About the connection between distance in Hilbert space and fidelity

Question

I have the feeling that there is a direct relationship between the distance in Hilbert space and the fidelity (similarity) between quantum states.

1. short version : the distance between two quantum states and is defined as the norm of their difference vector: d(ψ,ϕ)=∥ψ−ϕ∥=⟨ψ−ϕ∣ψ−ϕ⟩​

Now, as far as I undertstand, this norm/ distance in Hilbert space is a measure of how different two quantum states are, with fidelity providing a direct way to quantify their similarity.Higher fidelity thus indicates greater similarity between states, resulting in a smaller distance ; while lower fidelity (typically, with orthogonal states) indicates greater dissimilarity, resulting in a larger distance. Would you agree with that ?

2. Long version -- or an attempt to make a rigorous connection between the distance in Hilbert space and the notion of fidelity. At least, for pure states.

Let's take to pure states $$(\psi) and (\phi)$$ The fidelity is given by the absolute value of the inner product: $$ F(\psi, \phi) = |\langle \psi | \phi \rangle| $$

Now, the distance between the states is the norm of their difference: $$ d(\psi, \phi) = \|\psi - \phi\| = \sqrt{\langle \psi - \phi | \psi - \phi \rangle} $$

The, using the properties of inner products, we can expand the norm: $$ \|\psi - \phi\|^2 = \langle \psi - \phi | \psi - \phi \rangle = \langle \psi | \psi \rangle + \langle \phi | \phi \rangle - \langle \psi | \phi \rangle - \langle \phi | \psi \rangle $$

• For normalized states $$(psi\| = \|\phi\| = 1)$$: $$ \|\psi - \phi\|^2 = 2 - 2 \text{Re}(\langle \psi | \phi \rangle) $$ And since $$(\psi | \phi)$$ is a complex number, $$(\text{Re}(\langle \psi | \phi \rangle) \leq |\langle \psi | \phi \rangle|)$$: $$ \|\psi - \phi\|^2 = 2 - 2 |\langle \psi | \phi \rangle| = 2(1 - F(\psi, \phi)) $$ Therefore: $$ d(\psi, \phi) = \sqrt{2(1 - F(\psi, \phi))} $$

For pure states, the distance in Hilbert space is directly related to fidelity through the formula $$(d(\psi, \phi) = \sqrt{2(1 - F(\psi, \phi))})$$.

Again, what do you think of that ?

Answer 1

The connection between Fidelity and the Trace distance, defined as

$$ D(\rho,\sigma) = \frac 12 \operatorname{tr}(|\rho-\sigma|) $$

Is well known in literature. Check, for example, Nielsen, Michael A., and Isaac L. Chuang. Quantum computation and quantum information Chapter 9 section 2.3. pg 415-416. The bound you found is generalized to

$$ 1-F(\rho,\sigma)\leq D(\rho,\sigma)\leq \sqrt{1-F(\rho,\sigma)^2} $$

Which is proved first time in this article by Fuchs and van de Graaf.

There is a related question in other stack about to saturate this inequality here.