds, because a displacement lives only in the phase of φ. Moving k₀ is the mirror image.
Live results
- SPREAD Δx = a/√2
- 0.35 nm
- SPREAD Δk = 1/a
- 2.00 nm⁻¹
- PRODUCT Δx Δk (FLOOR ½)
- 0.71
- ELECTRON ⟨T⟩ = ħ²⟨k²⟩/2m
- 0.762 eV
Catch the common trap
The mistake this lesson is built to prevent.
Many students read Δx Δk ≥ ½ as a story about measurement: pin the particle down with a short-wavelength photon, the photon kicks it, and so the momentum becomes uncertain. The Fourier pair says something different. φ(k) is computed from ψ(x) by an integral before anything is measured, and Δx and Δk are the widths of |ψ|² and |φ|², two properties of one function. Prepare many identical copies, measure position on half of them and momentum on the other half, and the two histograms obey Δx Δp ≥ ħ/2 although no copy was measured twice, so none was disturbed. A rectangular radio pulse 1 ns long has its spectrum's first zeros 1 GHz either side of the carrier for the same mathematical reason, and nothing has touched it. The disturbance picture gets the scaling right and the cause wrong, and it wrongly suggests that a gentler measurement could beat the bound.
Test it: what changes?
- Δx = a/√2 grows
- Δk = 1/a shrinks
- Product Δx Δk stays 1/√2
- Mean kinetic energy falls toward ħ²k₀²/2m
- |ψ(x)|² unchanged
- |φ(k)|² slides, staying centred on k₀
- Δk stays 1/a while |⟨p⟩| = ħ|k₀| grows
- Mean kinetic energy ħ²(k₀² + 1/a²)/2m rises
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
An electron beam is prepared with the real momentum-space wavefunction φ(k) = e(−|k − k₀|/κ)/√κ, where k₀ = 5.0 nm⁻¹ and κ = 0.50 nm⁻¹. Check that φ is normalised, build ψ(x) from the synthesis integral, confirm Plancherel from the x side, and find ⟨p⟩, the mean speed, Δk and Δx.
View solution →
- Write q = k − k₀. Then ∫|φ|² dk = (1/κ) ∫ e(−2|q|/κ) dq = (1/κ)(κ) = 1, so φ is normalised. Units: κ is in m⁻¹, so 1/√κ carries m¹⁄², as a momentum-space wavefunction must.
- Synthesis: ψ(x) = (1/√(2πκ)) e(ik₀x) ∫ e(−|q|/κ) e(iqx) dq, the factor e(ik₀x) coming straight out of k = q + k₀. Splitting at q = 0 gives 1/(1/κ − ix) + 1/(1/κ + ix) = 2κ/(1 + κ²x²), so ψ(x) = √(2κ/π) e(ik₀x)/(1 + κ²x²).
- Plancherel from the x side: ∫|ψ|² dx = (2κ/π) ∫ dx/(1 + κ²x²)² = (2κ/π)(π/(2κ)) = 1, as in k. The peak density |ψ(0)|² = 2κ/π = 0.32 nm⁻¹ carries m⁻¹, as a probability density on a line must.
- |φ|² is symmetric about k₀, so ⟨k⟩ = k₀ and ⟨p⟩ = ħk₀ = (1.055 × 10⁻³⁴ J s)(5.0 × 10⁹ m⁻¹) = 5.3 × 10⁻²⁵ kg m s⁻¹. The classical speed ħk₀/m = 5.8 × 10⁵ m s⁻¹ is under 0.2% of c, so p = mv needs no relativistic correction.
- Spreads: ⟨(k − k₀)²⟩ = (1/κ) ∫ q² e(−2|q|/κ) dq = κ²/2, so Δk = κ/√2 = 0.35 nm⁻¹. In x, ⟨x²⟩ = (2κ/π) ∫ x² dx/(1 + κ²x²)² = 1/κ², so Δx = 1/κ = 2.0 nm. The product is 1/√2 = 0.71, above the floor ½: this is the exponential and Lorentzian pair again, with x and k exchanged.
- Limits: taking φ real put the packet at x = 0, and a factor e(−ikx₀) on φ would move it to x₀, carrying the peak density with it, without changing ⟨p⟩, Δk or Δx. As κ → 0, φ narrows to a spike and Δx = 1/κ grows without bound, the plane-wave limit that no normalised state reaches.
Answer. ψ(x) = √(2κ/π) e(ik₀x)/(1 + κ²x²), normalised in both spaces; ⟨p⟩ = 5.3 × 10⁻²⁵ kg m s⁻¹ (v = 5.8 × 10⁵ m s⁻¹), Δk = 0.35 nm⁻¹, Δx = 2.0 nm, Δx Δk = 0.71.
MediumExample 2
An electron is shared between two sites 2d apart: ψ(x) = [f(x − d) − f(x + d)]/N, where f(x) = e(−|x|/a)/√a has transform F(k) = √(2a/π)/(1 + a²k²), with a = 0.20 nm and d = 1.0 nm. Find φ(k) without a new integral, fix N by Plancherel, locate the zeros of |φ|², and compare the most probable |p| with ⟨p⟩.
View solution →
- The translation rule turns f(x ∓ d) into e(∓ikd) F(k). So φ(k) = F(k)(e(−ikd) − e(ikd))/N = −(2i/N) sin(kd) F(k). The separation of the two sites has become an oscillation in k, and since sin(kd) is a pure number φ keeps the m¹⁄² of F.
