Choi's theorem states that any completely positive map $\Phi(\cdot) : C^*_{n\times n} \rightarrow C^*_{m \times m}$ can be expressed as $\Phi(\rho) = \sum_{j=1}^r F_j^\dagger \rho F_j$, for some $n \times m$ matrices $F_j$. In order for the map to be trace preserving one needs to have $\sum_j F_j F_j^\dagger = I_n$.

How is this trace preserving condition derived? I couldn't see it.


3 Answers 3


This is a slight variation of the ideas behind the other answer.

Note that $\operatorname{Tr}(\Phi(\rho))=\operatorname{Tr}(\rho)$ for all states $\rho$ (read, all positive trace-1 operators) is equivalent to $\operatorname{Tr}(\Phi(X))=\operatorname{Tr}(X)$ holding for all operators $X$ (essentially because any operator can be written as a linear combination of positive operators, and the trace operation is linear).

Note that if $\Phi(X)=\sum_k A_k X A_k^\dagger$, then $\operatorname{Tr}(\Phi(X))=\operatorname{Tr}(X)$ is equivalent to $$\operatorname{Tr}\left[X\left(\sum_k A_k^\dagger A_k\right) \right]=\operatorname{Tr}(X).$$

Because this must hold for any operator $X$, we can see what it amounts to for $X=\lvert i\rangle\!\langle j\rvert$. Noting that $\operatorname{Tr}[|i\rangle\!\langle j| B]=\langle j|B| i\rangle$ and $\operatorname{Tr}[|i\rangle\!\langle j|]=\delta_{ij}$, this gives us $$\langle i|\sum_k A_k^\dagger A_k |j\rangle=\delta_{ij}.$$

This is nothing but the componentwise version of $\sum_k A_k^\dagger A_k=I$.

Note that here I used a slightly different notational convention for the Kraus decomposition than the one used in the original post. The reason is simply that I am more used to this one, but you might simply replace each $A_k\to A_k^\dagger$ switch between the two conventions.


Recall that the trace is both linear and invariant under cyclic permutation of the operators

$$ \mathrm{Tr}(\Phi(\rho))=\mathrm{Tr}\left(\sum_j F_j^\dagger \rho F_j\right)=\sum_j\mathrm{Tr}\left( F_j^\dagger \rho F_j\right)=\sum_j \mathrm{Tr}\left(F_jF_j^\dagger \rho \right)= \mathrm{Tr}\left(\sum_jF_jF_j^\dagger \rho \right)$$

You can clearly see that if $\sum_jF_jF_j^\dagger=I$ $\mathrm{Tr}(\Phi(\rho))=\mathrm{Tr}(\rho)$ for all $\rho$. To prove the converse, suppose $A$ is such that

$$\mathrm{Tr}(A\rho)=\mathrm{Tr}(\rho) $$ for all $\rho$, this corresponds to

$$ \sum_{ij}A_{ij}\rho_{ji}-\sum_i \rho_{ii}=0$$

By taking $\rho_{ij}=\delta_{ki}\delta_{lj}$

we get

$$A_{lk}-\delta_{lk}=0 $$

thus $A=I$.

  • $\begingroup$ Nice explanation. I couldn't follow where the $\delta$'s are coming from though. $\endgroup$
    – satya
    Commented Aug 1, 2019 at 2:47
  • $\begingroup$ sincethe trace preserving condition is valid for every matrix $\rho$, I picked some $\rho$ that allowed me to conclude that $A$ was the identity $\endgroup$ Commented Aug 1, 2019 at 6:47

This is not to hard to see when we rewrite the trace in terms of the Frobenius inner product on the space of matrices $M_{m,n}(K)$ when $K\in\{\mathbb{R},\mathbb{C}\}$ defined by letting $\langle A,B\rangle=\text{Tr}(AB^*)$.

We will use that fact that in an inner product space $V$, if $a,b\in V$, then $a=b$ if and only if $\langle x,a\rangle=\langle x,b\rangle$ for all $x\in V$.

Theorem: Suppose that $\Phi(X)=A_1XB_1^*+\dots+A_rXB_r^*$.

The following are equivalent:

  1. $\text{Tr}(\Phi(\rho))=\text{Tr}(\rho)$ for all trace 1 positive operators $\rho$.

  2. $\text{Tr}(\Phi(X))=\text{Tr}(X)$ for all operators $X$.

  3. $A_1^*B_1+\dots+A_r^*B_r=I$.


$2\rightarrow 1$ is clear since we simply restrict the equation to trace 1 positive operators. $1\rightarrow 2$ follows because the vector space of trace 1 positive operators spans the vector space of all positive operators.

$2\leftrightarrow 3.$ Using the cyclic invariance of the trace, we get

$$\text{Tr}(\Phi(X))=\text{Tr}(A_1XB_1^*+\dots+A_rXB_r^*) =\text{Tr}(XB_1^*A_1+\dots+XB_r^*A_r)=\langle X,A_1^*B_1+\dots+A_r^*B_r\rangle$$ while $\text{Tr}(X)=\langle X,I\rangle.$ Therefore, $\text{Tr}(X)=\text{Tr}(\Phi(X))$ if and only if $I=A_1^*B_1+\dots+A_r^*B_r$.



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