# projective measurement & POVM

Let us consider the following completely positive map $\mathcal{B}(\mathbb{C}^n)\ni\rho\mapsto L\rho L^\dagger$, where $L\in\mathcal{B}(\mathbb{C}^n)$ is any arbitrary operator (and can have rank $>1$). The above map is a completely positive Neumark's dilation theorem this can be extended to a projective measurement. Consider $A_1=L L^\dagger$ (and assume $A_1\leq1$), and choose suitable positive semidefinite operators $A_j$'s such that $\sum_jA_j=I$ (identity matrix). By Naimark, we shall get a projective measurement given by a projection operator $P_j\in\mathcal{B}(\mathbb{C}^n\otimes\mathcal{H}_{aux})$, where $\mathcal{H}_{aux}$ is the auxiliary space of minimal dimension for such extension. The relation is given (with respect to a fixed auxiliary state $\rho_{aux}$ as: \begin{equation} \langle m|A_j|n\rangle=\sum_{r,s}\langle m,r|P_j|n,s\rangle\langle s|\rho_{aux}|r\rangle \end{equation} (The above formula is taken from Quantum Theory: Concepts and Methods by Asher Peres)

$P_j$'s are not unique. They depend on the $A_j$'s we have chosen and also on the auxiliary $\rho_{aux}$. My question is the following: What will be the extension $\tilde{\rho}\in\mathcal{B}(\mathbb{C}^n\otimes\mathcal{H}_{aux})$ (of $\rho\in\mathcal{B}(\mathbb{C}^n)$) should be so that \begin{equation} Tr_2(P\tilde{\rho}P)=L'\rho L'^\dagger,\quad \text{where} \quad L'L'^\dagger=A_1. \end{equation} $Tr_2$ is the partial trace with respect to the second (auxiliary) system. (It will be nice if $L=L'$, but I guess it is too much to ask). My initial guess was $\tilde{\rho}=\rho\otimes\rho_{aux}$. But this does not seem to be true from a few examples. Please tell me, whether I am making any mistake as well.

UPDATE/OBSERVATION: If we put $\rho=I$ (the identity matrix) then the second equation should give us $A_1$.

• I am confused about the question. Are you asking for a $\tilde{\rho}$ and $L'$ such that given $\rho$, $P$ and $A_{1}$, you have $\mbox{Tr}_{2}\left(P\tilde{\rho}P\right)=L'\rho L'^{\dagger}$, $L'L'^{\dagger}=A_{1}$? – Alex L May 5 '13 at 15:34
• yes. i mean exactly what extension of $\rho$ (which I have written as $\tilde{rho}$) has to be taken such that the above relation holds. i do not want the right hand side to be 'splitted' as $\sum_k L_k'\rho L_k' ^\dagger$ such that $\sum_k L_k' L_k' ^\dagger=A_1$. – RSG May 5 '13 at 16:42