The time-dependent Schrödinger equation
Given V and Ψ now, every later Ψ follows
Every later lesson in this course is a solution of one equation, so this one reads it before anyone solves it: what each term measures, why there is a single time derivative and an explicit i, why solutions can be added, and how far a law fitted to plane waves can be trusted.
Build the model
One time derivative and one i: the present wavefunction fixes the future.
1The big idea
The time-dependent Schrödinger equation is the law of motion for everything in this course: supply the potential V(x, t) and the wavefunction at one instant, and it returns the wavefunction at every later instant. It can be motivated but not derived. Read a free plane wave e(ikx − iωt) through E = ħω and p = ħk, and ∂/∂t brings down −iE/ħ while ∂²/∂x² brings down −p²/ħ², so the classical E = p²/2m + V turns into iħ ∂Ψ/∂t = −(ħ²/2m) ∂²Ψ/∂x² + VΨ. Asserting that the same equation governs every state in every V(x), including potentials that change within a wavelength, is the postulate, and agreement with experiment is its only warrant. Three structural facts then do the work. It is first order in time, so Ψ(x,0) alone fixes the future. It is linear, so solutions combine with constant coefficients, and amplitudes, not probabilities, add. And the i is essential: without it the equation would describe diffusion, or diffusion run backwards, changing the size of every Fourier component and with it the total probability, while with it each component only turns in phase. The model stops where E = p²/2m does, once kinetic energies are no longer small beside mc², and it has no place for spin, for particles created or destroyed, or for what a measurement does to Ψ.
2In one sentence
The time-dependent Schrödinger equation, iħ ∂Ψ/∂t = −(ħ²/2m) ∂²Ψ/∂x² + VΨ, is the postulated law of motion for a wavefunction: given V and Ψ at one instant, it fixes Ψ at every later instant.
Example. For an electron ħ/2m = 5.79 × 10⁻⁵ m² s⁻¹, so a wavefunction curved over a = 0.10 nm changes on the time scale 2ma²/ħ = (1.0 × 10⁻²⁰ m²)/(5.79 × 10⁻⁵ m² s⁻¹) = 1.7 × 10⁻¹⁶ s, under a fifth of a femtosecond.
3Read every term as an energy times Ψ
In one dimension the equation reads iħ ∂Ψ/∂t = −(ħ²/2m) ∂²Ψ/∂x² + V(x, t)Ψ, and its right-hand side is abbreviated ĤΨ, the Hamiltonian acting on Ψ, which the next lesson takes apart. Check the units before trusting it. ħ is in J s, so iħ ∂Ψ/∂t is J s × s⁻¹ = J times the units of Ψ. ħ²/2m is J² s² kg⁻¹ = J m², and ∂²/∂x² supplies m⁻², so the kinetic term is J times Ψ as well; VΨ is J times Ψ by construction. Each term carries exactly one power of Ψ, so the equation cannot fix the size of Ψ: normalisation has to come from the Born rule, which gives Ψ units of m⁻¹⁄² in one dimension. The combination ħ/2m ties space to time. For an electron it is 5.79 × 10⁻⁵ m² s⁻¹, and curvature over 0.10 nm corresponds to ħ²/2ma² = 3.8 eV, the energy scale of chemistry.
4Plane waves suggest the equation; they do not derive it
Take the free plane wave Ψ = A e(ikx − iωt) and read it with Einstein's E = ħω and de Broglie's p = ħk. Differentiating in time brings down −iω, so iħ ∂Ψ/∂t = ħωΨ = EΨ; differentiating twice in space brings down −k², so −(ħ²/2m) ∂²Ψ/∂x² = (p²/2m)Ψ. The classical E = p²/2m has become a differential equation that every such wave satisfies, and so does any superposition of them. Adding VΨ for a constant V is still safe: every frequency shifts by V/ħ, a common phase e(−iVt/ħ) that no probability can detect. What no argument supplies is the case that matters, a V(x) varying over distances comparable to the wavelength, where no single k exists. Schrödinger postulated that the same equation holds there, for every state. Hydrogen's spectrum, measured tunnelling rates and the rest of this course are the evidence; none of it is a derivation.
5First order in time: Ψ(x,0) is the whole initial condition
A stretched string obeys ∂²y/∂t² = v²∂²y/∂x², second order in time, so predicting it takes both the initial shape y(x,0) and the initial velocity ∂y/∂t; Newton's second law likewise needs x and v. The Schrödinger equation has a single ∂/∂t, so knowing Ψ is knowing its rate of change: ∂Ψ/∂t = −(i/ħ)ĤΨ. Step forward by δt, recompute, repeat, and Ψ(x,0) alone fixes Ψ at every later time; numerical solvers do a refined version of exactly this, with δt far below ħ/E. The information a velocity would have carried lives in the phase of Ψ: e(ikx) and e(−ikx) have the same |Ψ|² everywhere, yet one moves right and the other left. The determinism is of Ψ, not of outcomes. The equation predicts the wavefunction exactly; what a position measurement returns is still only distributed as |Ψ|² dx.
