There is a well-known form of the unitary freedom of Kraus operators, which can be found in Nielsen's book, stating that two sets of Kraus operators describe the same physical process of the system(consider the whole space consist of the system and the environment) iff $\Pi_l=\sum_k u_{lk}F_k$, where $u_{lk}$ is the entry of the corresponding unitary matrix.

In the deduction of Kraus operator in Nielsen's book: $$\newcommand{\Tr}{{\operatorname{Tr}}} \begin{align} \Tr_{\rm env} \left[ U|\psi\rangle\otimes|0\rangle\langle\psi|\otimes\langle0|U^\dagger \right],\tag{1} \end{align} $$ the single $\Pi_l$ can be stated as $(I\otimes\langle l|)U(I\otimes|0\rangle)$. Refer to this question, we can write $U|\psi\rangle\otimes|0\rangle$ into $\sum_l\Pi_l|\psi\rangle\otimes|l\rangle$, so eq.(1) can be written into $$ \begin{align} \Tr_{\rm env} \left[ \left(\sum_l\Pi_l|\psi\rangle\otimes|l\rangle \right) \left(\sum_\tilde{l}\langle\psi|\Pi_\tilde{l}^\dagger\otimes\langle \tilde{l}| \right) \right],\tag{2} \end{align} $$ which have another form of unitary freedom: $$\Tr_{\rm env} \left[ (I\otimes U_{\rm env}) \left(\sum_l\Pi_l|\psi\rangle\otimes|l\rangle\right) \left(\sum_\tilde{l}\langle\psi|\Pi_\tilde{l}^\dagger\otimes\langle \tilde{l}|\right) (I\otimes U_{\rm env}^\dagger) \right], $$ from the perspective of eq.(1), this unitary freedom can be written as $F_l=(I\otimes\langle l|U_{\rm env})U|\psi\rangle\otimes|0\rangle$, while this one cannot be transformed into the form of $\Pi_l=\sum_k u_{lk}F_k$ easily.

To sum up, the question is that how to transform the 'new form' of unitary freedom $ F_l=(I\otimes\langle l|U_{\rm env})U|\psi\rangle\otimes|0\rangle$ into $\Pi_l=\sum_k u_{lk}F_k$?


1 Answer 1


The extra transformation $U_{\mathrm{env}}$ is equivalent to choosing a new basis for the environment states $$|l\rangle\to U_{\mathrm{env}}^\dagger |l\rangle.$$ This is the same freedom as the freedom to choose the basis for $|l\rangle$; equivalently, your question is whether there is additional freedom between the choices $$\Pi_l=(I\otimes \langle l|)U(I\otimes|0\rangle)\quad\mathrm{and}\quad P_k=(I\otimes \langle k|)U(I\otimes|0\rangle).$$ But this is exactly the unitary freedom captured by $\Pi_l=\sum_k u_{lk}F_k$, because we can always write $\langle l|=\sum_k u_{lk}\langle k|$, so there is no additional freedom here (we can identify $P_k=F_k$).

In fact, you have found the relationship between the components $u_{lk}$ and $U_{\mathrm{env}}$, which must satisfy $$\langle l|U_{\mathrm{env}}=\sum_{k}u_{lk}\langle k|\quad\Leftrightarrow\quad u_{lk}=\langle l |U_{\mathrm{env}}|k\rangle.$$


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