Given a Hilbert space $\mathbb{C}^N$ and the reduced dynamics $\Lambda(t)$ of the open quantum system, we can define the set of asymptotic states as

$$ \mathcal{A}=\left\{\tilde\rho \in \mathcal{S}(\mathbb{C}^N)\,|\,\exists\rho \in \mathcal{S}(\mathbb{C}^N)\; \text{such that}\; \lim_{t\rightarrow\infty} \Lambda(t)[\rho]=\tilde\rho \right\}\,, $$ where by $\mathcal{S}(\mathbb{C}^N)$ I mean the set of quantum states ($\mathrm{Tr}\,\rho=1, \rho^\dagger=\rho, \rho\ge0$).

If the open system dynamics is relaxing, then there is only one steady-state $\rho_{ss}$ and this is the unique asymptotic state, so $\mathcal{A}$ has one element. However, in general $\mathcal{A}$ may have $\infty$ elements.

For example, for a pure dephasing channel, we have $\infty$ asymptotic states in the z-axis (if the coherence plane is the xy-plane) in the Bloch sphere.

So my question is:

Are there any particular reduced dynamics $\Lambda(t)$ for which we can say, for example, that $\mathcal{A}$ has 2 and only 2 different elements?


  • $\begingroup$ If you think that pure dephasing dynamics has infinitely many asymptotic states, then the only possibilities are 1 or infinity. This is because of linearity: if $\rho_1$ and $\rho_2$ are asymptotic then so is $p\rho_1 +(1-p)\rho_2$ for all $0<p<1$. Therefore, the existence of more than 1 asymptotic state immediately implies “infinitely” many. I suggest therefore that a more interesting question is the number of linearly independent asymptotic states. $\endgroup$ – Mark Mitchison Dec 16 '18 at 18:10
  • $\begingroup$ Hi Mark! Thanks for you answer. Yes, you are right, I agree that the question you suggest would have been more interested. However, I don't understand why the only possibilities are 1 or infinity. I agree that by linearity of both $\Lambda(t)$ and $\lim$, if $\rho_1$ and $\rho_2$ are asymptotic, then there are infinity asymptotic states. Buy why is not possible that there are no asymptotic states at all? Is it not possible some kind of oscillation so the limit $\lim_{t\rightarrow\infty}\Lambda(t)$ does not exist? $\endgroup$ – Iyán Dec 16 '18 at 20:39

I will try to answer my question based on what Mark said.

If we assume that $\mathcal{A}$ has two elements, then by lineartiy of both $\lim$ and $\Lambda(t)$ we can find infinite different elements of $\mathcal{A}$.


By hypothesis, $\tilde\rho_1$, $\tilde\rho_2 \in \mathcal{A}$, so there must exist by definition of $\mathcal{A}$, $\rho_1$, $\rho_2$ such that: $$ \lim_{t\rightarrow\infty}\Lambda(t)[\rho_1]=\tilde\rho_1\quad\text{and}\quad\lim_{t\rightarrow\infty}\Lambda(t)[\rho_2]=\tilde\rho_2 $$

If we now consider a linear combination of $\rho_1$ and $\rho_2$ that also belongs to $\mathcal{S}(\mathbb{C}^N)$: $$ p\rho_1+(1-p)\rho_2\in\mathcal{S}(\mathbb{C}^N)\,,\quad p\in(0,1) $$ And we compute the limit when $t\rightarrow\infty$ we get: $$ \lim_{t\rightarrow\infty}\Lambda(t)[p\rho_1+(1-p)\rho_2]=\lim_{t\rightarrow\infty}\left(\Lambda(t)[p\rho_1]+\Lambda(t)[(1-p)\rho_2]\right)=\lim_{t\rightarrow\infty}\Lambda(t)[p\rho_1]+\lim_{t\rightarrow\infty}\Lambda(t)[(1-p)\rho_2]=p\tilde\rho_1+(1-p)\tilde\rho_2\in\mathcal{S}(\mathbb{C}^N)\quad\square $$

Geometrically this can be also be understand because linearity of $\Lambda(t)$ means that all points of the bloch sphere are transformed in the same way. In the particular case of $\mathcal{S}(\mathbb{C}^2)$, we can easily imagine that it is not possible to find a linear transformation that would lead to two asymptotic states.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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