Derivation

The Time-Dependent Schrödinger Equation

D-137 Home PU-202 Threads matter · waves · energy Depends on de-broglie-relation, planck-einstein-relation

Statement

For a single non-relativistic particle of mass m moving in a real scalar potential V(r,t), the complex wavefunction Ψ(r,t) obeys the linear first-order-in-time equation iℏ ∂_tΨ = −(ℏ²/2m)∇²Ψ + VΨ. This is the unique wave equation that (i) is linear in Ψ, (ii) is first order in time, and (iii) forces every free plane wave Ψ ∝ e^{i(k·r−ωt)} to satisfy the non-relativistic dispersion ℏω = ℏ²k²/2m built from the de Broglie relation p = ℏk and the Planck–Einstein relation E = ℏω.

Why it matters

The time-dependent Schrödinger equation (TDSE) is the equation of motion of non-relativistic quantum mechanics: given the state now, it fixes the state at every later time. Everything else — stationary states, tunnelling, spectra, scattering, the entire chemistry of atoms and molecules — is a consequence of solving it under boundary conditions.

Crucially the equation is not derived from classical mechanics; it is constructed so that its plane-wave solutions carry exactly the energy and momentum that de Broglie and Planck–Einstein already assigned to a matter wave. Seeing that construction — which choices are forced and which are postulates — is what separates understanding the equation from merely quoting it.

Assumptions

The particle is non-relativistic:we use E = p²/2m for kinetic energy; if dropped, the correct free dispersion is E² = p²c² + m²c⁴ and one is led to the Klein–Gordon or Dirac equations instead. The evolution is linear in Ψ:superpositions of solutions are solutions; if dropped, interference and the whole Hilbert-space structure collapse, and probabilities no longer combine by amplitude. The equation is first order in time:the state at one instant determines all later states, so Ψ(r,t₀) alone is a complete initial condition; if dropped (second order in t), one must also specify ∂_tΨ and the interpretation of |Ψ|² as a conserved probability density fails. The wavefunction is complex and the operator relation iℏ∂_t ↔ E, −iℏ∇ ↔ p holds:a genuinely complex phase e^−iEt/ℏ is required for a single travelling wave to have definite E and p; if Ψ were real, no single plane wave could be an eigenstate of both, and probability current would vanish identically. The potential V is real:this makes the Hamiltonian Hermitian and total probability conserved; if V has an imaginary part, norm is not preserved and the model describes absorption or particle loss rather than closed-system dynamics.

Derivation

Ψ(r,t) = A e^{i(k·r − ωt)}

Posit the simplest free-particle matter wave: a monochromatic plane wave of wavevector k and angular frequency ω, as demanded by de Broglie. A

p = ℏk, E = ℏω

Import the two prior results: de Broglie relation for momentum and Planck–Einstein relation for energy, so the wave carries definite p and E. A

Ψ = A e^{i(p·r − Et)/ℏ}

Substitute the physical variables p, E for the kinematic k, ω; the phase is now written entirely in dynamical quantities. A

∂_tΨ = −(iE/ℏ)Ψ ⇒ iℏ ∂_tΨ = EΨ

Differentiate once in time and rearrange: acting with iℏ∂_t extracts the eigenvalue E. This identifies the operator Ê = iℏ∂_t. B

−iℏ∇Ψ = pΨ ⇒ −ℏ²∇²Ψ = p²Ψ

Differentiate in space: −iℏ∇ extracts p, and applying it twice extracts p². This identifies the operator p̂ = −iℏ∇. B

E = p²/2m

Impose the non-relativistic free-particle energy–momentum relation. This is the physical input that selects the non-relativistic theory. A

EΨ = (p²/2m)Ψ

Multiply the dispersion relation by Ψ so each side can be replaced by its operator action from steps 4–5. A

iℏ ∂_tΨ = −(ℏ²/2m) ∇²Ψ

Substitute EΨ = iℏ∂_tΨ (step 4) and p²Ψ = −ℏ²∇²Ψ (step 5). The free TDSE now holds for a single plane wave. B

