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If $N\geq 2$, $a\in \mathbb{Z}_N$, and $a^r= 1$ for some $r$. Consider the operator $M_a$, which is related to order finding :

$M_a |x\rangle= |a \cdot x \pmod{N} \rangle $ if $x\in \mathbb{Z}_N$

What is an easy way to show that it's a unitary?

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    $\begingroup$ What are your thoughts on this? have you attempted anything to see why it's a unitary operation? $\endgroup$
    – FDGod
    Nov 19, 2023 at 20:14
  • $\begingroup$ I have no idea on what ket 3, for example, can possibly be. Or ket of the equiv. class 3 of $\mathbb{Z}_N$ $\endgroup$
    – metaUser
    Nov 19, 2023 at 20:34

1 Answer 1

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Assuming $\gcd(N,a)=1$, then the operator $M_a$ mapping $x\in\mathbb Z_N$ to $a\cdot x \pmod N$ is a permutation, which means that it’s reversible, ergo unitary.

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  • $\begingroup$ +1 The assumption is in fact not only sufficient, but also a necessary condition for unitarity of $M_a$. If $\gcd(N,a)=b>1$, then $N=ub$ and $a=vb$ for some non-zero $u,v\in\mathbb{Z}_N$ and $M_a|u\rangle=|0\rangle=M_a|0\rangle$, so $M_a$ fails to be injective. $\endgroup$ Nov 20, 2023 at 3:41
  • $\begingroup$ @AdamZalcman indeed, as another way to state the same, in the OP's framing $a$ must generate all of $\mathbb Z_N$, and hence $r$ must equal $N$. $\endgroup$ Nov 20, 2023 at 14:00

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