Skip to main content

Quantum core · institution-dependent · Interactive lesson

Building the Schrödinger equation

Everything you calculate for the rest of this course starts from one equation nobody can derive. What you can do, and what this topic does, is show that the wave relations plus the energy of a slow particle leave almost no other form standing — then name precisely what the surviving form assumes.

UPIV-06.01 · Lesson 1 of 9 in this unit. Position is not a completion score.
Starting model: Building the Schrödinger equationRe ψ for one mode evolved by the trial law ∂ψ/∂t = C ∂²ψ/∂x², with C = (ℏ/2m)exp(iθ). Solid curve: Re ψ at t = 0.20 fs for k = 1.0 × 10¹⁰ m⁻¹, with the dot on its leading crest. Dashed pair: the envelope |ψ|, now at 1.000 of the starting height marked by the faint pair and by the bar on the right.∂ψ/∂t = C ∂²ψ/∂x²C = (ℏ/2m) exp(iθ), θ = 90°turns 5.79, decays 0.00 (10¹⁵ s⁻¹)k = 1.0 × 10¹⁰ m⁻¹, t = 0.20 fsRe ψ solid, envelope dashed|ψ|only θ = 90° holds the envelope still
Start with a picture. Change the model in the lab below.

Quantum core · institution-dependent. Departments differ in how much quantum mechanics they include in Physics IV or teach in a dedicated course. Follow your university’s required mathematical depth and assessment; these numbered units are not automatically optional.

1

Build the model

You cannot derive it. You can show that almost nothing else would do.

The key idea

The time-dependent Schrödinger equation is the postulated law of motion for a quantum state: it fixes the rate of change of ψ in time, through a factor of i, from the curvature of ψ in space plus the potential energy at that point.

Example. For a free electron it collapses to ℏω = ℏ²k²/2m, so a 1.00 nm de Broglie wave (k = 6.28 × 10⁹ m⁻¹) must oscillate at ω = 2.29 × 10¹⁵ rad s⁻¹ — an energy of 1.50 eV, which is what a 1.50 V electron gun delivers.

The complete physical model and assumptions

Two experimental relations arrive from the matter-wave unit: E = ℏω and p = ℏk. Take them seriously and ask what differential equation a wave exp(i(kx − ωt)) would have to obey if it is to carry the classical energy of a slow free particle, E = p²/2m. In wave variables that reads ℏω = ℏ²k²/2m: one power of ω against two of k. A single time derivative brings down −iω and a double space derivative brings down −k², so the equation must be first order in time and second in position, and the surviving constant must be purely imaginary — a real one turns the same expression into a diffusion equation whose modes decay instead of oscillating. Demand also that superpositions of solutions be solutions, so interference survives evolution, and you are left with iℏ ∂ψ/∂t = −(ℏ²/2m) ∂²ψ/∂x² + Vψ. None of that is a derivation. Plane waves are not normalisable, and the substitution rule the argument leans on is the assumption doing the work rather than a consequence of anything earlier; the equation is a postulate warranted by the hydrogen spectrum, molecular vibrational ladders and tunnelling currents, and by nothing else. It also arrives with its price list visible: the classical E = p²/2m makes it non-relativistic, the single complex component makes it spinless, V is an external field rather than a partner in the dynamics, and V must be real or probability leaks away.

1Read the equation off a single plane wave

Start with the trial wave ψ = A exp(i(kx − ωt)) and the two relations E = ℏω, p = ℏk. Differentiating once in time brings down a factor −iω; differentiating twice in position brings down −k². The classical energy of a slow free particle, E = p²/2m, becomes ℏω = ℏ²k²/2m in wave variables — one power of ω set against two powers of k. That single count fixes the shape of the equation: first order in time, second in position. Now assemble it. Since ∂ψ/∂t = −iωψ, multiplying by iℏ gives ℏωψ; since ∂²ψ/∂x² = −k²ψ, multiplying by −ℏ²/2m gives (ℏ²k²/2m)ψ. Setting those two equal is exactly ℏω = ℏ²k²/2m, for every k at once. Add potential energy and the classical relation becomes E = p²/2m + V, so a term Vψ joins the right-hand side. Put a number on it: an electron at λ = 0.10 nm has k = 6.28 × 10¹⁰ m⁻¹, so ω = ℏk²/2m = 2.29 × 10¹⁷ rad s⁻¹ and E = 150 eV — the working energy of a low-energy electron diffraction beam, recovered from the wave relations alone.

