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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2020

Paper 3

Starting from $|0^n\rangle|0^n\rangle$, the first layer of <Hadamard gate>[Hadamard gates] gives
$$
2^{-n/2}\sum_{x\in B_n}|x\rangle|0^n\rangle.
$$
The <Boolean quantum oracle> then gives
$$
2^{-n/2}\sum_{x\in B_n}|x\rangle|f(x)\rangle.
$$
Using the <Walsh-Hadamard transform> identity
$$
H^{\otimes n}|x\rangle
=2^{-n/2}\sum_{y\in B_n}(-1)^{x\cdot y}|y\rangle,
$$
the final state is
$$
\boxed{2^{-n}\sum_{x,y\in B_n}(-1)^{x\cdot y}|y\rangle|f(x)\rangle.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 15C / b / i / Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 15C / b / i / Solution

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Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 15C / b / i / Solution

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