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Quantum Mechanics Core

Density Matrices & Mixed States

How to describe a qubit when you're not sure exactly what state it's in

By the end of this topic you'll be able to

Explain the difference between a pure state and a mixed state
Write the density matrix for a pure state and for a maximally mixed state
Explain why a qubit entangled with another system looks mixed on its own

Up to now we have always assumed we know exactly which state |ψ⟩ our system is in. Binney and Skinner point out how unrealistic this is even in principle: establishing a system's state this precisely requires that a measurement has just been made, collapsing it into a known eigenstate — a procedure that makes no allowance for the experimental error that is endemic in every real laboratory.

For a macroscopic object the assumption collapses entirely. How would one possibly determine the individual quantum states of the ~10²³ atoms in a diamond, given that measuring even one of them disturbs the rest?

So suppose instead that we admit we don't know the system's state, but believe it is in one of a complete set of states {|n⟩}, each with some probability pₙ. Crucially, this is not the same as saying the system is in the definite superposition Σₙ√pₙ|n⟩ — that would be a perfectly good pure state, and we are explicitly admitting we don't have one.

Binney and Skinner call this an impure state, reserving 'pure state' for an ordinary ket |ψ⟩; the field as a whole more often says mixed state, but the idea is identical either way: the object we need is the density operator, ρ ≡ Σₙ pₙ|n⟩⟨n|.

For an ordinary pure state, ρ reduces to the familiar projector |ψ⟩⟨ψ|. For a 50/50 classical toss between |0⟩ and |1⟩ — uncertainty about which was prepared, not a superposition of the two — ρ = ½|0⟩⟨0| + ½|1⟩⟨1| instead, a genuinely different mathematical object from the superposition (|0⟩+|1⟩)/√2 even though a casual glance might confuse them.

The terminology 'impure' is, as Binney and Skinner note, slightly unfortunate: a system described by such a ρ is in a perfectly good quantum state. What's impure is our knowledge of which state that is — not the state itself.

Density matrices become unavoidable, not merely convenient, for a qubit that is entangled with something else. Given the Bell state |Φ⁺⟩=(|00⟩+|11⟩)/√2, there is simply no ket that describes qubit A alone; tracing out B gives ρ_A = I/2, the maximally mixed state, and A looks on its own exactly like a fair coin, even though the full two-qubit system sits in a perfectly definite pure state.

|0⟩|1⟩
Purity (Bloch-vector length)1.00

At purity 1 the arrow sits on the surface — a pure state. Drag it down to 0 and the arrow shrinks to the center — the maximally mixed state I/2, indistinguishable from a fair coin no matter what you measure.

That apparent randomness in A is not fundamental uncertainty about A — it is the unavoidable price of describing A while deliberately ignoring B. The same formalism is the natural language for noise: any qubit that interacts with an uncontrolled environment ends up mixed, and the whole theory of decoherence and quantum error correction, coming up next, is written in terms of ρ rather than |ψ⟩.

Try It Yourself

Worked Example

Write the density matrix for the pure state |+⟩ = (|0⟩+|1⟩)/√2, then for a mixed state that's |0⟩ with 70% probability and |1⟩ with 30% probability. Compute Tr(ρ²) for each and explain what the difference means.

  1. 1Pure state: ρ = |+⟩⟨+| = (1/2)(|0⟩+|1⟩)(⟨0|+⟨1|) = (1/2)(|0⟩⟨0| + |0⟩⟨1| + |1⟩⟨0| + |1⟩⟨1|) = (1/2)[[1,1],[1,1]] in matrix form.
  2. 2Tr(ρ²) for this pure state: ρ² = (1/4)[[1,1],[1,1]][[1,1],[1,1]] = (1/4)[[2,2],[2,2]] = (1/2)[[1,1],[1,1]] = ρ itself (a property unique to pure-state projectors), so Tr(ρ²) = Tr(ρ) = 1/2+1/2 = 1.
  3. 3Mixed state: ρ = 0.7|0⟩⟨0| + 0.3|1⟩⟨1| = [[0.7, 0],[0, 0.3]] — diagonal, since there's no coherent superposition, just classical uncertainty about which basis state you have.
  4. 4Tr(ρ²) for this mixed state: ρ² = [[0.49, 0],[0, 0.09]], so Tr(ρ²) = 0.49 + 0.09 = 0.58.
Answer

The pure state gives Tr(ρ²) = 1 exactly; the mixed state gives Tr(ρ²) = 0.58 < 1. Tr(ρ²) = 1 is a general litmus test for purity — any ρ with Tr(ρ²) < 1 is a genuine statistical mixture, not a coherent superposition, no matter how it's written.

Reference

Pure state density matrixRank-1 matrix; Tr(ρ²) = 1
Mixed state density matrixWeighted mixture; Tr(ρ²) < 1
Maximally mixed qubitComplete uncertainty; looks like a fair coin on measurement
Partial traceDescribes subsystem A alone, ignoring B

Quick Check

What's the density matrix of qubit A alone, when the full 2-qubit system is in the Bell state |Φ⁺⟩?

What distinguishes a mixed state from a pure superposition state?