On the Incomparability of Operator Fidelity and Q-Power Means in the Near Order
Let $\mathbb{P}$ and $\mathbb{P}_0$ be the cones of $n\times n$ positive definite and positive semidefinite matrices over $\mathbb{C}$, respectively. We define
and call this the operator fidelity of $A$ and $B$. For $p\in\mathbb{R}$, define
with the identification that $Q_0(A,B)$ is the log-Euclidean mean of $A$ and $B$.
Let $A, B \in \mathbb{P}_0$. Then
- $A \# (\alpha B) = \alpha^{1/2}(A \# B)$ for every $\alpha > 0$.
- $\det(A \# B) = (\det A \, \det B)^{1/2}$.
Let $X$ be a $2\times 2$ positive definite matrix. Then $X \succeq I$ implies that $\operatorname{tr} X \le 1 + \det X$.
Proof. Let $\lambda_1, \lambda_2 \ge 0$ be the eigenvalues of $X$. The condition $X \succeq I$ is equivalent to saying $\lambda_1, \lambda_2 \ge 1$. This implies that $(\lambda_1 - 1)(\lambda_2 - 1) \ge 0$. This inequality expands to the required condition.
Let $X \in \mathbb{P}_0$ be a matrix of rank 1. Then $X^{1/2} = (\operatorname{tr} X)^{-1/2} X$.
Let $A \in \mathbb{P}$ and $B \in \mathbb{P}_0$ such that $B$ is of rank 1. Then
\[ A \# B = \left(\operatorname{tr} A^{-1}B\right)^{-1/2} B. \]Proof. Use Lemma 3 to get
\[ A\#B = A^{1/2}\!\left(A^{-1/2}BA^{-1/2}\right)^{1/2}\!A^{1/2} = \left(\operatorname{tr} A^{-1}B\right)^{-1/2}\!A^{1/2}\!\left(A^{-1/2}BA^{-1/2}\right)\!A^{1/2} = \left(\operatorname{tr} A^{-1}B\right)^{-1/2}\!B. \]There exist $A, B \in \mathbb{P}$ such that $F(A,B) \npreceq Q_p(A,B)$ for all $p > 0$.
Proof. Suppose $F(A,B) \preceq Q_p(A,B)$ for all $A, B$ and some $p \ge 0$; we use this assumption to produce a contradiction. When $A$ and $B$ are clear from context, write $F = F(A,B)$ and $Q = Q_p(A,B)$.
Notice that, just as choosing $A$ and $B$ fixes $F$, choosing $A$ and $F$ fixes $B$. That is, let $A, F \in \mathbb{P}$ and let $H = F^2$. With $B = A^{-1/2}HA^{-1/2}$, we have $F(A,B) = F$. Further, let $T = HA^{-1}$. Then $B$ is similar to $T$, as $B = A^{-1/2}TA^{1/2}$.
