University Physics IV · The Schrödinger Equation · 6.1
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.
Build the model
Connect the measurement to the mechanism.
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 physical content rather than an input; 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.
- Simple definition
- 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.
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.
Feed them a classical energy relation and a wave equation drops out. E = p²/2m + V gives the line above, and nothing further was assumed.
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.
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.
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.
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.
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.
Read 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.
The 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.
First 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.
Linearity 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.
Everything 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.
Why 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.
Change one variable at a time
Make the relationship visible.
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.
AMPLITUDE |ψ|, 1 AT t = 01.000
OSCILLATION RATE ω5.79 10¹⁵ rad s⁻¹
DECAY RATE0.00 10¹⁵ s⁻¹
ENERGY ℏω3.81 eV
Live interpretationAMPLITUDE |ψ|, 1 AT t = 0: 1.000. OSCILLATION RATE ω: 5.79 10¹⁵ rad s⁻¹. DECAY RATE: 0.00 10¹⁵ s⁻¹. ENERGY ℏω: 3.81 eV
Catch the common trap
Explain before calculating.
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?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyShow 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.
- 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.
AnswerIt 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.
MediumA free-particle law is proposed in the general form α ∂ψ/∂t = β ∂²ψ/∂x², with α and β complex constants. Require that every plane wave ψ = exp(i(kx − ωt)) obey ℏω = ℏ²k²/2m. Find β/α, show that it cannot be real, and evaluate the two rates a real and an imaginary ratio would predict for an electron with k = 1.0 × 10¹⁰ m⁻¹.
- Differentiate: ∂ψ/∂t = −iωψ and ∂²ψ/∂x² = −k²ψ, so the proposed law demands α(−iω) = β(−k²) for every k, giving ω = −i(β/α)k².
- Match the required dispersion ω = (ℏ/2m)k²: −i(β/α) = ℏ/2m, and since 1/(−i) = i this gives β/α = iℏ/2m. The ratio is purely imaginary, and rescaling α and β by a common complex factor leaves it alone — the textbook choice α = iℏ, β = −ℏ²/2m is one member of that family.
- Suppose instead the ratio were real, β/α = D. Then ω = −iDk² is purely imaginary, and ψ = exp(ikx) exp(−Dk²t) decays with no oscillation whatever. That is the diffusion equation, not a wave equation.
- Numbers for an electron: D = ℏ/2m = 5.79 × 10⁻⁵ m² s⁻¹ and k² = 1.0 × 10²⁰ m⁻², so Dk² = 5.79 × 10¹⁵ s⁻¹. A real ratio drops the mode to 1/e in 1/(Dk²) = 1.73 × 10⁻¹⁶ s = 0.17 fs.
- An imaginary ratio of the same magnitude instead holds the amplitude fixed and turns the phase through 2π every 2π/(Dk²) = 1.09 × 10⁻¹⁵ s = 1.09 fs. The size of the coefficient is set by the electron mass; only its phase decides between decay and rotation, and the existence of states of definite energy chooses rotation.
Answerβ/α = iℏ/2m, purely imaginary. A real ratio of the same size would kill an electron mode at k = 1.0 × 10¹⁰ m⁻¹ in 0.17 fs; the imaginary one rotates its phase once every 1.09 fs at fixed amplitude.
HardApply 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%.
- 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%.
AnswerThe 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.