Tools / Three-qubit state
A three-qubit state with one reference qubit
A three-qubit pure state is eight complex numbers, so it is eight arrows in one plane. Choose one qubit as the reference and the same state reads |0⟩|φ₀⟩ + |1⟩|φ₁⟩: two two-qubit states, one for each value of that qubit. Each branch has its own Schmidt bases and contributes four endpoints; two Schmidt axes span a plane, so a single Bloch sphere carries two planes — one per branch — and you can turn it to see them from any side.
Controls
Measure A, B, C or all three: the outcome appears here.No measurements yet.
The same state in two pictures
Amplitudes
Eight arrows in one plane
Σ|c|² = 1.000 — the arrows are normalized.
|000⟩
0.534122°
|001⟩
0.27917°
|010⟩
0.20670°
|011⟩
0.300244°
|100⟩
0.166163°
|101⟩
0.4058°
|110⟩
0.305347°
|111⟩
0.470289°
Arrow length is the magnitude, its angle the phase. A drag sets both; ←/→ turns an arrow and ↑/↓ resizes it.
Relative state
Two planes in one sphere
Drag the sphere to orbit it; with the sphere focused, ←/→ turns the view and ↑/↓ raises or lowers it. Each plane is filled where it cuts the sphere, and the stronger half is the half in front of the other plane.
|ψ⟩ = 0.703 |0⟩A|φ₀⟩ + 0.711 |1⟩A|φ₁⟩
| Quantity | |0⟩A | |1⟩A |
|---|---|---|
| Probability | 0.495 | 0.505 |
| Schmidt angle θ | 20° | 19° |
| cos²θ / sin²θ | 0.879 / 0.121 | 0.897 / 0.103 |
| Relative phase χ | 303° | 120° |
|φ₀⟩ = 0.938 |+b⟩₀|+c⟩₀ + e^{i303°} 0.348 |−b⟩₀|−c⟩₀
|φ₁⟩ = 0.947 |+b⟩₁|+c⟩₁ + e^{i120°} 0.320 |−b⟩₁|−c⟩₁
The planes meet at 87.5° along û = (−0.33, −0.79, 0.52); that diameter is the basis |+û⟩, |−û⟩, at polar 59° and azimuth 248°.
In the shared û basis all four amplitudes survive in each branch; the frame data left is λ₀ = 23°, μ₀ = 13°, λ₁ = 52°, μ₁ = 77°, and the twist α = 87.5° between the two meridians.
Two Schmidt planes, one per reference branch.
Why one qubit is the reference
Group the eight amplitudes by the value of one qubit and the state separates into two pieces:
\[|\psi\rangle = |0\rangle|\phi_0\rangle + |1\rangle|\phi_1\rangle .\]
Nothing is approximated — \(|\phi_0\rangle\) and \(|\phi_1\rangle\) are ordinary, unnormalised two-qubit states, and their lengths are the probabilities of the two values: \(w_0 = \langle\phi_0|\phi_0\rangle\) and \(w_1 = \langle\phi_1|\phi_1\rangle\), with \(w_0 + w_1 = 1\). The reference qubit contributes only its two kets; everything else lives in the other two qubits.
Each branch is then a problem this toolkit already solves. Divide \(|\phi_r\rangle\) by \(\sqrt{w_r}\) to normalise it, and it has a Schmidt form
\[|\phi_r\rangle = \sqrt{w_r}\left(\cos\theta_r\,|{+}p_r\rangle|{+}q_r\rangle + e^{i\chi_r}\sin\theta_r\,|{-}p_r\rangle|{-}q_r\rangle\right),\]
with its own axes for the two remaining qubits. Those axes are different in the two branches in general, which is the point of the picture: the same pair of qubits is described by a different pair of local bases depending on what the reference reads.
The two planes
Two axes through the origin span a plane, so each branch contributes one plane to the sphere. The plane is filled where it cuts the sphere — a disc seen at an angle — so the whole shared diameter lies visibly inside both planes rather than running off their edges. The disc is split along that diameter into the half nearer the camera and the half beyond it, and the nearer half is drawn stronger: where the two discs overlap, the stronger piece is the plane in front. Filling strength also tracks the branch probability, so the branch that is less likely fades.
The two planes generally differ, and that is the point of the picture: the same pair of qubits is described by a different pair of local bases depending on what the reference reads. A plane needs two independent directions, so when a branch's axes line up — GHZ, W, and any product branch — there is no plane to fill, and it is hidden rather than drawn as a stray line.
The intersection line
Both planes pass through the centre of the sphere, and two distinct planes through one point always meet in a line. With unit normals \(\hat n_r = \hat p_r \times \hat q_r\), the line runs along \(\hat u = \hat n_0 \times \hat n_1\), and the planes open at the dihedral angle \(\alpha = \arccos\lvert \hat n_0 \cdot \hat n_1\rvert\).
