Bounded and Unbounded Generators

For bounded Lindblad data, algebraic cancellation of the trace can be integrated along every state trajectory. For unbounded data, that step may fail because the formal expression has not specified the generator domain or excluded probability escaping through arbitrarily high levels. A pure-birth model makes the obstruction explicit: every finite-support calculation conserves probability, yet the minimal evolution can lose it.

Required background. The GKSL–Lindblad Theorem proves the bounded case. Generators of CP Semigroups defines domains and graph-norm cores.

Unbounded coefficients require a generator domain

Consider the symbolic expression

$$\mathcal L(\rho)=G\rho+\rho G^\dagger +\sum_jL_j\rho L_j^\dagger,\qquad G+G^\dagger+\sum_jL_j^\dagger L_j=0.$$

If $G$ or the total noise intensity is unbounded, these products need not exist as trace-class operators for every $\rho$. The second equality may hold only as a quadratic-form identity on a dense vector domain. It is then an infinitesimal balance identity, not yet a trace-preserving semigroup.

A construction must identify an operator or form domain, establish a closed generator or an appropriate weak evolution, and prove existence and conservation. The general minimal-semigroup approach constructs a CP trace-nonincreasing evolution from suitable dissipative data. Whether this minimal evolution is conservative is a further theorem; Chebotarev and Fagnola give useful sufficient criteria.

Even purely Hamiltonian evolution illustrates the domain issue. Self-adjoint $H$ defines $U_t\rho U_t^\dagger$ for every density operator, while the generator equals $-i[H,\rho]$ only where the derivative exists in trace norm. Finite-rank operators $|u\rangle\langle v|$ with $u,v\in D(H)$ are valid test elements, but a formal commutator written for an arbitrary state does not establish trace-norm differentiability.

An explosive birth process has zero formal trace derivative

Use basis states $|n\rangle$, $n\ge0$, and rates $\lambda_n=2^n$ in fixed inverse-time units. Let

$$L_n=\sqrt{\lambda_n}|n+1\rangle\langle n|, \qquad R=\sum_n\lambda_n|n\rangle\langle n|, \qquad G=-R/2.$$

Each individual $L_n$ is bounded, but their total intensity $R$ is unbounded. On finite linear combinations of matrix units, the formal generator is well defined and has zero trace. Its diagonal restriction is the pure-birth equation

$$\dot p_0=-\lambda_0p_0,\qquad \dot p_n=\lambda_{n-1}p_{n-1}-\lambda_np_n\quad(n\ge1).$$

For finitely supported $p$, summing the right-hand sides cancels exactly. The cancellation alone does not control the infinite-time sequence of jumps.

Starting in level zero, let the independent waiting times be $W_n$ with exponential rate $\lambda_n$. The time at which infinitely many jumps have occurred is

$$\zeta=\sum_{n=0}^\infty W_n,\qquad \mathbb E\zeta=\sum_{n=0}^\infty2^{-n}=2.$$

Since $\zeta$ is nonnegative and has finite expectation, it is finite almost surely. The minimal process terminates at explosion rather than assigning an additional state at infinity. Its finite-level probabilities therefore satisfy

$$\sum_n p_n(t)=\mathbb P(\zeta>t).$$

For example, Markov’s inequality gives $\mathbb P(\zeta>4)\le1/2$. At least half the probability has escaped the finite-level state space by this time, despite exact formal balance on every finite-support input.

The birth process has a normal CP extension

Let $p_{m|n}(t)$ be the minimal classical transition probability from $n$ to finite level $m\ge n$, with $p_{n|n}(t)=e^{-\lambda_nt}$. Set $S_t=\sum_ne^{-\lambda_nt/2}|n\rangle\langle n|$. Then define

$$\Phi_t(\rho)=S_t\rho S_t^\dagger +\sum_{m>n}p_{m|n}(t)\rho_{nn}|m\rangle\langle m|.$$

This is CP: besides $S_t$, its Kraus operators are $\sqrt{p_{m|n}(t)}|m\rangle\langle n|$ for $m>n$. Their completeness sum is diagonal with entries $\sum_{m\ge n}p_{m|n}(t)\le1$. Thus $\Phi_t$ is normal in the dual picture and trace nonincreasing.

The diagonal evolves by the classical transition semigroup. Off-diagonal entries evolve as $\rho_{mn}(t)=e^{-(\lambda_m+\lambda_n)t/2}\rho_{mn}(0)$. These two observations verify the semigroup law. Continuity follows first on finite matrix units and then, using contraction, on the whole trace class. Its generator agrees with the formal Lindblad expression on finite matrix units.

For the initial projector $|0\rangle\langle0|$, the trace at time $4$ is at most $1/2$. This is an explicit CP $C_0$ semigroup whose formal finite-support balance identity does not imply conservation.

A dense test space need not be a core

If trace cancellation held on the actual generator domain of a trace-nonincreasing $C_0$ semigroup, integration would prove conservation. If it held on a graph-norm core, continuity of trace would extend the cancellation to that full domain.

Therefore the finite-matrix-unit subspace in the explosive example, although dense in trace norm, cannot be a graph-norm core for the minimal generator. This is the precise mathematical location of the failed inference. One cannot choose a convenient dense test domain, perform a cancellation there, and silently identify its graph closure with the generator of a proposed evolution.

Finite cutoffs require similar care. Making the top level absorbing gives a conservative finite-level model. As the cutoff moves upward, escaped mass can accumulate at the moving boundary. Matrix elements at every fixed level may converge while the density operators fail to converge in trace norm. Conservation would pass through a trace-norm channel limit, so a limit that loses trace necessarily uses a weaker topology.

Exercises

Constant birth rates. Replace $\lambda_n=2^n$ by a constant $\lambda>0$. Does explosion occur?

Solution

The waiting times are independent identically distributed exponentials with positive mean $1/\lambda$. Their partial sums diverge almost surely by the law of large numbers. Equivalently, the number of births by finite time is Poisson distributed. The minimal process is conservative; there is no finite-time accumulation of infinitely many jumps.

Where the trace calculation stops. Why is $\sum_n\dot p_n=0$ not a valid proof of conservation in the explosive example?

Solution

Finite partial sums satisfy $d(\sum_{n=0}^Np_n)/dt=-\lambda_Np_N$. The boundary flux can survive when $N\to\infty$. Interchanging the infinite sum and derivative, or canceling separately divergent incoming and outgoing sums, requires justification absent from the formal calculation.

References

• Chebotarev, Alexander M., and Franco Fagnola. “Sufficient conditions for conservativity of minimal quantum dynamical semigroups.” Journal of Functional Analysis 153, 382–404 (1998). Preprint. Minimal CP evolutions, formal balance, and additional conservativity criteria.
• Engel, Klaus-Jochen, and Rainer Nagel. One-Parameter Semigroups for Linear Evolution Equations. Graduate Texts in Mathematics 194, Springer, 2000. doi:10.1007/b97696. Closed generators and the distinction between dense domains and cores.
• Lindblad, Göran. “On the generators of quantum dynamical semigroups.” Communications in Mathematical Physics 48, 119–130 (1976). doi:10.1007/BF01608499. The bounded theorem whose hypotheses exclude this unbounded-intensity example.