# Does the inequality $\mathrm{tr}(L^\dagger L \rho^2)-\mathrm{tr}(L^\dagger \rho L\rho )\geq 0$ hold generally?

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 $$L$$?

• What conditions are imposed on a Lindbladian operator? Commented May 6, 2023 at 9:30
• 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. Commented May 6, 2023 at 12:37
• $\rho=\sum^N_{i=1}\lambda_i|\phi_i><\phi_i|$ 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 $N\times N$. Commented May 6, 2023 at 17:49
• Are you saying that $L$ can be any $N \times N$ matrix? Commented May 6, 2023 at 17:54
• Yes. I consider that $\rho$ obeys the master equation $d\rho/dt=\mathcal{D}[L]\rho$ where $\mathcal{D}[L]\rho=L\rho L^\dagger -L^\dagger L\rho/2-\rho L^\dagger L/2$. Commented May 6, 2023 at 18:18

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 $$0 and $$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$$, both of which are possible.
No, it does not hold -- brute force numerics with random $$\rho$$ and $$L$$ finds counterexamples already for $$N=2$$.