---
schema_version: 1
id: PHYS-2026-08-01-01
date: 2026-08-01
updated_at: 2026-08-01
title: Liouvillian gaps, Jordan blocks, and observable relaxation
summary: "Why a finite Liouvillian gap fixes an asymptotic exponential scale but not the full relaxation curve, mixing time, or decay seen by a chosen observable."
language: en
entry_kind: daily
status: published
level: graduate-advanced
user_difficulty: unrated
domains:
- quantum-theory
- statistical-mechanics
- mathematical-physics
estimated_minutes: 60
---
## Key claim
**A nonzero Liouvillian spectral gap supplies an asymptotic exponential scale. It does not, by itself, specify the relaxation law or the time required to reach a given accuracy.**
Jordan blocks add polynomial factors. Nonorthogonal left and right eigenmodes alter prefactors. An initial state and an observable may also have zero overlap with the slowest mode.
## Theme
**Liouvillian spectra, exceptional points, and finite-accuracy mixing.**
## Guiding question
If every eigenvalue of a Markovian generator is known, what information is still missing from a prediction of observable relaxation?
## Setup
Consider a resonantly driven qubit with Markovian pure dephasing:
$$
\dot\rho=\mathcal L(\rho)
=-\ii\left[\frac{\Omega}{2}\sigma_x,\rho\right]
+\kappa(\sigma_z\rho\sigma_z-\rho),
\qquad \Omega,\kappa>0.
$$
This is a time-independent GKSL equation for the assumed effective model. It is not an exact statement about every microscopic bath that could produce dephasing.
Write
$$
\rho=\frac12(I+x\sigma_x+y\sigma_y+z\sigma_z).
$$
The Bloch equations are
$$
\dot x=-2\kappa x,
$$
and
$$
\frac{\dd}{\dd t}
\begin{pmatrix}y\\z\end{pmatrix}
=M\begin{pmatrix}y\\z\end{pmatrix},
\qquad
M=
\begin{pmatrix}
-2\kappa&-\Omega\\
\Omega&0
\end{pmatrix}.
$$
For $\Omega\ne0$, the unique stationary state is $\rho_{\rm ss}=I/2$.
> [!margin: Gap convention]
> For a finite-dimensional primitive semigroup, we use $\Delta_{\mathcal L}=\min_{\lambda\ne0}[-\operatorname{Re}\lambda]$. Primitivity excludes additional stationary or purely oscillatory sectors.
The Liouvillian spectrum contains $0$, the $x$-sector eigenvalue $-2\kappa$, and the two eigenvalues of $M$.
## Analysis
### Derivation: a finite-gap exceptional point
The characteristic polynomial of $M$ is
$$
\det(M-\lambda I)
=\lambda^2+2\kappa\lambda+\Omega^2.
$$
Hence
$$
\lambda_\pm=-\kappa\pm\sqrt{\kappa^2-\Omega^2}.
$$
This gives three regimes:
$$
\begin{array}{c|c}
\Omega<\kappa & \lambda_\pm\text{ real: overdamped}\\
\Omega=\kappa & \lambda_+=\lambda_-=-\kappa\text{: exceptional point}\\
\Omega>\kappa & \lambda_\pm=-\kappa\pm\ii\sqrt{\Omega^2-\kappa^2}
\end{array}
$$
At $\Omega=\kappa$,
$$
M=-\kappa I+N,
\qquad
N=\kappa
\begin{pmatrix}
-1&-1\\
1&1
\end{pmatrix},
\qquad
N^2=0,quad N\ne0.
$$
The repeated eigenvalue has only one eigenvector in this sector. The propagator is therefore
$$
\ee^{Mt}=\ee^{-\kappa t}\ee^{Nt}
=\ee^{-\kappa t}(I+tN).
$$
The gap is still
$$
\Delta_{\mathcal L}=\kappa.
$$
Yet a generic component in this sector contains
$$
(A+Bt)\ee^{-\kappa t},
$$
not a single exponential. Exceptional points of two-level Lindblad generators and their defective eigenmodes are analyzed explicitly by [Hatano](https://arxiv.org/abs/1903.04676).