2The i cannot be scaled away

Write the free law in its most general first-order form, ∂ψ/∂t = C ∂²ψ/∂x², and put a mode ψ = exp(ikx) f(t) into it. Then f′ = −Ck²f, so f = exp(−Ck²t). Split C into modulus and phase, C = |C| exp(iθ): the amplitude falls as exp(−|C|k²t cos θ) while the phase turns at |C|k² sin θ. Real C, meaning θ = 0, is the diffusion equation. Its modes decay and never oscillate, so there is no frequency to identify with E/ℏ, no state of definite energy, and no conserved total probability. Only θ = 90°, a purely imaginary C, gives pure rotation: |ψ| is untouched and ω = |C|k², which matches ℏω = ℏ²k²/2m when |C| = ℏ/2m = 5.79 × 10⁻⁵ m² s⁻¹. The two behaviours are not subtly different. For an electron with k = 1.0 × 10¹⁰ m⁻¹, a real coefficient would drop the mode to 1/e of its height in 0.17 fs, while the imaginary one holds every bit of it and turns the phase through a full cycle every 1.09 fs.

3First order in time is a claim about what a state is

Because only one time derivative appears, ψ(x, 0) alone determines ψ at every later instant. Compare what classical physics needs: Newton's law wants a position and a velocity, and the classical wave equation wants a displacement and its time derivative. Quantum mechanics gets away with one function because that function is complex and so is already carrying two real fields — |ψ| says where the particle is likely to be found and arg ψ says where it is going, since the local momentum is ℏ ∂(arg ψ)/∂x. That is what the complexity buys, and it is why 'the state' and 'the wave function at one instant' are the same object in this course. Go second order in time instead, as you must if you substitute into E² = p²c² + m²c⁴, and the bargain collapses: the Klein–Gordon equation needs ∂ψ/∂t at t = 0 as extra initial data, its plane waves come with ℏω of both signs, and the density it conserves is not |ψ|² and is not positive, so it cannot be read as a probability at all.

4Linearity is doing work, and it is not free

Every term contains exactly one power of ψ, so any combination aψ₁ + bψ₂ of solutions is a solution and the coefficients set at t = 0 are carried along untouched. Three things follow. An arbitrary initial state can be expanded on a convenient basis and evolved term by term, which is the method of the next four units. A two-path cross term survives the evolution, so interference is a consequence of the dynamics rather than an extra rule. And the map from ψ(0) to ψ(t) is a linear operator that, whenever V is real, also preserves the norm — the evolution is unitary. The cost is worth stating plainly: a linear, deterministic, norm-preserving law can never turn a spread-out superposition into one definite detector click, which is why the measurement postulate has to be a separate axiom rather than a theorem. Nor is a nonlinear patch free. A term like |ψ|²ψ makes the physics depend on how the state was normalised; such equations earn their place as mean-field approximations to many-body systems, never as the fundamental law.

5Everything the equation quietly assumes

Four assumptions sit on the face of it. It is non-relativistic, because E = p²/2m went in, and the mismatched derivative orders — one in time, two in space — mean it cannot be Lorentz covariant; a 1.5 eV electron has v/c = 2.4 × 10⁻³ and is safe, a 100 keV electron in a microscope has v/c = 0.55 and is not. It is spinless, because ψ is one complex number at each point with no internal index; an electron needs the two components of the Pauli equation or the four of Dirac's. It describes one particle in one dimension: three dimensions replace ∂²/∂x² with ∇², while N particles need a single wave function on 3N coordinates rather than N waves in space, which is where entanglement will come from. And V(x, t) is an external classical field, prescribed rather than solved for, with no back-reaction — and it must be real, since a complex V deliberately breaks norm conservation and is used precisely as a model of absorption.