For $\gamma \in (0,1)$ let
\[ A = \begin{pmatrix} \gamma^2 & 0 \\ 0 & 1 \end{pmatrix}. \]Due to Lemma 2 and Lemma 1, the inequality
\[ \gamma\,\operatorname{tr}\!\left[F^{-1}\# Q\right] \;\le\; \gamma + \gamma\left[\frac{\det Q}{\det F}\right]^{1/2} \tag{1} \]must be true for all $\gamma$. Thus the inequality must also hold in the limit $\gamma \to 0$. Note that $F$ is independent of $\gamma$, but $Q$ is not, so $\gamma$ can be collected with $Q$. To do this define
\[ L = \lim_{\gamma \to 0} \gamma^2 Q, \qquad \ell = \lim_{\gamma \to 0} \gamma^2 \det Q. \]Using again the properties of the geometric mean and determinants, inequality (1) in the limit $\gamma \to 0$, after squaring, becomes
\[ \operatorname{tr}^2\!\left(F^{-1}\# L\right) \;\le\; \frac{\ell}{\det F}. \tag{2} \]Our effort now will be in computing $L$ and $\ell$. Note that
\[ L = \lim_{\gamma \to 0} \gamma^2 Q_p(A,B) = \lim_{\gamma \to 0} Q_p(\gamma^2 A, \gamma^2 B) = Q_p\!\left(\lim_{\gamma \to 0}\gamma^2 A,\ \lim_{\gamma \to 0}\gamma^2 B\right). \]The limit of the first argument is $0$. And
\[ \lim_{\gamma \to 0}\gamma^2 B = \lim_{\gamma \to 0}\left(\gamma A^{-1/2}\right) H \left(\gamma A^{-1/2}\right) = \begin{pmatrix} h_{11} & 0 \\ 0 & 0 \end{pmatrix}, \]which is a rank 1 matrix. This immediately gives
\[ L = 2^{-1/p}\begin{pmatrix} h_{11} & 0 \\ 0 & 0 \end{pmatrix}. \]Therefore we apply Lemma 4 to evaluate the left-hand side of equation (2) as
\[ \operatorname{tr}^2\!\left(F^{-1}\# L\right) = \left(\operatorname{tr} FL\right)^{-1}\left(\operatorname{tr} L\right)^2 = \left(2^{-1/p}f_{11}h_{11}\right)^{-1}h_{11}^2\,2^{-2/p} = 2^{-1/p}\,\frac{h_{11}}{f_{11}}. \]Now let us focus on $\ell$. Note that
\[ \gamma^2\det Q_p(A,B) = \gamma^2\det Q_p(A,T) = \gamma^{-2}\det Q_p(\gamma^2A,\gamma^2T) = 2^{-2/p}\gamma^{-2}\Big(\det\!\left[(\gamma^2A)^p+(\gamma^2T)^p\right]\Big)^{1/p}. \]In the first equality we have used that $B$ is similar to $T$ via the transformation $A^{1/2}$, which keeps $A$ unchanged. Let $K = (\gamma^2T)^p$. Then we compute
\[ \det\!\left[(\gamma^2A)^p+K\right] = k_{11}\gamma^{2p}+k_{22}\gamma^{4p}+\gamma^{6p}+\det K = \gamma^{2p}\Big[k_{11}+k_{22}\gamma^{2p}+\gamma^{4p}+(\det H)^p\Big], \]using $\det K = \gamma^{4p}(\det H)^p$. So this gives
\[ \ell = 2^{-2/p}\lim_{\gamma \to 0}\Big[k_{11}+k_{22}\gamma^{2p}+\gamma^{4p}+(\det H)^p\Big]^{1/p}. \]Note here that $k_{11}$ and $k_{22}$ are functions of $\gamma$, but $H$ is not. We compute their limit using
\[ \lim_{\gamma\to0}K = \left(\lim_{\gamma\to0}\gamma^2T\right)^{\!p} = \begin{pmatrix} h_{11}^p & 0 \\ \ast & 0 \end{pmatrix}, \]leveraging the lower-triangular form of $\lim_{\gamma \to 0}\gamma^2T$ to easily calculate the $p$-th power. Finally we get
\[ \ell = 2^{-2/p}\left[h_{11}^p + (\det H)^p\right]^{1/p}. \]Thus inequality (2) reduces to
\[ 2^{-1/p}\,\frac{h_{11}}{f_{11}} \;\le\; 2^{-2/p}\,\frac{\left[h_{11}^p+(\det H)^p\right]^{1/p}}{\det F}, \]which can be further rearranged as
\[ 2\left(h_{11}\det F\right)^p \;\le\; f_{11}^p\left[h_{11}^p+(\det H)^p\right]. \tag{3} \]Our assumption is that this holds for every $F$ and $H = F^2$. But choose
\[ H = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} \qquad\text{and}\qquad F = \begin{pmatrix} \tfrac{2}{\sqrt5} & \ast \\ \ast & \ast \end{pmatrix}. \]Then inequality (3) says
\[ 1 \;\le\; \left(\frac{2}{\sqrt5}\right)^{p}, \]which is false for every $p > 0$. This is a contradiction.