In Quantum Metrology, the aim is to estimate some unknown parameters(I will talk about one parameter estimation in this post, while multiparameter is also available) as precise as possible. Without quantum resources, we can only realize the Standard Quantum Limit(SQL), normally referred to as the form $$var(\hat{\theta})=1/n$$, where $$\hat{\theta}$$ is the estimator(function of random variables to estimate the unknown parameter $$\theta$$) and $$n$$ is the number of experiments. While with quantum resources, we may reach the so-called Heisenberg's Limit, normally referred to as the form $$var(\hat{\theta})=1/n^2$$, a square enhancement in the precision. Here is an frequently used state example, stated that the GHZ state $$1/\sqrt{2}(|0\rangle^{\otimes n}+|1\rangle^{\otimes n})$$ after the evolution described by the unitary operator $$U = e^{-i\theta\sigma_z/2}\otimes e^{-i\theta\sigma_z/2}\otimes...$$ will become $$1/\sqrt{2}(|0\rangle^{\otimes n}+e^{in\theta}|1\rangle^{\otimes n})\tag{1}$$ ignoring the global phase. And we can estimate the value of the parameter by measuring this parameterized quantum state.

My question is, the HL will show its advantage only when the scale is $$1/n^2$$, while from eq.(1), we can easily see that there's a $$2\pi$$ period in an exponential function, no matter how small $$\theta$$ is, we cannot always enhance our precision by HL scale when $$n$$ pass some specific value. So, how to understand this paradox, did I miss something?

You are correct, but the SQL is a local limit, when you already have a very good idea what the value of $$\theta$$ is, so there is no contradiction.
Let's work through it. You measure some relative phase $$\Theta$$, and you infer that $$\Theta=n\theta+2\pi k,\quad k\in \mathbb{N}.$$ You work out that $$\theta=\frac{\Theta +2\pi k}{n},$$ where $$\Theta$$ and $$n$$ are known ($$n$$ is known because you set up your experiment) and $$k$$ is unknown. You now must determine what values of $$k$$ are possible, then you are finished.
Because the SQL is local, we know a priori that $$\theta_{\text{low}}\leq \theta\leq \theta_{\text{high}}$$ for some values of $$\theta_{\text{low}}$$ and $$\theta_{\text{high}}$$, so we deduce that $$\frac{n\theta_{\text{low}}-\Theta}{2\pi}\leq k\leq \frac{n\theta_{\text{high}}-\Theta}{2\pi}.$$ So as long as our initial knowledge has sufficiently small $$\theta_{\text{high}}-\theta_{\text{low}}$$, there will only be one possible value of $$k$$, and we will fully determine the value of $$\theta$$. If our initial knowledge is not precise enough, you are correct that using too large a value of $$n$$ will yield multiple possible results.
• @narip I mean exactly what I said, that the possible values of $\theta$ are restricted to a small region around the "true" value of $\theta$. See e.g. doi.org/10.1007/s00220-019-03433-4. This is why the SQL is an asymptotic limit, as you can only achieve that precision in the limit of doing lots of measurements. Nov 15 '21 at 15:05
• One more thing, why it's $2k\pi$ instead of $k\pi$? E.g. for states $\frac{1}{\sqrt{2}}\left( |0\rangle +e^{i\theta}|1\rangle \right)$, if we measure $\langle \sigma _x\rangle$ to get $\cos\theta$, and the value of $\theta\in[0,\pi]$ will be the same as $\theta\in[\pi,2\pi]$, which makes us unable to distinguish $\theta$ is in $[0,\pi]$ or in $[\pi,2\pi]$. Dec 8 '21 at 9:14
• Okay, $\langle \sigma_y \rangle= \sin\theta$, combine this will be enough to tell which $\theta$ is. Cheers. Dec 8 '21 at 9:23