6Why this is a postulate, and what pays for it

The plane-wave construction above is a plausibility argument, not a proof. It leans on states that are not normalisable, so they are not states of the theory at all, and the substitution rule E → iℏ ∂/∂t is the physical content rather than a step in the algebra. The history was messier still: Schrödinger reached the equation through a Hamilton–Jacobi and optical analogy, tried the relativistic version first, and set it aside because it gave the wrong fine structure for hydrogen — the missing ingredient being spin, which nothing in the argument had suggested. What warrants the equation is the ledger of what it then predicted correctly: the hydrogen spectrum with the measured Rydberg constant, the evenly spaced vibrational ladder of molecular spectra, tunnelling currents falling exponentially with barrier width closely enough that a scanning tunnelling microscope resolves single atomic steps, and the band structure of solids. Say in an exam that it is assumed and tested, never that it is proved.

2

Key equations

Read the units, assumptions and limits before substituting.

The equation itself

iℏ ∂ψ/∂t = −(ℏ²/2m) ∂²ψ/∂x² + V(x, t) ψ

Hand it ψ at one instant and a potential, and it returns ψ at every other instant. The entire dynamical content of the theory is this one line.

ψ(x, t) is complex, of units m⁻¹⁄² in one dimension; V in J, m in kg, ℏ = 1.055 × 10⁻³⁴ J s.

Wave relations read as substitution rules

E → iℏ ∂/∂t, p → −iℏ ∂/∂x

Feed them a classical energy relation and a wave equation drops out. E = p²/2m + V gives the line above — though the rule is itself a postulate, and linearity has to be demanded separately.

Both sides act on ψ. E = ℏω in J and p = ℏk in kg m s⁻¹, with ω in rad s⁻¹ and k in m⁻¹.

Free-particle dispersion

ω(k) = ℏk²/2m, so E = ℏ²k²/2m = p²/2m

The k² is the fingerprint of the second space derivative, and the check that the equation returns the kinetic energy you started from.

ω in rad s⁻¹, k in m⁻¹. Quadratic in k, unlike the linear ω = ck of a light wave in vacuum.

Group velocity carries the particle

vg = dω/dk = ℏk/m = p/m, vₚ = ω/k = vg/2

This fixes the order of the space derivative: only ω ∝ k² makes a packet travel at the classical p/m while its crests slide through at half that.

Both in m s⁻¹. The factor of two between them is a prediction of the equation, never an input to it.

Linearity, and what it protects

if ψ₁ and ψ₂ solve it, so does aψ₁ + bψ₂

Superposition survives evolution, so interference is built into the dynamics and any state may be expanded on a basis and evolved term by term.

a and b are complex constants, fixed once by the initial state and never touched again by the evolution.

The non-relativistic ceiling

E = √(p²c² + m²c⁴) ≈ mc² + p²/2m − p⁴/8m³c²

Names the domain: 0.01% at 100 eV but 9.8% at 100 keV, where this equation's de Broglie wavelength is already 4.8% too long.

The dropped term is a fraction K/2mc² of the kinetic energy; mc² = 511 keV for an electron.
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

1.0 10¹⁰ m⁻¹
0.4 to 1.6 10¹⁰ m⁻¹; steps of 0.1.
0.20 fs
0 to 0.5 fs; steps of 0.02.
90 °
0 to 90 °; steps of 5.

Try this. Leave θ at 90°, the Schrödinger case, and drag t: the crest slides right while the envelope never moves. Now drag θ to 0 and the same law becomes diffusion — at k = 1.0 the mode is down to 0.055 of its height by 0.50 fs, with no frequency left to call an energy.

Live results

AMPLITUDE |ψ|, 1 AT t = 0
1.000
OSCILLATION RATE ω
5.79 10¹⁵ rad s⁻¹
DECAY RATE
0.00 10¹⁵ s⁻¹
ℏω, ENERGY ONLY AT θ = 90°
3.81 eV

Results follow this lesson’s model and assumptions. Displayed values are rounded.

