Start here
A quantum particle in one dimension, already running. The coloured hump is where the
particle is likely to be found, the grey curve is the potential it lives in, and the colours
are the phase of its wave function: stripes mean it is moving. Below are the same wave
function in momentum space and a space-time picture that scrolls as time goes on.
Pick a potential on the right, then a starting wave function from its rows of
named states, and drag the sliders (x₀ , σ , k₀ for a packet). Click a thin
grey line on the plot to load that stationary state. Play pauses and resumes,
Restart goes back to the beginning, Backwards runs time in reverse (it pauses
at time zero, the starting state again; Play then goes on into negative time, the
state's past), and Speed slows a fast packet down.
Look for this: a stationary state only changes colour; anything else moves, spreads or
sloshes. The triangle and bar at the foot of the plot are the mean position and its spread,
and the readouts above the plot give the same numbers. Each experiment below sets up one
idea: press Set it up .
The time-dependent Schrödinger equation
$$i\hbar\frac{\partial\psi}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}+V(x,t)\,\psi$$
with $\hbar=m=1$ (the mass is a slider under Domain and numerics ). The filled curve
is $|\psi|^2$, the probability density, coloured by the
phase of $\psi$ (the colour wheel below); the grey region is $V(x)$ (an infinite wall is a grey
band), read on the right-hand energy axis, with the stationary states' energies as thin lines
and $\langle E\rangle$ dashed. The triangle at the foot is $\langle x\rangle$, and the bar
through it spans $\langle x\rangle\pm\sigma_x$. Under it is the wave function in momentum
space, $\phi(p)=\frac{1}{\sqrt{2\pi\hbar}}\int\psi(x)\,e^{-ipx/\hbar}\,dx$, drawn the same
way, with $\langle p\rangle\pm\sigma_p$.
Pick a potential and an initial state, then press Play . The initial states come in
four kinds, a row each: a Gaussian packet, a stationary state, an equal mix of two, and a
formula you type. Each row starts with its kind, then named starting points of that kind;
the highlighted button is the state loaded. Move one of its sliders and the highlight moves
to the kind: the state is now your own. Typed expressions may use any symbols: each new one
gets a slider. Change the potential while the clock runs, and the wave function feels it
suddenly.
Phase colours: red is positive real, cyan negative real,
yellow-green $+i$, violet $-i$. A wave $e^{ikx}$ has colours cycling along $x$: the faster they
cycle, the larger the momentum, and the direction of the cycling is the sign.
1. A free particle spreads
A narrow packet at rest. Its width grows as
$\sigma(t)=\sigma_0\sqrt{1+(\hbar t/2m\sigma_0^2)^2}$: the narrower it starts, the faster it
spreads, because a narrow packet is made of a wide range of momenta. Watch
$\sigma_x\sigma_p$ in the readout: it starts at its minimum, $\hbar/2$, and grows, while
$\sigma_p$ itself never changes (no force).
2. The phase is the momentum
A packet moving to the right has colours cycling along it, once every $2\pi/k_0$ (3.1 here,
for $k_0=2$): $k_0$ is the wavenumber, and the momentum is $\hbar k_0$. Drag the
$k_0$ slider through zero: the colour stripes stop, then cycle the other way, and the packet
goes left. The momentum panel shows the same thing: $|\phi(p)|^2$ is centred on
$\hbar k_0$.
3. Two packets interfere
Two packets moving towards each other. Where they overlap, $|\psi_1+\psi_2|^2$ has fringes
$|\psi_1|^2+|\psi_2|^2+2\,\mathrm{Re}\,\psi_1^*\psi_2$, spaced by $\pi/k_0$. They pass through
each other unchanged: the equation is linear.
4. Stationary states, and what moves
The second state of the infinite well. $|\psi|^2$ does not change at all; only the phase
turns, as $e^{-iE_2t/\hbar}$ (the colours cycle). Now choose 1 + 2 : a superposition of
two energies sloshes from side to side with period $2\pi\hbar/(E_2-E_1)$ (42 here). Only
superpositions of different energies move. The bars at the right edge show the state's
energy distribution $|c_n|^2$; they never change.
5. Quantum revival
A packet bouncing in a box. It spreads and seems to dissolve into a mess, yet at
$T_\text{rev}=4mL^2/\pi\hbar\approx127$ it reassembles exactly, and at fractions of
$T_\text{rev}$ it splits into several copies. The reason: the energies $n^2\pi^2/2L^2$ are all
whole multiples of one energy, so all the phases line up again. The space-time picture
(bottom) draws this as a "quantum carpet".
