
Physics V course sequence · Interactive lesson
Separation of variables and stationary states
The one substitution that turns quantum dynamics into linear algebra: solve Ĥ|n⟩ = Eₙ|n⟩ once, hang a phase clock e(−iEₙt/ħ) on each eigenvector, and the future of every initial state is a sum you already own. The price is hidden in the name — a lone eigenstate shows nothing moving at all.
UPV-04.04 · Lesson 4 of 9 in this unit. Position is not a completion score.Build the model
All quantum motion is interference between energy eigenstates.
The key idea
A stationary state is an eigenstate of a time-independent Hamiltonian, evolving only by the global phase e(−iEₙt/ħ), so every probability and expectation value it predicts is constant in time.
Example. For an electron in a 1.0 nm box, ψ₁ has E₁ = 0.376 eV: its phase turns once every h/E₁ = 11.0 fs while |ψ₁|² never moves; add ψ₂ and the density beats at (E₂ − E₁)/h = 273 THz.
The complete physical model and assumptions
Insert Ψ(x, t) = ψ(x)T(t) into iħ∂Ψ/∂t = ĤΨ and divide by ψT: the left side depends only on t, the right only on x, so both equal one constant E. That single move splits a partial differential equation on Hilbert space into a trivial clock, T = e(−iEt/ħ), and the eigenvalue problem Ĥψ = Eψ — and it is legal precisely because Ĥ is time-independent; give V a time dependence and no separation constant exists. The payoff is total: when the spectrum is purely discrete, as for the box or any potential that grows without bound, self-adjointness makes the eigenvectors |n⟩ a complete orthonormal basis, so any initial state expands as Σcₙ|n⟩ and evolves by each coefficient spinning at its own rate, cₙe(−iEₙt/ħ). Diagonalise Ĥ once — by hand for the box, with scipy.linalg.eigh on a grid — and the dynamics of every initial condition is solved. The cost is hidden in the name: a lone eigenstate is "stationary" because its unimodular phase cancels in every probability, so its density and every expectation value are frozen; nothing in the world of one eigenstate happens. All quantum motion is interference between eigenstates — relative phase winding at the Bohr frequencies (Eₘ − Eₙ)/ħ — which is why spectra measure energy differences and nothing else.
1Divide, and the variables fall apart
Let Ĥ be time-independent and try the product Ψ(x, t) = ψ(x)T(t) in iħ∂Ψ/∂t = ĤΨ. The time derivative hits only T, Ĥ hits only ψ, and dividing by ψT leaves iħT′(t)/T(t) = (Ĥψ)(x)/ψ(x). The left side is a function of t alone, the right of x alone, and two functions of independent variables can agree everywhere only if both equal one constant. Name it E: the single PDE has split into iħT′ = ET and Ĥψ = Eψ. Watch where the assumption earned its keep — if V depended on t, the right side would still contain t after the division and no separation constant would exist. Separation is not a trick that always works; it is a statement about closed, undriven systems.
2The clock integrates itself
iħT′ = ET is first-order and linear: T(t) = e(−iEt/ħ), a pure phase of modulus one, so the norm ⟨Ψ|Ψ⟩ never drifts — and hermiticity has already forced E real. The clock's rate is enormous: the ground state of an electron in a 1.0 nm box carries E₁ = 0.376 eV, so its phase turns at E₁/ħ = 5.7 × 10¹⁴ rad s⁻¹, one full cycle every 11.0 fs. And the rate is not even physically well defined: add a constant V₀ to the potential and every Eₙ shifts by V₀, respinning every clock together while changing nothing measurable. Absolute phase rates are gauge; only differences between them will ever appear in an experiment.
