2
$\begingroup$

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?

$\endgroup$

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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