6. A coherent state: almost classical
A Gaussian with the ground state's width, $\sigma=\sqrt{\hbar/2m\omega}=0.71$, displaced to
$x_0=3$ in a harmonic well. It swings back
and forth without changing shape (the bar $\langle x\rangle\pm\sigma_x$ keeps its length), with
$\langle x\rangle$ (the triangle) following the classical particle (the red dot, under
Show : Classical ) exactly: for a quadratic potential Ehrenfest's theorem
$\frac{d}{dt}\langle p\rangle=-\langle V'(x)\rangle$ is Newton's law. Try the V-shaped
potential with the same packet: there $\langle V'\rangle\ne V'(\langle x\rangle)$ and the two part
ways.
7. A squeezed state breathes
Narrower than the ground state ($\sigma=0.3$ against its 0.71) and at rest in the middle,
this packet stays put,
$\langle x\rangle=\langle p\rangle=0$, but its width oscillates at twice the oscillator's
frequency, trading position uncertainty for momentum uncertainty and back: the bars
$\pm\sigma_x$ and $\pm\sigma_p$ take turns to be long. $\sigma_x\sigma_p$ is back at its
minimum, $\hbar/2$, whenever $\sigma_x$ is at its largest or smallest.
8. Tunnelling
The packet's energy is below the top of the barrier ($\langle E\rangle$ dashed, under the
grey), yet part of it gets through. $P(x>0)$ in the readout is the probability on the far
side: once the packet has gone it is the transmission, 0.39 here (the plane-wave $T$ at this
energy is 0.39 too). Widen the barrier: the transmission falls roughly as $e^{-2\kappa a}$,
with $\kappa=\sqrt{2m(V_0-E)}/\hbar$. Inside the barrier the wave function decays without
oscillating: no colour stripes. Choose probability current under Show :
$j=\frac{\hbar}{m}\mathrm{Im}\,\psi^*\psi'$ is drawn blue where probability flows to the right
and red where it flows to the left; it pours through the barrier, and its integral over time
at a point is the probability that has passed that point. (A bound stationary state is real
up to a phase and carries no current at all.)
9. Reflection from a step, even going over it
Above the step, a classical particle would always go on; a wave is partly reflected, with
$R=\left(\frac{k-k'}{k+k'}\right)^2$, $k'=\sqrt{k^2-2mV_0/\hbar^2}$ the wavenumber over the
step. Here $k_0=2.4$ and $V_0=2$, so $k'=1.33$ and $R=0.083$; $P(x<0)$ ends a little higher,
at 0.094, because the slower parts of the packet reflect more. Choose Below the
step : everything comes back, after a short excursion into the forbidden region.
10. Bound states leak into the walls
A finite well has a finite number of bound states (here 3). Each one reaches into the
classically forbidden region beyond the walls, decaying as $e^{-\kappa|x|}$: the higher the
state, the farther. Click a level (a thin line) to load that state.
11. Passing over a well: resonance
A particle passing over the well is partly reflected, by the jump in $V$ at each edge.
Over the well its wavenumber is larger, $k'=\sqrt{k^2+2mV_0/\hbar^2}$, and the waves reflected
at the two edges cancel when the well is a whole number of half wavelengths wide,
$2ak'=n\pi$: then it all goes through. In general
$$T=\left[1+\frac{V_0^2\sin^2(2ak')}{4E(E+V_0)}\right]^{-1}.$$
This packet has $k_0=2.865$, so $2ak'=3\pi$, and passes almost entirely: $P(x>0)$ ends at
0.985. Now choose Off resonance : $k_0=1.19$, halfway to the resonance below
($2ak'=2.5\pi$), and more than two thirds comes back ($T=0.31$). The packets are wide ($\sigma=2.5$)
so that their spread in $k$, $1/2\sigma$, is narrow. Slow electrons crossing noble-gas atoms
almost unscattered (the Ramsauer–Townsend effect) are the same cancellation.
12. Tunnelling back and forth
A particle in the left well of a double well is the superposition
$(\psi_0+\psi_1)/\sqrt2$ of the two lowest states, which differ in energy by a tiny splitting
$\Delta E$. It tunnels to the right well and back with period $2\pi\hbar/\Delta E$ (here
$\Delta E=0.112$, so 56). Raise the
barrier: the splitting shrinks exponentially, and the tunnelling slows. This is the
ammonia molecule's inversion and the qubit's two states.
