# Why for $\Phi(\rho)=\sum_j F_j^\dagger\rho F_j$ to be trace preserving we need the condition $\sum_j F_j F_j^\dagger=I$?

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.

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 $$A=I$$.

• Nice explanation. I couldn't follow where the $\delta$'s are coming from though. Aug 1, 2019 at 2:47
• sincethe trace preserving condition is valid for every matrix $\rho$, I picked some $\rho$ that allowed me to conclude that $A$ was the identity 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. $$A_1^*B_1+\dots+A_r^*B_r=I$$.

Proof:

$$2\rightarrow 1$$ 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 $$I=A_1^*B_1+\dots+A_r^*B_r$$.

Q.E.D.