Fubini-Study Metric
The following are my notes from a small graduate talk I gave deriving the Fubini-Study Metric using quantum states.
Before we calculate the Fubini Study metric, we'll first examine the Fubini Study distance, defined as follows
\[ \begin{align*} \text{dist} _{FS} &= \arccos\sqrt{\frac{\braket{\psi}{\varphi}\braket{\varphi}{\psi}}{\braket{\psi}{\psi}\braket{\varphi}{\varphi}}}. \end{align*} \]Two quick test examples are the case where $\ket{\psi}$ and $\ket \varphi$ exist on the same ray, and the case where $\ket \psi$ and $\ket \varphi$ are orthogonal. First, if the two states exist on the same ray, then $\ket \varphi = re^{i\theta}\ket{\psi}$ and thus $\bra{\varphi} = re^{-i\theta}\bra{\psi}$. Therefore
\[ \begin{align*} \text{dist} _{FS} &= \arccos\sqrt{\frac{re^{i\theta}\braket{\psi}{\psi}re^{-i\theta}\braket{\psi}{\psi}}{\braket{\psi}{\psi}re^{i\theta}re^{-i\theta}\braket{\psi}{\psi}}} \\ &= \arccos \sqrt{\frac{r^2}{r^2}} \\ &= \arccos 1 \\ &= 0 \end{align*} \]Thus meaning that the distance (in the form of an angle) is 0 when measured between two states that differed by a scalar multiple. Which is exactly what we might expect.
If the two states are orthogonal, then $\braket{\psi}{\varphi} = 0$ and thus $\braket{\varphi}{\psi}=0$. Therefore,
\[ \begin{align*} \text{dist} _{FS} &= \arccos\sqrt{\frac{0}{\braket{\psi}{\psi}\braket{\varphi}{\varphi}}} \\ &= \arccos 0 \\ &= \frac{\pi}{2} \end{align*} \]Which is again, exactly what we might expect, since this says that the angle between orthogonal states is $\frac{\pi}{2} = 90^\circ$
To calculate the metric, we take the infinitesimal distance, squared. In this first calculation, we'll use a normalized state $\ket\psi$. That is, $\braket{\psi}=1$.
\[ \begin{align*} \dd s^2 &= \text{dist}_{FS}^2 (\psi,\psi + \dd \psi) \\ &=\arccos^2\sqrt{\frac{\braket{\psi}{\psi + \dd \psi}\braket{\psi + \dd \psi}{\psi}}{\braket{\psi}{\psi}\braket{\psi + \dd \psi}{\psi + \dd \psi}}} \\ &= \arccos^2{\sqrt{\frac{\big(\braket{\psi}{\psi} +\braket{\psi}{\dd \psi}\big)\big(\braket{\psi}{\psi} +\braket{\dd \psi}{\psi}\big)}{\braket{\psi}{\psi}\big(\braket{\psi}{\psi}+ \braket{\psi}{\dd \psi} + \braket{\dd \psi}{\psi} + \braket{\dd \psi}{\dd \psi}\big)}}} \\ &= \arccos^2{\sqrt{\frac{\big(1+ \braket{\psi}{\dd \psi}\big)\big(1+ \braket{\dd \psi}{\psi}\big)}{\big(1+ \braket{\psi}{\dd \psi} + \braket{\dd \psi}{\psi} + \braket{\dd \psi}{\dd \psi}\big)}}} \\ &= \arccos^2\sqrt{\frac{(1+x)(1+y)}{1+x+y+z}} \\ &= \arccos^2 \sqrt{\frac{1+x+y+xy}{1+x+y+z}} \\ &= \arccos^2 \sqrt{1 + \frac{xy-z}{1+x+y+z}} \\ &= \arccos^2 \sqrt{1+(xy-z)(1-(x+y+z))} \\ &= \arccos^2 \sqrt{1 + (xy-z)} \\ &= \arccos^2 \left(1 + \frac{1}{2}(xy-z)\right) \end{align*} \]Letting $-\ep = \frac{1}{2}(xy-z)$ we have:
\[ \begin{align*} g &= \arccos(1-\ep) \\ \implies \cos(g) &= 1-\ep \\ \implies 1 - \frac{g^2}{2} &= 1 - \ep \\ \implies\ep &= \frac{g^2}{2} \\ \implies g&= \sqrt{2\ep} \\ \implies \arccos^2(1-\ep) &= 2\ep \\ \iff \arccos^2(1+\frac{1}{2}(xy-z) &= -(xy-z) = z-xy \end{align*} \]Therefore, re-exchanging our variables, we have that the Fubini Study metric is
