Given A ∈ F 2 n × n {\displaystyle A\in \mathbb {F} _{2}^{n\times n}} (an upper- triangular binary matrix of size n × n {\displaystyle n\times n} ) and b ∈ F 2 n {\displaystyle b\in \mathbb {F} _{2}^{n}} (a binary vector of length n {\displaystyle n} ),
define a function q : F 2 n → Z 4 {\displaystyle q:\mathbb {F} _{2}^{n}\to \mathbb {Z} _{4}} :
and
There exists a z ∈ F 2 n {\displaystyle z\in \mathbb {F} _{2}^{n}} such that
Find z {\displaystyle z} .4
With 3 registers; the first holding A {\displaystyle A} , the second containing b {\displaystyle b} and the third carrying an n {\displaystyle n} -qubit state, the circuit has controlled gates which implement U q = ∏ 1 < i < j < n C Z i j A i j ⋅ ⨂ j = 1 n S j b j {\displaystyle U_{q}=\prod _{1<i<j<n}CZ_{ij}^{A_{ij}}\cdot \bigotimes _{j=1}^{n}S_{j}^{b_{j}}} from the first two registers to the third.
This problem can be solved by a quantum circuit, H ⊗ n U q H ⊗ n ∣ 0 n ⟩ {\displaystyle H^{\otimes n}U_{q}H^{\otimes n}\mid 0^{n}\rangle } , where H is the Hadamard gate, S is the S gate and CZ is CZ gate. It is solved by this circuit because with p ( z ) = | ⟨ z | H ⊗ n U q H ⊗ n | 0 n ⟩ | 2 {\displaystyle p(z)=\left|\langle z|H^{\otimes n}U_{q}H^{\otimes n}|0^{n}\rangle \right|^{2}} , p ( z ) > 0 {\displaystyle p(z)>0} iff z {\displaystyle z} is a solution.5
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Watts, Adam Bene; Kothari, Robin; Schaeffer, Luke; Tal, Avishay (June 2019). "Exponential separation between shallow quantum circuits and unbounded fan-in shallow classical circuits". Proceedings of the 51st Annual ACM SIGACT Symposium on Theory of Computing. Vol. 362. Association for Computing Machinery. pp. 515–526. arXiv:1906.08890. doi:10.1145/3313276.3316404. ISBN 9781450367059. S2CID 195259496. 9781450367059 ↩