For the initial condition $z(0)=1$, $y(0)=0$, eliminate $y$ using $\dot z=\Omega y$. This gives
$$
\ddot z+2\kappa\dot z+\Omega^2z=0.
$$
At the exceptional point,
$$
\boxed{z(t)=(1+\kappa t)\ee^{-\kappa t}}.
$$
The logarithmic decay rate still approaches the gap:
$$
-\lim_{t\to\infty}\frac{1}{t}\ln|z(t)|=\kappa.
$$
The polynomial changes the relaxation curve and produces a subleading $\ln\ln(1/\varepsilon)$ correction to the time needed to reach accuracy $\varepsilon$.
### Scope and assumptions: gap, mixing time, and mode geometry
For a diagonalizable finite-dimensional generator, write its decaying part as
$$
\ee^{t\mathcal L}-\mathcal P_{\rm ss}
=\sum_{n\ge1}\ee^{\lambda_nt}\mathcal P_n,
$$
where
$$
\mathcal P_n=|R_n\rangle\!\rangle\langle\!\langle L_n|
$$
for biorthogonally normalized right and left eigenmatrices. An observable deviation is
$$
\delta\langle O(t)\rangle
=\sum_{n\ge1}\ee^{\lambda_nt}
\langle\!\langle O|R_n\rangle\!\rangle
\langle\!\langle L_n|\delta\rho(0)\rangle\!\rangle.
$$
The eigenvalues fix the available exponential factors. The two overlaps decide which factors appear in a particular experiment.
If the coefficient of every mode with real part $-\Delta_{\mathcal L}$ vanishes, the observable relaxes faster than the global gap suggests. Oscillatory eigenvalues add phases, and Jordan chains add powers of $t$.
> [!margin: Observable gap]
> One may define an effective decay rate for a chosen pair $(O,\rho_0)$ from the slowest mode with nonzero overlap. This is not a new spectral gap of $\mathcal L$.
Non-normality matters even away from an exact exceptional point. The norms of the projectors $\mathcal P_n$ can be large when left and right eigenvectors are nearly parallel.
Large modal contributions may then cancel at one time and reinforce at another. This changes intermediate-time behavior and sensitivity to perturbations, although a completely positive trace-preserving map remains contractive in trace distance between states.
Define the worst-case trace-distance mixing time by
$$
t_{\rm mix}(\varepsilon)
=\inf\left\{t:\sup_{\rho}
\frac12\left\|\ee^{t\mathcal L}(\rho)-\rho_{\rm ss}\right\|_1
\le\varepsilon\right\}.
$$
A bound of the schematic form
$$
\sup_\rho\frac12\left\|\ee^{t\mathcal L}(\rho)-\rho_{\rm ss}\right\|_1
\le C(1+t^{p})\ee^{-\Delta_{\mathcal L}t}
$$
requires more than $\Delta_{\mathcal L}$. The constant $C$ depends on eigenmode geometry and norm conversion; $p$ records the largest relevant Jordan block.
Consequently,
$$
t_{\rm mix}(\varepsilon)
\lesssim
\frac{1}{\Delta_{\mathcal L}}
\left[\ln C+\ln\frac1\varepsilon+p\ln t_{\rm mix}\right].
$$
The gap fixes the leading exponential scale at fixed finite dimension. It does not determine $C$, $p$, or the finite-accuracy time.
> [!margin: Many-body scaling]
> For a family $\mathcal L_L$, a size-independent gap need not give size-independent trace-norm mixing. Prefactors can grow with Hilbert-space dimension or with properties of $\rho_{\rm ss}$. Logarithmic Sobolev constants provide stronger convergence control in many settings.
Quantum logarithmic Sobolev inequalities and their relation to convergence bounds are developed by [Kastoryano and Temme](https://arxiv.org/abs/1207.3261). A 2026 preprint further proposes mode trace norms as predictors beyond the gap [Zhou](https://arxiv.org/abs/2601.06256).
The last source is recent and should be read as a current proposal, not as the basis of the finite-dimensional Jordan-block result above.
