Plane waves, and why no plane wave is a state
Sharp momentum, flat density, and no finite norm.
Separate the free equation and out comes e(i(kx − ωt)), a wave with one sharp momentum that the rest of this unit leans on. Before you trust it, learn exactly what it is not: no amplitude makes it a probability distribution, no finite sum of such waves does either, and yet genuine states can mimic its shape and its momentum as closely as you like.
Build the model
The first solution of the free equation fails the one test every state must pass.
1The big idea
Separate the free Schrödinger equation and the first solution to appear is ψ(x, t) = A e(i(kx − ωt)) with ω = ħk²/(2m): one wavenumber, hence one momentum ħk, and an exact solution for every complex A. A state must also satisfy ∫|ψ|² dx = 1, and there the plane wave fails. Because the travelling solution is forced to be complex, its modulus is |A| at every x and every t, so a stretch of length L holds probability |A|²L and the whole line holds an infinite amount for any A other than zero. Normalising rescales a finite integral; it cannot rescue an infinite one, and adding finitely many plane waves only redistributes a density that never falls off. What survives is a limit. Wrap the same oscillation in a smooth envelope of width σ and normalise: every member of that family is a genuine state, its shape matches the plane wave over any fixed region as σ grows, and its wavenumber spread 1/(2σ) shrinks to zero. But its density falls as 1/σ everywhere, so the normalised family converges to zero, and the plane wave reappears only once normalisation is abandoned. The idealisation still earns its keep, supplying the energy, the wavelength and a finite probability current ħk|A|²/m that describes a steady beam. It stops working the moment a question needs one particle in a finite region.
2In one sentence
A plane wave is the free-particle solution ψ(x, t) = A e(i(kx − ωt)) with ω = ħk²/(2m); it has the sharp momentum ħk, but its probability density |A|² is the same everywhere, so it cannot be normalised and is not a physical state.
Example. Choose |A|² = 1.0 × 10⁶ m⁻¹ so that one micrometre of the line holds probability 1. The next micrometre holds another 1, a metre holds 1.0 × 10⁶, and the whole line holds infinitely much; making A smaller only moves the failure further out.
3Why the travelling solution has to be complex
Try ψ = A e(i(kx − ωt)) in iħ ∂ψ/∂t = −(ħ²/2m) ∂²ψ/∂x². The time derivative brings down −iω, so the left side is ħωψ; two space derivatives bring down (ik)² = −k², so the right side is (ħ²k²/2m)ψ. They agree at every x and t exactly when ħω = ħ²k²/(2m), for any complex A. For an electron with k = 1.0 × 10¹⁰ m⁻¹, λ = 0.63 nm, that gives E = 6.1 × 10⁻¹⁹ J = 3.8 eV and ω = 5.8 × 10¹⁵ rad s⁻¹. Now try the real wave A cos(kx − ωt), which serves perfectly well on a string. The left side becomes iħωA sin(kx − ωt) and the right side (ħ²k²/2m)A cos(kx − ωt): an imaginary sine against a real cosine, equal nowhere. The equation is first order in time with an i in front, so it accepts only the dependence e(−iωt) with ħω = E, and a real cosine smuggles in e(+iωt) as well. The travelling solution must be complex, and that one fact decides what follows.
4A density with no crests and no edges
Write A = |A| e(iα). Then Re ψ = |A| cos(kx − ωt + α) and Im ψ = |A| sin(kx − ωt + α): two waves a quarter-cycle apart, sliding along +x together. The density is ψ*ψ = |A|²(cos² + sin²) = |A|², the same number at every x, at every t and for every k. There is no crest where the particle prefers to be and nothing moves, as befits a stationary solution of the single energy ħω. The density carries no trace of the momentum ħk it was built with; that information lives entirely in the phase kx − ωt. The standing wave A cos(kx) e(−iωt), built from +k and −k with equal weight, does have structure, |A|² cos²(kx), with nodes half a wavelength apart. But it never decays either, and averaged over any whole number of half-wavelengths it still holds |A|²/2 per unit length. With crests or without, neither density falls off as |x| grows, and that is the whole problem.
5No amplitude repairs the integral
A state needs ∫|ψ|² dx = 1 over the whole line. For the plane wave any stretch of length L holds |A|²L, wherever the stretch sits, so for any nonzero A a stretch 1/|A|² long already holds probability 1 and the rest of the line holds infinitely more. Shrinking A only moves the failure outward: halve A and that stretch becomes four times longer; divide A by 1000 and it becomes a million times longer, still finite, with an infinite remainder beyond it. The only finite choice is A = 0, which describes no particle at all. Compare the bound states normalised earlier in the course: there the integral was finite before A was chosen, and normalising meant dividing by that finite number. Rescaling can turn 7.3 into 1; it cannot turn an infinite integral into 1. Waiting does not help, because |A|² never changes in time. The plane wave is at least bounded, |ψ| = |A| everywhere, which is why it is kept as a limit rather than discarded like the negative-energy solutions e(±κx), each of which grows without bound in one direction.
