advanced30 min studyLesson 938 in Physics

Numerical Lindblad evolution and positivity

Distinguish a physical finite master equation from a numerical update that only preserves trace.

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What you’ll understand

  • State the finite GKSL generator with nonnegative rates and declared operator units.
  • Compute exact amplitude-damping populations and coherence.
  • Use density eigenvalues to expose a nonphysical Euler update.

An open quantum system is generally described by a density matrix rather than one pure wavefunction. Its evolution need not conserve the purity or energy of the subsystem, but a valid unconditional state must remain positive and have unit trace. A Lindblad-form generator can supply those properties under its mathematical assumptions. A numerical method applied to that generator does not automatically inherit them, and convergence to the chosen equation still does not validate the physical approximations used to obtain it.

Specify a finite completely positive evolution model

For a finite-dimensional system, take Hermitian H, dimensionless jump operators L-r, and nonnegative rates gamma-r with inverse-time units. The GKSL equation combines the Hamiltonian commutator with jump and anticommutator terms. With fixed coefficients this generates a completely positive trace-preserving semigroup. Complete positivity matters because the map must remain positive when the system is considered alongside an untouched ancillary system, not merely for a few tested density matrices.

A physical derivation may invoke weak coupling, short environmental correlations and a secular or other controlled approximation. Those assumptions are not implied by observing that a fitted generator has a convenient matrix form. Strongly correlated initial system-environment states, memory effects or an inadequate Hilbert cutoff can require a different model. Identify whether the equation is stipulated phenomenology, a derived approximation or a benchmark, and vary numerical controls separately from those physical assumptions.

ρ˙=−iℏ[H,ρ]+∑rγr(LrρLr†−12{Lr†Lr,ρ}),γr≥0\dot\rho=-\frac{i}{\hbar}[H,\rho]+\sum_r\gamma_r\left(L_r\rho L_r^\dagger-\tfrac12\{L_r^\dagger L_r,\rho\}\right),\qquad \gamma_r\ge0
Rho is dimensionless, H has energy units and each gamma has inverse-time units. The finite fixed-coefficient GKSL assumptions support the stated semigroup result.↔ Scroll wide equations horizontally.

Solve an amplitude-damping benchmark analytically

Choose zero Hamiltonian and a jump operator taking excited state one to ground state zero. The excited population decays with the rate gamma, while an off-diagonal coherence decays at half that rate. The ground population follows from unit trace. A coherent initial pure state generally becomes mixed at intermediate times before reaching the ground state. Matching only its populations would miss part of the benchmark.

For a Hermitian two-by-two trace-one matrix, nonnegative diagonal entries and nonnegative determinant ensure positive semidefiniteness. Its purity is another useful diagnostic, but purity alone does not replace the eigenvalue test for general matrices. The Bloch vector of a mixed qubit lies inside the unit sphere; a pure-state-only display can reproduce its populations while having too much coherence. Such a display must not be interpreted as a dissipative evolution solver.

ρ11(t)=e−γtρ11(0),ρ01(t)=e−γt/2ρ01(0),ρ00(t)=1−ρ11(t)\rho_{11}(t)=e^{-\gamma t}\rho_{11}(0),\qquad \rho_{01}(t)=e^{-\gamma t/2}\rho_{01}(0),\qquad \rho_{00}(t)=1-\rho_{11}(t)
These exact formulas apply to the declared zero-Hamiltonian two-state amplitude-damping generator. Off-diagonal entries remain complex conjugates.↔ Scroll wide equations horizontally.

Audit the discrete map and its refinement

A generic ODE integrator can preserve trace and Hermiticity while generating a negative eigenvalue at a finite timestep. Forward Euler makes this visible immediately for a sufficiently large amplitude-damping step. Smaller steps can reduce a particular violation, but positivity for a few initial states does not establish complete positivity of the map for every system-and-ancilla state. Exponential or specially constructed channel updates offer stronger structure when their implementation is controlled.

