# Saturating an inequality relating the operator norm and the total variation distance

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 $$n$$?

• I guess not really your question, but $U=V$? Commented Jul 27, 2023 at 13:39
• Fair enough, though let's say nontrivially saturate it Commented Jul 27, 2023 at 13:40
• 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, $$\|p_u - p_v\| = \sum_x |p_u(x)-p_v(x)| \leq \sum_x |p_u(x)| + |p_v(x)| = \sum_x p_u(x) + p_v(x) = 2$$ Commented Jul 27, 2023 at 15:40
• @Rammus That should be an answer I think Commented May 22 at 15:09