6Linear: amplitudes add, and interference follows
If Ψ₁ and Ψ₂ solve the equation for the same V, substitute c₁Ψ₁ + c₂Ψ₂. Every term is linear in Ψ, so the result is c₁ times one true equation plus c₂ times another, and it holds. The coefficients must be constants: a c₂(t) leaves a stray iħ c₂′(t)Ψ₂ on the left. What adds is the amplitude, so the density gains a cross term: |Ψ|² = |c₁Ψ₁|² + |c₂Ψ₂|² + 2Re(c₁*c₂Ψ₁*Ψ₂). Release two electron packets 0.20 nm wide at x = ±1.0 nm. Each spreads, and by 4 fs, now 1.17 nm wide, they overlap. There the cross term repeats every 1.50 nm, and near x = 0 the density swings between twice what adding the two densities would give and almost nothing, with a bright fringe at x = 0 where the two contributions arrive in phase. Adding densities instead gives one smooth hump. The fringes are the signature of an equation linear in Ψ, not in |Ψ|².
7Without the i, probability would not be conserved
Write the free equation as ∂Ψ/∂t = (iħ/2m) ∂²Ψ/∂x² and delete the i: it becomes ∂Ψ/∂t = (ħ/2m) ∂²Ψ/∂x², the diffusion equation, with D = 5.79 × 10⁻⁵ m² s⁻¹ for an electron. Feed it a Fourier component e(ikx) and it decays as e(−Dk²t): fine structure is erased first, ∫|Ψ|² dx falls, and running time backwards blows up. Deleting the i from iħ ∂Ψ/∂t instead gives exactly that backwards equation, with every component growing as e(+Dk²t). With the i the same component becomes e(ikx − iωt) with ω = ħk²/2m; its modulus never changes, only its phase turns, which is why the total probability can stay at 1 (the continuity equation proves it later in this unit). The i also forces Ψ to be complex. Write Ψ = R + iI with R and I real and V real: the equation splits into ħ ∂R/∂t = ĤI and ħ ∂I/∂t = −ĤR. A wavefunction that starts real acquires an imaginary part at once, and R and I drive each other the way position and momentum do in an oscillator.
8Where the equation stops being the right one
The kinetic term is p²/2m, so the equation is non-relativistic by construction. The exact relation is p²/2m = K(1 + K/2mc²), with K the true kinetic energy, so the equation overstates the energy of a given wave by the fraction K/2mc²: 1.0% for a 10 keV electron and 20% for the 200 keV electrons of a transmission microscope, whose wavelengths are always computed relativistically. Feed the plane-wave argument E² = p²c² + m²c⁴ instead and it returns a different equation, the Klein–Gordon equation, second order in time. The Schrödinger equation also has no room for spin, which the Pauli equation adds by hand and the Dirac equation produces; it keeps the particle number fixed, so it cannot describe pair creation above 2mc² = 1.02 MeV; and it treats V as a given external field. Finally, it is silent about measurement: it evolves Ψ smoothly between measurements, and what a measurement does to Ψ is a separate postulate, met in Unit 4.
Key equations
What each relation says, and the units it is written in.
The equation, in one dimension
iħ ∂Ψ/∂t = −(ħ²/2m) ∂²Ψ/∂x² + V(x, t)Ψ
Every term is an energy times Ψ. The right-hand side is written ĤΨ, the Hamiltonian acting on Ψ, unpacked in the next lesson.
ħ = 1.055 × 10⁻³⁴ J s; m in kg; x in m; t in s; V in J; Ψ in m⁻¹⁄²
Free-particle dispersion it demands
ω = ħk²/2m
A plane wave e(ikx − iωt) with V = 0 solves the equation only at this ω, which is E = p²/2m read through E = ħω and p = ħk.
ω in rad s⁻¹, k in m⁻¹; for an electron ħ/2m = 5.79 × 10⁻⁵ m² s⁻¹
Superposition
Ψ = c₁Ψ₁ + c₂Ψ₂
Any such combination is again a solution. Amplitudes add, so |Ψ|² gains the interference term 2Re(c₁*c₂Ψ₁*Ψ₂).
