University Physics V · The Schrödinger Equation · 4.4
Separation of Variables & 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.
Build the model
Connect the measurement to the mechanism.
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: 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.
- Simple definition
- 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.
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.
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, with the |n⟩ complete.
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.
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.
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.
Divide, 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.
The 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.
One 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 an atom parked in an energy eigenstate does not radiate: its charge density is a static distribution, and static distributions launch no waves. Nothing observable in the world of one eigenstate ever happens.
Two 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.
The eigenbasis is the general solution
Self-adjointness delivers more than real eigenvalues: the |n⟩ form a complete orthonormal basis, 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.T @ ψ0, 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.
Where 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.
Change one variable at a time
Make the relationship visible.
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.
BEAT ν = (Eₙ−E₁)/h273 THz
BEAT PERIOD h/(Eₙ−E₁)3.67 fs
MEAN ENERGY ⟨H⟩0.94 eV
P(E₁) — FIXED FOR ALL t0.50
Live interpretationBEAT 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
Catch the common trap
Explain before calculating.
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?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyShow 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.
- 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.
AnswerBoth 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.
MediumThe 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.
- 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.
AnswerP(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).
HardThe 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.
- 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.
AnswerExactly 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.