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Let $U$ be an $n$-qubit unitary, and let $p_U(x) = |\langle x | U | 0\rangle |^2$ be the probability of obtaining $x \in \{0,1\}^n$ on the all zero input. Given two $n$-qubit unitaries $U$ and $V$, it is straightforward to show (see, e.g., Nielsen and Chuang, page 194) that for all $x \in \{0,1\}^n$, $|p_U(x) - p_V(x)| \leq 2 \|U - V\|$, where the righthand side is the operator norm between $U$ and $V$. Therefore, summing over all $x$ implies $\| p_U - p_V \|_{\text{TVD}} \leq 2^n \|U - V\|$, where the LHS is the total variation distance.

I am interested if this bound can ever be saturated. If not, are there $U$ and $V$ such that $\| p_U - p_V \|_{\text{TVD}}$ is exponential in $n$?

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    $\begingroup$ I guess not really your question, but $U=V$? $\endgroup$
    – DaftWullie
    Commented Jul 27, 2023 at 13:39
  • $\begingroup$ Fair enough, though let's say nontrivially saturate it $\endgroup$ Commented Jul 27, 2023 at 13:40
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    $\begingroup$ The total variational distance, as you defined it, is bounded above by 2. So no, you cannot make it exponential in $n$. To see it, $$ \|p_u - p_v\| = \sum_x |p_u(x)-p_v(x)| \leq \sum_x |p_u(x)| + |p_v(x)| = \sum_x p_u(x) + p_v(x) = 2 $$ $\endgroup$
    – Rammus
    Commented Jul 27, 2023 at 15:40
  • $\begingroup$ @Rammus That should be an answer I think $\endgroup$ Commented May 22 at 15:09

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