|q⟩ Bad Qubits

intermediate · Physics · Density Matrices & Mixed States

Time Evolution of ρ

States evolve. For a closed quantum system the density operator evolves by a rule that follows directly from the Schrödinger equation, and it has a clean differential form.

Unitary evolution of ρ

If a pure state evolves as ∣ψ(t)⟩=U(t)∣ψ(0)⟩|\psi(t)\rangle = U(t)|\psi(0)\rangle with a unitary U(t)U(t), then its density operator ρ=∣ψ⟩⟨ψ∣\rho = |\psi\rangle\langle\psi| evolves as

ρ(t)=U(t) ρ(0) U†(t).\rho(t) = U(t)\,\rho(0)\,U^{\dagger}(t).

By linearity this holds for any mixture too: each ∣ψi⟩⟨ψi∣|\psi_i\rangle\langle\psi_i| transforms the same way, and the probabilities pip_i are untouched. The "sandwich" UρU†U\rho U^{\dagger} is the universal rule for closed-system evolution.

The von Neumann equation

Differentiating ρ(t)=Uρ0U†\rho(t) = U\rho_0 U^{\dagger} with U(t)=e−iHt/ℏU(t) = e^{-iHt/\hbar} gives the von Neumann equation (also called the quantum Liouville equation):

iℏ dρdt=[H,ρ]=Hρ−ρH.i\hbar\,\frac{d\rho}{dt} = [H, \rho] = H\rho - \rho H.

This is the density-operator counterpart of the Schrödinger equation. Note the sign and ordering: it is the opposite of the Heisenberg equation of motion for operators, because ρ\rho is a state, not an observable.

Properties preserved by the flow

Unitary evolution preserves everything that makes ρ\rho a valid, equally-pure state:

That last point is important: turning a pure state into a mixed one requires either entanglement with an environment (then take a partial trace) or a non-unitary process. Pure unitary dynamics alone moves a state around the Bloch sphere without changing ∣r⃗∣|\vec r|.

Try it

Apply a ZZ gate to ρ=(3/41/41/41/4)\rho = \begin{pmatrix}3/4 & 1/4\\ 1/4 & 1/4\end{pmatrix} via ρ′=ZρZ†\rho' = Z\rho Z^{\dagger} (with Z†=ZZ^{\dagger} = Z), and return the coherence ρ01′\rho'_{01}. The ZZ gate flips the sign of the off-diagonal term, sending 1/4→−1/41/4 \to -1/4 — a phase flip, with the populations unchanged.

Run your code to see the quantum state.

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