Ψ = ∫ d³k Φ(k) e^{i(k·r − ω(k)t)}

Because steps 4, 5 and 8 involve only linear operators, an arbitrary superposition of plane waves also satisfies step 8. Linearity promotes the single-mode result to a general free wavefunction. C

E = p²/2m + V(r,t)

Restore interactions via the classical Hamiltonian: total energy is kinetic plus potential. Promoting E → iℏ∂_t, p → −iℏ∇, with V acting by multiplication, gives the full equation. C

Result

iℏ ∂Ψ/∂t = −(ℏ²/2m) ∇²Ψ + V(r,t)Ψ = ĤΨ

Reading. The rate of change of the wavefunction’s phase and amplitude (left) is set by the total-energy operator, the Hamiltonian Ĥ = −(ℏ²/2m)∇² + V (right). The factor i makes evolution unitary rather than diffusive: probability is transported and interferes, not smeared away. Given Ψ now, the equation dictates Ψ at every later instant.

Units check. [ℏ∂_tΨ] = (J·s)(s⁻¹)[Ψ] = J·[Ψ]. [ℏ²∇²Ψ/m] = (J·s)²(m⁻²)/kg ·[Ψ] = (kg²m⁴s⁻²)(m⁻²)(kg⁻¹)[Ψ] = kg·m²·s⁻²·[Ψ] = J·[Ψ]. [VΨ] = J·[Ψ]. All three terms carry units of energy × [Ψ], so the equation is dimensionally homogeneous for any units of Ψ.

Limiting cases

Free particle (V = 0). Recovers step 8; plane waves obey ℏω = ℏ²k²/2m, i.e. the non-relativistic dispersion the equation was built to reproduce.
Time-independent V(r). Separation Ψ = ψ(r)e^−iEt/ℏ gives the time-independent Schrödinger equation Ĥψ = Eψ for stationary states.
Slowly varying, large action (ℏ → 0). Writing Ψ = e^iS/ℏ and keeping leading order yields the classical Hamilton–Jacobi equation ∂_tS + (∇S)²/2m + V = 0: the classical limit.
Constant V = V₀. Only shifts the global phase by e^{−iV₀t/ℏ}; no observable is changed, consistent with energy being defined up to a constant.

Breaks when

Speeds approach c. The kinetic input E = p²/2m (step 6) is the non-relativistic approximation; at high energy one needs E² = p²c² + m²c⁴, giving the Klein–Gordon/Dirac equations. Spin then also emerges and cannot be added by hand here.
Particle number changes. A single fixed-mass wavefunction cannot describe pair creation, decay, or photon emission/absorption; those require quantum field theory, where Ψ is promoted to an operator field.
Open or dissipative systems. If the particle exchanges energy incoherently with an environment, unitary evolution fails; one needs a density-matrix master (Lindblad) equation rather than the TDSE for a pure state.
Strong, fast fields where the dipole/scalar-potential picture fails. Minimal coupling p → p − qA must replace the bare p²/2m; a plain scalar V then misrepresents magnetic and gauge effects.

Failure modes

Sign/factor slips in the operators. Using p̂ = +iℏ∇ or dropping the i in iℏ∂_t reverses the direction of phase evolution and breaks the correspondence ℏω = ℏ²k²/2m.
Treating the TDSE as second order in time “like a wave equation.” It is first order; assuming you also need ∂_tΨ(t₀) as data is wrong and doubles the (spurious) solution space.
Forcing Ψ to be real. A cosine is a sum of two counter-propagating exponentials with opposite p; it is not a momentum eigenstate and has zero net current.
Plugging E = ½mv² numbers into ℏω while using group velocity v_g as phase velocity. For matter waves v_phase = ω/k = v/2 ≠ v_g = dω/dk = v; confusing them mislocates the particle.
Assuming stationary states are static. ψ(r)e^−iEt/ℏ has time-dependent phase; only |Ψ|² is static. Superposing two energies restores visible time dependence.
Adding a complex V casually. Doing so silently violates norm conservation; legitimate only as an effective absorbing model, never for a closed system.

