Let $H$ be a Hilbert space, which we interpret as a space of quantum states.

- If $U(t):H\to H$ is a unitary
**norm-continuous**one-parameter group with $U(0)=I$, (essentially)*Cauchy's functional equation*says it has the form $$U(t)=e^{itA}$$ for some**bounded**self-adjoint Hamiltonian $A$. - If $U(t):H\to H$ is a unitary
**strongly-continuous**one-parameter group with $U(0)=I$,*Stone's theorem*says it has the form $$U(t)=e^{itA}$$ for some**unbounded**self-adjoint Hamiltonian $A$.

Now let $\mathcal{S}_1$ be the set of trace-class operators of trace $1$ on $H$, which we interpret as states in an open quantum system.

- If $\mathcal{T}(t):\mathcal{S}_1\to\mathcal{S}_1$ is a trace-preserving completely positive
**norm-continuous**one-parameter group with $\mathcal{T}(0)=I$,*Lindblad's theorem*says it has the form $$\mathcal{T}(t)=e^{t\mathcal{L}}$$ where $$\mathcal{L}(\rho)=-i[A,\rho]+\sum_{k=0}^{\infty}L_k\rho L_k^*+L_k^*L_k\rho$$ is a Lindbladian with**bounded**$A$, $L_k$ where $A$ is self-adjoint and $\sum_{k=0}^{\infty}L_k^*L_k$ is strongly summable.

My question is:

**What can we say in this case when $\mathcal{T}(t)$ is only strongly continuous?**