3One eigenstate: every statistic stands still
Put the system in a single Ψₙ = ψₙe(−iEₙt/ħ). The density is Ψₙ*Ψₙ, and the phase meets its conjugate: |Ψₙ|² = |ψₙ|², frozen. The same cancellation runs through every expectation value — for any time-independent operator A, ⟨A⟩ = ⟨ψₙ|A|ψₙ⟩ with no t left anywhere, and ΔH = 0 exactly. This is why "stationary", and it does not mean at rest: the box ground state has ⟨p²⟩ = (πħ/L)² > 0, plenty of kinetic energy with no motion of any probability. It is also why, in this semiclassical picture, an atom parked in an energy eigenstate does not radiate: its charge density is a static distribution, and static distributions launch no waves; the spontaneous decay of real excited levels comes from coupling to the quantised field, which this Ĥ leaves out. Nothing observable in the world of one eigenstate ever happens.
4Two eigenstates: the Bohr frequency appears
Now superpose: Ψ = aψ₁e(−iE₁t/ħ) + bψ₂e(−iE₂t/ħ) with a, b real. Squaring gives |Ψ|² = a²ψ₁² + b²ψ₂² + 2abψ₁ψ₂ cos[(E₂ − E₁)t/ħ]. The diagonal terms are the frozen stationary densities; all motion lives in the cross term, oscillating at the single frequency ω₂₁ = (E₂ − E₁)/ħ — a difference, immune to the V₀ gauge. For the 1.0 nm box the gap is E₂ − E₁ = 3E₁ = 1.13 eV, so the density sloshes at ν = ΔE/h = 273 THz, and with a = b = 1/√2 the mean position swings as ⟨x⟩ = L/2 − 0.180L cos ω₂₁t. That sloshing charge is exactly what radiates: the frequencies a system emits are its level gaps over h, which is why spectroscopy measures differences and never absolute energies.
5The eigenbasis is the general solution
Self-adjointness delivers more than real eigenvalues: for a purely discrete spectrum, as in the box, the |n⟩ form a complete orthonormal basis (a continuous part adds an integral over generalised eigenstates, which are not vectors in the Hilbert space), so any initial state expands as |Ψ(0)⟩ = Σcₙ|n⟩ with cₙ = ⟨n|Ψ(0)⟩, and linearity evolves each term separately: |Ψ(t)⟩ = Σcₙe(−iEₙt/ħ)|n⟩. Each |cₙ|² — the probability that an energy measurement returns Eₙ — is constant forever: energy statistics are conserved even while the density churns. Numerically this is the spectral method: discretise Ĥ on a grid, call E, U = scipy.linalg.eigh(H), project c = U.conj().T @ ψ0 (U†, the conjugate transpose), and the state at any t is U @ (np.exp(−1j*E*t/ℏ)*c). One diagonalisation buys every time at once, with no step-size error accumulating — t = 10 ps costs the same as 10 fs.
6Where separation stops working
Two boundaries mark the method's edge. Drive the system — V = V(x, t), a laser field, an oscillating gate — and the very first step fails: no separation constant exists, there are no stationary states, and evolution needs the time-ordered propagator of the previous topic rather than one phase per level. Degeneracy is a softer edge: if E₂ = E₃, any combination of ψ₂ and ψ₃ is again an eigenstate, its cross term's Bohr frequency is zero, and that whole superposition is itself stationary — "the" stationary states are no longer unique. What survives everywhere is the moral: a superposition across distinct energies is never stationary, and its motion is a sum of cosines at the gaps (Eₘ − Eₙ)/ħ and nothing else.
Key equations
Read the units, assumptions and limits before substituting.
Separation ansatz
Ψ(x, t) = ψ(x) e(−iEt/ħ)
Turns iħ∂ₜΨ = ĤΨ, a PDE, into one ODE the clock solves at once and one eigenvalue problem carrying all the physics.
Legal only when ∂Ĥ/∂t = 0. E is the separation constant, in J or eV; ħ = 1.055 × 10⁻³⁴ J s.The eigenvalue problem
Ĥ|n⟩ = Eₙ|n⟩
Solve it once — by hand or with scipy.linalg.eigh — and the future of every initial state is already determined.
Why "stationary"
|Ψₙ(x, t)|² = |ψₙ(x)|², d⟨A⟩/dt = 0
One eigenstate predicts frozen statistics. Stationary means unchanging, not at rest: the box ground state keeps ⟨p²⟩ = (πħ/L)².