### Physical interpretation: critical damping, slow sectors, and memory
The $y$-$z$ equations are identical to a classical damped oscillator:
$$
\ddot z+2\kappa\dot z+\Omega^2z=0.
$$
The Liouvillian exceptional point is the critical-damping point. The term $t\ee^{-\kappa t}$ comes from a repeated defective root, not from a closing gap.
In the strong-dephasing regime $\Omega\ll\kappa$,
$$
\lambda_+
=-\kappa+\kappa\sqrt{1-\frac{\Omega^2}{\kappa^2}}
\simeq-\frac{\Omega^2}{2\kappa}.
$$
Thus
$$
\tau_{\rm slow}\simeq\frac{2\kappa}{\Omega^2}.
$$
Increasing the microscopic dephasing rate can slow the population dynamics. Frequent destruction of phase coherence suppresses the coherent transfer that changes $z$, producing a quantum-Zeno-like slow mode.
A cluster of slow eigenvalues separated from the rest can create a metastable window. The system first approaches a low-dimensional manifold and only later reaches its unique stationary state.
Identifying that manifold requires the slow eigenmatrices, not only the smallest nonzero real part. This spectral construction is developed by [Macieszczak, Guță, Lesanovsky, and Garrahan](https://arxiv.org/abs/1512.05801).
> [!margin: Markovian boundary]
> If the reduced evolution contains a memory kernel, a single constant $\mathcal L$ need not organize the full dynamics. Exceptional points of an approximate Bloch generator may then miss features of the microscopic evolution.
For example, a time-nonlocal equation may take the form
$$
\dot\rho(t)=\int_0^t\dd t'\,K(t-t')\rho(t').
$$
The comparison by [Seshadri, Li, and Galperin](https://arxiv.org/abs/2401.04011) shows how the approximations leading to a Bloch master equation can limit Liouvillian-exceptional-point reasoning in a driven two-level model.
## False claim to diagnose
> If a primitive finite-dimensional Lindbladian has gap $\Delta>0$, every observable deviation must equal $C\ee^{-\Delta t}$ at late times.
This statement contains three errors.
First, a Jordan block at the gap edge gives $t^m\ee^{-\Delta t}$. Second, complex eigenvalues add oscillations. Third, the initial state or observable may have zero overlap with all gap-edge modes.
The defensible statement is narrower:
> At fixed finite dimension, the gap identifies the slowest available asymptotic exponential rate. The realized decay law also depends on generalized eigenspaces and experimental overlaps.
## What follows — and what does not
| Statement | Status |
| --- | --- |
| The qubit model has $\lambda_\pm=-\kappa\pm\sqrt{\kappa^2-\Omega^2}$. | Exact within the stated GKSL model |
| $\Omega=\kappa$ is a second-order exceptional point. | Exact matrix result |
| The gap at that point is $\kappa$. | Exact with the stated gap convention |
| Every component decays as a pure $\ee^{-\kappa t}$. | False; a Jordan chain gives $t\ee^{-\kappa t}$ |
| A finite gap guarantees a unique stationary state. | False without primitivity or equivalent assumptions |
| The gap alone fixes $t_{\rm mix}(\varepsilon)$. | False; prefactors, Jordan order, and norms enter |
| A chosen observable must decay at the global gap. | False when its slow-mode overlap vanishes |
| A size-independent many-body gap implies uniform rapid mixing. | Not without further bounds |
| The same spectral analysis applies unchanged to non-Markovian dynamics. | False in general |
## Exercise
For
$$
M=
\begin{pmatrix}
-2\kappa&-\Omega\\
\Omega&0
\end{pmatrix},
$$
complete the following tasks.
1. Derive $\lambda_\pm$ and classify the three damping regimes.
2. At $\Omega=\kappa$, derive $z(t)$ for $z(0)=1$, $y(0)=0$ without diagonalizing $M$.
3. Compute $y(t)$ and the trace distance from $I/2$.
4. Estimate the small-$\varepsilon$ time at which $|z(t)|\le\varepsilon$.
5. For $\Omega\ll\kappa$, derive the slow rate and explain its dependence on $\kappa$.
Hint 1
Use $\dot z=\Omega y$ to obtain a second-order equation. Critical damping gives a repeated root.
