# How asymptotically efficient is quantum state tomography of a flat qubit?

Suppose you receive $n$ copies of a qubit rotated by an unknown angle. That is to say, you're given the state:

$$T(\theta) = \left(\sin(\theta) \left|0\right\rangle + \cos(\theta) \left|1\right\rangle\right)^{\otimes n}$$

and you want to estimate $\theta$ by measuring the state in clever ways. How accurately can that be done?

I think the answer is somewhere around $O(\frac{1}{\sqrt{n}})$, but I'm not sure. The reasoning I used to get that guess is based on the amount of orthogonality between $T(\theta_1)$ and $T(\theta_2)$. It needs to be bounded below some constant for $\theta_1$ to be usually-distinguishable from $\theta_2$, so:

$T(\theta_1) \cdot T(\theta_2)^\dagger$

$= \left(\sin(\theta_1) \left|0\right\rangle + \cos(\theta_1) \left|1\right\rangle\right)^{\otimes n} \cdot \left(\sin(\theta_2) \left\langle0\right| + \cos(\theta_2) \left\langle1\right|\right)^{\otimes n}$

$= \left( \sum_{i=0}^n \left| {}^n_i\right\rangle \sin(\theta_1)^{n-i} \cos(\theta_1)^i \right) \cdot \left( \sum_{i=0}^n \left\langle {}^n_i\right| \sin(\theta_2)^{n-i} \cos(\theta_2)^i \right)$

$= \sum_{i=0}^n {n \choose i} (\sin \theta_1 \sin \theta_2)^{n-i} (\cos \theta_1 \cos \theta_2)^i$

$= \left(\sin \theta_1 \sin \theta_2 + \cos \theta_1 \cos \theta_2 \right)^n$

$= cos^n(\theta_1 - \theta_2)$

$\approx (1 - \frac{\theta_d^2}{2})^n$

and in order for that to converge to a constant overlap, you need $n \propto \theta_d^{-2}$ so that you get $(1 - \frac{c}{n})^n \approx e^{-c} \in O(1)$. So it seems like we can't do better than $\theta_d \in O(\frac{1}{\sqrt{n}})$ because of the large amount of overlap between the possible states as $\theta$ varies.

This bound is surprising to me because just directly measuring the bits and taking the average already gives an $O(\frac{1}{\sqrt{n}})$ estimate. So I suspect that I made an error in reasoning or approximation when computing the overlap.