6Adding a few plane waves does not help
Perhaps two plane waves can cancel each other far away. Take equal amplitudes at k₁ = 1.00 × 10¹⁰ m⁻¹ and k₂ = 1.02 × 10¹⁰ m⁻¹. At t = 0 the density is |A|² |e(ik₁x) + e(ik₂x)|² = 2|A|²[1 + cos((k₂ − k₁)x)], a beat pattern swinging between 0 and 4|A|² with period 2π/(k₂ − k₁) = 31 nm, and at later times the pattern only slides along. Its zeros are genuine, but it repeats forever, so every 31 nm holds the same probability and the integral over L again grows like 2|A|²L. Any finite sum Σ Aⱼ e(ikⱼx) with distinct kⱼ behaves the same way: the cross terms oscillate and average to zero over long stretches, leaving a mean density Σ|Aⱼ|² that never falls off. Interference that is destructive everywhere far from the centre needs a continuous spread of wavenumbers, a whole function of k integrated over k. That is the wave packet, built properly later in this unit; the point here is that no finite superposition of plane waves is a state.
7A limit of states, not a state
What does converge is a family of genuine states. Put the plane wave under a smooth envelope and normalise: ψσ(x) = (2πσ²)⁻¹⁄⁴ e(−x²/(4σ²)) e(ikx). Its density is a Gaussian of rms width σ, so ∫|ψσ|² dx = 1 for every σ, and its wavenumber spread is Δk = 1/(2σ). For an electron with k = 1.0 × 10¹⁰ m⁻¹, widening σ from 1.0 μm to 1.0 mm cuts Δk/k from 5.0 × 10⁻⁵ to 5.0 × 10⁻⁸, and over any fixed stretch the oscillation looks ever more like e(ikx). The price is a ceiling that no point can exceed: the peak density 1/(σ√(2π)) falls from 4.0 × 10⁵ m⁻¹ to 4.0 × 10² m⁻¹, and a 1.0 nm detector at the centre drops from probability 4.0 × 10⁻⁴ to 4.0 × 10⁻⁷. As σ → ∞ the normalised states tend to zero at every x. Only the rescaled function (2πσ²)¹⁄⁴ ψσ tends to e(ikx), and it gets there by giving up normalisation. Exactly sharp momentum and total probability exactly 1 are two limits that cannot be taken together.
8What a plane wave is still good for
Plane waves stay the everyday tool of the continuum under one rule: ask only questions whose answer does not depend on the infinite norm. The energy ħ²k²/(2m), the wavelength 2π/k and the dispersion ω(k) belong to k alone. The probability current j = (ħ/m) Im(ψ* ∂ψ/∂x) = ħk|A|²/m is finite, and it is the classical flux n v once |A|² is read as particles per metre of a steady beam. At k = 1.0 × 10¹⁰ m⁻¹ an electron moves at ħk/m = 1.16 × 10⁶ m s⁻¹, so |A|² = 1.0 × 10³ m⁻¹ carries 1.16 × 10⁹ electrons per second, a current of 0.19 nA. Ratios of currents, such as the fraction of a beam reflected at a step, are what scattering calculations report, and A cancels from them. A question about one particle, where it is or whether a detector of finite size clicks, needs a normalised packet instead. The next topic keeps the plane waves and replaces the failed normalisation with one built for the continuum: delta normalisation.
Key equations
What each relation says, and the units it is written in.
The free plane wave
ψ(x, t) = A e(i(kx − ωt)) with ω = ħk²/(2m)
Substitution turns both sides of the free equation into multiples of ψ, ħω on the left and ħ²k²/(2m) on the right. Every complex A and every real k works, which is where the trouble starts.
A: complex constant; k: wavenumber in m⁻¹; ω: angular frequency in rad s⁻¹; m: mass in kg; ħ = 1.055 × 10⁻³⁴ J s
A density with no features
|ψ(x, t)|² = ψ*ψ = |A|²
Re ψ = |A| cos(kx − ωt + α) and Im ψ = |A| sin(kx − ωt + α) both oscillate, with α = arg A, but cos² + sin² = 1 erases every trace of position and time.
|A|² would be probability per metre, in m⁻¹; the same value at every x and t, whatever k is
Probability in a stretch of length L
∫ |ψ|² dx from x₀ to x₀ + L = |A|² L
It grows without bound as L → ∞ unless A = 0. Setting |A|² = 1/L normalises only a cut-off wave, and then A → 0 as L → ∞.
