Majorana Stars

On the Bloch sphere a qubit is one arrow: two real numbers (θ, φ) for a state with two complex amplitudes, once you drop the norm and the global phase. A spin-j state has 2j+1 amplitudes — the readout above lists them — so it needs 4j real numbers, and one arrow cannot hold them. Two angles per point, 2j points: that is exactly the count. Majorana noticed in 1932 that the state itself does the bookkeeping. Write the 2j+1 amplitudes as the coefficients of a polynomial of degree 2j; its 2j roots, dropped onto the sphere by stereographic projection, are the stars you see.

Every star is a legitimate Bloch vector. For the symmetric states of n qubits — GHZ, W, Dicke, or n copies of the same qubit — the picture is literal: the state is the symmetrised product of n single qubits, and the stars are those n qubits’ Bloch vectors. Pick Symmetric qubits and Product: all the stars sit stacked on one point, because the qubits really are all in the same state. Pick GHZ: they spread into a regular polygon on the equator. Pick W: n−1 at |0⟩ and one at |1⟩ — the W state is, in spin language, simply |n/2, n/2−1⟩.

Drag the Rotation angle slider. Whatever shape the constellation has, it turns as a rigid body — the hull keeps its shape, the pairwise angles between stars never change, and the trails are concentric circles about the axis you chose. That is what an SU(2) rotation is: the same physical rotation applied to every one of the 2j underlying qubits at once. The single-qubit Bloch sphere only ever shows you this case, which is why it makes every operation look like a rotation.

Now set the rotation back to zero and drag Twist χ instead. This applies e^{−iχJ_z²}, the one-axis-twisting interaction that squeezes spin in atomic clocks and, on qubits, is a pairwise Z⊗Z coupling — an entangling operation. The stars do not turn; they separate. A tiny χ already scatters a stacked coherent state into a ring, while the ⟨J⟩ arrow has barely shortened: most of the state is still pointing where it was, but the constellation announces immediately that the state has left the set of “n identical qubits”. Keep going and the arrow shrinks toward zero as the stars spread over the whole sphere. That shrinking is the honest measure of how far from classical the state has gone, and a single arrow was never going to show you the rest.

Try the twist at j = ½. Nothing happens: J_z² is a constant there, so the twist is a global phase. One qubit has no room to squeeze. The moment a second point appears on the sphere, it does.

Keep the twist going and the stars come back. For a half-integer j the constellation returns to its starting shape at χ = π; for an integer j the trip takes 2π, but halfway, at χ = π, the phases e^{−iπm²} = (−1)^m are exactly a rotation by π about z: a stacked coherent state that was scattered over the sphere re-collapses to a single point on the far side. Halfway to that, at χ = π/2, it passes through a cat state — the coherent state superposed with its own π-rotated twin — whose stars form a regular ring in the plane perpendicular to the original direction, just as GHZ’s do. The slider spans exactly one period, so what you see at its right-hand end is the start again.

All stars on one point — a spin coherent state, the closest a big spin gets to a classical arrow; |⟨J⟩| = j exactly, and the hull collapses to nothing. For qubits this is a product state: no entanglement at all.

Stars split between the poles — the |j, m⟩ states: j+m at the top, j−m at the bottom. The hull is a single line through the centre of the sphere.

A regular polygon on the equator — the cat / GHZ family. ⟨J⟩ vanishes: the amplitudes sit at m = ±j, which no single J± step connects, so there is no direction the spin “points”, only a shape. States whose stars are spread as evenly as possible over the sphere (Platonic solids, for the right j) are the most non-classical spin states there are, and the ones best suited to sensing rotations about every axis at once.

The polynomial’s roots are found numerically, and a root that appears 2j times over is inherently harder to pin down than 2j separate roots — so a coherent state’s stacked stars may sit a fraction of a degree apart at j = 3. That is a property of polynomials, not of the physics, and it disappears the instant the stars actually separate.