# Show $I+\tau\mathcal{L}$ is completely positive when $\tau \leq 1/\lambda_{\mathrm{max}}(\mathcal{L})$

I am not very well-versed when in comes to open quantum systems which is why I need some help. In a paper, I encountered the following situation:

Let $$\mathcal{L}$$ be a Lindbladian so the time evolution of a system is described by the master equation $$\frac{d\rho(t)}{dt}=\mathcal{L}\rho(t)$$

Now, the paper argues that while $$e^{\tau \mathcal{L}}$$ is a proper channel meaning CPTP map, its first order approximation $$I + \tau \mathcal{L}$$

is not necessarily CPTP. In particular, they argue that $$\tau$$ has to be chosen sufficiently small in order for $$I + \tau \mathcal{L}$$ to also be CP. That's the part I want to understand.

So, let $$\mathcal{L}$$ be a Lindbladian and denote by $$\lambda_{\mathrm{max}}(\mathcal{L})$$ the maximum magnitude eigenvalue of the Choi state corresponding to $$\mathcal{L}$$, i.e.

$$J(\mathcal{L}) = (\mathcal{L} \otimes I)(|\Omega\rangle\langle\Omega|)\\ \lambda_{\mathrm{max}}(\mathcal{L})=\mathrm{max}_{\lambda\in\mathrm{spec}(J(\mathcal{L}))}|\lambda|$$

(Here, I am confused already: Because they talk about magnitudes, it seems that $$J(\mathcal{L})$$ need not be positive semi-definite, i.e. it can have negative eigenvalues. But I should be Hermitian, shouldnt it? In general, what properties does $$\mathcal{L}$$ have?)

Then, I want to convince myself of the following statement: The map $$\Phi:=I+\tau\mathcal{L}$$ is also completely positive, whenever $$\tau\leq 1/\lambda_{\mathrm{max}}(\Phi)$$.

My approach: The Choi state corresponding to the map $$\Phi'$$ is $$J(\Phi') = |\Omega\rangle\langle\Omega| + \tau J(\mathcal{L})$$

Now, the spectrum of $$\tau J(\mathcal{L})$$ has magnitude $$\leq 1$$ by construction but how can I argue about the eigenvalues of the sum?

Also, why is $$I+\tau\mathcal{L}$$ trace preserving? Again, I guess this comes down to my lack of knowledge about the properties of $$\mathcal{L}$$...

The relation you are asking about is true in general for any Hermitian operator. This is just because the eigenvalues of $$I + \tau L$$ are $$1 + \tau\lambda_i$$, where $$\lambda_i$$ are the eigenvalues of $$L$$.

So if $$1 - \tau|\lambda_{min}|$$ is positive then all the eigenvalues of $$I + \tau L$$ are positive.

(This is a weaker condition than $$\tau < 1/|\lambda|_{max}$$, since by definition $$1/|\lambda|_{max} \le 1/|\lambda_{min}|$$ )

• Right... So, is the Lindbladian $\mathcal{L}$ always Hermitian? (I really dont know anything about master equations yet...) Commented Feb 1, 2022 at 21:44
• The density operator is Hermitian, so your first equation implies that $\mathcal L$ is Hermitian as well Commented Feb 1, 2022 at 22:21
• No, $\mathcal L$ is not Hermitian. The master equation shows that it is Hermitian-preserving which is different from the operator itself being Hermitian. Indeed, every normal (hence every Hermitian) channel is necessarily identity-preserving and, similarly, any Hermitian $\mathcal L$ necessarily satisfies $\mathcal L({\bf1})=0$ showing that the Hermitian Lindbladians are a strict subset of all Lindbladians. Indeed one can give a counterexample showing that the conjectured inequality is wrong. Commented May 3 at 7:17

If the authors of the paper in question include the case $$\tau=0$$ in their formulation "sufficiently small", then this is correct. In other words:

There exist Lindbladians $$\mathcal L$$ such that $${\rm id}+\tau\mathcal L$$ is completely positive if and only if $$\tau=0$$.

We will give a counterexample below but first a word on the simpler condition of trace preservation, which you also asked about. Because $$e^{t\mathcal L}$$ is trace preserving for all $$t$$, differentiating the corresponding condition $${\rm tr}(e^{t\mathcal L}(X))={\rm tr}(X)$$ at $$t=0$$ shows that $${\rm tr}(\mathcal L(X))=0$$ for all $$X$$, that is, $$\mathcal L$$ has to be trace annihilating. But this shows that $${\rm id}+\tau\mathcal L$$ is trace-preserving for all $$\tau\in\mathbb C$$: $${\rm tr}(({\rm id}+\tau\mathcal L)(X))={\rm tr}(X)+\tau{\rm tr}(\mathcal L(X))={\rm tr}(X)$$

Thus, complete positivity really is the problem here; in fact it occurs whenever $$\mathcal L$$ has eigenvalues on the imaginary axis. For example, consider the valid generator $$\mathcal L(X):=-i[\sigma_z,X]$$ in which case $${\rm id}+\tau\mathcal L$$ for all $$\tau\in\mathbb R$$ gives rise to the following Choi matrix: $$J({\rm id}+\tau\mathcal L)=|\Omega\rangle\langle\Omega|+\tau\begin{pmatrix}0&0&0&-2i\\0&0&0&0\\0&0&0&0\\2i&0&0&0\end{pmatrix}= \begin{pmatrix}1&0&0&1-2i\tau\\0&0&0&0\\0&0&0&0\\1+2i\tau&0&0&1\end{pmatrix}$$ But the smallest (non-zero) eigenvalue of this matrix is $$1-\sqrt{1+4\tau^2}$$ which is non-negative if and only if $$\tau=0$$. In other words $${\rm id}+\tau\mathcal L$$ is completely positive if and only if $$\tau=0$$. In particular this shows that your statement about the maximal eigenvalue cannot hold: in this example, $$|\lambda_{\rm max}(\mathcal L)|=1$$ but $$\tau\leq 1/|\lambda_{\rm max}(\mathcal L)|=1$$ is not enough to guarantee complete positivity.