Looking for a proof of a trace identity and (possibly) relationship to Kraus operators

Question

$\DeclareMathOperator\PSU{PSU}\DeclareMathOperator\U{U}\DeclareMathOperator\SU{SU}\DeclareMathOperator\O{O}\DeclareMathOperator\SO{SO}\DeclareMathOperator\GL{GL}\DeclareMathOperator\ASL{ASL}\DeclareMathOperator\SL{SL}\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\Spin{Spin}\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\PGL{PGL}\DeclareMathOperator\G{G}\DeclareMathOperator\F{F}\DeclareMathOperator\E{E}\DeclareMathOperator\Tr{Tr}$Let $$ \langle A,B \rangle =\Tr(A^\dagger B) $$ be the Hilbert-Schmidt inner product on the space of $ N \times N $ complex matrices. Let $ \mathscr{E} $ be an orthonormal error basis with respect to the Hilbert-Schmidt inner product. What is a simple proof that $$ \sum_{E \in \mathscr{E}} E A E^\dagger= \Tr(A) \, I $$ where $ I $ is the identity matrix? Is there a relationship between this fact and Kraus operators? Can this equation be seen as just a rotation of the Kraus operators?

Answer 1

One pretty simple way to prove this is to compute the Choi representations for the two maps in question: $$ \Phi(A) = \sum_{E\in\mathscr{E}} E A E^{\dagger} \quad\text{and}\quad \Psi(A) = \operatorname{Tr}(A)I. $$

For the first map, we have $$ J(\Phi) = \sum_{E\in\mathscr{E}} \operatorname{vec}(E) \operatorname{vec}(E)^{\dagger} = I \otimes I, $$ where $\operatorname{vec}(E)$ refers to the vectorization of $E.$ The first equality follows from a general relationship that links Kraus and Choi representations through the vectorization map and the second equality follows from the fact that $\mathscr{E}$ is an orthonormal error basis. For the second map, we have $J(\Psi) = I\otimes I$ by evaluating the Choi representation.

Choi representations are faithful, so the two maps are equal.

Answer 2

The $(p,q)^{\text{th}}$ entry of $\sum_k E_k A E_k^\dagger$ is $$ \sum_k \sum_r \sum_s (E_k)_{pr} A_{rs} (E_k^\dagger)_{sq} = \sum_k \sum_r \sum_s (E_k)_{pr} A_{rs} (\overline{E_k})_{qs}. $$ Rearranging the sums to $$ \sum_r \sum_s A_{rs} \sum_k (E_k)_{pr} (\overline{E_k})_{qs} $$ and using completeness of the basis simplifies the inner sum over $k$ and we get $$ \sum_r \sum_s A_{rs} \delta_{pq} \delta_{rs} $$ which is only non zero when $p=q$ and $r=s$ (and is equal to $ \sum_r A_{rr} = \text{tr}(A) $ when it is non-zero).

Justifying the simplification of the inner sum over $k$: we have the following identity for the error basis $\mathcal{E}$ \begin{equation*} \text{for all } X,Y, \quad \operatorname{Tr}(X^\dagger Y) = \sum_k \operatorname{Tr}(X^\dagger E_k) \operatorname{Tr}(E_k^\dagger Y). \end{equation*} Choose $X = |p\rangle\langle r|, Y = |q\rangle\langle s|$, where $\{|i\rangle\}$ is an orthonormal basis for $\mathbb{C}^N$. Then:

1. $\operatorname{Tr}(X^\dagger Y) = \operatorname{Tr}\bigl( (|p\rangle\langle r|)^\dagger (|q\rangle\langle s|) \bigr) = \delta_{pq}\delta_{rs}.$

2. $\operatorname{Tr}(X^\dagger E_k) = \operatorname{Tr}(|r\rangle\langle p| E_k) =(E_k)_{pr}.$

3. $\operatorname{Tr}(E_k^\dagger Y) = \operatorname{Tr}(E_k^\dagger |q\rangle\langle s|) = (E_k^\dagger)_{sq}.$

The identity above for the error basis $\mathcal{E}$ becomes

\begin{equation*} \delta_{pq}\delta_{rs} = \sum_{k} (E_k)_{pr} (E_k^\dagger)_{sq} = \sum_{k} (E_k)_{pr} \overline{(E_k)}_{qs}. \end{equation*}