I have a question regarding Kraus operators. Any quantum channel can be written in terms of Kraus operators as $E(\rho)= \sum_{i=0}^n K_i \rho K_i^{\dagger}$ where $\rho$ is the initial density matrix. So you can propagate any initial density matrix through this quantum channel with Kraus operators.

Does this mean that if we have the output density matrix, we can decompose it in terms of Kraus operators and find the exact Kraus operators of the channel? In related questions, Choi representation is transformed to Kraus operators but that approach is not clear to me. Does anyone have a simple example to understand it or a simple way to obtain the Kraus operators?

  • $\begingroup$ what do you mean with "decompose the final density matrix in terms of Kraus operators"? The map/channel is decomposed with Kraus ops, not the density matrix $\endgroup$
    – glS
    Commented Aug 28, 2021 at 9:24
  • $\begingroup$ @gIS If you don't know what quantum channel U that you have created and you only know the final state U \rho_{initial}U^{dag} can you get the expression of K\rhoK^{dag} for any initial density matrix that you want? Have I misunderstood something? $\endgroup$ Commented Aug 28, 2021 at 10:06
  • $\begingroup$ so you are asking for the Kraus decomposition of a channel sending some arbitrary initial state to the output $\rho$? The problem with this is that there will in general be many channels, with potentially different Kraus decompositions, sending any given $\rho_{\rm in}$ to your $\rho$. As a trivial example, the identity channel, $\Phi(\rho)=\rho$ for each $\rho$, and the replacement channel $\Phi(\sigma)={\rm Tr}(\sigma)\rho$, both give you the output state $\rho$ for some input, but are completely different channels $\endgroup$
    – glS
    Commented Aug 28, 2021 at 10:08
  • $\begingroup$ Yes. Is there a way to restrict that? Do you know any literature that can help with that? So I guess there is not a way to get something like that ... $\endgroup$ Commented Aug 28, 2021 at 10:11
  • $\begingroup$ what kind of "restrictions" are you referring to? $\endgroup$
    – glS
    Commented Aug 28, 2021 at 10:14

1 Answer 1


Finding Kraus operators

The method of determining the Kraus operators of a quantum channel $\Phi: L(\mathcal{X})\to L(\mathcal{Y})$ from the knowledge of its action on a set of inputs is called quantum process tomography. See section 8.4.2 on page 389 in Nielsen & Chuang for details. In particular, see equation $(8.168)$ on page 392 for how Kraus operators are computed. In general, the method requires the knowledge of the action of $\Phi$ on a basis of the input space $L(\mathcal{X})$. For one qubit, this means knowing the action of $\Phi$ on at least four input states.

Recovering $\Phi$ from knowledge of its action on single input

As @gIS points out in the comments, it is in general impossible to recover $\Phi$ from the knowledge of its action on a single input density matrix $\rho\in D(\mathcal{X})\subset L(\mathcal{X})$. However, as you anticipate in your responses, it is possible to place certain restrictions on $\Phi$ that do make such determination possible.

Below, I describe two such sets of restrictions. The first one is motivated by basic linear algebra and is not particularly surprising. The second one comes from quantum information science and is perhaps more unexpected.

Action of $\Phi$ on a basis

Let $d=\dim\mathcal{X}$ and let $X_1, X_2, \dots, X_{d^2}\in L(\mathcal{X})$ denote a basis of the input space $L(\mathcal{X})$. Suppose that in addition to knowing the action of $\Phi$ on $\rho$, we also know the action of $\Phi$ on $X_1, X_2, \dots, X_{d^2-1}$ where $\rho$ is linearly independent from $X_1, X_2, \dots, X_{d^2-1}$. In this case, $\rho, X_1, X_2, \dots, X_{d^2-1}$ is itself a basis of $L(\mathcal{X})$ and so we can recover $\Phi$.

A special case of this occurs when $d=1$ and $\Phi$ is a state preparation channel which is indeed fully described by its sole output state.

Action of $\Phi$ on a full rank bipartite input

It turns out that with the help of an auxiliary system, a quantum channel can be recovered from its action on a single input. More precisely, if we know the action of $\Phi\otimes I$ on a joint state $\rho\in D(\mathcal{X}\otimes\mathcal{X})$ whose Schmidt rank is $d^2$ then we can recover $\Phi$ from the knowledge of $(\Phi\otimes I)(\rho)$ alone.

The method is called entanglement-assisted process tomography (EAPT) or ancilla-assisted process tomography (AAPT). In the former, $\rho$ is entangled, in the latter it is not. The latter is possible because, perhaps somewhat surprisingly, there do exist unentangled states of full Schmidt rank. See description of figure 1 c in this paper for an example.

  • $\begingroup$ Thank you so much!!! This is so helpful!!!! $\endgroup$ Commented Aug 28, 2021 at 17:47
  • $\begingroup$ You're welcome! I'm glad the answer is helpful :-) $\endgroup$ Commented Aug 28, 2021 at 19:44
  • $\begingroup$ I guess I misunderstood the question: I thought it was about assuming $\rho$ was an output of the channel. Here you characterise what happens when you know $\Phi(\rho)$ for some $\rho$ (i.e. you know how a specific input evolves through the channel) $\endgroup$
    – glS
    Commented Aug 28, 2021 at 20:18
  • $\begingroup$ Well, the equation for $E(\rho)$ in the question does use $\rho$ as input. I agree with you that the question was somewhat unclear which usually happens when some confusion exists, so I've decided to describe the task of channel determination in general hoping that it would clarify the problem for the OP. Also, the question did suggest that they were considering finding the channel based on just a single output, so I described when such a feat is and isn't possible. BTW: Your comments under the question helped clarify it enough for me to take a stab at it. Thanks! $\endgroup$ Commented Aug 28, 2021 at 20:30
  • $\begingroup$ It is my fault . I wasn't clear from the beggining. Sorry if I confused you. Zalcman's answer was helpful because it make some things clearer. I am new here. I will try to ask to the point questions. $\endgroup$ Commented Aug 28, 2021 at 21:01

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