🧭Stern–Gerlach Experiment Simulator - spin measurement, analyser chains, Bell test

How to use Spin & the Stern–Gerlach experiment

1. 1Pick spin 1/2 or spin 1 and load a classic sequence (e.g. Z+ → X+ → Z), or build your own with Add analyser
2. 2Set each analyser angle and the beam it passes; on the canvas you can also drag a magnet up or down to rotate it, or click a blocked beam to pass it
3. 3Press Fire to watch atoms fly one by one, or fire 10000 at once, and compare measured fractions with the quantum prediction in the table
4. 4Open Classical vs quantum for the screen patterns and a deflection estimate, then Entangled pairs to tune four angles and read the CHSH S value

FAQ

Why does Z+ → X+ → Z split into two beams again?

The first magnet selects atoms with spin up along z. To an x measurement that state is an equal superposition of +x and −x, and after the measurement it is +x, so the z information is gone. Measuring +x along z gives +z and −z with probability cos²(45°) = 1/2 each, so the third magnet splits the beam evenly. Spin components along different axes cannot both be definite.

Why is there a half angle in cos²(Δθ/2)?

A spin-1/2 state picks up amplitudes in θ/2 under rotation (a 360° rotation flips its sign; only 720° restores it). The overlap between the + state along θ₁ and the + state along θ₂ is cos((θ₂−θ₁)/2), and the probability is its square. So analysers 90° apart give 50%, and Z against −Z (180°) gives 0. Light polarisation follows cos²θ (Malus's law) instead, because the photon is a spin-1 particle with only two transverse states.

Why did classical physics predict a continuous band?

If each atom carried a classical magnetic needle in a random direction, its component along the field gradient, μ cos α, could take any value between −μ and +μ, and so could the deflection. For isotropic directions cos α is uniform on [−1, 1], which gives an even vertical band. Stern and Gerlach saw two separate spots in 1922, showing that the angular-momentum component is quantised.

Silver has no orbital angular momentum, so where does the moment come from?

Silver has a single 5s valence electron (l = 0) outside closed shells. The moment comes from that electron's spin s = 1/2, whose component m_s can only be ±1/2, giving μ_z ≈ ∓μ_B and exactly two beams. At the time it was taken as proof of Bohr's orbital quantisation; the right explanation only came with electron spin in 1925.

What does the CHSH S value tell us?

Any theory in which outcomes are fixed in advance by local hidden variables obeys |S| ≤ 2. For the singlet, quantum mechanics predicts E(a,b) = −cos(a−b), giving |S| = 2√2 ≈ 2.83 at a = 0°, a′ = 90°, b = 45°, b′ = 135°. Experiments, including the loophole-free tests of 2015, observe the violation, so nature cannot be described by local hidden variables.