
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.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.
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.
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.Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
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.
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 °
This experiment is kept only while this page is open. It does not change saved course progress.
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.
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.
- Differentiate the trial wave: ∂ψ/∂t = −iωψ and ∂²ψ/∂x² = −k²ψ.
- Substitute: iℏ(−iω)ψ = −(ℏ²/2m)(−k²)ψ. Since i(−i) = 1, this is ℏωψ = (ℏ²k²/2m)ψ, and ψ is nowhere zero, so ω = ℏk²/2m. In energy terms E = ℏω = ℏ²k²/2m = p²/2m with p = ℏk — the wave solves the equation exactly when it carries the classical kinetic energy.
- Numbers: ω = (1.055 × 10⁻³⁴ × 1.00 × 10²⁰) ÷ (2 × 9.109 × 10⁻³¹) = 5.79 × 10¹⁵ rad s⁻¹.
- Energy: E = ℏω = 1.055 × 10⁻³⁴ × 5.79 × 10¹⁵ = 6.10 × 10⁻¹⁹ J = 3.81 eV.
- Velocities: vₚ = ω/k = 5.79 × 10¹⁵ ÷ 1.00 × 10¹⁰ = 5.79 × 10⁵ m s⁻¹, while vg = dω/dk = ℏk/m = 1.16 × 10⁶ m s⁻¹, exactly twice vₚ. The particle speed is the group velocity: p/m = ℏk/m returns the same 1.16 × 10⁶ m s⁻¹, and v/c = 3.9 × 10⁻³ confirms the non-relativistic form is safe here.
Answer. It solves the equation only for ω = ℏk²/2m. At k = 1.00 × 10¹⁰ m⁻¹: ω = 5.79 × 10¹⁵ rad s⁻¹, E = 3.81 eV, vₚ = 5.79 × 10⁵ m s⁻¹ and vg = 1.16 × 10⁶ m s⁻¹ — the crests move at half the particle's speed.
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.
- Put ψ = A exp(i(kx − ωt)) in. The left side gives iℏ(−iω)ψ = ℏωψ; the right gives −iℏu(ik)ψ = ℏukψ. So ℏω = ℏuk and the dispersion is linear, ω = uk — the dispersion of light in vacuum, not of a massive particle.
- Fix u by the one relation the law is supposed to reproduce. Setting ℏuk₀ = ℏ²k₀²/2m gives u = ℏk₀/2m = 5.79 × 10⁵ m s⁻¹, and at k₀ the energy does come out right: E = ℏuk₀ = 3.81 eV.
- It is already wrong at that wavenumber. Here vg = dω/dk = u = 5.79 × 10⁵ m s⁻¹, while the particle the law is meant to describe moves at p/m = ℏk₀/m = 1.16 × 10⁶ m s⁻¹ — the packet crawls at half the speed of its own particle. Worse, vg = u whatever k is, so a fast electron and a slow one would travel at the same speed, and vₚ = ω/k = u makes the crests keep pace with the packet instead of running through it at half speed.
- Move off k₀ and the energy goes too. At k = 2.00 × 10¹⁰ m⁻¹ the law gives E = ℏuk = 7.62 eV where E = ℏ²k²/2m demands 15.24 eV, a factor of two out. One constant u cannot track a k², so the law can be tuned to agree at a single wavenumber and nowhere else.
- The symmetry argument closes it. ω = uk is odd in k, so reversing the direction of travel sends E = ℏω to −E and a left-moving free electron would carry negative energy; the law has also singled out one direction along an empty line where nothing distinguishes one. The classical E = p²/2m is even in p and asks for ω ∝ k², and a k² can only come from two space derivatives.
Answer. One space derivative gives ω = uk. Matching the energy at k₀ = 1.00 × 10¹⁰ m⁻¹ needs u = ℏk₀/2m = 5.79 × 10⁵ m s⁻¹, but the group velocity is then 5.79 × 10⁵ m s⁻¹ against the classical 1.16 × 10⁶ m s⁻¹, the energy at k = 2.00 × 10¹⁰ m⁻¹ comes out 7.62 eV instead of 15.24 eV, and ℏω goes negative for a left-moving electron. The space derivative has to be second order.
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.
- Substituting into E²ψ = (p²c² + m²c⁴)ψ gives (iℏ ∂/∂t)²ψ = c²(−iℏ ∂/∂x)²ψ + m²c⁴ψ, that is −ℏ² ∂²ψ/∂t² = −ℏ²c² ∂²ψ/∂x² + m²c⁴ψ — the Klein–Gordon equation.
- First structural break: it is second order in time, so ψ(x, 0) no longer determines the future and ∂ψ/∂t at t = 0 must be supplied as well. Its plane waves also come in two branches, ℏω = ±√(ℏ²c²k² + m²c⁴).
- Second break: the density this equation conserves is not |ψ|² and is not positive definite, so it cannot be read as a probability. Repairing that means abandoning the single-particle reading altogether, and even then the equation describes spin 0, while an electron needs Dirac's.
- Now the numerical domain. Expanding E = √(p²c² + m²c⁴) = mc² + p²/2m − p⁴/8m³c² + …, the first dropped term is a fraction p²/4m²c² = K/2mc² of the kinetic energy, with mc² = 511 keV.
- For wavelengths, λ = h/p with the exact p, so λₙᵣ/λᵣₑₗ = √(1 + K/2mc²). Setting that ratio to 1.01 gives K/2mc² = 0.0201, so K = 0.0201 × 1.022 MeV = 20.5 keV.
- Check it against a standard electron-microscope setting: at K = 100 keV, √(1 + 100/1022) = 1.048, so the Schrödinger wavelength of 3.88 pm overstates the true 3.70 pm by 4.8%.
Answer. The Klein–Gordon equation, −ℏ² ∂²ψ/∂t² = −ℏ²c² ∂²ψ/∂x² + m²c⁴ψ: second order in time, two frequency branches, and no positive conserved density. The 1% wavelength error arrives near 20 keV and reaches 4.8% at 100 keV.
Quick check
Select an answer and explain it before checking.
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.