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On pg. 31 in Nielsen and Chuang, it's said that:

This procedure can easily be generalized to functions on an arbitrary number of bits, by using a general operation known as the Hadamard transform, or sometimes the Walsh– Hadamard transform

Aren't we trying to generalize to an arbitrary number of qubits? If it's bits, why isn't it qubits?

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  • $\begingroup$ can you also add (in the post, not in the comments) what "procedure" is the text talking about? $\endgroup$
    – glS ♦
    Sep 15, 2021 at 11:14
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    $\begingroup$ I would guess that this means simply the following. To parallelize a boolean function of one bit you need one qubit+one ancilla (to store the result). To parallelize a function of n bits you would need $n$ qubits + one ancilla. So to me bits and qubits could be used interchangeably here. $\endgroup$
    – Nikita Nemkov
    Sep 15, 2021 at 13:18

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