Mixed States and the Density Matrix

A statevector |ψ⟩ describes a qubit you know everything about. The density matrix ρ = Σ pᵢ |ψᵢ⟩⟨ψᵢ| describes the case where you do not: the qubit is |ψ₁⟩ with probability p₁, |ψ₂⟩ with probability p₂, and so on. For a single pure state it collapses to ρ = |ψ⟩⟨ψ| and says nothing the ket did not.

Read it like this. The diagonal is the populations — ρ₀₀ is P(0), ρ₁₁ is P(1), and they sum to 1 because Tr ρ = 1 for any state. The off-diagonal is the coherence, ρ₀₁ = ρ₁₀*. For |ψ⟩ = cos(θ/2)|0⟩ + e^(iφ)sin(θ/2)|1⟩ it is ½ sin θ e^(−iφ): it carries the phase, and it is the part that interferes. ρ is always Hermitian and never has a negative eigenvalue. The tests in lib/quantum/densityMatrix.test.ts check the first two for every state this page can produce.

For one qubit, ρ = (I + r·σ)/2 where r is exactly the Bloch vector. Squaring and taking the trace gives Tr ρ² = (1 + |r|²)/2, so |r| = √(2 Tr ρ² − 1). A pure state has Tr ρ² = 1 and sits on the surface. The maximally mixed state I/2 has Tr ρ² = ½ and sits at the centre, with no arrow at all. Everything in between is inside the ball.

In the first scene, drag λ from 0 to 100% and watch three things move together: the teal cells go to zero, the purity readout falls from 1 toward ½, and the arrow drops onto the z axis. The diagonal does not move, so P(0) is the same before and after. That is what the arrow on the decoherence page was doing: T2 shrinks ρ₀₁ by e^(−t/T2), and this page is the same picture with the numbers showing.

Set the second scene to measure along Z. The superposition (|0⟩ + |1⟩)/√2 gives 50/50. So does a coin flip between |0⟩ and |1⟩. Looking only at Z, you cannot tell them apart, and people conclude from this that a superposition “is just” a state you do not know yet.

Now measure along X. The superposition gives |+⟩ with certainty; the mixture is still 50/50 — and is 50/50 along Y, and along every other axis, because I/2 has no direction. Turn the phase φ and the superposition’s X and Y statistics trade places while the mixture’s do nothing. The two matrices share a diagonal; the whole difference is the off-diagonal. A superposition is a definite state with a definite phase. A mixture is ignorance, and ignorance has no phase.

One thing worth knowing and not visible here: I/2 is also 50% |+⟩ and 50% |−⟩, and 50% of any other antipodal pair. The decomposition of a mixed state into pure states is not unique, and no measurement can recover which one “really” happened. ρ is the whole story.

The third scene starts with the Bell state Φ⁺ = (|00⟩ + |11⟩)/√2, which is pure — its 4×4 ρ has Tr ρ² = 1. Now suppose you hold one qubit and lose the other. What is the state of the qubit you have? Take the partial trace: sum over the outcomes of the qubit you gave up, which is what the smaller matrix on the right does to the larger one on the left.

The answer is I/2, the maximally mixed state. Not approximately — exactly, and for either qubit. The arrow disappears into the centre of the ball, the purity reads 0.500, and the entropy reads one full bit. The pair as a whole has zero entropy and each half has the maximum a qubit can have. That is what entanglement is, said in terms of ρ: information that lives in the correlations and in neither part.

Compare with |+⟩|0⟩. It is also pure, but trace out either qubit and what remains is still pure — the arrow stays on the surface. Then pick cos α|00⟩ + sin α|11⟩ and drag α from 0 to π/4. The reduced purity runs from 1 down to ½ following cos⁴α + sin⁴α, and the arrow slides down the z axis into the centre. How mixed a part is measures how entangled the whole was.

Nothing here evolves in time. The dephasing slider applies a channel once; the rates, T1, T2 and the bound T2 ≤ 2T1 are on the decoherence page. There is no amplitude damping in the first scene either, so the diagonal never changes there — a real qubit’s would. Measurement updates ρ in a way this page does not show (it reports probabilities, not post-measurement states), the entropy readout is for one qubit only, and the two-qubit states are fixed presets rather than anything you can build gate by gate — the simulator is for that.