University Physics IV · The Schrödinger Equation · 6.4
Stationary States & Superpositions
Solve the eigenvalue problem once and evolution becomes bookkeeping. This lesson shows why a single energy level is a frozen picture, how to project any starting state onto the levels with one integral each, and why two levels together make the probability density swing at the gap between them.
Build the model
Connect the measurement to the mechanism.
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.
- Simple definition
- 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.
|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 problem
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 = 1
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 phases
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⁻¹
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 s
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 nm
Why 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.
Projecting 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.
Only 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.
The 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, which is why a statistical mixture of the same two levels shows no oscillation at all. 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.
Numbers: 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.
What 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.
Change one variable at a time
Make the relationship visible.
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.
BEAT PERIOD h/ΔE0.917 fs
RELATIVE PHASE / 2π0.00 turns
MEAN POSITION ⟨x⟩0.160 nm
MEAN ENERGY ⟨H⟩3.76 eV
Live interpretationBEAT PERIOD h/ΔE: 0.917 fs. RELATIVE PHASE / 2π: 0.00 turns. MEAN POSITION ⟨x⟩: 0.160 nm. MEAN ENERGY ⟨H⟩: 3.76 eV
Catch the common trap
Explain before calculating.
An electron in an infinite well of width 0.50 nm is prepared as Ψ(x,0) = (√3 ψ₁ + ψ₂)/2, with E₁ = 1.50 eV and E₂ = 6.02 eV. How often does ⟨x⟩ return to its starting value?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn 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.
- 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 1.46 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 (1.46 cycles) in 1.00 fs with no observable consequence.
MediumAn 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.
- 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.
AnswerA = √(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.
HardThe 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.
- 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.