# How to go from finite to infinitesimal form of the Fubini-Study metric?

As mentioned e.g. in the Wikipedia page, given a pair of pure states $$\psi,\phi\in\mathbb{CP}^{N-1}$$, the geodesic distance between them is $$\gamma(\psi,\phi) = \arccos\sqrt{\frac{\langle\psi|\phi\rangle\langle\phi|\psi\rangle}{\langle\psi|\psi\rangle\langle\phi|\phi\rangle}},$$ while the corresponding infinitesimal form, obtained taking $$\phi=\psi+\delta\psi$$, reads $$ds^2 = \frac{\langle \delta\psi|\delta\psi\rangle}{\langle\psi|\psi\rangle} - \frac{\langle\delta\psi|\psi\rangle \langle\psi|\delta\psi\rangle}{\langle\psi|\psi\rangle^2}.$$

This should be a simpler case of the more general calculation for the Bures metric discussed in this answer, while a related discussion is also found here. What's a direct way to perform this calculation?

Quite amusingly, I just realised pretty much this exact question was previously asked on math.SE, and an answer was given there a few years ago... by myself. I completely forgot. I'll copy-paste the answer I gave there and keep the post up for future reference.

Start by considering the following identities, obtained via basic complex algebra and Taylor expansions: $$|\langle \psi,\psi+\delta\psi\rangle|^2 = \|\psi\|^4\left\lvert 1+ \frac{\langle\psi,\delta\psi\rangle}{\|\psi\|^2}\right\rvert^2 = \|\psi\|^4 \left(1 +\frac{2\operatorname{Re}\langle\psi,\delta\psi\rangle}{\|\psi\|^2} + \frac{|\langle\psi,\delta\psi\rangle|^2}{\|\psi\|^4}\right)$$

$$\|\psi+\delta\psi\|^2=\|\psi\|^2 + \|\delta\psi\|^2 + 2\operatorname{Re}\langle\psi,\delta\psi\rangle$$

$$\frac{1}{\|\psi+\delta\psi\|^2}\simeq\frac{1}{\|\psi\|^2} \left( 1 - \frac{2\operatorname{Re}\langle\psi,\delta\psi\rangle}{\|\psi\|^2} - \frac{\|\delta\psi\|^2}{\|\psi\|^2} + \frac{4[\operatorname{Re}\langle\psi,\delta\psi\rangle]^2}{\|\psi\|^4} \right)$$

$$\arccos\sqrt{1-x} \simeq \sqrt x, \quad\text{for } x\to 0^+.$$

Our goal is to compute the quantity $$\gamma(\psi,\psi+\delta\psi) = \arccos\sqrt{\frac{|\langle \psi,\psi+\delta\psi\rangle|^2}{\|\psi\|^2 \|\psi+\delta\psi\|^2} }.$$

Assuming $$\|\psi\|^2=1$$ for simplicity (we can keep the $$\|\psi\|$$ terms around but nothing substantial changes in the expressions) we have $$\frac{\lvert\langle\psi,\psi+\delta\psi\rangle\rvert^2}{\|\psi+\delta\psi\|^2} \simeq 1 + \lvert\langle\psi,\delta\psi\rangle\rvert^2 - \|\delta\psi\|^2,$$ therefore $$\gamma(\psi,\psi+\delta\psi)\simeq\arccos\sqrt{1+\lvert\langle\psi,\delta\psi\rangle\rvert^2-\|\delta\psi\|^2} \simeq \sqrt{\|\delta\psi\|^2-\lvert\langle\psi,\delta\psi\rangle\rvert^2},$$ and finally $$ds^2 \equiv \gamma(\psi,\psi+\delta\psi)^2 \simeq \|\delta\psi\|^2-\lvert\langle\psi,\delta\psi\rangle\rvert^2,$$ which is the expression we were looking for.