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I'm working through the Nielsen and Chuang book and am trying to understand the Deutsch-Jozsa algorithm. I think I understand it as it's presented, however I'm curious what would happen if we used $|1^n\rangle$ instead of $|0^n\rangle$. At the summation steps I've been told the innermost sum will go to $0$, but I'm having trouble figuring out why this would be the case.

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    $\begingroup$ Which step are you asking about? Are you asking to (1) perform the first $n$ Hadamard transforms on the first register of $n$ qubits initially prepared in $\vert 1\rangle$, then (2) evaluate the oracle onto the second register, then (3) perform the other Hadamard transform on the first register and measure? Traditionally step (1) is done with qubits initially prepared in $\vert 0\rangle$. $\endgroup$
    – Mark S
    Mar 28 at 0:14

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