L in m, so |A|²L is dimensionless; x₀ can be anywhere because the density is flat
Probability current of a plane wave
j = (ħ/m) Im(ψ* ∂ψ/∂x) = ħk|A|²/m
Finite and constant. Read |A|² as particles per metre of a steady beam and j is the classical flux n v. Waves at +k and −k with equal |A| superpose to j = 0.
j in s⁻¹ in one dimension; ħk/m is the classical speed p/m in m s⁻¹
A normalised state that approaches it
ψσ(x) = (2πσ²)⁻¹⁄⁴ e(−x²/(4σ²)) e(ikx)
As σ grows its shape matches e(ikx) over any fixed stretch ever more closely, and its wavenumber spread 1/(2σ) shrinks toward zero.
σ: envelope width in m, the rms width of |ψσ|²; the prefactor, in m⁻¹⁄², gives ∫|ψσ|² dx = 1 for every σ
What the normalised family converges to
|ψσ(x)|² ≤ 1/(σ√(2π)) → 0 as σ → ∞
The normalised states tend to zero at every x. Only the rescaled (2πσ²)¹⁄⁴ ψσ tends to e(ikx): sharp momentum and unit norm cannot be had together.
bound in m⁻¹; a window of width d in m therefore holds probability at most d/(σ√(2π))
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Drag σ from 1 nm to 8 nm: across the frame the curves look more and more like a plane wave and Δk falls eightfold, yet the peak density sinks by the same factor and the chance of |x| ≤ 1 nm drops from 0.683 to 0.099. Then move k: Re ψ and Im ψ change, |ψ|² does not.
Live results
- PEAK DENSITY |ψ(0)|²
- 0.199 nm⁻¹
- P(|x| ≤ 1 nm)
- 0.383
- WAVENUMBER SPREAD Δk
- 0.250 nm⁻¹
- RELATIVE SPREAD Δk/k
- 0.167
Catch the common trap
The mistake this lesson is built to prevent.
The tempting repair is to make A small: take |A|² = 1/L, with L the length of the apparatus or even of the universe, and call the plane wave normalised. That answers a different question. A e(ikx) extends over the whole line, so its integral is |A|² times an infinite length, and for any fixed nonzero A a finite stretch 1/|A|² long already holds probability 1. Setting |A|² = 1/L normalises some other function: either the wave cut off at ±L/2, whose wavenumber distribution has a central peak of half-width 2π/L rather than zero width, or the wave on a ring of circumference L, a different space in which k must be a multiple of 2π/L. Both are legitimate computing devices. Neither shows that the plane wave is a state, because sending L → ∞ drives A to zero and the limit of those normalised functions is zero everywhere.
Test it: what changes?
- Wavenumber spread 1/(2σ) shrinks
- Peak density 1/(σ√(2π)) falls
- Probability within |x| ≤ 1 nm drops
- Shape looks more like a plane wave locally
- |ψ|² and peak density unchanged
- Re ψ and Im ψ oscillate faster
- Δk unchanged, so Δk/k falls
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
A student models an electron in a 4.0 nm region as ψ = A e(ikx) with k = 5.0 × 10⁹ m⁻¹, choosing |A|² = 2.5 × 10⁸ m⁻¹ so that the region holds probability 1. Keeping the same function on the whole x axis, find the probability it assigns to the 4.0 nm region, to a 1.0 mm stretch and to the whole line. What did the student actually normalise?
View solution →
- The density is |ψ|² = |A|² = 2.5 × 10⁸ m⁻¹ at every x, whatever k is, so the probability in any stretch is |A|² times its length. Dimensions: m⁻¹ × m is a pure number, as a probability must be.
- 4.0 nm region: 2.5 × 10⁸ m⁻¹ × 4.0 × 10⁻⁹ m = 1.0, as the student intended.
- 1.0 mm stretch: 2.5 × 10⁸ m⁻¹ × 1.0 × 10⁻³ m = 2.5 × 10⁵, which no probability can be.
- Whole line: |A|² times an infinite length diverges. Only A = 0 keeps it finite, and zero describes no particle.
- The student normalised the wave cut off outside the 4.0 nm region. Its wavenumber distribution has a central peak of half-width 2π/L = 2π/(4.0 × 10⁻⁹ m) = 1.6 × 10⁹ m⁻¹, about 31% of k, so it is far from a state of sharp momentum.
Answer. 1.0 in the region, 2.5 × 10⁵ in 1.0 mm, and infinity on the whole line. The normalised object is a cut-off wave whose wavenumber is uncertain by about 1.6 × 10⁹ m⁻¹, about 31% of k.
MediumExample 2
A steady 50 eV electron beam carries a current of 2.0 nA. Far from the source, model it as ψ = A e(i(kx − ωt)). Find k and the speed ħk/m, the value of |A|² for which the probability current equals the particle flux, and the mean number of electrons in a 1.0 cm length of beam. Check the speed against classical mechanics.
