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I've seen a lot of analyses on quantum circuit error bound based on the norm difference $\Vert U - V \Vert$.

On the other hand, I've also seen a lot of papers that use the gate fidelity $\frac{1}{2^n}\lvert \text{tr}(U^\dagger V) \rvert$.

Is there a precise relationship between the two, e.g. does one upper bound the other?

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  • $\begingroup$ Do you have a reference for p=1? $\endgroup$ Mar 17 at 5:09
  • $\begingroup$ @NorbertSchuch reference for exactly what are you looking for? $\endgroup$
    – FDGod
    Mar 17 at 7:36
  • $\begingroup$ @FDGod Where someone would use the $\|\cdot\|_1$ norm to measure the difference between unitaries. This seems neither correct nor smart: Operationally, it is the wrong norm (you want the $\|\cdot\|_\infty$ norm), it is no easier to compute (as e.g. the fidelity would be), and it is larger than the $\|\cdot\|_\infty$-norm, and thus harder to bound (or you will get worse bounds). $\endgroup$ Mar 17 at 14:50
  • $\begingroup$ One comment on the relationship of the measures: The fidelity (as you define it) is independent of a phase choice of $U$, while the norm distance isn't -- to relate them, you want to at least include a minimization over the relative phase of U and V. $\endgroup$ Mar 17 at 14:51
  • $\begingroup$ @NorbertSchuch got it. And yeah, my bad, I didn’t realise the mistake while writing my answer. $\endgroup$
    – FDGod
    Mar 17 at 16:47

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The gate fidelity can be expressed with the Frobenius norm as follows: \begin{align*} \min_{\varphi\in\mathbb R}\frac1{2^n}\|U-e^{i\varphi}V\|_2^2&=\min_{\varphi\in\mathbb R} \frac1{2^n}{\rm tr}((U-e^{i\varphi}V)^\dagger (U-e^{i\varphi}V)) \\ &=\min_{\varphi\in\mathbb R}\frac1{2^n}{\rm tr}(U^\dagger U-e^{i\varphi}U^\dagger V-e^{-i\varphi}V^\dagger U+V^\dagger V)\\ &=\min_{\varphi\in\mathbb R}\frac2{2^n}{\rm tr}({\bf1}_{2^n})-\frac1{2^n}{\rm tr}(e^{i\varphi}U^\dagger V+e^{-i\varphi}V^\dagger U)\\ &=2\Big(1-\frac1{2^n}\max_{\varphi\in\mathbb R}{\rm Re}(e^{i\varphi}{\rm tr}(U^\dagger V))\Big)=2\Big(1-\frac1{2^n}|{\rm tr}(U^\dagger V)|\Big)\,. \end{align*} Using the upper bound $\|X\|_2\leq\sqrt m\|X\|_\infty$ which is valid for all $X\in\mathbb C^{m\times m}$ the above identity can be related to the operator norm distance via $$ \frac1{2^n}|{\rm tr}(U^\dagger V)|=1-\frac1{2^{n+1}}\min_{\varphi\in\mathbb R}\|U-e^{i\varphi}V\|_2^2\geq 1-\frac{(\sqrt{2^n})^2}{2^{n+1}}\min_{\varphi\in\mathbb R}\|U-e^{i\varphi}V\|_\infty^2 $$ which altogether yields the following relation: $$ \boxed{\frac{|{\rm tr}(U^\dagger V)|}{2^n}\geq 1-\frac12\min_{\varphi\in\mathbb R}\|U-e^{i\varphi}V\|_\infty^2} $$

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  • $\begingroup$ Same comment as above at the question: What you want to do (or should do) is to include a minimization over the relative phase of U and V in the 2-norm. Then, the 2-norm and the fidelty should become equal. (And this optimization is, generally, what you want -- unless these are in fact controlled operations. (But even in that case you can easily fix the phase with a rotation on the control, so you still care about the optimal choice.)) $\endgroup$ Mar 17 at 14:53
  • $\begingroup$ You're of course right, I adjusted my answer accordingly. $\endgroup$ Mar 17 at 17:51

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