A little curiosity goes a long way
Learn by looking.
9 small questions. Explore each answer in the lab.
Probability has a shape
Where might you find an electron?
An orbital describes a quantum state. Its probability density tells you how likely you are to find the electron in a small region of space. A lobe is part of that density pattern; it is not an electron’s path.
Probability density = |ψ|²
Try it
Start with the two-lobed 2p state. Open Adjust and change ℓ from 1 to 0 while keeping n = 2. Rotate each shape to compare them.
What to notice
Surface mode draws one density level. The wavefunction continues beyond that surface; it is not a hard boundary around an electron.
Open in the labSame density, new colors
What do phase colors tell you?
The wavefunction has a magnitude and a phase. Squaring its magnitude gives probability density, so different phase values can share the same density. The colors reveal information that density alone hides.
ψ = |ψ| exp(iφ)
Try it
Rotate this nonzero-m state. In Adjust, switch Phase colors off and on. The density shape stays the same while its coloring changes.
What to notice
These are complex spherical-harmonic states. The colors are not physical colors, and a stationary phase pattern is not a movie of an electron moving.
Open in the labClimb the energy levels
Does a bigger n mean tighter binding?
In this hydrogen model, the energy depends only on n. The energy is negative because the electron is bound. As n increases, the energy approaches zero and the state becomes less tightly bound.
Eₙ = −13.6057 eV / n²
Try it
Read the energy for n = 2, then use Adjust to select n = 4. Predict the new energy before checking the readout.
What to notice
The magnitude falls to one quarter of its previous value: about −3.40 eV becomes −0.85 eV. Selecting a state does not simulate absorption or a transition.
Open in the labFollow the photon’s energy
How much energy can light give away?
A photon scattering from a free electron initially at rest transfers some of its energy into the electron’s kinetic energy. For a fixed incoming energy, a larger scattering angle transfers more energy.
Electron kinetic energy = E − E′
Try it
Start with a 511 keV photon at 90°, where the outgoing photon and electron each carry about 255.5 keV. In Adjust, compare 0° and 180°.
What to notice
At 0° there is no energy transfer in this ideal model. At 180° the transfer is largest. Arrows represent momentum vectors, not particle trajectories.
Open in the labOne event, two diagrams
Why draw the same process twice?
The two lowest-order QED diagrams attach the incoming and outgoing photons to the electron line in different orders. Both contribute to the same scattering process. Add their complex amplitudes before taking the squared magnitude.
𝓜 = 𝓜ₛ + 𝓜ᵤ
Try it
Open Explain in the lab and inspect the photon and electron lines. Compare the two diagrams: the external particles are the same; their attachments to the electron line differ.
What to notice
The internal line is a propagator, not an observed intermediate path. Diagram spacing is bookkeeping, not a measured distance or a sequence of frames.
Open in the labTurn the question around
What changes when you rotate a measurement?
Prepare spin up along z. An ideal analyzer gives only two possible outcomes, +ℏ/2 and −ℏ/2 along its own axis. Rotating that axis changes their probabilities, even though every particle starts in the same prepared state.
P(+) = cos²(θ / 2) for a z+ preparation
Try it
Predict the split at 90°, then send a batch of particles. Compare the observed counts with the model’s probabilities. Try 0°, 90°, and 180°.
What to notice
At 90° each outcome has a 50% probability; a finite batch need not divide exactly in half. Spin is intrinsic angular momentum, not a tiny ball rotating in space.
Open in the labAsk the same question twice
Will a second measurement agree?
An ideal spin measurement prepares the state associated with its outcome. Immediately measuring the same axis again repeats that outcome, provided nothing changes the state in between.
P(same outcome | same axis, unchanged state) = 1
Try it
Send particles through two analyzers aligned along z. Compare their outcomes, then rotate both analyzers to x and repeat with the same z+ source.
What to notice
Two x measurements of a z+ preparation can start with either sign, but the second sign matches the first for each particle. Agreement does not mean the first outcome was predetermined.
Open in the labA measurement changes what follows
Can inserting an analyzer change the result?
A z+ preparation measured directly along z always gives +. Insert an x measurement first and the final z outcomes become equally likely. Keeping both x outcomes still changes the ensemble reaching the final analyzer.
z+ → measure x → measure z: P(+) = P(−) = ½
Try it
Compare the z result with and without the intermediate x analyzer. Send a batch for each arrangement, then rotate the intermediate analyzer back to z.
What to notice
Both branches continue; no particles are filtered out. The displayed routes are a schematic of recorded outcomes. This model excludes coherent recombination and evolution between measurements.
Open in the labPreview illustrations are schematic. Each lab includes its model assumptions and references in Explain.