University Physics V · The Schrödinger Equation · 4.2
The Time-Dependent Equation in Dirac Notation
Everything a quantum system does between measurements is this one line. Learn to read it the way you read F = ma: which operator drives the motion, why the ket right now is the entire initial data, and how the i and the dagger between them hold total probability at exactly one.
Build the model
Connect the measurement to the mechanism.
Quantum dynamics is a single postulate: the state vector of a closed system obeys iħ d|Ψ⟩/dt = Ĥ|Ψ⟩, with the Hamiltonian of the last topic recast as the generator of time translation. Nothing derives this law — Schrödinger guessed it in 1926, and every interference fringe and atomic spectrum since has voted for it. Its structure does all the work.
First order in time: |Ψ(0)⟩ alone fixes |Ψ(t)⟩ forever, forwards and backwards, with no second initial condition and no "velocity of the state". Linear: solutions superpose with constant coefficients, so the superposition principle of Unit 3 survives the motion untouched. And the factor of i conspires with Ĥ† = Ĥ to make the flow norm-preserving — energy components turn their phases at rates Eₙ/ħ without ever changing length, which is why total probability holds at 1 with no one enforcing it.
The price is printed on the label: the law is non-relativistic (first order in time, second in space), knows nothing of spin until ℂ² is tensored in, applies only while the system stays closed, and contains no measurement — evolution under this equation is smooth and deterministic, and the Born rule's randomness enters by a different postulate entirely.
- Simple definition
- The time-dependent Schrödinger equation iħ d|Ψ(t)⟩/dt = Ĥ|Ψ(t)⟩ is the postulate that a closed system's state vector is driven through Hilbert space by its Hamiltonian, the self-adjoint energy operator.
- Example
- Feed it an eigenstate with E = 2.0 eV and it returns |Ψ(t)⟩ = e(−iEt/ħ)|E⟩: the phase turns at ω = E/ħ = 3.0 × 10¹⁵ rad s⁻¹, once around every 2.1 fs, while every probability stands still.
First order in t: |Ψ(0)⟩ alone is complete initial data. No state 'velocity' is needed — unlike x and ẋ in Newton, or shape and velocity profile for a string.
ħ = 1.055 × 10⁻³⁴ J s = 6.582 × 10⁻¹⁶ eV s; Ĥ self-adjoint, in J; |Ψ(t)⟩ a unit ket
One abstract equation, many concrete forms: choose a basis and the ket law becomes the PDE that grid codes discretise.
Ψ(x, t) = ⟨x|Ψ(t)⟩ in m⁻¹ᐟ²; m in kg, V in J — the same ket law hit with ⟨x|
What a computer integrates: N coupled linear first-order ODEs on a complex vector, one per basis ket — solveᵢᵥₚ territory.
Hₖⱼ = ⟨k|Ĥ|j⟩ in J, cₖ dimensionless; comes from inserting Σⱼ|j⟩⟨j| = 𝟙
Normalise once, normalised forever. The i and the dagger cancel jointly — drop either one and total probability leaks.
zero exactly when Ĥ† = Ĥ on its domain; in ⟨x| language the dagger is an integration by parts
Only energy differences are observable — set the zero of V wherever the algebra is cleanest.
E₀ any real constant in J; the phase is global, so no |⟨a|Ψ⟩|² moves
Conservation laws for free: [Ĥ, Â] = 0 freezes ⟨Â⟩ for every state — commutators are where constants of the motion live.
for  with no explicit time dependence; [Ĥ, Â] carries J times the units of Â
First order in time: the ket is the whole initial data
Newton's second law is second order: launching a projectile takes a position and a velocity. The classical wave equation is second order too — a string needs its shape and its velocity profile. The Schrödinger equation is first order: d|Ψ⟩/dt is already determined by |Ψ⟩ itself, so a single ket at a single instant fixes the entire history, forward and backward. There is no independent 'velocity of the state' to prescribe and no freedom left once |Ψ(0)⟩ is given. That makes quantum evolution between measurements perfectly deterministic — given the ket at noon, the equation names the ket at any other time. Reversibility is built in as well: complex-conjugate the equation and send t → −t and you hold an equally valid solution, which is why any irreversibility in physics must be smuggled in from outside this law.
