1
$\begingroup$

Is it true that a generic operator that is annihilated by the Lindblad superoperator (with both Hamiltonian and dissipative parts of the dynamics) has to be annihilated separately by both the Hamiltonian dynamics and the dissipative dynamics, or are there cases where it is only the sum of the two terms that vanishes, but not each of them separately?

$\endgroup$
1
  • $\begingroup$ They don't have to be annihilated by both individually. You can always absorb some terms from one into the other (they are not unique) so this can't be required. $\endgroup$ Commented Mar 29 at 3:41

1 Answer 1

1
$\begingroup$

To give an explicit example, consider $H=\sigma_x$, $V=|0\rangle\langle 1|$, and $$\mathcal L:=-i[H,\cdot]+V(\cdot)V^\dagger-\frac12V^\dagger V(\cdot)-(\cdot)\frac12V^\dagger V\,.$$ The (in fact unique) steady state of the corresponding dynamics then is $$ \sigma=\begin{pmatrix}\frac59&\frac{2i}9\\-\frac{2i}9&\frac49\end{pmatrix} $$ (so $\mathcal L(\sigma)=0$), but $$ -i[H,\sigma]=\begin{pmatrix} -\frac{4}{9} & \frac{i}{9} \\ -\frac{i}{9} & \frac{4}{9} \end{pmatrix}\neq 0\tag1 $$ (and $V\sigma V^\dagger-\frac12V^\dagger V\sigma-\sigma \frac12V^\dagger V\neq 0$ as it is of course the negative of (1)).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.