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  1. We investigate the problem of predicting the output behavior of unknown quantum channels. Given query access to an -qubit channel and an observable , we aim to learn the mapping to within a small error for most sampled from a distribution . Previously, Huang, Chen, and Preskill proved a surprising result that even if is arbitrary, this task can be solved in time roughly , where is the target prediction error. However, their guarantee applied only to input distributions invariant under all single-qubit Clifford gates, and their algorithm fails for important cases such as general product distributions over product states . In this work, we propose a new approach that achieves accurate prediction over essentially any product distribution , provided it is not "classical" in which case there is a trivial exponential lower bound. Our method employs a "biased Pauli analysis," analogous to classical biased Fourier analysis. Implementing this approach requires overcoming several challenges unique to the quantum setting, including the lack of a basis with appropriate orthogonality properties. The techniques we develop to address these issues may have broader applications in quantum information. 
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  2. Gowers, Tim (Ed.)
    We prove that for $$d\geq 0$$ and $$k\geq 2$$, for any subset $$A$$ of a discrete cube $$\{0,1\}^d$$, the $k-$higher energy of $$A$$ (i.e., the number of $2k-$tuples $$(a_1,a_2,\dots,a_{2k})$$ in $$A^{2k}$$ with $$a_1-a_2=a_3-a_4=\dots=a_{2k-1}-a_{2k}$$) is at most $$|A|^{\log_{2}(2^k+2)}$$, and $$\log_{2}(2^k+2)$$ is the best possible exponent. We also show that if $$d\geq 0$$ and $$2\leq k\leq 10$$, for any subset $$A$$ of a discrete cube $$\{0,1\}^d$$, the $k-$additive energy of $$A$$ (i.e., the number of $2k-$tuples $$(a_1,a_2,\dots,a_{2k})$$ in $$A^{2k}$$ with $$a_1+a_2+\dots+a_k=a_{k+1}+a_{k+2}+\dots+a_{2k}$$) is at most $$|A|^{\log_2{ \binom{2k}{k}}}$$, and $$\log_2{ \binom{2k}{k}}$$ is the best possible exponent. We discuss the analogous problems for the sets $$\{0,1,\dots,n\}^d$$ for $$n\geq2$$. 
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