QiLing Tools

🔬1D Schrödinger Equation Solver - Potential Well Energy Levels, Quantum Tunneling Animation, Superposition

Solve the one-dimensional Schrödinger equation numerically in your browser. By default the particle is an electron, with lengths in nm and energies in eV; you can switch to atomic units (ħ = mₑ = 1). The bound-state panel discretises the Hamiltonian with finite differences into a symmetric tridiagonal matrix, finds the lowest levels by Sturm-sequence bisection and the normalised wavefunctions by inverse iteration, draws them on the potential and compares every level with the analytic result for the infinite well, harmonic oscillator and finite well; you can also draw any piecewise potential straight onto the plot. The wave-packet panel evolves a Gaussian packet with the Crank–Nicolson scheme as it hits a barrier, step, well or double barrier, animating Re ψ, Im ψ and |ψ|², tracking reflection and transmission, and comparing with an exact transfer-matrix T(E). The superposition panel shows two eigenstates sloshing at ΔE/ħ, with ⟨x⟩ against time.

Free · No sign-up · Runs in your browserLast updated:

Loading tool...

How to use Quantum Mechanics Simulator (Schrödinger Equation)

  1. 1Pick a panel: bound states, wave-packet tunneling or superposition
  2. 2Choose a potential and set the width, depth, ħω, barrier height and width, or draw the potential on the plot
  3. 3Click a level to see its wavefunction and probability density, play the packet to measure transmission, or pick two states to watch them oscillate

FAQ

What are the energy levels of an infinite square well, and how big are they for an electron in a 1 nm box?
Eₙ = n²π²ħ²/(2ma²) for n = 1, 2, 3…, so the levels grow as n², with wavefunctions ψₙ = √(2/a)·sin(nπx/a). For an electron ħ²/2mₑ ≈ 0.0381 eV·nm², so a 1 nm box gives E₁ ≈ 0.376 eV, E₂ ≈ 1.50 eV and E₃ ≈ 3.38 eV. Halving the width multiplies every level by four, which is why smaller quantum dots glow bluer.
How do you calculate the tunneling probability through a barrier?
For energy E below a square barrier of height V₀ and width a, T = [1 + V₀² sinh²(κa) / (4E(V₀ − E))]⁻¹ with κ = √(2m(V₀ − E))/ħ. For thick barriers T ≈ 16(E/V₀)(1 − E/V₀)e^(−2κa), falling exponentially with width. A 0.8 eV electron meeting a 1 eV, 0.5 nm barrier has T ≈ 0.24. Above the barrier T is not automatically 1: transmission is complete only when k′a = nπ.
Why does the wave packet's transmission differ from the plane-wave T(E₀)?
A packet of finite width contains a spread of momenta; a Gaussian of width σ has a momentum spread of ħ/(2σ). Because T(E) changes steeply with energy, roughly exponentially when tunneling, the faster components contribute more, so the measured transmission is the momentum-weighted average ⟨T⟩, usually above T(E₀). Wider packets have sharper momentum and approach the plane-wave value. The tool shows both alongside the simulated result.
Why use the Crank–Nicolson method?
It discretises time implicitly about the midpoint: (1 + iHΔt/2ħ)ψⁿ⁺¹ = (1 − iHΔt/2ħ)ψⁿ. This Cayley form is unitary, so total probability is conserved exactly at every step, and it is stable for any time step; in 1D each step only needs a tridiagonal solve, so the cost scales linearly with the grid size. A simple explicit Euler step instead makes the norm grow until the solution blows up.
Why does a superposition of two levels slosh back and forth, and how is that related to light emission?
For Ψ = c₁ψ₁e^(−iE₁t/ħ) + c₂ψ₂e^(−iE₂t/ħ) the density contains the cross term 2c₁c₂ψ₁ψ₂cos(ΔE·t/ħ), so the distribution oscillates at ω = ΔE/ħ with period h/ΔE. If the two states have opposite parity, ⟨x⟩ oscillates too, giving an oscillating dipole that radiates light at ν = ΔE/h, which is Bohr's frequency condition. States of equal parity have zero dipole matrix element, which is why such dipole transitions are forbidden.