Interactive Physics IV modelRe ψ for one mode evolved by the trial law ∂ψ/∂t = C ∂²ψ/∂x², with C = (ℏ/2m)exp(iθ). Solid curve: Re ψ at t = 0.20 fs for k = 1.0 × 10¹⁰ m⁻¹, with the dot on its leading crest. Dashed pair: the envelope |ψ|, now at 1.000 of the starting height marked by the faint pair and by the bar on the right.∂ψ/∂t = C ∂²ψ/∂x²C = (ℏ/2m) exp(iθ), θ = 90°turns 5.79, decays 0.00 (10¹⁵ s⁻¹)k = 1.0 × 10¹⁰ m⁻¹, t = 0.20 fsRe ψ solid, envelope dashed|ψ|only θ = 90° holds the envelope still
Re ψ for one mode evolved by the trial law ∂ψ/∂t = C ∂²ψ/∂x², with C = (ℏ/2m)exp(iθ). Solid curve: Re ψ at t = 0.20 fs for k = 1.0 × 10¹⁰ m⁻¹, with the dot on its leading crest. Dashed pair: the envelope |ψ|, now at 1.000 of the starting height marked by the faint pair and by the bar on the right.

Predict → change one variable → check

Plan a change before running it. Every other input stays at the saved baseline.

Saved baseline. Wavenumber k: 1.0 10¹⁰ m⁻¹; Elapsed time t: 0.20 fs; Phase θ of the coefficient C: 90 °

Baseline 1.0; choose 0.4 to 1.6, in steps of 0.1.
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

The i is not bookkeeping to be dropped at the end. Students arriving from circuits and classical waves use complex exponentials as a convenience and take the real part when they finish; here that is fatal, because the i sits inside the equation. Write ψ = u + iv and separate the parts: ℏ ∂u/∂t = −(ℏ²/2m) ∂²v/∂x² + Vv, and ℏ ∂v/∂t = (ℏ²/2m) ∂²u/∂x² − Vu. Each real field is driven by the curvature of the other, so u alone solves nothing — Re ψ is not a solution of the equation Re ψ came from. And there is nothing left to take the real part of at the end: the prediction is |ψ|² = u² + v², already real, and constant in time for a plane wave, whereas Re ψ swings between ±A. A real formulation does exist, but it needs both fields and this same coupling — the complex equation wearing a disguise.

5

Practice & worked examples

3 authored examples. Make an attempt, then reveal the reasoning one step at a time.

EasyExample 1

Show that ψ(x, t) = A exp(i(kx − ωt)) satisfies the free Schrödinger equation iℏ ∂ψ/∂t = −(ℏ²/2m) ∂²ψ/∂x² only when ω = ℏk²/2m, then evaluate ω, the energy in eV, the phase velocity and the group velocity for an electron with k = 1.00 × 10¹⁰ m⁻¹.

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

Feed the same substitution rule a different classical guess: E = up, the massless relation with c replaced by a fixed speed u. That gives a law with one space derivative, iℏ ∂ψ/∂t = −iℏu ∂ψ/∂x. Find its dispersion, choose u so the law returns the correct energy for an electron at k₀ = 1.00 × 10¹⁰ m⁻¹, then show it already fails at k₀ and fails worse at k = 2.00 × 10¹⁰ m⁻¹, and give the symmetry reason no choice of u can rescue 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.

HardExample 3

Apply the same substitution rule, E → iℏ ∂/∂t and p → −iℏ ∂/∂x, to the relativistic relation E² = p²c² + m²c⁴. Write the equation it produces, give two structural reasons it is a different theory rather than a corrected Schrödinger equation, and find the electron kinetic energy above which the Schrödinger form's de Broglie wavelength is wrong by more than 1%.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 6 steps revealed.

6

Quick check

Select an answer and explain it before checking.

A student proposes the free-particle law ∂ψ/∂t = (ℏ/2m) ∂²ψ/∂x², identical in structure to the Schrödinger equation except that the factor of i has been dropped. Substituting a mode ψ = exp(ikx) f(t), what actually goes wrong?
7

Review & continue

The time-dependent Schrödinger equation is the postulated law of motion for a quantum state: it fixes the rate of change of ψ in time, through a factor of i, from the curvature of ψ in space plus the potential energy at that point.

  • 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.