\[ \begin{align*} \dd s^2 &= \braket{\dd \psi}{\dd \psi} - \braket{\dd \psi}{\psi}\braket{\psi}{\dd \psi} \end{align*} \]If we repeat the calculation for a non-normalized state, then we instead divide the numerator and denominator inside the square root by $\braket{\psi}{\psi}\braket{\psi}{\psi}$ and so
\[ \begin{align*} \dd s^2 &= \text{dist}_{FS}^2 (\psi,\psi + \dd \psi) \\ &=\arccos^2\sqrt{\frac{\braket{\psi}{\psi + \dd \psi}\braket{\psi + \dd \psi}{\psi}}{\braket{\psi}{\psi}\braket{\psi + \dd \psi}{\psi + \dd \psi}}} \\ &= \arccos^2{\sqrt{\frac{\big(\braket{\psi}{\psi} +\braket{\psi}{\dd \psi}\big)\big(\braket{\psi}{\psi} +\braket{\dd \psi}{\psi}\big)}{\braket{\psi}{\psi}\big(\braket{\psi}{\psi}+ \braket{\psi}{\dd \psi} + \braket{\dd \psi}{\psi} + \braket{\dd \psi}{\dd \psi}\big)}}} \\ &= \arccos^2{\sqrt{\frac{\big(1+ \frac{\braket{\psi}{\dd \psi}}{\braket{\psi}{\psi}}\big)\big(1+ \frac{\braket{\dd \psi}{\psi}}{\braket{\psi}{\psi}}\big)}{\big(1+ \frac{\braket{\psi}{\dd \psi}}{\braket{\psi}{\psi}} + \frac{\braket{\dd \psi}{\psi}}{\braket{\psi}{\psi}} + \frac{\braket{\dd \psi}{\dd \psi}}{\braket{\psi}{\psi}}\big)}}} \\ &= \arccos^2\sqrt{\frac{(1+x)(1+y)}{1+x+y+z}} \\ &= \arccos^2 \sqrt{\frac{1+x+y+xy}{1+x+y+z}} \\ &= \arccos^2 \sqrt{1 + \frac{xy-z}{1+x+y+z}} \\ &= \arccos^2 \sqrt{1+(xy-z)(1-(x+y+z))} \\ &= \arccos^2 \sqrt{1 + (xy-z)} \\ &= \arccos^2 \left(1 + \frac{1}{2}(xy-z)\right) \\ &\dots\\ &= z-xy \\ &= \frac{\braket{\dd \psi}{\dd \psi}}{\braket{\psi}{\psi}} - \frac{\braket{\psi}{\dd \psi}\braket{\psi}{\dd \psi}}{\braket{\psi}{\psi}\braket{\psi}{\psi}} \\ &= \frac{\braket{\dd \psi}{\dd \psi}}{\braket{\psi}{\psi}} - \frac{\braket{\psi}{\dd \psi}\braket{\psi}{\dd \psi}}{\big| \braket{\psi}{\psi}\big|^2} \end{align*} \]Alternatively, let $\ket \psi$ be some normalized state, the we can decompose $\ket{\dd \psi}$ into parallel and perpendicular components as follows
\[ \begin{align*} \ket{\dd \psi_\perp} = \ket{\dd \psi} - \braket{\psi}{\dd \psi}\ket{\psi} \end{align*} \]As a check, $\braket{\psi}{\dd \psi_\perp}$ should equal zero. So
\[ \begin{align*} \braket{\psi}{\dd \psi_\perp} &= \bra{\psi}\big(\ket{\dd \psi}-\braket{\psi}{\dd \psi}\ket{\psi}\big) \\ &= \braket{\psi}{\dd \psi}- \braket{\psi}{\dd \psi} \braket{\psi}{\psi} \\ &= \braket{\psi}{\dd \psi}- \braket{\psi}{\dd \psi} \\ &= 0 \end{align*} \]Continuing our calculation,
\[ \begin{align*} \dd s^2 &= \braket{\dd \psi _\perp}{\dd \psi_\perp} \\ &= \big(\bra{\dd \psi} - \braket{\dd \psi}{\psi}\bra{\psi}\big) \big(\ket{\dd \psi} - \braket{\psi}{\dd \psi}\ket{\psi}\big) \\ &= \braket{\dd \psi}{\dd \psi} - \braket{\psi}{\dd \psi} \braket{\dd \psi}{\psi} - \braket{\dd \psi}{\psi} \braket{\psi}{\dd \psi} + \braket{\dd \psi}{\psi} \braket{\psi}{\dd \psi}\braket{\psi}{\psi} \\ &= \braket{\dd \psi}{\dd \psi} - \braket{\psi}{\dd \psi} \braket{\dd \psi}{\psi} - \braket{\dd \psi}{\psi} \braket{\psi}{\dd \psi} + \braket{\dd \psi}{\psi} \braket{\psi}{\dd \psi} \\ &= \braket{\dd \psi}{\dd \psi} - \braket{\psi}{\dd \psi}\braket{\dd \psi}{\psi} \end{align*} \]Which is the same metric that we got from examining the infinitesimal Fubini Study distance.