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In Breuer and Petruccione's book 'The theory of open quantum systems', section 2.4.3, 'representation theorem of quantum operations', a quantum operation $\phi_m$ corresponding to a measurement outcome m is defined as a map from one density matrix to another with following properties-

1. $0 \leq tr [\phi_m(\rho)] \leq 1$
2. Convex linearity: $\phi _m \left ( \Sigma a_i \rho _i\right )=\Sigma a_i \phi_m(\rho_i)$
3. The operator $(\phi_m \otimes I)$ is Completely postivite, where this new operator is defined as-

\begin{equation}(\phi_m \otimes I) \left ( \Sigma A_i \otimes B_i\right )\equiv \Sigma \phi_m(A_i) \otimes B_i \tag{1}\label{1}\end{equation}

My question is- in \eqref{1}, do these matrices $A_i$s and $B_i$s need to be density matrices? Because, I think it is not always possible to partition a density matrix in this way in a given basis. For example, lets take a maximally entangled density matrix

$$|\psi\rangle \langle \psi |= \frac {1}{2}\begin{pmatrix} 1 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 1 \end{pmatrix}$$

We can write $$|\psi\rangle \langle \psi |=\frac{1}{2} \left [ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \otimes \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} + \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \otimes \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \otimes \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \right ]$$

Then, $$(\phi_m \otimes I) \equiv \frac{1}{2} \left [ \phi_m \left [ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \right ] \otimes \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + \phi_m \left [ \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \right ] \otimes \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} + \phi_m \left [ \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \right ] \otimes \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} + \phi_m \left [ \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \right ] \otimes \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \right ]$$

But the problem is, $ \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$ is not a desnity matrix. So is $\phi_m \left [ \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \right ]$ defined?

Comments: If we think of $\phi_m$ in terms of Kraus operator representation, as suggested by @flippiefanus in comments, then this map is sure defined. The reason I was not doing that is because in the textbook I mentioned in the beginning, the Kraus operators are calculated using this map on maximally entangled qubits. So, if the solution suggested is true, and if we want to find the Kraus operators corresponding to a given dynamical map, then it seems we need to know how this map maps any given matrix, not necessarily density matrix.

In Breuer and Petruccione's book 'The theory of open quantum systems', section 2.4.3, 'representation theorem of quantum operations', a quantum operation $\phi_m$ corresponding to a measurement outcome m is defined as a map from one density matrix to another with following properties-

1. $0 \leq tr [\phi_m(\rho)] \leq 1$
2. Convex linearity: $\phi _m \left ( \Sigma a_i \rho _i\right )=\Sigma a_i \phi_m(\rho_i)$
3. The operator $(\phi_m \otimes I)$ is Completely postivite, where this new operator is defined as-

\begin{equation}(\phi_m \otimes I) \left ( \Sigma A_i \otimes B_i\right )\equiv \Sigma \phi_m(A_i) \otimes B_i \tag{1}\label{1}\end{equation}

My question is- in \eqref{1}, do these matrices $A_i$s and $B_i$s need to be density matrices? Because, I think it is not always possible to partition a density matrix in this way in a given basis. For example, lets take a maximally entangled density matrix

$$|\psi\rangle \langle \psi |= \frac {1}{2}\begin{pmatrix} 1 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 1 \end{pmatrix}$$

We can write $$|\psi\rangle \langle \psi |=\frac{1}{2} \left [ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \otimes \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} + \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \otimes \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \otimes \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \right ]$$

Then, $$(\phi_m \otimes I) \equiv \frac{1}{2} \left [ \phi_m \left [ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \right ] \otimes \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + \phi_m \left [ \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \right ] \otimes \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} + \phi_m \left [ \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \right ] \otimes \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} + \phi_m \left [ \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \right ] \otimes \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \right ]$$

But the problem is, $ \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$ is not a desnity matrix. So is $\phi_m \left [ \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \right ]$ defined?

Comments: If we think of $\phi_m$ in terms of Kraus operator representation, as suggested by @flippiefanus in comments, then this map is sure defined. The reason I was not doing that is because in the textbook I mentioned in the beginning, the Kraus operators are calculated using this map on maximally entangled qubits. So, if the solution suggested is true, and if we want to find the Kraus operators corresponding to a given dynamical map, then it seems we need to know how this map maps any given matrix, not necessarily density matrix.