- Normalise: ∫|φ|² dk = (4/N²) ∫ sin²(kd) |F|² dk = (2/N²)[1 − ∫ cos(2kd) |F|² dk], using ∫|F|² dk = 1. Plancherel identifies the last integral as the overlap S = ∫ f(x − d) f(x + d) dx = (1 + 2d/a) e(−2d/a) = 11 e(−10) = 5.0 × 10⁻⁴.
- So N² = 2(1 − S) = 1.9990 and N = 1.414, within 0.03% of √2: the lumps barely overlap, and N → √2 as d/a grows, the limit of two separately normalised pieces.
- |φ(k)|² = (4/N²) sin²(kd) (2a/π)/(1 + a²k²)². It vanishes wherever kd = nπ, at k = 0, ±3.14, ±6.28 nm⁻¹ and so on: fringes spaced π/d = 3.1 nm⁻¹ under an envelope about 1/a = 5.0 nm⁻¹ wide. The site separation 2d = 2π/(fringe spacing) can be read off, as a slit separation is read from a double-slit pattern.
- ψ is real, so |φ|² is even and ⟨p⟩ = 0, yet |φ(0)|² = 0: the probability density vanishes exactly at the mean momentum. Maximising sin²(kd)/(1 + a²k²)² gives d cot(kd) = 2a²k/(1 + a²k²), solved by k = 1.46 nm⁻¹, so the most probable |p| is ħ(1.46 nm⁻¹) = 1.5 × 10⁻²⁵ kg m s⁻¹.
- Compare the symmetric combination f(x − d) + f(x + d): sin becomes cos and the most probable momentum becomes zero. The two |ψ|² differ only through the tiny overlap term, yet the momentum pictures are opposite: the relative sign of the lumps lives almost entirely in φ.
Answer. φ(k) = −(2i/N) sin(kd) √(2a/π)/(1 + a²k²) with N = 1.414; |φ|² vanishes at k = nπ/d, every 3.1 nm⁻¹, including k = 0; ⟨p⟩ = 0 while the most probable |p| is 1.5 × 10⁻²⁵ kg m s⁻¹.
HardExample 3
A student models an electron held against a hard wall at x = 0 by ψ(x) = √(2/a) e(−x/a) for x ≥ 0 and ψ = 0 for negative x, with a = 0.30 nm. Find φ(k) and check it by Plancherel; find the probabilities that |k| ≤ 1/a and that |k| ≥ 10/a; show that ⟨p²⟩ is infinite and say what in ψ causes it; recover ⟨x⟩ from φ alone; and judge whether the model is physical.
View solution →
- φ(k) = (1/√(2π)) √(2/a) ∫₀^∞ e(−x(1/a + ik)) dx = (1/√(πa)) × 1/(1/a + ik) = √(a/π)/(1 + iak). Checks: √a carries m¹⁄²; φ(0) = √(a/π) is the area √(2a) under ψ divided by √(2π); and φ(−k) = φ*(k), as a real ψ requires.
- |φ|² = (a/π)/(1 + a²k²), and ∫|φ|² dk = (a/π)(π/a) = 1, matching ∫|ψ|² dx = (2/a)(a/2) = 1. |φ|² is even, so ⟨p⟩ = 0.
- P(|k| ≤ K) = (2/π) atan(aK). At K = 1/a this is (2/π)(π/4) = 0.50: half the probability lies within |p| ≤ ħ/a = 3.5 × 10⁻²⁵ kg m s⁻¹. At K = 10/a it is only 0.937, leaving 0.063 beyond |p| = 3.5 × 10⁻²⁴ kg m s⁻¹. For the two-sided e(−|x|/a)/√a the same tail holds just 4 × 10⁻⁴.
- ⟨k²⟩ = (a/π) ∫ k² dk/(1 + a²k²). The integrand tends to 1/(πa) at large |k|, so the integral grows linearly with any cutoff and diverges: ⟨p²⟩ and the kinetic energy are infinite. A cutoff at |k| = 100/a would already give about 27 eV, and each tenfold raise of the cutoff multiplies that by about ten.
- The cause is the jump. ψ leaps from 0 to √(2/a) at x = 0, so dψ/dx contains √(2/a) δ(x), and ⟨p²⟩ = ħ² ∫|dψ/dx|² dx then contains ∫δ(x)² dx, which is infinite: x-space gives the same verdict. A jump gives |φ|² ∝ k⁻²; the kink of the two-sided exponential gives k⁻⁴ and a finite ⟨p²⟩.
- Position from k-space, with x → i d/dk: i dφ/dk = √(a/π) a/(1 + iak)², so φ* i dφ/dk = (a²/π)(1 − iak)/(1 + a²k²)². The odd imaginary part integrates to zero, leaving ⟨x⟩ = (a²/π)(π/(2a)) = a/2 = 0.15 nm, equal to ∫ x|ψ|² dx. Verdict: a hard wall demands ψ(0) = 0, so a physical state such as x e(−x/a) has a kink, not a jump, and finite energy.
Answer. φ(k) = √(a/π)/(1 + iak), normalised; P(|k| ≤ 1/a) = 0.50, i.e. |p| ≤ 3.5 × 10⁻²⁵ kg m s⁻¹, and P(|k| ≥ 10/a) = 0.063; ⟨p²⟩ diverges because ψ jumps at the wall; ⟨x⟩ = a/2 = 0.15 nm; the model is unphysical.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
φ(k) is ψ in another basis; |φ|² dk is a probability.
Compute φ(k) = (1/√(2π)) ∫ ψ e(−ikx) dx, check φ(0) and ∫|φ|² dk = ∫|ψ|² dx, then find ⟨pⁿ⟩ = ∫ (ħk)ⁿ |φ|² dk. ψ must be square-integrable, and k-space simplifies only operators built from p̂; a potential V(x) becomes an integral operator there.