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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.
Starting model: Separation of variables and stationary statesAn electron in a 1.0 nm box with weight 0.50 on ψ₁ and 0.50 on level n = 2. Solid: |Ψ(x,t)|² at t = 0.0 fs; dashed: its time average, fixed by the |cₙ|² alone; the dot tracks ⟨x⟩. Δφ = (Eₙ − E₁)t/ħ is the relative phase of the two clocks. Everything that moves does so at the one Bohr frequency (Eₙ − E₁)/h = 273 THz.|Ψ(x, t)|², 1.0 nm boxsolid: now · dashed: mean · dot: ⟨x⟩Δφ = 0.0 radx = 0x = L
Start with a picture. Change the model in the lab below.
1

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.

2

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.

Ĥ self-adjoint on its stated domain; every Eₙ real; ⟨m|n⟩ = δₘₙ; for a purely discrete spectrum, as in the box, the |n⟩ are complete.

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.
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

0.50
0 to 1; steps of 0.05.
2
2 to 4; steps of 1.
0.0 fs
0 to 12 fs; steps of 0.1.

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.

Interactive Physics V modelAn electron in a 1.0 nm box with weight 0.50 on ψ₁ and 0.50 on level n = 2. Solid: |Ψ(x,t)|² at t = 0.0 fs; dashed: its time average, fixed by the |cₙ|² alone; the dot tracks ⟨x⟩. Δφ = (Eₙ − E₁)t/ħ is the relative phase of the two clocks. Everything that moves does so at the one Bohr frequency (Eₙ − E₁)/h = 273 THz.|Ψ(x, t)|², 1.0 nm boxsolid: now · dashed: mean · dot: ⟨x⟩Δφ = 0.0 radx = 0x = L
An electron in a 1.0 nm box with weight 0.50 on ψ₁ and 0.50 on level n = 2. Solid: |Ψ(x, t)|² at t = 0.0 fs; dashed: its time average, fixed by the |cₙ|² alone; the dot tracks ⟨x⟩. Δφ = (Eₙ − E₁)t/ħ is the relative phase of the two clocks. Everything that moves does so at the one Bohr frequency (Eₙ − E₁)/h = 273 THz.

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

Baseline 0.50; choose 0 to 1, in steps of 0.05.
4. Compared with the saved baseline, this result will…

This experiment is kept only while this page is open. It does not change saved course progress.

4

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.

5

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.

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.

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.

6

Quick check

Select an answer and explain it before checking.

An electron in a 1.0 nm infinite well is prepared in (ψ₁ + ψ₂)/√2, with E₁ = 0.38 eV and E₂ = 1.50 eV. Each component solves the time-independent equation. What does the probability density |Ψ(x, t)|² do as time runs?
7

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.

Extra challenging questions

Connect several ideas, explore supplied data, and compare your reasoning with a worked solution.