Ψ₁, Ψ₂ solutions for the same V; c₁, c₂ complex constants, fixed in time
One step forward in time
Ψ(x, t + δt) ≈ Ψ(x, t) − (iδt/ħ) ĤΨ(x, t)
First order in time: Ψ at one instant supplies its own rate of change, so no initial velocity is needed or allowed.
δt in s, small against ħ/E: ħ/E = 0.22 fs when E = 3 eV
Shifting the zero of V
V → V + V₀ gives Ψ → Ψ e(−iV₀t/ħ)
A matter wave's absolute frequency E/ħ depends on where V = 0 is put, so only energy differences are observable.
V₀ a constant energy in J; |Ψ|² and every probability are unchanged
The non-relativistic price
p²/2m = K(1 + K/2mc²)
Exact: the equation's kinetic energy runs high by K/2mc², 1.0% for a 10 keV electron and 20% at 200 keV.
K the true, relativistic kinetic energy in J; mc² = 511 keV for an electron
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Hold d at 1.0 nm and step t up from 0.5 fs: the gap between the packets stays nearly empty until each has spread to about half of d, then fringes grow that the dashed curve never shows. Raise d and the fringes crowd closer. The picture depends only on t/m, so a proton would take 1836 times as long.
Live results
- WIDTH OF EACH PACKET σ(t)
- 1.17 nm
- CROSS-TERM PERIOD AT x = 0
- 1.50 nm
- DE BROGLIE ESTIMATE ht/2md
- 1.45 nm
- DENSITY AT x = 0
- 0.473 nm⁻¹
Catch the common trap
The mistake this lesson is built to prevent.
Students who met complex exponentials in AC circuits or classical waves bring a habit with them: write e(ikx − iωt) for convenience, take the real part at the end, and treat Re Ψ as the physical wave and the i as bookkeeping. Here that fails twice. First, the equation contains i explicitly, so its real and imaginary parts are coupled, ħ ∂R/∂t = ĤI and ħ ∂I/∂t = −ĤR, and neither part is a solution on its own. Second, the Born rule uses both parts. For a plane wave |Ψ|² = |A|² is the same everywhere, as a particle of definite momentum must be, whereas (Re Ψ)² would be a train of moving stripes with zeros where the particle could never be found. The complex Ψ is the physics, not a convenience.
Test it: what changes?
- Each packet spreads wider
- Packets overlap and fringes appear near x = 0
- de Broglie estimate ht/2md grows with t
- Fringe period approaches ht/2md from above
- Fringes crowd closer together
- Width of each packet unchanged
- de Broglie estimate ht/2md falls as 1/d
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
In a region where V = 0, an electron is offered two candidate wavefunctions with k = 5.0 × 10⁹ m⁻¹: the real wave A cos(kx − ωt) and the complex wave A e(ikx − iωt). Substitute each into the time-dependent Schrödinger equation. Which can satisfy it, and at what ω?
View solution →
- Real candidate: iħ ∂Ψ/∂t = iħωA sin(kx − ωt) is purely imaginary, while −(ħ²/2m) ∂²Ψ/∂x² = (ħ²k²/2m)A cos(kx − ωt) is purely real. They agree at every x and t only if A = 0, so no value of ω rescues it.
- Complex candidate: ∂/∂t brings down −iω, so the left side is iħ(−iω)Ψ = ħωΨ; ∂²/∂x² brings down −k², so the right side is (ħ²k²/2m)Ψ. Both are multiples of the same Ψ, and they match when ω = ħk²/2m.
- Numbers: ω = (ħ/2m)k² = (5.79 × 10⁻⁵ m² s⁻¹)(2.5 × 10¹⁹ m⁻²) = 1.45 × 10¹⁵ rad s⁻¹. Units: m² s⁻¹ × m⁻² = s⁻¹.
- Check against classical mechanics: ħω = 1.526 × 10⁻¹⁹ J = 0.953 eV, and with p = ħk = 5.27 × 10⁻²⁵ kg m s⁻¹, p²/2m = 1.526 × 10⁻¹⁹ J too. The equation has rebuilt the classical kinetic energy.
Answer. Only the complex wave works, at ω = 1.45 × 10¹⁵ rad s⁻¹, so E = ħω = 0.953 eV = p²/2m. The real cosine fails for every ω.
MediumExample 2
At t = 0 an electron in a region where V = 0 has the real wavefunction Ψ(x,0) = A e(−x²/2a²) with a = 0.25 nm. Using only the equation, find ∂Ψ/∂t at x = 0, estimate when Im Ψ(0, t) reaches 5% of A, and say how fast |Ψ(0, t)|² is changing at t = 0.
