# How to extend a quantum operation to an auxiliary Hilbert space?

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.

• Perhaps you can express $\phi_m$ in terms of Kraus operators. That would allow you to operator on the elements in the matrix. Feb 15, 2021 at 13:11
• @flippiefanus do you mean, I take reduced density Matrix of the system qubit, apply the map on it (directly or using krauss operators) and then take the tensor product of the result with the environment qubit reduced density Matrix? Feb 15, 2021 at 13:18
• Edit: but this does not make sense in one scenario- suppose i have a qubit maximally entangled with another qubit, and i apply an identity map to my qubit. Then the total operator will map an entangled state to a uncorrelated tensor product state, even though i did not disturb the system. Feb 15, 2021 at 13:30
• No, I meant it in a more abstract mathematical sense. The point is the model $\phi_m$ in terms of a set of operators that you can apply on opposite sides of the density operator. Feb 16, 2021 at 3:36
• @flippiefanus suppose $A_i$'s are the kraus operator of $\phi_m$. Are the Kraus operators for $\phi_m \otimes I$ given by $A_i \otimes I$? In that case, the map $\phi_m$ can be defined even if the input Matrix is not a density Matrix, and i guess the map $\phi_m \otimes I$ can be defined for maximally entangled qubits as well. Feb 16, 2021 at 5:25

$$\Phi_m$$ is just a linear map. It acts on any linear operator in its input space, not just density matrices.