Discussion

The derivation is really a construction under constraints. Demanding linearity, first-order time evolution, and consistency with p = ℏk, E = ℏω for free waves leaves essentially no freedom: the operator dictionary E → iℏ∂_t, p → −iℏ∇ is forced, and applying it to the classical energy relation E = p²/2m + V writes the equation for you. What cannot be derived this way is the interaction structure — that we borrow the classical Hamiltonian is a genuine postulate, justified after the fact by agreement with experiment (hydrogen spectrum, tunnelling rates, and so on).

The imaginary unit is not cosmetic. It converts a would-be diffusion equation (∂_tΨ = D∇²Ψ, which damps) into a unitary one whose solutions oscillate and interfere. Continuity ∂_t|Ψ|² + ∇·j = 0 with j = (ℏ/m)Im(Ψ*∇Ψ) then guarantees that total probability is conserved — a direct consequence of V being real and Ĥ Hermitian.

Structurally, the TDSE says iℏ∂_tΨ = ĤΨ generates a one-parameter unitary group Û(t) = e^−iĤt/ℏ (for time-independent Ĥ), so Ĥ is the generator of time translations exactly as momentum generates space translations. This places the equation inside the general framework in which observables are Hermitian generators of symmetries; the specific −ℏ²∇²/2m + V is just one Hamiltonian, and the same evolution law governs spin, many-body, and field systems once Ĥ is chosen appropriately. Nothing in the derivation privileges the position representation; it is the Schrödinger-picture, coordinate-basis face of the abstract law iℏ d|Ψ⟩/dt = Ĥ|Ψ⟩.

Common misconceptions. The TDSE is not derived from Newton’s laws — it is a new axiom that reduces to them in the ℏ → 0 limit. And Ψ is not a physical wave in space like sound or light; it is a complex probability amplitude whose modulus-squared, not the field itself, is observable.

Worked examples

Free electron: verify the dispersion. An electron is prepared as a plane wave with de Broglie wavelength λ = 0.50 nm. Find ω from the TDSE dispersion and compare E with the kinetic energy.

Symbols: k = 2π/λ, ℏω = ℏ²k²/2m ⇒ ω = ℏk²/2m, E = ℏω. A

k = 2π/(0.50×10⁻⁹ m) = 1.257×10¹⁰ m⁻¹

Insert λ. A

ω = ℏk²/2m = (1.055×10⁻³⁴)(1.257×10¹⁰)² / (2×9.11×10⁻³¹)

Insert ℏ = 1.055×10⁻³⁴ J·s, m = 9.11×10⁻³¹ kg. A

ω = (1.055×10⁻³⁴)(1.580×10²⁰)/(1.822×10⁻³⁰) = 9.15×10¹⁵ rad·s⁻¹

Arithmetic. A

ω ≈ 9.15×10¹⁵ rad/s, E = ℏω ≈ 9.65×10⁻¹⁹ J ≈ 6.0 eV

Reading. Cross-check via kinetic energy: E = ℏ²k²/2m = p²/2m with p = ℏk = 1.326×10⁻²⁴ kg·m/s gives E = (1.326×10⁻²⁴)²/(1.822×10⁻³⁰) = 9.65×10⁻¹⁹ J. The frequency demanded by the TDSE and the kinetic energy agree exactly, as the construction guarantees.

Two-level phase beating. A particle is Ψ = (1/√2)(ψ₁e^{−iE₁t/ℏ} + ψ₂e^{−iE₂t/ℏ}) with E₂ − E₁ = 2.00 eV. Find the oscillation period of |Ψ|².