The basis the line defines
Every diameter of the Bloch sphere is a basis: its two ends are the \(\pm 1\) eigenstates of \(\hat u \cdot \vec{\sigma}\), so \(\hat u\) names the orthonormal pair \(\{|{+}\hat u\rangle, |{-}\hat u\rangle\}\). This one sphere carries both non-reference qubits, so that is a single basis read on either of them — the one basis the picture adapts to both branches at once. Lying in both planes is what buys it:
\[\hat u \in \Pi_0 \cap \Pi_1 \quad\Longleftrightarrow\quad \hat p_r, \hat q_r \in \operatorname{span}(\hat u, \hat m_r) \quad (r = 0, 1).\]
Read it in the frame whose \(z\) axis is \(\hat u\). No Schmidt axis then winds around \(\hat u\): each branch's two axes lie on one meridian, and inside a branch they differ only in how far they lean away from \(\hat u\). All the azimuthal disagreement in the picture therefore collapses into a single meridian angle per branch, and the gap between those two meridian angles is exactly the dihedral angle. The readout under the sphere names the two lean angles \(\lambda_r\) and \(\mu_r\) for each branch, and the twist \(\alpha\).
That is the line's significance for the three-qubit state. The reference cut produces two branches with two Schmidt frames, and those frames disagree in exactly two ways: by their lean along a common axis, and by a twist around it. The line is that common axis — the only direction on which both branches agree — and \(\alpha\) is the residue of the twist, the part no change of frame removes. Rotating about \(\hat u\) by \(\alpha\) carries one Schmidt plane exactly onto the other, so \(\hat u\) is the axis that relates them. The two frames coincide only when \(\alpha = 0\) — the product, GHZ and W cases, where the planes already lie on top of each other.
The line also ignores which of the two non-reference qubits you call the first: exchanging them swaps \(\hat p_r\) with \(\hat q_r\) and so flips both normals, and flipping both vectors leaves a cross product unchanged. The axes and the normals move under that exchange; \(\hat u\), and with it \(\alpha\), does not.
Look along û turns the camera down the shared axis. Seen end on, the two planes flatten into two straight lines through the centre, crossing at the dihedral angle: the meridians themselves.
Three qualifications. When the planes coincide, the intersection is the whole shared plane, so no single diameter is distinguished and every basis inside that plane qualifies. A maximally entangled branch \((\theta_r = 45^\circ)\) has no unique Schmidt bases, so its plane, and the line through it, inherit that freedom of convention. And the line belongs to this picture, which identifies the Bloch spheres of the two non-reference qubits: a local unitary on either qubit turns the planes and the line, while the branch weights and the Schmidt angles stay put.
Both branches in the shared basis
Write both branches in the basis the diameter names, \(\{|{+}\hat u\rangle, |{-}\hat u\rangle\}\) on each of the two remaining qubits. Neither branch keeps two terms: all four amplitudes survive, because \(\hat u\) is not a Schmidt axis of either qubit unless the leans vanish. Nothing physical is lost — a change of local basis is a local unitary, so the Schmidt values and every entanglement measure stay exactly where they were — but the Schmidt coefficients no longer sit on the diagonal.
What the shared basis does remove is azimuthal freedom. In the frame whose \(z\) axis is \(\hat u\), a branch's two Schmidt axes lie on one great circle through \(\hat u\) — the same meridian, up to relabelling \(|{\pm}\hat q\rangle\), which swaps those two basis vectors and nothing more. So each local frame is a rotation about the single axis \(\hat k_r = \hat u \times \hat m_r\), by the lean \(\lambda_r\) on the first qubit and \(\mu_r\) on the second, and between branches that axis turns by \(\alpha\) about \(\hat u\). Two branches that started with four axis angles each are left with four leans and one twist.
This is also why the line is a basis and not a Schmidt basis. A plane needs two independent axes, so \(\hat u\) could be the Schmidt axis of both qubits at once only when both axes collapse onto it — a product branch, with no plane left to draw. A product basis in which both branches are Schmidt-diagonal therefore exists only where the planes coincide, and there the diameter stops being distinguished in the first place.
What this cut does not show
Three qubits have more than one way to be entangled, and a single reference sees only one cut. Choose A and the branches are states of B and C; choose B and the same amplitudes regroup into states of A and C. The three cuts carry different information — GHZ is entangled across every cut, while W concentrates its entanglement differently — so switching the reference is not a cosmetic change but a different question asked of the same state.
Things to try
- Start on Two planes: the two branches carry different Schmidt frames, so the two planes sit at an angle. Drag the sphere and the view turns with it.
- Switch the Reference qubit to B without touching the state: the same eight amplitudes are regrouped, the branch probabilities change, and both planes move.
- Load GHZ: each branch is a product state along the same axis, so the four endpoints are collinear and the planes vanish. The reference qubit is perfectly correlated with the other two, which is what a plane with no area means here.
- Load W: the branches have probabilities 1/3 and 2/3, so the legend and the table show unequal weights even before any plane appears.
- Press Look along û: the camera turns down the shared axis, and both planes flatten into straight lines crossing at the dihedral angle.
- Rotate until a plane is edge on: it thins to a line, then widens again on the far side.
- Drag one of the eight arrows, then Normalize: the branch probabilities and both planes follow the amplitudes you set.
This model follows pure states only, and measurement uses your browser's random number generator. A mixed three-qubit state needs a density matrix and a different notion of Schmidt form, which this tool does not attempt.