The i is load-bearing, not decoration
Delete the i and watch what breaks: dΨ/dt = −(1/ħ)ĤΨ is a diffusion-type equation, and expanding in energy eigenvectors makes each component decay as e(−Eₙt/ħ). The norm drains away and, after a long time, only the lowest-energy component survives, because every excited fraction dies faster. Restore the i and the same expansion gives e(−iEₙt/ħ): each component keeps its length and only turns its phase, at angular rate Eₙ/ħ. Rotation instead of decay is the entire difference between quantum evolution and heat flow, and it hangs on one factor. Numerical physicists exploit the broken version on purpose: propagate in 'imaginary time' t → −iτ, renormalising as you go, and any starting guess relaxes onto the ground state — a standard SciPy exercise, and a reminder of exactly what the i was doing in the equation you meant to solve.
Hermiticity turns the crank on norm conservation
Differentiate ⟨Ψ|Ψ⟩ and use the equation twice. The ket side gives d|Ψ⟩/dt = −(i/ħ)Ĥ|Ψ⟩; daggering the whole equation gives d⟨Ψ|/dt = +(i/ħ)⟨Ψ|Ĥ†. Adding the two contributions, d⟨Ψ|Ψ⟩/dt = (i/ħ)⟨Ψ|(Ĥ† − Ĥ)|Ψ⟩, which vanishes exactly when Ĥ† = Ĥ. A self-adjoint Hamiltonian therefore conserves total probability with no side condition: normalise once, normalised forever. Both hypotheses earn their keep. In the position representation the dagger move is an integration by parts, so 'self-adjoint' includes the boundary terms — a wavefunction that fails to fall off fast enough can leak norm through the edge of the calculation. And the failure mode is a tool: add a complex absorbing potential −iW(x) near the grid boundary and the norm decays deliberately, swallowing outgoing flux before a periodic box wraps it back around.
Hit the law with a bra: components a computer can integrate
Dirac notation earns its keep the moment you need numbers. Apply ⟨k| to the equation and insert Σⱼ|j⟩⟨j| = 𝟙: iħ ċₖ = Σⱼ Hₖⱼ cⱼ, with cₖ = ⟨k|Ψ⟩ and Hₖⱼ = ⟨k|Ĥ|j⟩. One abstract law has become N coupled first-order ODEs — precisely what scipy.integrate.solveᵢᵥₚ eats. The smallest honest case is ammonia's two inversion states, nitrogen above or below the H₃ plane, with Ĥ = E₀𝟙 − Δσ̂ₓ: iħċ₁ = E₀c₁ − Δc₂ and iħċ₂ = −Δc₁ + E₀c₂. Starting in |1⟩, the populations trade as P₂ = sin²(Δt/ħ), and with Δ = 4.9 × 10⁻⁵ eV that oscillation runs at 2Δ/h ≈ 24 GHz — the microwave line the ammonia maser rides. The same two-line reduction, with the basis grown from 2 kets to a spatial grid, is how every wave-packet movie in this course is computed.
Linearity: superpositions never tangle
The equation contains |Ψ⟩ to the first power and nothing else, so if |Ψ₁(t)⟩ and |Ψ₂(t)⟩ both solve it, α|Ψ₁(t)⟩ + β|Ψ₂(t)⟩ solves it with the same constants α and β at every time. Superposition is not an initial-condition trick that the dynamics erodes; it is preserved exactly, which is why interference survives evolution and a two-slit pattern can be computed one path at a time and summed at the screen. Linearity also forbids state-space chaos: unitarity keeps the inner product of two nearby kets constant, so no trajectory in Hilbert space peels away from a neighbour exponentially — quantum 'chaos' has to hide in the spectrum of Ĥ instead. A practical corollary: adding E₀𝟙 to Ĥ multiplies every solution by the one global phase e(−iE₀t/ħ), so only energy differences ever reach an experiment, and the zero of potential energy is yours to place.