View solution →
- E = 50 eV = 50 × 1.602 × 10⁻¹⁹ J = 8.01 × 10⁻¹⁸ J, and k = √(2mE)/ħ = √(2 × 9.109 × 10⁻³¹ kg × 8.01 × 10⁻¹⁸ J) / 1.055 × 10⁻³⁴ J s = 3.62 × 10¹⁰ m⁻¹, a wavelength of 0.173 nm.
- Speed: ħk/m = (1.055 × 10⁻³⁴ × 3.62 × 10¹⁰)/(9.109 × 10⁻³¹) = 4.19 × 10⁶ m s⁻¹. Classically v = √(2E/m) = √(2 × 8.01 × 10⁻¹⁸ / 9.109 × 10⁻³¹) = 4.19 × 10⁶ m s⁻¹, the same, and only 1.4% of c, so the non-relativistic equation is fine.
- Particle flux: I/e = 2.0 × 10⁻⁹ A / 1.602 × 10⁻¹⁹ C = 1.25 × 10¹⁰ electrons per second.
- Set j = ħk|A|²/m = v|A|² equal to the flux: |A|² = 1.25 × 10¹⁰ s⁻¹ / 4.19 × 10⁶ m s⁻¹ = 2.98 × 10³ m⁻¹. Units: s⁻¹ ÷ m s⁻¹ = m⁻¹, a number per metre.
- In 1.0 cm: 2.98 × 10³ m⁻¹ × 0.010 m = 29.8, about 30 electrons, a mean spacing of 0.34 mm. Here |A|² is a beam density, and an infinitely long beam rightly contains infinitely many electrons.
Answer. k = 3.62 × 10¹⁰ m⁻¹, v = 4.19 × 10⁶ m s⁻¹ (matching √(2E/m)), |A|² = 2.98 × 10³ m⁻¹, about 30 electrons per centimetre of beam.
HardExample 3
A neutron is modelled by the normalised state ψσ(x) = (2πσ²)⁻¹⁄⁴ e(−x²/(4σ²)) e(ikx) with k = 1.2 × 10¹⁰ m⁻¹. Confirm the normalisation. Choose σ so that |ψσ| stays within 1% of its central value over a region 1.0 mm wide, then find Δk/k, using Δk = 1/(2σ), and the probability that a 1.0 μm detector at x = 0 fires. What fraction of the probability lies in the 1% region, and how does it depend on σ?
View solution →
- |ψσ|² = (2πσ²)⁻¹⁄² e(−x²/(2σ²)). With ∫ e(−x²/(2σ²)) dx = σ√(2π), the total is (2πσ²)⁻¹⁄² × σ√(2π) = 1 for every σ; m⁻¹ × m is dimensionless, as it must be.
- |ψσ(x)|/|ψσ(0)| = e(−x²/(4σ²)) ≥ 0.99 requires x² ≤ −4σ² ln 0.99, so |x| ≤ 2σ√(−ln 0.99) = 0.2005σ. A region 0.4010σ wide equal to 1.0 mm gives σ = 2.49 mm.
- Δk = 1/(2σ) = 1/(2 × 2.494 × 10⁻³ m) = 200 m⁻¹, so Δk/k = 200/(1.2 × 10¹⁰) = 1.7 × 10⁻⁸: the momentum is sharp to about 2 parts in 10⁸.
- Across the detector the density changes by a fraction of only x²/(2σ²) = (0.5 × 10⁻⁶)²/(2 × (2.494 × 10⁻³)²) = 2.0 × 10⁻⁸, so P = d/(σ√(2π)) = 1.0 × 10⁻⁶ m / (2.494 × 10⁻³ m × 2.507) = 1.6 × 10⁻⁴.
- The 1% region is |x| ≤ 0.2005σ, which holds erf(0.2005/√2) = erf(0.1418) = 0.159 of the probability for every σ. Widening σ tenfold widens the plane-wave-like region tenfold, spreads the same 15.9% over it and cuts the detector's probability tenfold; as σ → ∞ the region covers any fixed stretch while the probability per metre inside it falls to zero.
Answer. σ = 2.49 mm, Δk/k = 1.7 × 10⁻⁸, and the detector fires with probability 1.6 × 10⁻⁴. The 1% region always holds 15.9% of the probability, whatever σ is.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
A plane wave solves the equation but is never a state.
|A e(i(kx − ωt))|² = |A|² everywhere, so ∫|ψ|² dx diverges for any A ≠ 0, and no finite sum of plane waves fixes it. Use plane waves for E, λ and the current ħk|A|²/m of a steady beam; one particle in a finite region needs a normalised packet.