In Breuer and Petruccione's book 'The theory of open quantum systems', section 2.4.3, 'representation theorem of quantum operations', a quantum operation $\phi_m$ corresponding to a measurement outcome m is defined as a map from one density matrix to another with following properties-

1. $0 \leq tr [\phi_m(\rho)] \leq 1$
2. Convex linearity: $\phi _m \left ( \Sigma a_i \rho _i\right )=\Sigma a_i \phi_m(\rho_i)$
3. The operator $(\phi_m \otimes I)$ is Completely postivite, where this new operator is defined as-

\begin{equation}(\phi_m \otimes I) \left ( \Sigma A_i \otimes B_i\right )=\Sigma \phi_m(A_i) \otimes B_i \tag{1}\label{1}\end{equation}

My question is regarding the definition of the operator $(\phi_m \otimes I)$ (on this auxiliary larger Hilbert space). For example, how will this operator act on a maximally entangled bipartite qubit state given by

$$|\psi\rangle = \frac {1}{\sqrt{2}}\left (|00\rangle + |11\rangle \right )$$

The desnity matrix is given as,

$$|\psi\rangle \langle \psi |= \frac {1}{2}\begin{pmatrix} 1 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 1 \end{pmatrix}$$

Now, in order for us to use the definition \eqref{1}, we need a decomposition of $|\psi\rangle \langle \psi |$ given as

$$|\psi\rangle \langle \psi |= \Sigma A_i \otimes B_i$$

But I don't think it is possible to express $|\psi\rangle \langle \psi |$ in this form in this basis because the two qubits are entangled. That is, I think a maximally entangled qubit cannot be written as classical mixture of unentangled qubits.

Then how do we find $ (\phi_m \otimes I)(|\psi\rangle \langle \psi |)$?

In Breuer and Petruccione's book 'The theory of open quantum systems', section 2.4.3, 'representation theorem of quantum operations', a quantum operation $\phi_m$ corresponding to a measurement outcome m is defined as a map from one density matrix to another with following properties-

1. $0 \leq tr [\phi_m(\rho)] \leq 1$
2. Convex linearity: $\phi _m \left ( \Sigma a_i \rho _i\right )=\Sigma a_i \phi_m(\rho_i)$
3. The operator $(\phi_m \otimes I)$ is Completely postivite, where this new operator is defined as-

\begin{equation}(\phi_m \otimes I) \left ( \Sigma A_i \otimes B_i\right )\equiv \Sigma \phi_m(A_i) \otimes B_i \tag{1}\label{1}\end{equation}

My question is- in \eqref{1}, do these matrices $A_i$s and $B_i$s need to be density matrices? Because, I think it is not always possible to partition a density matrix in this way in a given basis. For example, lets take a maximally entangled density matrix

$$|\psi\rangle \langle \psi |= \frac {1}{2}\begin{pmatrix} 1 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 1 \end{pmatrix}$$

We can write $$|\psi\rangle \langle \psi |=\frac{1}{2} \left [ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \otimes \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} + \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \otimes \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \otimes \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \right ]$$

Then, $$(\phi_m \otimes I) \equiv \frac{1}{2} \left [ \phi_m \left [ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \right ] \otimes \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + \phi_m \left [ \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \right ] \otimes \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} + \phi_m \left [ \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \right ] \otimes \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} + \phi_m \left [ \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \right ] \otimes \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \right ]$$

But the problem is, $ \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$ is not a desnity matrix. So is $\phi_m \left [ \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \right ]$ defined?

Comments: If we think of $\phi_m$ in terms of Kraus operator representation, as suggested by @flippiefanus in comments, then this map is sure defined. The reason I was not doing that is because in the textbook I mentioned in the beginning, the Kraus operators are calculated using this map on maximally entangled qubits. So, if the solution suggested is true, and if we want to find the Kraus operators corresponding to a given dynamical map, then it seems we need to know how this map maps any given matrix, not necessarily density matrix.