Holds for any time-independent A; the phase e(−iEₙt/ħ) has modulus 1 and cancels against its conjugate.General solution on the eigenbasis
|Ψ(t)⟩ = Σₙ cₙ e(−iEₙt/ħ)|n⟩, cₙ = ⟨n|Ψ(0)⟩
The whole dynamics with no time-stepping: one phase per level. This is the spectral method a diagonalisation hands you.
|cₙ|² is the probability of measuring Eₙ — constant in time; Σ|cₙ|² = 1 throughout.Bohr beat frequency
ωₘₙ = (Eₘ − Eₙ)/ħ, νₘₙ = (Eₘ − Eₙ)/h
Only differences are observable: shift V by a constant and every Eₙ/ħ changes while every ωₘₙ survives — spectra measure gaps.
ΔE in J gives ω in rad s⁻¹ and ν in Hz; a 1 eV gap beats at 242 THz.Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Park the weight at 0 or 1 and drag t: nothing moves — a lone eigenstate's clock is unobservable. Set it to 0.50 for the strongest beat, then raise n from 2 to 3: the density still beats but the dot freezes at L/2, because x₁₃ = 0 — parity silences ⟨x⟩ while the density keeps time.
Live results
- BEAT ν = (Eₙ−E₁)/h
- 273 THz
- BEAT PERIOD h/(Eₙ−E₁)
- 3.67 fs
- MEAN ENERGY ⟨H⟩
- 0.94 eV
- P(E₁) — FIXED FOR ALL t
- 0.50
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. Weight |cₙ|² on upper level: 0.50; Upper level n: 2; Time t: 0.0 fs
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
Students learn that "overall phase is unobservable" and conclude the factors e(−iEₙt/ħ) can be dropped wherever they appear — which would freeze all of quantum mechanics. The rule licenses removing one global factor from the whole ket. A lone stationary state has only that global factor, so its clock really is invisible; a superposition carries one clock per component, and after factoring out e(−iE₁t/ħ) the relative phases e(−i(Eₙ−E₁)t/ħ) remain — and they are measurable, because they are the beat in |Ψ|². The genuine invariance is different: shifting V by a constant respins every clock at the same extra rate, so no absolute Eₙ/ħ appears in any observable, while every difference (Eₘ − Eₙ)/ħ does. Drop the phases component by component and you predict a universe with energy levels and no light.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
Show that Ψₙ(x, t) = ψₙ(x)e(−iEₙt/ħ) solves the time-dependent Schrödinger equation whenever Ĥψₙ = Eₙψₙ, then find how long the ground-state phase of an electron in a 1.0 nm infinite well (E₁ = 0.376 eV) takes to complete one full turn — and state what an observer can see of it.
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.
- Left side: iħ∂ₜΨₙ = iħψₙ(−iEₙ/ħ)e(−iEₙt/ħ) = Eₙψₙe(−iEₙt/ħ).
- Right side: ĤΨₙ = (Ĥψₙ)e(−iEₙt/ħ) = Eₙψₙe(−iEₙt/ħ), since Ĥ acts on x alone and the phase is a constant to it. The two sides agree for every t, so the ansatz solves the equation.
- The phase completes 2π when E₁t/ħ = 2π, i.e. t = 2πħ/E₁ = h/E₁.
- E₁ = 0.376 eV = 6.02 × 10⁻²⁰ J, so t = 6.626 × 10⁻³⁴ / 6.02 × 10⁻²⁰ = 1.10 × 10⁻¹⁴ s = 11.0 fs.
- An observer sees none of it: |Ψ₁|² = |ψ₁|² and every ⟨A⟩ is constant. The clock is real in the mathematics and invisible in any experiment performed on this state alone.
Answer. Both sides equal Eₙψₙe(−iEₙt/ħ); the ground-state phase turns once every h/E₁ = 11.0 fs, and no measurement on the lone eigenstate can detect it.
