# Can one turn a non-Trace Preserving map into one that is Trace Preserving?

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?

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.
• What do we mean by matrix inequality $\sum_{i=1}^n A_i^\dagger A_i \le \mathbb{I}$?