
Quantum core · institution-dependent · Interactive lesson
Stationary states and superpositions
Nothing inside a single energy level ever moves, yet the particle it describes is not at rest, and two levels together run a clock good to the femtosecond. This lesson settles all three claims for an electron in a half-nanometre box, then names the assumption the whole method quietly borrows.
UPIV-06.04 · Lesson 4 of 9 in this unit. Position is not a completion score.Quantum core · institution-dependent. Departments differ in how much quantum mechanics they include in Physics IV or teach in a dedicated course. Follow your university’s required mathematical depth and assessment; these numbered units are not automatically optional.
Build the model
An eigenstate holds still; two of them together keep time.
The key idea
A stationary state is a single energy eigenfunction carrying the phase e(−iEt/ħ); its probability density and all its expectation values are frozen, so only a superposition of different energies can make anything move.
Example. For an electron in a 0.50 nm infinite well, ψ₁ alone gives a density fixed for all time; add ψ₂ with equal weight and ⟨x⟩ swings between 0.160 nm and 0.340 nm every 0.917 fs while ⟨H⟩ stays at 3.76 eV.
The complete physical model and assumptions
Once the time-independent equation has handed you the pairs (Eₙ, ψₙ), the entire dynamics collapses into one rule: attach e(−iEₙt/ħ) to each eigenfunction and add. Each half of that rule costs something. A single eigenstate picks up a phase of unit modulus, so |Ψ|² and every expectation value of a time-independent observable are frozen — "stationary" describes the picture, not the particle, which still carries kinetic energy ⟨p²⟩/2m. Motion appears only when two different energies are present, because in the sum the moduli |cₙ| stay locked and only the relative phases turn: for two levels the interference term rides on cos(ΔE⋅t/ħ), so the density sloshes at the gap and at nothing else. Getting the cₙ costs one overlap integral each, cₙ = ∫ψₙ*Ψ(x,0)dx, and the whole scheme rests on a completeness claim — that the eigenfunctions really do span every admissible state. That claim is assumed here, delivered by Sturm–Liouville theory for a well-behaved bound problem in one dimension, and false as stated the moment the spectrum acquires a continuous part.
1Why the picture holds still
Put Ψₙ(x, t) = ψₙ(x)e(−iEₙt/ħ) into Born's rule. The density is Ψₙ*Ψₙ = ψₙ*ψₙ e(+iEₙt/ħ)e(−iEₙt/ħ) = |ψₙ|², because a real energy makes the phase factor a unit complex number. The same cancellation runs through any sandwich integral ⟨A⟩ = ∫Ψ*ÂΨ dx whose operator carries no explicit t, so ⟨x⟩, ⟨p⟩, ⟨H⟩ and every variance are constants of the motion. The probability current is static too, and for a real bound eigenfunction it is zero everywhere. What "stationary" does not mean is at rest: in the infinite well ⟨p⟩ = 0 but ⟨p²⟩ = 2mEₙ, so an electron in the n = 2 state of a 0.50 nm well carries 6.02 eV of kinetic energy while its picture never twitches. The description is frozen; the energy is not zero.
2Projecting the start onto the ladder
To evolve an arbitrary start, first write it on the eigenbasis: Ψ(x,0) = Σ cₙψₙ(x) with cₙ = ∫ψₙ*Ψ(x,0)dx. That integral is an inner product, and it isolates one coefficient only because the ψₙ are orthonormal — multiply the sum by ψₘ*, integrate, and every term but m collapses. Take the parabolic start Ψ(x,0) = A x(L−x) in an infinite well. Normalising with ∫₀L x²(L−x)²dx = L⁵/30 gives A = √(30/L⁵), and the standard integral ∫₀L x(L−x)sin(nπx/L)dx = 2L³[1−(−1)ⁿ]/n³π³ then gives cₙ = 4√60/n³π³ for odd n and exactly zero for even n. So c₁ = 0.9993 and |c₁|² = 0.9986: the parabola is 99.86% ground state. The even levels vanish for a reason worth naming — Ψ(x,0) is symmetric about L/2 while ψ₂ is antisymmetric, so their overlap cancels term by term.
