Yes, even more, $S_i$ and $T_i$ can be chosen the same. This is is mapping is know as a Krauss decomposition: it is is the generalisation of a unitary transformation that includes decoherence. The Krauss decomposition is also known to be a completely-positive trace-preserving map, meaning it maps density matrices to density matrices.

The Lindblad equation is the continuous version of this trace-preserving map; for an infinitesimal timestep you could have for example $S_i=\sqrt{dt}L_i$ and $S_0=1-dt\sum_j L_i^\dagger L_i$.

After a finite evolution time, the result is still a completely positive trace preserving maps, so the decomposition should exist in principle, although I don't have an immediate clean expression for it from the top of my head.

CAVEAT: I have assumed Open quantum systems means a Markovian one, where the Lindblad equation is actually valid. More general cases of open quantum systems, with an environment that is not memory-free, exist too and their situation is more complicated.

EDIT: I missed that $S_i$ and $T_i$ have to be unitary. This is not necessarily the case for the Krauss decomposition, so choosing them the same may not be allowed in general.

Yes, even more, $S_i$ and $T_i$ can be chosen the same. This mapping is known as a Kraus decomposition: it is the generalisation of a unitary transformation that includes decoherence. The Kraus decomposition is also known to be a completely positive trace-preserving map, meaning it maps density matrices to density matrices.

The Lindblad equation is the continuous version of this trace-preserving map; for an infinitesimal timestep you could have for example $S_i=\sqrt{dt}L_i$ and $S_0=1-dt\sum_j L_i^\dagger L_i$.

After a finite evolution time, the result is still a completely positive trace-preserving map, so the decomposition should exist in principle, although I don't have an immediate clean expression for it from the top of my head.

CAVEAT: I have assumed open quantum systems means a Markovian one, where the Lindblad equation is actually valid. More general cases of open quantum systems, with an environment that is not memory-free, exist too and their situation is more complicated.

EDIT: I missed that $S_i$ and $T_i$ have to be unitary. This is not necessarily the case for the Kraus decomposition, so choosing them the same may not be allowed in general.

Yes, even more, $S_i$ and $T_i$ can be chosen the same. This is is mapping is know as a Krauss decomposition: it is is the generalisation of a unitary transformation that includes decoherence. The Krauss decomposition is also known to be a completely-positive trace-preserving map, meaning it maps density matrices to density matrices.

The Lindblad equation is the continuous version of this trace-preserving map; for an infinitesimal timestep you could have for example $S_i=\sqrt{dt}L_i$ and $S_0=1-dt\sum_j L_i^\dagger L_i$.

After a finite evolution time, the result is still a completely positive trace preserving maps, so the decomposition should exist in principle, although I don't have an immediate clean expression for it from the top of my head.

CAVEAT: I have assumed Open quantum systems means a Markovian one, where the Lindblad equation is actually valid. More general cases of open quantum systems, with an environment that is not memory-free, exist too and their situation is more complicated.

Yes, even more, $S_i$ and $T_i$ can be chosen the same. This is is mapping is know as a Krauss decomposition: it is is the generalisation of a unitary transformation that includes decoherence. The Krauss decomposition is also known to be a completely-positive trace-preserving map, meaning it maps density matrices to density matrices.

The Lindblad equation is the continuous version of this trace-preserving map; for an infinitesimal timestep you could have for example $S_i=\sqrt{dt}L_i$ and $S_0=1-dt\sum_j L_i^\dagger L_i$.

After a finite evolution time, the result is still a completely positive trace preserving maps, so the decomposition should exist in principle, although I don't have an immediate clean expression for it from the top of my head.

CAVEAT: I have assumed Open quantum systems means a Markovian one, where the Lindblad equation is actually valid. More general cases of open quantum systems, with an environment that is not memory-free, exist too and their situation is more complicated.

EDIT: I missed that $S_i$ and $T_i$ have to be unitary. This is not necessarily the case for the Krauss decomposition, so choosing them the same may not be allowed in general.