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 $\mathcal{L}$...


2 Answers 2


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 $1/|\lambda|_{max} \le 1/|\lambda_{min}|$ )

  • $\begingroup$ Right... So, is the Lindbladian $\mathcal{L}$ always Hermitian? (I really dont know anything about master equations yet...) $\endgroup$
    – Marsl
    Commented Feb 1, 2022 at 21:44
  • $\begingroup$ The density operator is Hermitian, so your first equation implies that $\mathcal L$ is Hermitian as well $\endgroup$
    – user341440
    Commented Feb 1, 2022 at 22:21
  • $\begingroup$ 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. $\endgroup$ 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.


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