3Only the relative phases turn
Because the moduli |cₙ| are fixed, the probability of each energy is fixed, and ⟨H⟩ = Σ|cₙ|²Eₙ never changes — that is energy conservation written in this language. For the parabola above, ⟨H⟩ = E₁Σodd 0.9986/n⁴ = 0.9986 × (π⁴/96) E₁ = 1.0132E₁, about 1.3% above the ground-state energy and constant forever. What does change is phase, and only relative phase matters. Factor e(−iE₁t/ħ) out of the whole sum: it multiplies every term equally, and a global phase cancels in Ψ*Ψ and in every expectation value, so it is not physics. What survives is exp[−i(Eₙ−E₁)t/ħ] riding on each higher term. Two states differing by a global phase are the same state; two states differing by a relative phase are different states with different densities.
4The cross term is the whole of the motion
Square a two-term sum and three pieces appear. With real eigenfunctions and cₙ = |cₙ|e(iφₙ), |Ψ|² = |c₁|²ψ₁² + |c₂|²ψ₂² + 2|c₁||c₂|ψ₁ψ₂cos(ω₂₁t − δ), where ω₂₁ = (E₂−E₁)/ħ and δ = φ₂ − φ₁. The first two terms are the frozen densities of the separate levels; every bit of the motion lives in the third. Three consequences follow. The beat carries the energy difference and nothing else — add 10 eV to both levels and the density is unchanged, because the shift is a global phase. Kill either coefficient and the cross term dies with it. A statistical mixture of the same two levels fails for a different reason: it keeps both weights but holds no definite relative phase, so averaging over that phase erases the third term and leaves a density that never moves. Preparing the pair coherently or statistically gives identical energy statistics and different densities, and the beat is the measurement that tells them apart. And the cross term integrates to zero over x by orthogonality, so the norm stays at 1 while the density sloshes: probability is moved around, never created.
5Numbers: an electron in a half-nanometre well
Take L = 0.50 nm. E₁ = h²/8mL² = 2.41×10⁻¹⁹ J = 1.50 eV and Eₙ = n²E₁, so E₂ = 6.02 eV and ΔE = 3E₁ = 4.51 eV. Then ω₂₁ = ΔE/ħ = 6.86×10¹⁵ rad s⁻¹ and the beat period is T = h/ΔE = 0.917 fs, a sloshing frequency of 1.09 PHz — fast enough that only attosecond pulses can strobe it. With equal weights on ψ₁ and ψ₂, ⟨x⟩ = L/2 − (16L/9π²)cos(ω₂₁t) runs from 0.160 nm to 0.340 nm about the midpoint 0.250 nm, while ⟨H⟩ = 2.5E₁ = 3.76 eV sits still throughout. And because every Eₙ here is an integer multiple of E₁, all the phases realign together: the whole wavefunction, not merely the 1–2 pair, revives exactly after Tᵣₑᵥ = h/E₁ = 2.75 fs, three beats later. That exact revival is a peculiarity of the n² spectrum; a general potential gives only partial ones.
6What is assumed, and where it fails
Three assumptions are doing work here. Completeness: that every admissible Ψ(x,0) really is Σcₙψₙ. Sturm–Liouville theory delivers it for a bound problem on a finite interval, but it is imported, not proved, and it fails as stated once the spectrum has a continuous part — free and scattering states need ∫dE φ(E)ψE, an integral, not a sum. A time-independent potential: if V depends on t there are no stationary states to expand on and the scheme is void from the first line. And a resolved basis inside degenerate subspaces: when two independent eigenfunctions share an energy, any combination of them is also an eigenfunction, so you must first choose an orthonormal pair — Gram–Schmidt will do — before the overlap integral means anything. One practical caveat too: the series is truncated in practice, and a start with a corner in it converges only as slowly as its coefficients decay.
Key equations
Read the units, assumptions and limits before substituting.
Stationary state
Ψₙ(x, t) = ψₙ(x) e(−iEₙt/ħ)
|e(−iEₙt/ħ)| = 1 for real Eₙ, so the density and every expectation value stand still.
Eₙ in J; ħ = 1.055×10⁻³⁴ J s; ψₙ is a solution of the eigenvalue problemExpansion coefficient
cₙ = ∫ ψₙ*(x) Ψ(x, 0) dx
One integral per level projects the start onto the ladder, and |cₙ|² is the probability of measuring Eₙ.
cₙ complex, dimensionless when ψₙ carries m(−1/2); Σ|cₙ|² = 1Evolving a superposition
Ψ(x, t) = Σₙ cₙ ψₙ(x) e(−iEₙt/ħ)
No PDE to re-solve: evolution is a phase turned on each term at its own rate.