Each term evolves by its own phase (TDSE, stationary states); interference term goes as cos[(E₂−E₁)t/ℏ]. B

ω_beat = (E₂ − E₁)/ℏ, T = 2πℏ/(E₂−E₁) = h/(E₂−E₁)

Rearrange for period before numbers. A

ΔE = 2.00 eV = 2.00×1.602×10⁻¹⁹ J = 3.204×10⁻¹⁹ J

Convert to SI. A

T = (6.626×10⁻³⁴ J·s)/(3.204×10⁻¹⁹ J)

Insert h = 6.626×10⁻³⁴ J·s. A

T = 2.07×10⁻¹⁵ s ≈ 2.07 fs

Reading. Although each stationary state has a static |ψ_i|², their superposition’s probability density oscillates at the Bohr frequency set by the energy gap — the physical clock behind atomic transitions. Units: J·s / J = s. A 2 eV gap (visible-light scale) gives femtosecond beating, consistent with optical periods.

Problems

Show explicitly that Ψ = A e^i(kx−ωt) solves the 1-D free TDSE only if ω = ℏk²/2m.
Solution
Compute iℏ∂_tΨ = iℏ(−iω)Ψ = ℏωΨ and −(ℏ²/2m)∂_x²Ψ = −(ℏ²/2m)(ik)²Ψ = (ℏ²k²/2m)Ψ. Equality for all x,t requires ℏω = ℏ²k²/2m, i.e. ω = ℏk²/2m. This is exactly E = p²/2m under E=ℏω, p=ℏk.
For a matter wave show the phase velocity is v_p = ω/k and the group velocity v_g = dω/dk, and evaluate their ratio. Which equals the classical particle speed?
Solution
With ω = ℏk²/2m: v_p = ω/k = ℏk/2m = p/2m = v/2. v_g = dω/dk = ℏk/m = p/m = v. Ratio v_p/v_g = 1/2. The group velocity v_g = v equals the classical particle speed; the phase velocity is unobservable and half as large.
A proton (m = 1.673×10⁻²⁷ kg) has kinetic energy 1.00 keV. Find its de Broglie wavelength and the angular frequency the TDSE assigns to its plane wave.
Solution
E = 1.00 keV = 1.602×10⁻¹⁶ J. p = √(2mE) = √(2×1.673×10⁻²⁷×1.602×10⁻¹⁶) = √(5.360×10⁻⁴³) = 7.32×10⁻²² kg·m/s. λ = h/p = 6.626×10⁻³⁴/7.32×10⁻²² = 9.05×10⁻¹³ m (0.905 pm). ω = E/ℏ = 1.602×10⁻¹⁶/1.055×10⁻³⁴ = 1.52×10¹⁸ rad/s.
Starting from the 3-D TDSE with time-independent V(r), use separation of variables Ψ = ψ(r)T(t) to derive the time-independent Schrödinger equation and the form of T(t).
Solution
Substitute: iℏψT′ = [−(ℏ²/2m)∇²ψ + Vψ]T. Divide by ψT: iℏT′/T = [−(ℏ²/2m)∇²ψ + Vψ]/ψ. LHS depends only on t, RHS only on r, so both equal a constant E. Then iℏT′ = ET ⇒ T(t) = e^−iEt/ℏ, and −(ℏ²/2m)∇²ψ + Vψ = Eψ, the TISE. Hence Ψ = ψ(r)e^−iEt/ℏ and |Ψ|² = |ψ|² is stationary.
Derive the continuity equation ∂_tρ + ∇·j = 0 from the TDSE with real V, identifying ρ and j. Explain why a complex V breaks it.
Solution
Let ρ = Ψ*Ψ. Then ∂_tρ = Ψ*∂_tΨ + Ψ∂_tΨ*. From the TDSE, ∂_tΨ = (1/iℏ)[−(ℏ²/2m)∇²Ψ + VΨ] and its conjugate (with V* = V). Substituting, the V terms cancel and ∂_tρ = (iℏ/2m)(Ψ*∇²Ψ − Ψ∇²Ψ*) = −∇·j with j = (ℏ/2mi)(Ψ*∇Ψ − Ψ∇Ψ*) = (ℏ/m)Im(Ψ*∇Ψ). If V = V_R + iV_I, the V terms leave a residue ∂_tρ + ∇·j = (2V_I/ℏ)ρ ≠ 0, so total probability is not conserved — the model gains or loses particles.