Monitor trace, Hermiticity, the smallest eigenvalue, observable convergence and solver residuals together. Clipping negative eigenvalues or renormalizing after a step changes the method and can hide rather than quantify its error. Adaptive tolerances and Hilbert dimension must be varied until the target observable stabilizes; positivity at each finite size does not certify that the chosen bath model or truncated operators describe the intended experiment. Numerical stability and physical validity remain separate questions.

ρn+1=ρn+Δt L(ρn),tr⁡ρn+1=1 ⟹̸ ρn+1≥0\rho_{n+1}=\rho_n+\Delta t\,\mathcal L(\rho_n),\qquad \operatorname{tr}\rho_{n+1}=1\ \not\Longrightarrow\ \rho_{n+1}\ge0
The trace implication fails for a generic discrete update even when the exact generator is GKSL. A numerical negative eigenvalue must be interpreted with precision and convergence evidence rather than erased silently.↔ Scroll wide equations horizontally.

Damping a coherent plus state

Start from plus-state density and evolve with the declared amplitude-damping equation until γt=ln⁡2\gamma t=\ln2. Find populations, coherence, determinant and purity. Does the result remain pure?

  1. The excited population starts at one half and is reduced by one half, giving ρ11=1/4\rho_{11}=1/4 and ground population 3/43/4. The coherence is reduced by one over square root two, giving ρ01=1/(22)\rho_{01}=1/(2\sqrt2).

  2. The determinant is (3/4)(1/4)−[1/(22)]2=3/16−1/8=1/16(3/4)(1/4)-[1/(2\sqrt2)]^2=3/16-1/8=1/16. Nonnegative populations and determinant confirm a positive two-state density matrix.

  3. The purity is (3/4)2+(1/4)2+2[1/(22)]2=7/8(3/4)^2+(1/4)^2+2[1/(2\sqrt2)]^2=7/8, below one. The state has become mixed, with unit trace, under this exact subsystem evolution.

The result

The exact state has populations three quarters and one quarter, coherence one over twice square-root-two, positive determinant one sixteenth and purity seven eighths.

Explore the model

Interactive lab

Interactive lab · quantum

A pure display cannot reproduce a mixed state completely

Reset and set mixing angle sixty degrees, phase zero; predict excited probability one quarter and Bloch x square-root-three over two. Compare a fresh Reset run at sixty degrees, phase 180 degrees; predict the same populations and opposite Bloch x. These are pure states with different coherences, not the mixed damping state or a Lindblad evolution.

A pure display cannot reproduce a mixed state completelyA normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Current readings: Probability |0⟩: 71.1%; Probability |1⟩: 28.9%; Bloch (x, y, z): 0.78, 0.45, 0.42. All parameters are adjustable with the labeled controls below.xyz|0⟩|1⟩Basis probabilitiesP(0)71.1%P(1)28.9%
Bloch vector / P(0)P(1)Phase latitude
Probability |0⟩
71.1%
Probability |1⟩
28.9%
Bloch (x, y, z)
0.78, 0.45, 0.42
t = 0 s

A normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forMatching populations in a pure-qubit instrument does not reproduce density-matrix coherence or mixedness.

Check your understanding

What does unit trace and Hermiticity of a numerical density matrix establish about positivity?

Try it yourself

Apply one forward-Euler step of the zero-Hamiltonian amplitude-damping equation to the initial excited state, with γΔt=1.2\gamma\Delta t=1.2. Compute the two populations and eigenvalues. Compare the excited population with the exact same-time solution.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The initial excited population has derivative minus gamma, and the ground population has the opposite derivative.

Compare with the solution

Euler gives excited population 1−1.2=−0.21-1.2=-0.2 and ground population 1.2. The diagonal matrix is Hermitian with trace one, but its eigenvalues 1.2 and minus 0.2 violate positivity. Exact evolution instead gives excited population e−1.2≈0.3012e^{-1.2}\approx0.3012 and ground population about 0.6988. This finite step fails as a state update despite using a valid GKSL generator; it does not refute the positivity of the exact equation.

Take it with you

  • Exact GKSL structure and a discrete integrator have different guarantees.
  • Populations, coherences and eigenvalues test different state properties.
  • Convergence and positivity cannot by themselves validate environmental or truncation assumptions.

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