ωₙ = Eₙ/ħ in rad s⁻¹; the moduli |cₙ| never change, only the phasesTwo-level density and its cross term
|Ψ|² = |c₁ψ₁|² + |c₂ψ₂|² + 2|c₁c₂|ψ₁ψ₂ cos(ω₂₁t − δ)
Only the third term moves, and it moves at the gap alone — the mean energy is irrelevant to it.
real ψₙ; δ = arg c₂ − arg c₁; ω₂₁ = (E₂ − E₁)/ħ in rad s⁻¹Beat period set by the gap
T = 2π/ω₂₁ = h/ΔE
Turns a level spacing into a clock: wider gaps beat faster, and ΔE = 0 means no beat at all.
ΔE = 4.51 eV gives T = 0.917 fs, using h = 4.1357×10⁻¹⁵ eV sMean position, infinite well, levels 1 and 2
⟨x⟩ = L/2 − (16L/9π²) cos(ω₂₁t)
Gives the beat a quotable amplitude, and it collapses to zero whenever the two levels share a parity.
equal weights; 16/9π² = 0.1801, so L = 0.50 nm swings ±0.090 nm about 0.250 nmInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Drag time with n = 2: the dashed curve is frozen while the solid one sloshes, and the dot returns every 0.917 fs. Push |c₂|² to 0 or to 1 and the beat dies — interference needs both terms. Then set n = 3: the density still breathes, but ⟨x⟩ stops moving.
Live results
- BEAT PERIOD h/ΔE
- 0.917 fs
- RELATIVE PHASE / 2π
- 0.00 turns
- MEAN POSITION ⟨x⟩
- 0.160 nm
- MEAN ENERGY ⟨H⟩
- 3.76 eV
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Time t since preparation: 0.00 fs; Weight |c₂|² on the upper level: 0.50; Upper level n: 2
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The costly error is to hand a superposition a single energy — usually its mean — and evolve it with one factor e(−i⟨H⟩t/ħ). Do that and every term turns at the same rate, no relative phase ever accumulates, the cross term freezes, and you predict a density that never moves: precisely the behaviour of the eigenstate you were trying to leave behind. A superposition of two energies does not have an energy; it has two, with probabilities |c₁|² and |c₂|², and its motion is the difference between their phase rates, not their average. The mirror error is reading "stationary" as "at rest". An eigenstate has a static density, but ⟨p²⟩ = 2mEₙ ≠ 0 in the infinite well — the picture stands still, the kinetic energy does not vanish.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
An electron sits in the n = 2 state of an infinite well of width L = 0.50 nm, so Ψ(x, t) = ψ₂(x)e(−iE₂t/ħ). Show that the probability density does not change with time, then find E₂, the angular frequency E₂/ħ, and the phase the wavefunction accumulates in 1.00 fs.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- Born's rule: |Ψ|² = Ψ*Ψ = ψ₂*(x)e(+iE₂t/ħ) · ψ₂(x)e(−iE₂t/ħ). The two exponentials multiply to e⁰ = 1 because E₂ is real, so |Ψ(x, t)|² = |ψ₂(x)|² for every t.
- Energies: E₁ = h²/8mL² = (6.626×10⁻³⁴)² ÷ [8 × 9.109×10⁻³¹ × (0.50×10⁻⁹)²] = 2.41×10⁻¹⁹ J = 1.50 eV, and E₂ = 4E₁ = 9.64×10⁻¹⁹ J = 6.02 eV.
- ω₂ = E₂/ħ = 9.64×10⁻¹⁹ J ÷ 1.055×10⁻³⁴ J s = 9.14×10¹⁵ rad s⁻¹.
- In t = 1.00 fs the phase advances ω₂t = 9.14 rad, which is 9.14/2π = 1.45 full turns of the complex phase — and nothing measurable has changed, because an overall phase cancels in every |Ψ|² and every ⟨A⟩.
Answer. |Ψ(x, t)|² = |ψ₂(x)|² at all times. E₂ = 6.02 eV, ω₂ = 9.14×10¹⁵ rad s⁻¹, and the phase turns through 9.14 rad (1.45 cycles) in 1.00 fs with no observable consequence.
