 Solution

Apply one Hadamard gate independently to each qubit. These gates act in parallel, so the quantum circuit implements $H^{\otimes n}$ with $n$ gates and depth one.

For a computational basis state, factorization givesThe exponent is the dot product over the finite field ${\mathbb{F}}_2$, because multiplying the separate signs adds their exponents modulo two. This is the Walsh-Hadamard transform, equivalently the quantum Fourier transform on the finite Abelian group $({\mathbb{Z}}_2{)}^n$.

In particular, the all-zero input becomes the uniform coherent superposition $2^{-n/2}{\sum }_y|y⟩$. It is a useful initial operation because an oracle can then act on all amplitudes in one call, and subsequent interference can reveal a global property. Measurement alone does not read all function values: the later interference step is what turns the preparation into an algorithmic advantage.