Read the fine print on the postulate
Every symbol carries an assumption. The Ĥ of the previous topic is p̂²/2m + V(x̂): non-relativistic, since the equation is first order in time but second order in space, and no Lorentz transformation preserves that mismatch — an electron near c needs the Dirac equation. The state space is L²(ℝ), so spin is absent until you tensor on ℂ² and add the Pauli term. The system must be closed: couple it to an environment you do not track and the ket description itself fails, replaced by a density matrix whose Lindblad equation carries precisely the non-unitary terms this law forbids. And nothing in this smooth, deterministic flow ever selects a measurement outcome — the Born rule and projection of Unit 3 are separate postulates. Whenever a problem seems to need the state to 'jump', the jump is being imported from that other postulate, not from anything iħ d|Ψ⟩/dt = Ĥ|Ψ⟩ can do.
Change one variable at a time
Make the relationship visible.
Slide Δ up and the populations trade faster — the first full flip lands at πħ/(2Δ). Then raise Γ: the equation stays linear, but Ĥ† ≠ Ĥ now, and the dashed norm peels off 1 as e(−Γt/ħ). Hermiticity, not linearity, is what holds total probability at one.
P₁ AT MARKER t0.54
P₂ AT MARKER t0.46
NORM ⟨Ψ|Ψ⟩ AT t1.00
FLIP TIME πħ/(2Δ)1.96 ħ/E₀
Live interpretationP₁ AT MARKER t: 0.54. P₂ AT MARKER t: 0.46. NORM ⟨Ψ|Ψ⟩ AT t: 1.00. FLIP TIME πħ/(2Δ): 1.96 ħ/E₀
Catch the common trap
Explain before calculating.
To model an unstable level, a student replaces Ĥ by Ĥ − i(Γ/2)𝟙, with Γ = 4.1 × 10⁻⁸ eV real and Ĥ still self-adjoint. According to iħ d|Ψ⟩/dt = (Ĥ − i(Γ/2)𝟙)|Ψ⟩, what happens to the norm ⟨Ψ|Ψ⟩ of a state normalised at t = 0?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn electron occupies an energy eigenstate |E⟩ of some Hamiltonian, with E = 2.0 eV. Solve the time-dependent equation for |Ψ(t)⟩, find the angular frequency of the phase and the time for one full turn, and state what an experiment can see of this motion.
- On an eigenstate the operator equation collapses to one scalar ODE: writing |Ψ(t)⟩ = c(t)|E⟩ and using Ĥ|E⟩ = E|E⟩ gives iħ dc/dt = E c.
- Integrate: c(t) = e(−iEt/ħ), so |Ψ(t)⟩ = e(−iEt/ħ)|E⟩. The modulus of c never changes — only its argument turns.
- Angular rate: ω = E/ħ = (2.0 × 1.602 × 10⁻¹⁹ J)/(1.055 × 10⁻³⁴ J s) = 3.0 × 10¹⁵ rad s⁻¹.
- One full turn takes T = 2π/ω = h/E = (4.136 × 10⁻¹⁵ eV s)/(2.0 eV) = 2.1 × 10⁻¹⁵ s — about 2 fs.
- A global phase is unobservable: every probability |⟨a|Ψ⟩|² is phase-free, so nothing measurable moves. The turning phase only becomes visible against a second energy component — the beat of the next topic.
Answer|Ψ(t)⟩ = e(−iEt/ħ)|E⟩ with ω = 3.0 × 10¹⁵ rad s⁻¹, one turn every 2.1 fs; no observable of the lone eigenstate changes.
