Show the equivalence between two definitions of the Bures (/Fisher/Wasserstein) metric

Question

I would like to understand the connection between these two, very different, definitions of the Bures distance.

1. Through the traces of the density matrices $\rho_1$, $\rho_2$:

\begin{equation} d(\varrho_1, \varrho_2):=\sqrt{\mbox{tr}\varrho_1 + \mbox{tr} \varrho_2 -2\mbox{tr}\left(\varrho_1^{1/2}\,\varrho_2\, \varrho_1^{1/2}\right)^{1/2}}\,. \end{equation}

2. Through the purifications of the density matrices $\varrho_1$ and $\varrho_2$:

\begin{equation} d(\varrho_1\,, \varrho_2)= \min_{over~purification} ||\psi_{\varrho_1} - \psi_{\varrho_2}||\,, \end{equation}

where the minimum is again taken over all possible purifications $|\psi_{\varrho_1}\rangle$ and - $|\psi_{\varrho_2}\rangle$, of the states $\varrho_1$ and $\varrho_2$.

How to see that these two definitions are equivalent? What apparatus is needed for that?

Answer 1

I'll assume that everything is finite-dimensional. We start by noting the following facts:

1. Consider a linear operator $X\in \operatorname{Lin}(\mathcal X,\mathcal Y)$. Let $\operatorname{vec}(\cdot):\operatorname{Lin}(\mathcal X,\mathcal Y)\to\mathcal Y\otimes\mathcal X$ denote the vectorisation operation, and let $\operatorname{unvec}(\cdot):\mathcal Y\otimes\mathcal X\to\operatorname{Lin}(\mathcal X,\mathcal Y)$ denote its inverse. These are definable via their action on a basis as: $$\operatorname{vec}(e_i e_j^\dagger) = e_i\otimes e_j, \qquad \operatorname{unvec}(e_i \otimes e_j) = e_i e_j^\dagger.$$ Note also that if $\psi\in\mathcal X\otimes\mathcal Y$ satisfies $\psi=\operatorname{vec}(A)$ for some $A\in\mathrm{Lin}(\mathcal Y,\mathcal X)$, then $$\operatorname{Tr}_2[\psi\psi^\dagger] = AA^\dagger.$$

2. Note that the Hilbert-Schmidt norm of $X$, $\|X\|_2\equiv \sqrt{\operatorname{Tr}(X^\dagger X)}$, is connected with the standard Euclidean norm (also denoted with $\|\cdot\|_2$) of its vectorisation, via $$\|X\|_2 = \|\mathrm{vec}(X)\|_2.$$

3. Note that the set of purifications of a given state (density matrix) $\rho\in\mathrm{Pos}(\mathcal X)$, can be characterised as the set of state vectors $$\{\operatorname{vec}(\sqrt\rho U^\dagger) : \,\, U\in\mathrm U(\mathcal X,\mathcal Z)\},$$ where $\mathrm U(\mathcal X,\mathcal Z)$ denotes the set of isometries $\mathcal X\to\mathcal Z$, where $\mathcal Z$ is some auxiliary space with $\dim(\mathcal Z)\ge \operatorname{rank}(\rho)$. We can assume $\mathcal Z=\mathcal X$ for simplicity, so that $U$ is unitary and $U^\dagger=U^{-1}$.

You can have a look e.g. at Watrous' book for more details on this formalism.

Let then $\psi_\rho\equiv\operatorname{vec}(\sqrt{\rho}U^\dagger)$ and $\psi_\sigma\equiv\operatorname{vec}(\sqrt\sigma V^\dagger)$ be purifications of $\rho$ and $\sigma$, respectively. Then $$\|\psi_\rho - \psi_\sigma\|_2^2 = \|\sqrt\rho U^\dagger-\sqrt\sigma V^\dagger\|_2^2 = \operatorname{Tr}(\rho)+\operatorname{Tr}(\sigma) - \operatorname{Tr}[U\sqrt\rho\sqrt\sigma V^\dagger + V\sqrt\sigma\sqrt\rho U^\dagger] \\ = \operatorname{Tr}(\rho)+\operatorname{Tr}(\sigma) - {\rm Tr}(\sqrt\rho\sqrt\sigma W) - {\rm Tr}(W^\dagger \sqrt\sigma \sqrt\rho) \\ = \operatorname{Tr}(\rho)+\operatorname{Tr}(\sigma) - 2\operatorname{Re}{\rm Tr}(\sqrt\rho\sqrt\sigma W),$$ where $W\equiv V^\dagger U$. Minimising the above expression with respect to the purifications amounts to maximising the last two terms with respect to the unitaries, that is, maximising with respect to $W$ the expression $$\operatorname{Re}{\rm Tr}(\sqrt\rho\sqrt\sigma W).$$ The result is almost there: maximising $|\operatorname{Tr}(\sqrt\rho\sqrt\sigma W)|$ over the set of possible unitaries $W$ gives $\operatorname{Tr}|\sqrt\rho\sqrt\sigma|\equiv \|\sqrt\rho\sqrt\sigma\|_1$, where $\|A\|_1$ is the trace norm defined as $\|A\|_1 \equiv \operatorname{Tr}\sqrt{A^\dagger A}.$ You thus get $$\operatorname{Tr}|\sqrt\rho\sqrt\sigma| = \operatorname{Tr}\sqrt{\sqrt\sigma\rho\sqrt\sigma}.$$ Upon changing $W$ with a simple global phase, we can see that maximising $|\operatorname{Tr}(\sqrt\rho\sqrt\sigma W)|$ gives the same result as maximising $\operatorname{Re}\operatorname{Tr}(\sqrt\rho\sqrt\sigma W)$, and thus we get the result.