Period of phase leads the advantage of Heisenberg's Limit disappear?

Question

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(HL), normally referred to as the form $var(\hat{\theta})=1/n^2$, a square enhancement in the precision. Here is a frequently used state example, stating 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?

Answer 1

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.