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The CZ gate is known to be symmetric : $CZ(a;b)=CZ(b;a)$;

what about $CCZ(a,b;t) \stackrel{?}{=} CCZ(a,t;b) \stackrel{?}{=} CCZ(t,a;b) \cdots $

same question for $CCCZ$ gates..

I think the answer is yes, but I'd like to see a reference or better yet a nice way to prove it.

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A $\mathsf C Z$ gate rotates the phase of $|11\rangle$, and does nothing to the three other basis states.

A $\mathsf C^{n-1}Z$ gate will rotate the phase of the $|11\ldots 1\rangle$ state, and will do nothing to the $2^n-1$ other states.

If any of the $n$ qubits are $|0\rangle$, no phasing occurs.

Therefore such gates are symmetric and invariant under permutation of the qubit indices. For $n=2$, we often emphasize this by circuit diagrams having the $\mathsf C Z$ gate be symmetric.

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  • $\begingroup$ seems so obvious now...nice argument $\endgroup$
    – unknown
    Aug 1 at 16:27

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