MediumExample 2
An electron in an infinite well of width L is prepared as Ψ(x,0) = A x(L − x). Normalise it, expand it on the eigenbasis ψₙ = √(2/L) sin(nπx/L), and find |c₁|², |c₃|² and the mean energy ⟨H⟩ in units of E₁. Use ∫₀L x(L−x)sin(nπx/L) dx = 2L³[1 − (−1)ⁿ]/n³π³.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Normalise: ∫₀L A²x²(L−x)²dx = A²L⁵/30 = 1, so A = √(30/L⁵).
- cₙ = ∫₀L ψₙ*Ψ(x,0)dx = √(2/L) · √(30/L⁵) · 2L³[1−(−1)ⁿ]/n³π³ = 2√60[1−(−1)ⁿ]/n³π³. Even n gives 1−(−1)ⁿ = 0: Ψ(x,0) is symmetric about L/2, ψ₂ is antisymmetric, and the overlap cancels.
- Odd n gives cₙ = 4√60/n³π³ = 0.9993/n³. So c₁ = 0.9993, |c₁|² = 0.9986, and |c₃|² = 0.9986/3⁶ = 1.37×10⁻³. The parabola is 99.86% ground state.
- ⟨H⟩ = Σ|cₙ|²Eₙ = E₁ Σodd (0.9986/n⁶)⋅n² = 0.9986 E₁ Σodd 1/n⁴ = 0.9986 × (π⁴/96) E₁ = 1.0132E₁, which is exactly 10E₁/π².
- Because the |cₙ| never change, 1.0132E₁ is also the mean energy at every later time: the state moves, its energy statistics do not.
Answer. A = √(30/L⁵); |c₁|² = 0.9986, |c₃|² = 1.37×10⁻³, all even coefficients vanish by parity; ⟨H⟩ = 10E₁/π² = 1.0132E₁, constant in time.
HardExample 3
The same 0.50 nm well is prepared as Ψ(x,0) = (ψ₁ + ψ₂)/√2. Using ⟨n|x|n⟩ = L/2 and ⟨1|x|2⟩ = −16L/9π², find ⟨x⟩(t), its two turning values, the beat period, and ⟨H⟩. Then say when the whole wavefunction — not just ⟨x⟩ — repeats itself.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- Ψ(x, t) = (1/√2)[ψ₁e(−iE₁t/ħ) + ψ₂e(−iE₂t/ħ)]. Factor out e(−iE₁t/ħ): it is a global phase and cancels in every expectation value, leaving only the relative phase ω₂₁t with ω₂₁ = (E₂−E₁)/ħ.
- ⟨x⟩ = ½⟨1|x|1⟩ + ½⟨2|x|2⟩ + 2⋅½⋅⟨1|x|2⟩cos(ω₂₁t) = L/2 − (16L/9π²)cos(ω₂₁t).
- Amplitude: 16/9π² = 0.1801, so 0.1801 × 0.50 nm = 0.0901 nm. ⟨x⟩ runs from 0.250 − 0.090 = 0.160 nm at t = 0 to 0.250 + 0.090 = 0.340 nm half a period later.
- ΔE = E₂ − E₁ = 3E₁ = 4.51 eV = 7.23×10⁻¹⁹ J, so ω₂₁ = 6.86×10¹⁵ rad s⁻¹ and T = 2π/ω₂₁ = h/ΔE = 0.917 fs — a sloshing frequency of 1.09 PHz.
- ⟨H⟩ = ½E₁ + ½E₂ = 2.5E₁ = 3.76 eV, and it is fixed: the position oscillates, the energy statistics do not.
- Every Eₙ = n²E₁ is an integer multiple of E₁, so all the phases e(−iEₙt/ħ) realign together after Tᵣₑᵥ = h/E₁ = 2.75 fs — exactly three beats of the 1–2 pair. Ψ itself, not merely ⟨x⟩, is restored.
Answer. ⟨x⟩ = L/2 − (16L/9π²)cos(ω₂₁t), swinging between 0.160 nm and 0.340 nm with T = 0.917 fs; ⟨H⟩ = 2.5E₁ = 3.76 eV, constant; the full state revives after 2.75 fs.
Quick check
Select an answer and explain it before checking.
Review & continue
A stationary state is a single energy eigenfunction carrying the phase e(−iEt/ħ); its probability density and all its expectation values are frozen, so only a superposition of different energies can make anything move.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.