In a standard Quantum Sensing (QS) task one aims at estimating an unknown parameter $\theta$, encoded into an $n$-qubit probe state, via measurements of the system. The success of this task hinges on the ability to correlate changes in the parameter to changes in the system response $\mathcal{R}(\theta)$ (i.e., changes in the measurement outcomes). For simple cases the form of $\mathcal{R}(\theta)$ is known, but the same cannot be said for realistic scenarios, as no general closed-form expression exists. In this work we present an inference-based scheme for QS. We show that, for a general class of unitary families of encoding, $\mathcal{R}(\theta)$ can be fully characterized by only measuring the system response at $2n+1$ parameters. In turn, this allows us to infer the value of an unknown parameter given the measured response, as well as to determine the sensitivity of the sensing scheme, which characterizes its overall performance. We show that inference error is, with high probability, smaller than $\delta$, if one measures the system response with a number of shots that scales only as $\Omega(\log^3(n)/\delta^2)$. Furthermore, the framework presented can be broadly applied as it remains valid for arbitrary probe states and measurement schemes, and, even holds in the presence of quantum noise. We also discuss how to extend our results beyond unitary families. Finally, to showcase our method we implement it for a QS task on real quantum hardware, and in numerical simulations.
翻译:在标准的量子感测(QS)任务中,任务一旨在估算一个未知的参数$\theta$,通过系统测量将它编码成一个以美元为单位的量位探测器状态。任务的成功取决于将参数的变化与系统响应$\mathcal{R}(\theta)$(即测量结果的变化)的变化联系起来的能力。简单的例子中,美元(mathcal{R}(\theta)$)的形式是已知的,但对于现实的假设情况,无法这样说,因为不存在一般的封闭式表达。在这项工作中,我们为QS提出了一个基于推论的系统状态。我们显示,对于一个总编码的统一组合类别,$\mathcal{R}(\\\theta)$($)$的值变化,只能用2n+1美元的参数来测量系统的反应。反过来,让我们推算一个未知的参数的价值,根据测量的响应量值,以及确定测得的数值,甚至可以用来判断测测测算其总体性能说明其表现的精确度。我们整体性 QQQ3。我们从一个数值算出一个数值的数值的数值的数值值值值值值值值值值值值值值值值值值值值值值值值值值值值值值值值值值值值值值值值值值值值值值值值值值值值值值值值值值值值值值,如果在一个比值里,如果在一比值里算算算算算算出一个比值里算出一个比值里算出一个比值的值的值里值的值的值的值的值,如果在一个比值的值的值的值值值的值的值的值的值,在一次的值里值里值里值里值里值里值里值里值里值里值里值里, 。