13. The delta function
An attractive $V=-\alpha\,\delta(x)$ has exactly one bound state, $\psi=\sqrt\alpha\,e^{-\alpha|x|}$
with $E=-\alpha^2/2$. Here the delta is a narrow Gaussian well of width $w$ and area
$\alpha$, and its level sits a little higher, at $-0.465$: the true bound state has a cusp at the
delta, made of very short wavelengths, which a well of any width blunts. A packet sent at it is transmitted with
probability close to $T=k^2/(k^2+\alpha^2)$: one half for $k=\alpha$ (0.51 here, for the
same reason: watch $P(x>0)$ once the packet has gone). The slower the packet, the less it can tell the well from a true delta.
14. The quantum bouncer
A ball dropped onto a floor in uniform gravity, $V=gx$. The red dot is the classical ball.
The quantum one keeps time for a few bounces, then dissolves, and later partly revives. The
stationary states are Airy functions. Choose Typed : $xe^{-bx}$ is the textbook
variational guess for the lowest one. Drag $b$ and watch $\langle E\rangle=\hbar^2b^2/2m+3g/2b$:
it is least, 1.966, at $b=1.145$, just above the true ground state, 1.856 (the lowest level), as
the variational principle promises.
15. Bloch oscillations
An electron in a crystal ($V_0\cos 2\pi x/a$) pulled by a constant force (the $Fx$ in $V$:
the force is $-F$, to the left) does not accelerate forever: it oscillates, with period
$2\pi\hbar/Fa$ (25 here, swinging about 16 to the left), because its momentum runs to
the edge of the Brillouin zone and is Bragg-reflected. Set $F=0$ to see it simply spread.
A little of the packet leaks to higher bands each time (Zener tunnelling); a smaller $V_0$
leaks more.
16. Driving at resonance
The ground state of a harmonic oscillator pushed by $F\cos\omega t$. At resonance
($\omega=1$) the packet swings ever wider without changing shape: it is a coherent state
whose amplitude grows linearly. Off resonance it beats.
17. A sudden change
The ground state of a narrow box. While it runs, drag the box's width $L$: the walls move
and the wave function, unchanged at that instant, is now a superposition of the new box's
states (watch the bars at the right). Move the slider slowly instead and it stays in the
ground state: the adiabatic theorem.
18. Measurement
Press Position under Make a measurement : a position is chosen at random with
probability $|\psi|^2$, and the wave function collapses to a narrow packet there ($\sigma=0.05$,
a two-hundredth of the box), which then spreads. Press Energy : one stationary state
is chosen with probability $|c_n|^2$
and the wave function becomes it, motionless. Measure the energy twice: the same answer.
19. Running time backwards
A spread-out mess: a packet hit a barrier a while ago, and part went through, part came back.
Backwards is on: press Play , and the reflected and transmitted parts come back
together into the original packet, and the clock pauses at $t=0$. The
Schrödinger equation is time-reversal symmetric (the absorbing walls are not, so they are
off here). Press Play again to go on into negative time, the packet's past under the
same Hamiltonian: it moves back to the left and is wider, so at $t=0$ it was at its narrowest
(a Gaussian with no chirp spreads the same way into the past as into the future,
$\sigma(t)=\sigma_0\sqrt{1+(\hbar t/2m\sigma_0^2)^2}$). In this box its past soon includes a bounce
off the left wall; with absorbing walls the past would be wrong, as what they take can't come
back.
How it is computed
The split-operator method: each step multiplies $\psi$ by $e^{-iV\Delta t/2\hbar}$ in position
space, by $e^{-i\hbar k^2\Delta t/2m}$ in momentum space (one FFT there and one back), and by
$e^{-iV\Delta t/2\hbar}$ again. Every factor has magnitude 1, so probability is conserved
exactly, the kinetic energy is exact, and the error per step is of order $\Delta t^3$. The
stationary states are found separately, by a finite-difference eigenvalue solver. Walls are
infinitely high; ring joins the ends; absorbing edges swallow what reaches them.
Perfectly sharp steps are an idealisation the method pays for with tiny time steps, so a
potential with jumps loses its Fourier components above $k=16$ (reflection and tunnelling for
$k<5$ are those of the sharp step), and walls taller than 30 are rounded over 0.05. The time
step adapts to the potential where the wave function is.
Rewritten in 2026 from the BrainFlux Schrodinger1D Java applet.
Source: github.com/gallicchio/brainflux ,
MIT licence.