
Physics V course sequence · Interactive lesson
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.
UPV-04.02 · Lesson 2 of 9 in this unit. Position is not a completion score.Build the model
One first-order, linear, norm-preserving law generates all closed-system quantum motion.
The key idea
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.
The complete physical model and assumptions
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.
1First 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: for a real Hamiltonian such as p̂²/2m + V(x̂), complex-conjugate the position-space 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.
2The 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 scale as the real exponential e(−Eₙt/ħ). The norm is no longer conserved and, after a long time, only the lowest-energy component survives, because every excited fraction dies relative to it. 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 with a nonzero ground-state overlap 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.
3Hermiticity 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.
4Hit 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.
5Linearity: 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 together with unitarity 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.
6Read 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 master equation, of Lindblad form in the Markovian limit, 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.
Key equations
Read the units, assumptions and limits before substituting.
The law of motion
iħ d|Ψ(t)⟩/dt = Ĥ|Ψ(t)⟩
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 ketPosition representation
iħ ∂Ψ/∂t = −(ħ²/2m) ∂²Ψ/∂x² + V(x)Ψ
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|Components in any basis
iħ ċₖ = Σⱼ Hₖⱼ cⱼ, cₖ = ⟨k|Ψ⟩
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| = 𝟙Norm conservation
d⟨Ψ|Ψ⟩/dt = (i/ħ)⟨Ψ|(Ĥ† − Ĥ)|Ψ⟩ = 0
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 partsEnergy offset = global phase
Ĥ → Ĥ + E₀𝟙 ⇒ |Ψ(t)⟩ → e(−iE₀t/ħ)|Ψ(t)⟩
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 movesMotion of an expectation value
d⟨Â⟩/dt = (i/ħ)⟨[Ĥ, Â]⟩
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 ÂInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. 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.
Live results
- P₁ 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/ε
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. Coupling Δ: 0.8 ε; Anti-Hermitian leak Γ: 0.00 ε; Marker time t: 3.00 ħ/ε
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
Because everyone calls Ψ a wave, students carry over the classical wave equation's habits: two initial conditions, real solutions, and a pitch ω proportional to k. All three fail. The Schrödinger equation is first order in time, so |Ψ(0)⟩ alone fixes the future — supplying ∂Ψ/∂t as extra data over-determines the problem. Its free-particle dispersion is ω = ħk²/2m, quadratic, which is why packets spread as they travel. And the i makes complexity compulsory, not cosmetic: start Ψ real and iħ ∂Ψ/∂t = ĤΨ forces ∂Ψ/∂t to be imaginary, so the state leaves the real line immediately — Ψ = cos(kx − ωt) on its own solves no Schrödinger equation. Structurally this is a diffusion equation with an imaginary diffusion constant, and that i is exactly what trades decay for rotation.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
An 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.
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.
- 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 taken up with stationary states two topics on.
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.
MediumExample 2
In 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⟩.
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.
- 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.
Answer. P₂(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.
HardExample 3
For 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).
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.
- 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.
Quick check
Select an answer and explain it before checking.
Review & continue
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.
- 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.