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Does the inequality $$\mathrm{tr}(L^\dagger L \rho^2)-\mathrm{tr}(L^\dagger \rho L\rho )\geq 0$$ hold for any density matrix $\rho$ and any non-Hermitian Lindblad operator $L$?

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    $\begingroup$ What conditions are imposed on a Lindbladian operator? $\endgroup$
    – Rammus
    Commented May 6, 2023 at 9:30
  • $\begingroup$ To follow up, if you just naively apply Cauchy-Schwarz you find that $\mathrm{tr}[L^\dagger \rho L \rho] \leq \mathrm{tr}[\sqrt{\rho L^\dagger L \rho}]$ and the latter quantity can be expressed in terms of the eigenvalues of $L^\dagger L \rho^2$ which appears in your question. In particular if all the eigenvalues of the latter operator are larger than $1$ then your result would follow from Cauchy Schwarz as then $\sqrt{\rho L^\dagger L \rho} \leq \rho L^\dagger L \rho$. However, if there are no additional restrictions on $L$ (it is just non-Hermitian) then the result is not true. $\endgroup$
    – Rammus
    Commented May 6, 2023 at 12:37
  • $\begingroup$ $\rho=\sum^N_{i=1}\lambda_i|\phi_i><\phi_i|$ is the $N\times N$ density matrix, where $\lambda_i$ and $|\phi_i>$ are the i-th eigenvalue and i-th component state, respectively. $L$ is the $N\times N$ Lindblad operator operator representing the decoherence process. No conditions are imposed on $L$, except for $N\times N$. $\endgroup$
    – Kochan
    Commented May 6, 2023 at 17:49
  • $\begingroup$ Are you saying that $L$ can be any $N \times N$ matrix? $\endgroup$
    – Rammus
    Commented May 6, 2023 at 17:54
  • $\begingroup$ Yes. I consider that $\rho$ obeys the master equation $d\rho/dt=\mathcal{D}[L]\rho$ where $\mathcal{D}[L]\rho=L\rho L^\dagger -L^\dagger L\rho/2-\rho L^\dagger L/2$. $\endgroup$
    – Kochan
    Commented May 6, 2023 at 18:18

2 Answers 2

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TL;DR: No. The inequality fails for example when $\rho=\mathrm{diag}(\frac23,\frac13)$ and $L=|0\rangle\langle 1|$.

It's not hard to see that the inequality is satisfied if $\rho$ is pure or maximally mixed, so let's set $\rho=\mathrm{diag}(p, q)$ where $0<p<1$ and $q=1-p$. A natural candidate for $L$ is the annihilation operator $L=|0\rangle\langle 1|$. We have $$ \mathrm{tr}(L^\dagger L\rho^2)=\langle 1|\rho^2|1\rangle=q^2\tag1 $$ and $$ \mathrm{tr}(L^\dagger \rho L\rho)=\langle 0|\rho|0\rangle\langle 1|\rho|1\rangle=pq\tag2 $$ so \begin{align} \mathrm{tr}(L^\dagger L\rho^2)-\mathrm{tr}(L^\dagger \rho L\rho)=q(q-p).\tag3 \end{align} Clearly, this expression is positive if $q>p$ and negative if $p>q$, both of which are possible.

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No, it does not hold -- brute force numerics with random $\rho$ and $L$ finds counterexamples already for $N=2$.

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