Hint 2
At the exceptional point, set $u=\kappa t$. Solving $(1+u)\ee^{-u}=\varepsilon$ iteratively gives $u=\ln(1/\varepsilon)+\ln u+o(1)$.
**Oral check 1.** Why can a polynomial prefactor change the mixing time without closing the Liouvillian gap?
**Oral check 2.** What two overlaps decide whether a slow Liouvillian mode appears in a measured signal?
Solution
The characteristic equation is
$$
\lambda^2+2\kappa\lambda+\Omega^2=0,
$$
so
$$
\lambda_\pm=-\kappa\pm\sqrt{\kappa^2-\Omega^2}.
$$
The roots are distinct and real for $\Omega<\kappa$, repeated and defective for $\Omega=\kappa$, and complex conjugates for $\Omega>\kappa$.
Eliminating $y$ gives
$$
\ddot z+2\kappa\dot z+\Omega^2z=0.
$$
At $\Omega=\kappa$,
$$
z(t)=(A+Bt)\ee^{-\kappa t}.
$$
The conditions $z(0)=1$ and $\dot z(0)=\kappa y(0)=0$ give $A=1$ and $B=\kappa$:
$$
z(t)=(1+\kappa t)\ee^{-\kappa t}.
$$
Since $y=\dot z/\kappa$,
$$
y(t)=-\kappa t\ee^{-\kappa t}.
$$
Here $x(t)=0$. The trace distance from $I/2$ is
$$
D\!\left(\rho(t),\frac I2\right)
=\frac12\sqrt{y(t)^2+z(t)^2}
=\frac{\ee^{-\kappa t}}{2}
\sqrt{(\kappa t)^2+(1+\kappa t)^2}.
$$
For the simpler condition $|z(t)|\le\varepsilon$, let $u=\kappa t$. At small $\varepsilon$,
$$
u=\ln\frac1\varepsilon+\ln\ln\frac1\varepsilon+o(1).
$$
Thus the gap supplies the coefficient $1/\kappa$, while the Jordan block adds the double-logarithmic correction.
Finally,
$$
\lambda_+
=-\kappa+\kappa\sqrt{1-\frac{\Omega^2}{\kappa^2}}
\simeq-\frac{\Omega^2}{2\kappa}.
$$
The slow time is $2\kappa/\Omega^2$. Strong dephasing suppresses the coherent process that transfers population between the $\sigma_z$ states.
## Check your understanding
Two primitive Lindbladians have the same spectrum, including multiplicities, but one is diagonalizable and the other has a Jordan block at the gap edge.
Which asymptotic quantity must agree, and which features of a finite-time relaxation experiment can differ?
You may also reply with “deeper,” “too easy,” “too hard,” or your derivation.
## Further Reading
- [A short introduction to the Lindblad Master Equation](https://arxiv.org/abs/1906.04478)
- [Exceptional points of the Lindblad operator of a two-level system](https://arxiv.org/abs/1903.04676)
- [Quantum logarithmic Sobolev inequalities and rapid mixing](https://arxiv.org/abs/1207.3261)
- [Towards a theory of metastability in open quantum dynamics](https://arxiv.org/abs/1512.05801)
- [Liouvillian exceptional points of an open driven two-level system](https://arxiv.org/abs/2401.04011)
- [Universal Predictors for Mixing Time more than Liouvillian Gap](https://arxiv.org/abs/2601.06256)
## Connections and next step
- Generator level: the spectrum supplies exponential rates in operator space.
- Geometric level: projectors, Jordan chains, and pseudospectral sensitivity shape finite-time dynamics.
- Experimental level: state preparation and observable choice select which modes are visible.
- Many-body level: rapid mixing requires uniform control beyond a finite-size gap.
- Approximation level: a constant Lindbladian presupposes time-local Markovian reduction.
- Next step: compare spectral-gap bounds with logarithmic Sobolev and pseudospectral bounds for local Lindbladians.
- Revisit: connect the slow strong-dephasing mode to the quantum-Zeno limit and to metastable manifolds.