What is the intuition behind Bures and angle metrics?

Question

I am reading Distance measures to compare real and ideal quantum processes and it is explained the motivation behind Bures metric and angle metric.

Bures metric is defined as:

$$B(\rho,\sigma)=\sqrt{2-2 F(\rho,\sigma)}$$

Angle metric is defined as:

$$A(\rho,\sigma)=\arccos(\sqrt{F(\rho,\sigma)})$$

Where $F(\rho,\sigma)$ is the fidelity between $\rho$ and $\sigma$ density matrices. He says that we can understand such motivation on pure states: we would see it comes from the usual euclidian distance.

If I do such calculations, I would define the euclidian distance as:

$$d(X,Y)=||X-Y||=\sqrt{\langle X-Y | X-Y \rangle}=\sqrt{2-2 Re(\langle X | Y \rangle)} $$

To find the Bure metric I have to assume $\langle X | Y \rangle \geq 0$.

But why would it be the case ? For instance if I consider:

$$|\psi \rangle = | a \rangle + |b \rangle $$

I cannot change the relative phase between $|a \rangle$ and $|b \rangle$ as I want (because it would change the physical state $|\psi \rangle$). Thus if $\langle a | b \rangle $ is not a positive number I guess there is nothing much I can do for that.

How to understand the intuition behind such metric then? Should I actually consider it as an "abstract" definition on which I verify that it satisfies the axioms of a metric? But it would be weird in the way the paper explains the motivation behind.

Similar question for the angle metric.

[edit]: I think it might come from the fact we want to define a distance between physical states. Considering $|\Phi \rangle$ and $| \Psi \rangle$ two physical state, their global phase do not matter. Thus, to have a simple formula we can choose their phases $\phi_{\Psi}, \phi_{\Phi}$ so that $\langle \Psi | \Phi \rangle \geq 0$ which correspond to the upper bound: $\sup_{\phi_{\Psi}, \phi_{\Phi}}(Re[\langle \Psi | \Phi \rangle])=\langle \Psi | \Phi \rangle$. It somehow makes sense because we are interested into distance between physical and not mathematical states. We can thus fix the global phases of the two states as we would like.

Does that make sense ?

Answer 1

Filling out a number of details for the sake of a complete answer —

Starting from the linked article, Distance measures to compare real and ideal quantum processes [arXiv:quant-ph/0408063], the definition of fidelity is given in Eqn. (4) as $$ F(\rho,\sigma) = \mathrm{tr}\Bigl( \!\sqrt{\sqrt{\rho} \!\phantom|\sigma \phantom|\!\!\sqrt{\rho}\phantom|}\Bigr)^2$$ — which might look a bit intimidating, but demonstrates two important things about fidelity: that it is defined in general on density operators (not just state vectors), and that it is always a non-negative real number. If you want to compute it for pure states, the definition above ends up being equivalent to $$ F(\lvert \psi\rangle\! \langle \psi\rvert,\lvert \phi\rangle\! \langle \phi\rvert) = \langle\psi\vert \phi\rangle\! \langle\phi\vert \psi\rangle = \bigl\lvert \langle\psi\vert \phi\rangle \bigr\rvert^2$$ which is always a non-negative real, and in particular, which does not depend on any global phases that you might consider for either the state $\lvert \psi \rangle$ or $\lvert \phi \rangle$ (which is not physical information about the state).

The Bures metric (from the second column of page 4) is then $$ B(\rho,\sigma) = \sqrt{2 - 2\sqrt{F(\rho,\sigma)}} $$ which for pure states simplifies to $$\begin{aligned} B(\lvert \psi\rangle\! \langle \psi\rvert,\lvert \phi\rangle\! \langle \phi\rvert) &= \sqrt{2 - 2\sqrt{F(\lvert \psi\rangle\! \langle \psi\rvert,\lvert \phi\rangle\! \langle \phi\rvert)}} \\&= \sqrt{2 - 2\bigl\lvert \langle\psi\vert \phi\rangle \bigr\rvert} \\&= \sqrt{2 - 2 \max \langle\psi'\vert \phi'\rangle},\end{aligned} $$ where the maximum is taken over unit vectors $\lvert \psi'\rangle \propto \lvert \psi\rangle$ and $\lvert \phi'\rangle \propto \lvert \phi\rangle$.

You ask (not unreasonably) why, for pure states, you would take the absolute value $\lvert \langle \psi \vert \phi \rangle \rvert$, instead of the real part $\mathrm{Re}\,\langle \psi \vert \phi \rangle$ as you would if you were dealing directly with the inner products of vectors $\lvert \psi \rangle$ and $\lvert \phi \rangle$. The answer is that, because we are interested in the states and not actually in particular vectors which represent those states, working directly with the state vectors won't necessarily provide a sensible answer. For a state $\lvert \phi' \rangle \propto \lvert \phi \rangle$, the values of $\mathrm{Re}\,\langle \psi \vert \phi \rangle$ and $\mathrm{Re}\,\langle \psi \vert \phi' \rangle$ usually won't be the same — but whether we use $\lvert \phi' \rangle$ or $\lvert \phi \rangle$ to represent the state should be a purely arbitrary choice with no impact either on the physics or on our the analysis of the physics. Any choice of formula should be stable under such arbitrary choices, and furthermore (for a metric) should yield the value $0$ if we were to consider different ways $\lvert \phi' \rangle$ and $\lvert \phi \rangle$ to represent the same state.

Bear in mind that, at the end of the day, their remark about simplifying to the Euclidean metric is likely to have been a quick attempt to provide intuition, rather than a serious attempt to provide a formal statement. However, there is a sense in which taking the absolute value (or if you prefer, the maximum inner product among equivalent states up to global phases) is the correct approach to considering the connection to the "Euclidean distance" between "states", and I expect that this is what they have in mind.