When Does Adaptivity Help for Quantum State Learning?

We consider the classic question of state tomography: given copies of an unknown quantum state $\rho \in \mathbb{C}^{d \times d}$, output $\widehat{\rho}$ which is close to $\rho$ in some sense, e.g. trace distance or fidelity. When one is allowed to make coherent measurements entangled across all copies, $\Theta\left(d^{2} / \varepsilon^{2}\right)$ copies are necessary and sufficient to get trace distance $\varepsilon$ [18], [29]. Unfortunately, the protocols achieving this rate incur large quantum memory overheads that preclude implementation on near-term devices. On the other hand, the best known protocol using incoherent (single-copy) measurements uses $O\left(d^{3} / \varepsilon^{2}\right)$ copies [24], and multiple papers have posed it as an open question to understand whether or not this rate is tight [6], [18]. In this work, we fully resolve this question, by showing that any protocol using incoherent measurements, even if they are chosen adaptively, requires $\Omega\left(d^{3} / \varepsilon^{2}\right)$ copies, matching the upper bound of [24]. We do so by a new proof technique which directly bounds the “tilt” of the posterior distribution after measurements, which yields a surprisingly short proof of our lower bound, and which we believe may be of independent interest. While this implies that adaptivity does not help for tomography with respect to trace distance, we show that it actually does help for tomography with respect to infidelity. We give an adaptive algorithm that outputs a state which is $\gamma$-close in infidelity to $\rho$ using only $\widetilde{O}\left(d^{3} / \gamma\right)$ copies, which is optimal for incoherent measurements. In contrast, it is known [18] that any nonadaptive algorithm requires $\Omega\left(d^{3} / \gamma^{2}\right)$ copies. While it is folklore that in 2 dimensions, one can achieve a scaling of $O(1 / \gamma)$, to the best of our knowledge, our algorithm is the first to achieve the optimal rate in all dimensions.