MediumExample 2
The same electron (1.0 nm box, E₁ = 0.376 eV, E₂ = 4E₁ = 1.504 eV) is prepared in Ψ(0) = (ψ₁ + ψ₂)/√2. Find the probability of each energy outcome at any later time, ⟨H⟩ and ΔH, and the frequency and amplitude of the oscillation of ⟨x⟩, given the box matrix element x₁₂ = −16L/(9π²) = −0.180L.
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.
- The coefficients only rotate: c₁(t) = e(−iE₁t/ħ)/√2 and c₂(t) = e(−iE₂t/ħ)/√2, so |c₁|² = |c₂|² = ½ forever — P(E₁) = P(E₂) = 0.5 at every t.
- ⟨H⟩ = ½E₁ + ½E₂ = ½(0.376 + 1.504) = 0.940 eV, constant. From ⟨H²⟩ = ½(E₁² + E₂²), the spread is ΔH = (E₂ − E₁)/2 = 0.564 eV, also constant.
- The density's cross term beats at ν = (E₂ − E₁)/h: ΔE = 3E₁ = 1.128 eV = 1.807 × 10⁻¹⁹ J, so ν = 1.807 × 10⁻¹⁹ / 6.626 × 10⁻³⁴ = 2.73 × 10¹⁴ Hz = 273 THz, period 3.67 fs.
- ⟨x⟩(t) = L/2 + 2⋅(½)⋅x₁₂ cos(ω₂₁t) = L/2 − 0.180L cos(ω₂₁t): the mean position sloshes between 0.32L and 0.68L at 273 THz.
- Energy statistics frozen while position swings — exactly the division of labour stationary-state interference promises.
Answer. P(E₁) = P(E₂) = ½ for all t; ⟨H⟩ = 0.940 eV and ΔH = 0.564 eV, both constant; ⟨x⟩ oscillates about L/2 with amplitude 0.180L at 273 THz (period 3.67 fs).
HardExample 3
The electron is prepared instead in Ψ(0) = (ψ₁ + ψ₂ + ψ₃)/√3, with Eₙ = n²E₁ and E₁ = 0.376 eV. List every frequency at which |Ψ(x, t)|² oscillates, give each in THz, and find the earliest time at which the state returns to Ψ(0) up to a global phase.
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.
- Ψ(t) = (ψ₁e(−iE₁t/ħ) + ψ₂e(−i4E₁t/ħ) + ψ₃e(−i9E₁t/ħ))/√3. Squaring gives three static diagonal terms and three moving cross terms.
- Each cross term beats at one Bohr frequency: (E₂−E₁)/h = 3E₁/h, (E₃−E₂)/h = 5E₁/h, (E₃−E₁)/h = 8E₁/h. Nothing else appears — no sums of energies, no absolute Eₙ/h.
- E₁/h = 6.02 × 10⁻²⁰ J / 6.626 × 10⁻³⁴ J s = 9.09 × 10¹³ Hz = 90.9 THz, so the motion's spectrum is 273, 455 and 727 THz — in the ratio 3 : 5 : 8.
- Revival: factor out e(−iE₁t/ħ). The relative phases e(−i3E₁t/ħ) and e(−i8E₁t/ħ) both return to 1 only when E₁t/ħ is a multiple of 2π, since 3 and 8 share no common factor: tᵣₑᵥ = 2πħ/E₁ = h/E₁.
- tᵣₑᵥ = 6.626 × 10⁻³⁴ / 6.02 × 10⁻²⁰ = 1.10 × 10⁻¹⁴ s = 11.0 fs — the box revival time 4mL²/(πħ); every Bohr frequency above is an integer multiple of 1/tᵣₑᵥ.
Answer. Exactly three frequencies — 273, 455 and 727 THz, the Bohr gaps 3, 5 and 8 × E₁/h. First return to Ψ(0) up to global phase at t = h/E₁ = 11.0 fs.
Quick check
Select an answer and explain it before checking.
Review & continue
A stationary state is an eigenstate of a time-independent Hamiltonian, evolving only by the global phase e(−iEₙt/ħ), so every probability and expectation value it predicts is constant in time.
- 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.