Wave Function Calculator
Calculate quantum wave function properties including probability density, momentum, energy, and phase/group velocities.
About this calculator
This calculator models a free particle (an electron by default) as a plane wave, psi(x,t) = A * e^(i(kx - omega*t)), and reads off its measurable properties. The wave number k = 2*pi/wavelength sets the spatial oscillation via the de Broglie relation, and the angular frequency omega = h-bar*k^2/(2m) follows from the free- particle Schrodinger dispersion relation (equivalent to E = p^2/(2m) written in terms of h-bar and k). From there, momentum p = h-bar*k and energy E = h-bar*omega are the standard de Broglie relations.
Two velocities fall out of the same wave: the phase velocity v_phase = omega/k, at which the wave's oscillation pattern moves, and the group velocity v_group = d(omega)/dk = h-bar*k/m, which equals the particle's actual classical velocity p/m. For this free-particle dispersion relation the two are not equal -- phase velocity works out to exactly half of group velocity -- a well-known and slightly counterintuitive feature of matter waves that has no counterpart in everyday wave intuition built from sound or water waves. The probability density, |psi|^2 = A^2, is constant everywhere and at all times for this idealized single plane wave: it doesn't depend on position or time at all, which is the textbook illustration of why a pure plane wave is not physically normalizable (a real particle's wave function is a localized wave packet, a superposition of many wavelengths, not a single infinite plane wave).
Inputs
Results
Probability Density |ψ|²
1
Real Part
1
Imaginary Part
0
How to Use This Calculator
- Enter Amplitude of the wave function — higher amplitude increases probability density at that location.
- Enter Wavelength (m) — the de Broglie wavelength; smaller wavelength means higher momentum.
- Enter Position (m) — the spatial point where you want to evaluate the wave function.
- Enter Time (s) to see how the wavefunction evolves; at t = 0 you get the initial state.
- Read Probability Density |ψ|² — the measurable probability of finding the particle at that position, plus Momentum and Energy from the de Broglie relations.
How the result changes with Amplitude
| Amplitude | Probability Density |ψ|² | Real Part | Imaginary Part |
|---|---|---|---|
| 0.5 | 0.25 | 0.5 | 0 |
| 0.75 | 0.563 | 0.75 | 0 |
| 1.5 | 2.25 | 1.5 | 0 |
| 2.5 | 6.25 | 2.5 | 0 |
What each input means
- Amplitude
- Wave function amplitude
- Wavelength
- De Broglie wavelength
- Position
- Position in space
- Time
- Time parameter
How this is calculated
Formula
ψ(x,t) = A × e^(i(kx - ωt))Worked example, using the default values
- Identify Input Parameters4 parametersAmplitude = 1, Wavelength = 1e-10, Position = 0, Time = 0 = 4 input(s) provided
- Calculate Probability Density |ψ|²Probability Density |ψ|²1 = 1
- Calculate Real PartReal Part1 = 1
- Calculate Imaginary PartImaginary Part0 = 0
- Calculate MomentumMomentum = Number(momentum.toExponential(3))6.626e-24 = 6.626e-24
- Calculate EnergyEnergy = Number(energy.toExponential(3))2.41e-17 = 2.41e-17
Engine last updated . Checked against 3 independently-derived tests — how we verify calculators. Built by Paul Gunder, a software engineer, not a licensed financial, medical, or legal professional.
Frequently Asked Questions
Why doesn't the probability density change with position or time?
A single plane wave psi(x,t) = A*e^(i(kx-omega*t)) has a constant magnitude everywhere -- only its phase (the argument of cos/sin) depends on x and t, not its amplitude. Since probability density is |psi|^2 = A^2, it comes out completely uniform: this is the standard textbook reason a pure plane wave isn't a physically realistic, localized particle state, and why real wave packets are built from a superposition of many wavelengths instead.
Why are phase velocity and group velocity different here?
For this free-particle dispersion relation, omega is proportional to k^2, so phase velocity (omega/k) and group velocity (d(omega)/dk) scale differently with k -- working out to v_phase = v_group / 2. Group velocity is the one that matches the particle's actual classical velocity (p/m); phase velocity, the speed of the wave crests themselves, is a real but less physically intuitive quantity for matter waves.
What determines the wave's momentum and energy?
Both follow directly from the wavelength through the de Broglie relations: momentum p = h-bar*k where k = 2*pi/wavelength, and energy E = h-bar*omega using the free-particle dispersion relation for omega. A shorter wavelength means a larger wave number k, which means higher momentum and higher energy -- the same relationship that underlies electron diffraction and other matter-wave phenomena.
Is this the actual wave function of a real particle like an electron in an atom?
No -- this models an idealized free particle with no potential energy acting on it, using a single infinite plane wave. A bound particle (an electron in an atom, for example) has a wave function shaped by the Schrodinger equation with a specific potential, producing quantized energy levels and a spatially localized, non-uniform probability density -- quite different from this calculator's uniform free-particle case.
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