A trace non-preserving quantum channel $\mathcal{A}$ takes a state $\rho$ to $\rho^\prime$, i.e., $\sum_{i=1}^{n} A_i \rho A_i^\dagger = \rho^\prime$, with $\sum_{i}^{n} A_i^\dagger A_i \ne \mathbf{I}$. Can one make this map trace-preserving?

One way, say by rewriting the Kraus operators $B_i = A_i/\sqrt{{\rm Tr} \rho^\prime}$, such that $\sum_{i}^{n} B_i^\dagger B_i = \mathbf{I}$ and problem seems to be solved. Does the set $\{B_i \}$ define a valid quantum channel?


1 Answer 1


Firstly, in your example the operators $B_i$ depend on the input state to the channel which makes the channel ill-defined. E.g., if $A_0 = |0\rangle\langle 0 |$ and $A_1 = \frac12 |1 \rangle \langle 1 |$ then you have a trace non-increasing CP map but the trace of the channel acting on $\rho = |0\rangle\langle 0|$ is different from it acting on $\rho=|1\rangle \langle 1|$.

Now back to the question, suppose the Kraus operators for your original map satisfy $$ \sum_{i=1}^n A_i^{\dagger} A_i \leq \mathbb{I} $$ so the map is trace non-increasing. Then let $M = \mathbb{I} - \sum_{i=1}^n A_i^{\dagger}A_i$, note that $M \geq 0$ and so we can define a unique positive sqaure root $M^{1/2}$. Now let $B_i = A_i$ for $i \in \{1,2,\dots, n\}$ and let $B_{n+1} = M^{1/2}$. Then you have a CP map defined by Kraus operators $\{B_i\}_{i=1}^{n+1}$ which satisfies $$ \begin{aligned} \sum_{i=1}^{n+1} B_{i}^{\dagger}B_{i} &= \sum_{i=1}^n A_{i}^{\dagger} A_i + \left(\sqrt{\mathbb{I} - \sum_{i=1}^n A_i^{\dagger}A_i}\right)^{\dagger} \left(\sqrt{\mathbb{I} - \sum_{i=1}^n A_i^{\dagger}A_i}\right)\\ &= \sum_{i=1}^n A_{i}^{\dagger} A_i + \left(\mathbb{I} - \sum_{i=1}^n A_i^{\dagger}A_i\right) \\ &= \mathbb{I} \end{aligned} $$ and so it is a valid quantum channel.

For the case where $\sum_{i=1}^n A_i^{\dagger}A_i \nleq \mathbb{I}$ you can just define $$ B_i = \frac{A_i}{\|\sum_{i=1}^n A_i^{\dagger}A_i\|} $$ where in this context $\|\sum_{i=1}^n A_i^{\dagger}A_i\|$ is just the largest eigenvalue of the operator $\sum_{i=1}^n A_i^{\dagger}A_i$. Then $\sum_{i=1}^n B_i^{\dagger}B_i \leq \mathbb{I}$ and you can follow the above construction again.

  • $\begingroup$ Thanks @Rammus. Could you suggest some reference where I can find this procedure? $\endgroup$
    – seeker
    Commented Aug 10, 2022 at 14:09
  • $\begingroup$ No sorry, it's just a construction I came up with. Probably someone has done something similar before but I can't give you an example. $\endgroup$
    – Rammus
    Commented Aug 10, 2022 at 15:58
  • $\begingroup$ What do we mean by matrix inequality $\sum_{i=1}^n A_i^\dagger A_i \le \mathbb{I}$? $\endgroup$
    – Rob
    Commented Feb 3, 2023 at 12:37

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