Trace norm and Bures distance I am given that the Bures distance between two density matrices (i.e. semi positive-definite self-adjoint maps which have trace equal to unity) $\rho_1,\rho_2$ is: $D_B(\rho_1,\rho_2) = 2(1-\|\rho_1^{1/2} \rho_{2}^{1/2}\|)$ where $\|\cdot\|$ is the $L^1$-norm defined by $\|A\| = \mathrm{Tr}(\sqrt{A^\star A})$,
I try to prove that $D_B(\rho_1,\rho_2) =0 \implies \rho_1 = \rho_2$.
I have already proven that $0\leq D_B(\rho_1,\rho_2)\leq 2$. My attempt is to write using that $\mathrm{Tr}(\rho_1) = \mathrm{Tr}(\rho_2) = 1$:
$$1 = \|\rho_1^{1/2}\rho_{2}^{1/2}\| = \|\rho_1\| = \|\rho_2\|,$$
however does it follow from this that $\rho_1 = \rho_2$ should hold?
 A: Conceptually this is very simple -- just write this out in a particularly nice way and apply Cauchy Schwarz.
It suffices the show $\|\rho_1^{1/2} \rho_{2}^{1/2}\|=\|\rho_1^{1/2} \rho_{2}^{1/2}\|_{S_1}\leq 1$ with equality iff $\rho_{1}=\rho_{2}$. I include the $S_1$ since this is a Schatten 1 norm.
It is also convenient to quasi-linearize the Schatten 1 norm by writing it as
$\Big \Vert A \Big \Vert_{S_1}=\max_U \Big \vert \text{trace}\big(U^*A\big)\Big\vert$ with $U\in U_n$.  This comes the from Polar Decomposition of $A$ (with optional use (i) of von Neumann Trace Inequality to verify the maximum or (ii) diagonalize the HPSD matrix in A's Polar Decomposition and apply triangle inequality when writing out the trace).
Proof
$\|\rho_1^{1/2} \rho_{2}^{1/2}\|_{S_1}$
$=\max_U \Big \vert \text{trace}\Big(U^*\rho_1^\frac{1}{2} \rho_2^\frac{1}{2}\Big) \Big \vert $
$=\max_U \Big \vert \text{trace}\Big(\big(U^*\rho_1^\frac{1}{2}\big) \rho_2^\frac{1}{2}\Big) \Big \vert $
$=\max_U \Big \vert \text{trace}\Big(\big(\rho_2^\frac{1}{2}\big)\big(U^*\rho_1^\frac{1}{2}\big)\Big) \Big \vert $
$=\max_U \Big \vert \text{trace}\Big(\big(\rho_2^\frac{1}{2}\big)^*\big(U^*\rho_1^\frac{1}{2}\big)\Big) \Big \vert $
$\leq\Big \Vert \rho_2^\frac{1}{2}\Big \Vert_F\cdot \Big \Vert U^*\rho_1^\frac{1}{2}\Big \Vert_F$
$= \text{trace}\big(\rho_2\big)^\frac{1}{2}\cdot \text{trace}\big(\rho_1\big)^\frac{1}{2}$
$= 1 \cdot 1$
$=1$
with equality iff $\rho_1^\frac{1}{2} = \rho_2^\frac{1}{2}$ i.e. iff $\rho_1 = \rho_2$ by uniqueness of the square-root
further explanation:
examining the equality conditions of Cauchy Schwarz, which imply $\eta \cdot U^*\rho_1^\frac{1}{2} = \rho_2^\frac{1}{2}$
$1=\vert \eta\vert^2\cdot \Big\Vert U^*\rho_1^\frac{1}{2}\Big \Vert_F^2=\vert \eta\vert^2\cdot\text{trace}\big(\rho_1\big)= \text{trace}\big(\rho_2\big)=1\implies \eta$ is on the unit circle.
Thus we can say
$Q\rho_1^\frac{1}{2} = \big(\eta \cdot U^*\big)\rho_1^\frac{1}{2} = \rho_2^\frac{1}{2}$
if $\rho_2$ was PD, it would be immediate that $Q=I$ since Polar Decomposition is unique for invertible matrices.
Since these are only PSD, we take a somewhat uglier finish by proving $\rho_1^\frac{1}{2}=Q\rho_1^\frac{1}{2}$.
$Q\rho_1^\frac{1}{2} =\rho_2^\frac{1}{2}$ means that the Schatten 1 norm is given by taking the trace.  Using the earlier quasi-linear representation of the Schatten 1 norm, this implies
$ \max_V \Big \vert \text{trace}\big(V^*Q\rho_1^\frac{1}{2}\big)\Big\vert=\Big\Vert Q\rho_1^\frac{1}{2}\Big\Vert_{S_1}=\text{trace}\big(Q\rho_1^\frac{1}{2}\big) \leq \text{trace}\big(\rho_1^\frac{1}{2}\big)=\big \vert\text{trace}\big(\rho_1^\frac{1}{2}\big)\big \vert$
(with the upper bound by von Neumann trace or earlier mentioned triangle inequality argument) and
$\big \vert\text{trace}\big(\rho_1^\frac{1}{2}\big)\big \vert\leq  \max_V \Big \vert \text{trace}\big(V^*Q\rho_1^\frac{1}{2}\big)\Big\vert$
(by selection $V:=Q$)
$\implies\text{trace}\big(\rho_1^\frac{1}{2}\big)=\text{trace}\big(Q\rho_1^\frac{1}{2}\big)$
and one more application of Cauchy-Schwarz gives
$\text{trace}\Big(Q\rho_1^\frac{1}{2}\Big)=\text{trace}\Big(\big(Q\rho_1^\frac{1}{4}\big)\rho_1^\frac{1}{4}\Big)\leq \Big \Vert Q\rho_1^\frac{1}{4}\Big \Vert_F\cdot \Big \Vert \rho_1^\frac{1}{4}\Big \Vert_F=\text{trace}\Big(\rho_1^\frac{1}{2}\Big)$
which is met with equality $\implies \gamma \cdot Q \rho_1^\frac{1}{4}=\rho_1^\frac{1}{4}$ (where $\gamma$ is on the unit circle, by checking Frobenius norms) and multiplying each side on the right by $\rho_1^\frac{1}{4}\implies \gamma \cdot Q \rho_1^\frac{1}{2}=\rho_1^\frac{1}{2}\implies \gamma^{-1} \rho_1^\frac{1}{2}=Q \rho_1^\frac{1}{2}= \rho_2^\frac{1}{2}$ and the RHS is PSD thus the LHS is well, $\implies \gamma = 1$