MediumIn the basis (|1⟩, |2⟩) of the nitrogen atom sitting above or below the plane of NH₃, the Hamiltonian is Ĥ = E₀𝟙 − Δσ̂ₓ with Δ = 4.9 × 10⁻⁵ eV. The molecule starts in |1⟩. Solve the component equations, find P₂(t), and give the first time the molecule is certain to be found in |2⟩.
- Drop E₀𝟙 first: it multiplies every solution by the global phase e(−iE₀t/ħ) and moves no probability.
- Components from Hₖⱼ = ⟨k|Ĥ|j⟩: the only surviving entries are the off-diagonal −Δ, so iħċ₁ = −Δc₂ and iħċ₂ = −Δc₁.
- With c₁(0) = 1, c₂(0) = 0 the solution is c₁ = cos(Δt/ħ), c₂ = i sin(Δt/ħ). Substitute back to verify, and check the norm: cos² + sin² = 1 at every t, as Ĥ† = Ĥ demands.
- P₂(t) = |c₂|² = sin²(Δt/ħ), first reaching 1 when Δt/ħ = π/2, i.e. t = πħ/(2Δ).
- t = π × 6.582 × 10⁻¹⁶ eV s/(2 × 4.9 × 10⁻⁵ eV) = 2.1 × 10⁻¹¹ s, and the population oscillates at 2Δ/h = 9.8 × 10⁻⁵/4.136 × 10⁻¹⁵ = 23.7 GHz.
AnswerP₂(t) = sin²(Δt/ħ); certainty first arrives at t = πħ/(2Δ) = 21 ps, and the population oscillates at 2Δ/h ≈ 24 GHz — the ammonia maser's microwave line.
HardFor the same ammonia Hamiltonian Ĥ = −Δσ̂ₓ with Δ = 4.9 × 10⁻⁵ eV, use d⟨Â⟩/dt = (i/ħ)⟨[Ĥ, Â]⟩ to find the motion of ⟨σ̂z⟩ and ⟨σ̂y⟩ starting from |1⟩, and check the result against the amplitude solution P₂ = sin²(Δt/ħ).
- Pauli algebra needed: [σ̂ₓ, σ̂z] = −2iσ̂y and [σ̂ₓ, σ̂y] = 2iσ̂z.
- d⟨σ̂z⟩/dt = (i/ħ)(−Δ)⟨[σ̂ₓ, σ̂z]⟩ = (i/ħ)(−Δ)(−2i)⟨σ̂y⟩ = −(2Δ/ħ)⟨σ̂y⟩, and likewise d⟨σ̂y⟩/dt = +(2Δ/ħ)⟨σ̂z⟩.
- These are precession equations: the Bloch vector turns about the x axis at Ω = 2Δ/ħ. With ⟨σ̂z⟩(0) = 1 and ⟨σ̂y⟩(0) = 0 for |1⟩: ⟨σ̂z⟩ = cos(2Δt/ħ), ⟨σ̂y⟩ = sin(2Δt/ħ).
- Consistency: ⟨σ̂z⟩ = P₁ − P₂ = 1 − 2 sin²(Δt/ħ) = cos(2Δt/ħ) — the double angle is the amplitude frequency doubled, because probabilities are squared moduli.
- Ω = 2Δ/ħ = 9.8 × 10⁻⁵ eV/6.582 × 10⁻¹⁶ eV s = 1.5 × 10¹¹ rad s⁻¹; meanwhile [Ĥ, σ̂ₓ] = 0, so ⟨σ̂ₓ⟩ is a constant of the motion and stays at its initial 0.
Answer⟨σ̂z⟩(t) = cos(2Δt/ħ), ⟨σ̂y⟩(t) = sin(2Δt/ħ): precession about x at Ω = 2Δ/ħ = 1.5 × 10¹¹ rad s⁻¹, with ⟨σ̂ₓ⟩ frozen at 0 by [Ĥ, σ̂ₓ] = 0 — squaring the amplitudes reproduces it exactly.