The Exercise

Let $\rho$ and $\sigma$ be states of a quantum system with a finite-dimensional state space, and let $P$ be an invertible positive operator on that space, an observable with strictly positive eigenvalues. Then $\operatorname{tr}(\rho P)$ is the mean value of $P$ in the state $\rho$ , and $\operatorname{tr}(\sigma P^{-1})$ is the mean value of its inverse in $\sigma$ . The exercise is to show that the infimum of the product of these two mean values, over all such $P$ , is the square of the fidelity between $\rho$ and $\sigma$ .

The library states this for any two positive operators, and adds a pair of qubit states for which the square cannot be dropped.

The Alberti Product

Let $E$ be a finite-dimensional complex Hilbert space. For a positive operator $\rho$ on $E$ , $\sqrt{\rho}$ is its positive square root, and $P > 0$ says that the operator $P$ is positive and invertible.

For operators $\rho$ , $\sigma$ and an invertible operator $P$ on $E$ , the Alberti product of $\rho$ and $\sigma$ at $P$ is the real number $A(\rho,\sigma;P) = \operatorname{Re}\operatorname{tr}(\rho P)\, \operatorname{Re}\operatorname{tr}(\sigma P^{-1}).$ (1)^✦

The trace norm $\|T\|_1$ of a linear map $T$ between finite-dimensional inner product spaces is the sum of its singular values $s_1(T), \dots, s_d(T)$ , with $d$ the dimension of the space $T$ acts on: $\|T\|_1 = \sum_{i=1}^{d} s_i(T).$ (2)^✦

Two Qubit States

The second claim is about real $2\times 2$ matrices. For real square matrices $\rho$ and $\sigma$ , the fidelity $F(\rho,\sigma)$ is the number $F(\rho,\sigma) = \operatorname{tr}\sqrt{\sqrt{\rho}\,\sigma\sqrt{\rho}},$ (3)^✦ and for an invertible real matrix $P$ the Alberti product is $A(\rho,\sigma;P) = \operatorname{tr}(\rho P)\,\operatorname{tr}(\sigma P^{-1}).$ (4)^✦

The two states are the qubit density matrices $\rho_0 = \begin{pmatrix} 9/10 & 0 \\ 0 & 1/10 \end{pmatrix}, \qquad \sigma_0 = \begin{pmatrix} 1/10 & 0 \\ 0 & 9/10 \end{pmatrix},$ (5) diagonal in the computational basis with eigenvalues $9/10$ and $1/10$ in opposite order: $\rho_0$ is mostly $|0\rangle$ , and $\sigma_0$ is mostly $|1\rangle$ .

The Claim

Theorem 1. Let $\rho$ and $\sigma$ be positive operators on a finite-dimensional complex Hilbert space $E$ . The infimum of their Alberti products (1) over the invertible positive operators $P$ on $E$ is the square of the trace norm of $\sqrt{\rho}\sqrt{\sigma}$ : $\inf_{P > 0} A(\rho,\sigma;P) = \bigl\|\sqrt{\rho}\,\sqrt{\sigma}\bigr\|_1^2.$ (6)^✦

Theorem 2. Let $\rho_0$ and $\sigma_0$ be the qubit states (5). There is a positive definite real $2\times 2$ matrix $P$ whose Alberti product (4) lies below the fidelity (3): $A(\rho_0,\sigma_0;P) < F(\rho_0,\sigma_0).$ (7)^✦

For states $\rho$ and $\sigma$ , where $\|\sqrt{\rho}\sqrt{\sigma}\|_1$ is the fidelity $F(\rho,\sigma)$ , the first theorem is the exercise.

What Is Checked

The first theorem is stated over the Alberti product (1) of operators and the trace norm (2); the second over the fidelity (3) and the Alberti product (4) of real matrices, at the states (5). For a positive operator $\rho$ , a positive operator $\sigma$ , and an invertible positive operator $P$ , the two mean values entering the Alberti product are real and nonnegative.