Posted on 4th February 2025
A little mathematics: I have claimed the following in a number of talks, hence perhaps a proof is called for. Given two Density Matrices \( \rho_1, \rho_2 \) (so positive-semidefinite, trace one matrices) we say that the associated quantum states are distinguishable when \( \newcommand{\Tr}{\operatorname{Tr}}\Tr(\rho_1 \rho_2) = 0 \). Let \( T \) be a quantum channel; we are interested in when \( T(\rho_1) \) and \( T(\rho_2) \) are distinguishable.
Claim: \( T(\rho_1) \) and \( T(\rho_2) \) are distinguishable if and only if \( \Tr\big(T(|\xi\rangle\langle\xi|) T(|\eta\rangle\langle\eta|) \big)=0 \) for each \( \newcommand{\im}{\operatorname{Im}}\xi \in \im(\rho_1) \) and \( \eta \in \im(\rho_2) \).
We can expand a density matrix \( \rho_1 \) as \( \rho_1 = \sum_i |\xi_i \rangle \langle \xi_i| \) for some orthogonal family \( (\xi_i) \) (this is just diagonalisation of positive matrices).
Similarly let \( \rho_2 = \sum_j |\eta_j\rangle\langle\eta_j| \). Then \[ \Tr(T(\rho_1)T(\rho_2))=0 \quad\Leftrightarrow\quad \sum_{i,j} \Tr\big( T(|\xi_i \rangle \langle \xi_i|) T(|\eta_j \rangle \langle \eta_j|) \big) = 0 \qquad\qquad (\dagger). \] Given positive \( x,y \) we see that \( \Tr(xy) = \Tr(y^{1/2} x^{1/2} x^{1/2} y^{1/2}) = \Tr(z^*z) \geq 0 \) say, where \( z = x^{1/2} y^{1/2} \). So each summand in the right-hand-side of \( (\dagger) \) is positive, and hence \( (\dagger) \) is equivalent to \( \Tr\big( T(|\xi_i \rangle \langle \xi_i|) T(|\eta_j \rangle \langle \eta_j|) \big) = 0 \) for each \( i,j \). This proves the "if" part of our claim, as each \( \xi_i \) is in the image of \( \rho_1 \), and similarly for \( \eta_j \).
Recall that \( \rho_1 \) and \( \rho_1^{1/2} \) have the same image (diagonalise, for example). Let \( \xi \in \im(\rho_1) \), so \( \xi = \rho_1^{1/2}(\xi') \) for some \( \xi' \). Then, for any \( \psi \), \[ (\psi | |\xi\rangle\langle\xi| \psi) = (\psi|\xi)(\xi|\psi) = (\psi|\rho_1^{1/2}(\xi')) (\rho_1^{1/2}(\xi')|\psi) = (\rho_1^{1/2}(\psi)|\xi')(\xi'|\rho_1^{1/2}(\psi)). \] Using Cauchy-Schwarz, this is \[ \leq \|\xi'\|^2 \|\rho_1^{1/2}(\psi)\|^2 = \|\xi'\|^2 (\psi|\rho_1(\psi)). \] This holds for all \( \psi \) and so \( |\xi\rangle\langle\xi| \leq \|\xi'\|^2 \rho_1 \). Set \( k_1 = \|\xi'\|^2 \). Similarly, for \( \eta\in\im(\rho_2) \) there is \( k_2\geq0 \) with \( |\eta\rangle\langle\eta| \leq k_2 \rho_2 \). Then \begin{align*} \Tr\big(T(|\xi\rangle\langle\xi|) T(|\eta\rangle\langle\eta|) \big) &= \Tr\big(T(|\eta\rangle\langle\eta|)^{1/2} T(|\xi\rangle\langle\xi|) T(|\eta\rangle\langle\eta|)^{1/2} \big) \\ &\leq k_1 \Tr\big(T(|\eta\rangle\langle\eta|)^{1/2} T(\rho_1) T(|\eta\rangle\langle\eta|)^{1/2} \big) \\ &= k_1 \Tr\big(T(\rho_1)^{1/2} T(|\eta\rangle\langle\eta|) T(\rho_1)^{1/2} \big) \\ &\leq k_1 k_2 \Tr\big(T(\rho_1)^{1/2} T(\rho_2)) T(\rho_1)^{1/2} \big) = k_1 k_2 \Tr\big( T(\rho_1) T(\rho_2) \big). \end{align*} So, if \( T(\rho_1) \) and \( T(\rho_2) \) are distinguishable, then also \( \Tr\big(T(|\xi\rangle\langle\xi|) T(|\eta\rangle\langle\eta|) \big)=0 \), showing the "only if" part of the claim.
Notice that we only used that \( T \) was positive (so preserves the semi-definite ordering).