# What is the difference between the “Fubini-Study distances” $\arccos|\langle\psi|\phi\rangle|$ and $\sqrt{1-|\langle\psi|\phi\rangle|}$?

I sometimes see the "Fubini-Study distance" between two (pure) states $$|\psi\rangle,|\phi\rangle$$ written as $$d(\psi,\phi)_1=\arccos(|\langle\psi|\phi\rangle|),$$ for example in the Wikipedia page. Other sources (e.g. this paper in pag. 16), use the definition $$d(\psi,\phi)_2=\sqrt{1-|\langle\psi|\phi\rangle|}.$$

What is the difference between these two definitions? Is one preferred over the other?

• The first one is giving you a path that connects two points so distance in the sense of geometry. The second is more of an analysis flavor where you talk about various norms on vector spaces. The difference of two vectors gives the intuition of the path being a straight line in the above vector space. It does not talk directly about giving any geometric path in the manifold. – AHusain Jan 24 at 17:05
• @AHusain but what confuses me is that they are both called "Fubini-Study distance". Then when someone refers to the "Fubini-Study metric/distance" without explicitly giving the expression, which of the two expression should we think they are referring to? Also, if they are both distances, they both should come with a geometric interpretation, no? – glS Jan 24 at 17:11
• I'd only call the first one a distance. The difficulty comes from the coincidences of R^n. So many possible notions coincide leading to different definitions with the same name. – AHusain Jan 24 at 17:55
• The first distance is the length of the geodesic (corresponding to the Fubiny-Study metric) on the complex projective space $CP^{n-1}$ connecting the two rays (defined by the two vectors). It can be computed, for instance, by solving the Hamilton-Jacobi equation. – David Bar Moshe Jan 27 at 13:33
• @DavidBarMoshe if you could expand on that, it would make for a great answer I think – glS Jan 28 at 15:01

Recall the law of cosines for two unit vectors $$\mathbf{u}$$ and $$\mathbf{v}$$ in $$\mathbb R^2$$: $$\|\mathbf{u}-\mathbf{v}\|^2 = 2-2\cos\theta,$$ where $$\theta$$ is the angle between the vectors. Similarly, you'll recall the definition of the inner product, $$\langle \mathbf u|\mathbf {v}\rangle = \cos\theta.$$ So, $$\|\mathbf{u}-\mathbf{v}\| = \sqrt{2}\sqrt{1- \langle\mathbf{u}|\mathbf{v}\rangle}.$$ For these unit vectors in $$\mathbb R^2$$, $$\theta$$ is the arc length on the unit circle intersecting these vectors. It is a distance. In your notation, $$d(\mathbf{u},\mathbf{v})_1$$ is the arc length and $$d(\mathbf{u},\mathbf{v})_2$$ is the Euclidean distance. Before we generalize to quantum states, you should note that nothing said so far would suggest one is preferred over the other.
For complex vectors we can do something analogous. Define the "Hermitian angle" of two complex vectors $$\psi$$ and $$\phi$$ in $$\mathbb C^n$$ to be $$\cos\theta = |\langle\psi|\phi\rangle|.$$ Now consider $$\|\psi -\phi\|^2 = 2-2\Re[\langle\psi|\phi\rangle] \geq 2-2|\langle\psi|\phi\rangle|.$$ So you can see that $$d(\psi,\phi)_1$$ and $$d(\psi,\phi)_2$$ are analogous, but not quite, the arc length and distance between the quantum states.