View solution →
- The equation is first order in time, so Ψ(x,0) alone gives the rate: ∂Ψ/∂t = (iħ/2m) ∂²Ψ/∂x² when V = 0. No initial velocity is needed, or even allowed.
- Differentiate twice: ∂²Ψ/∂x² = A(x²/a⁴ − 1/a²) e(−x²/2a²), which at x = 0 is −A/a², negative because the peak curves downward.
- So ∂Ψ/∂t(0,0) = −iħA/(2ma²) = −iγA, with γ = (ħ/2m)/a² = (5.79 × 10⁻⁵ m² s⁻¹)/(6.25 × 10⁻²⁰ m²) = 9.26 × 10¹⁴ s⁻¹ (m² s⁻¹ ÷ m² = s⁻¹). The rate is purely imaginary, so a real Ψ cannot stay real.
- To first order Im Ψ(0, t) = −γAt, which reaches 0.05A in magnitude at t = 0.05/γ = 5.4 × 10⁻¹⁷ s, about 0.054 fs.
- The density's rate is ∂|Ψ|²/∂t = 2Re(Ψ* ∂Ψ/∂t) = 2Re(A × (−iγA)) = 0 at t = 0, so |Ψ(0, t)|² starts changing only at second order. The exact solution, worked in Unit 5, agrees: at t = 0.054 fs the peak density has fallen by just 0.50%.
Answer. ∂Ψ/∂t(0,0) = −i(9.26 × 10¹⁴ s⁻¹)A; Im Ψ(0, t) reaches 5% of A after about 0.054 fs, while ∂|Ψ(0, t)|²/∂t = 0 at t = 0.
HardExample 3
Two electron packets, each the real Gaussian (2πσ₀²)⁻¹⁄⁴ exp[−(x ∓ d)²/4σ₀²] with σ₀ = 0.20 nm, start at rest at x = ±d with d = 1.2 nm and V = 0. Alone, each becomes a Gaussian of width σ(t) = σ₀√(1 + τ²), τ = ħt/(2mσ₀²), with phase τ(x ∓ d)²/(4σ(t)²) plus a phase common to both. Find the period of the interference near x = 0 at t = 6.0 fs and compare it with ht/(2md).
View solution →
- Linearity: each packet solves the free equation, so Ψ = C(Ψ₊ + Ψ₋) does too, with C fixed once at t = 0. The overlap integral is e(−d²/2σ₀²) = e(−18) = 1.5 × 10⁻⁸, so C = 1/√2 to that accuracy, and the norm then stays 1.
- Amplitudes add, so |Ψ|² = ½(|Ψ₊|² + |Ψ₋|²) + Re(Ψ₊*Ψ₋). The common phase cancels in the cross term, which oscillates as cos Δφ with Δφ = τ[(x + d)² − (x − d)²]/(4σ(t)²) = τxd/σ(t)².
- Numbers: τ = (ħ/2m)t/σ₀² = (5.79 × 10⁻⁵ m² s⁻¹)(6.0 × 10⁻¹⁵ s)/(4.0 × 10⁻²⁰ m²) = 8.68, so σ(t) = 0.20 nm × √76.4 = 1.75 nm. That exceeds d, so the packets overlap strongly and fringes are visible.
- Δφ grows by 2π over Δx = 2πσ(t)²/(τd) = 2π(1.748 nm)²/(8.68 × 1.2 nm) = 1.84 nm. The falling envelope pulls the visible maxima of |Ψ|² slightly inward, to ±1.76 nm, but the cross term's period is 1.84 nm.
- Classical correspondence: an electron reaching x from ∓d in time t moves at (x ± d)/t, so its de Broglie wavenumbers are m(x ± d)/(ħt). They differ by 2md/(ħt), giving fringes ht/(2md) = (6.63 × 10⁻³⁴ J s × 6.0 × 10⁻¹⁵ s)/(2 × 9.11 × 10⁻³¹ kg × 1.2 × 10⁻⁹ m) = 1.82 nm. The exact period is larger by (1 + τ²)/τ² = 1.013.
Answer. The interference term repeats every 1.84 nm, 1.3% above the de Broglie estimate ht/(2md) = 1.82 nm; the gap is the factor (1 + τ²)/τ², which tends to 1 as t grows.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
Ψ now fixes Ψ later; amplitudes, not probabilities, add.
Supply V(x, t) and Ψ(x,0) alone; no initial velocity is needed. Combine solutions only with constant complex coefficients, so amplitudes add and interference appears. Trust the equation while kinetic energies are small beside mc², with no spin or particle creation, and not for what a measurement does to Ψ.