Suppose we are given two maps $\Phi$ and $\Psi$ such that $$\|\Phi-\Psi\|_{\diamond}\leqslant\varepsilon.$$ What can we say about $\left\|\Phi^{\otimes t}-\Psi^{\otimes t}\right\|_{\diamond}$? Is it upper-bounded by $t\varepsilon$? Is the result different if $\Phi$ and $\Psi$ are unitary?

The fact that the upper bound grows linearly with the number of copies seems a bit too good to be true to me. I tried to think about some maps that are indistinguishable using a single query but highly distinguishable when using two queries but haven't found one such pair yet.


1 Answer 1


TL;DR: If $\Phi$ and $\Psi$ are quantum channels (unitary or otherwise), then things are "too good to be true".

Proposition. If $\|\Phi-\Psi\|_\diamond\le\varepsilon$ and $m=\max(\|\Phi\|_\diamond, \|\Psi\|_\diamond)\leqslant 1$, then for all $t\in\mathbb{N}$

$$ \|\Phi^{\otimes t}-\Psi^{\otimes t}\|_\diamond\leqslant t\varepsilon.\tag1 $$

Proof. Assume that $\|\Phi^{\otimes t}-\Psi^{\otimes t}\|_\diamond\leqslant t\varepsilon$. Then $$ \begin{align} \|\Phi^{\otimes(t+1)}-\Psi^{\otimes(t+1)}\|_\diamond&=\|\Phi\otimes\Phi^{\otimes t}-\Phi\otimes\Psi^{\otimes t}+\Phi\otimes\Psi^{\otimes t}-\Psi\otimes\Psi^{\otimes t}\|_\diamond\tag2\\ &\leqslant\|\Phi\otimes\Phi^{\otimes t}-\Phi\otimes\Psi^{\otimes t}\|_\diamond+\|\Phi\otimes\Psi^{\otimes t}-\Psi\otimes\Psi^{\otimes t}\|_\diamond\tag3\\ &=\|\Phi\|_\diamond\|\Phi^{\otimes t}-\Psi^{\otimes t}\|_\diamond+\|\Phi-\Psi\|_\diamond\|\Psi^{\otimes t}\|_\diamond\tag4\\ &\leqslant mt\varepsilon+m^t\varepsilon\tag5\\ &\leqslant (t+1)\varepsilon\tag6 \end{align} $$ which completes proof by induction.$\square$

Now, if $\Omega$ is a quantum channel, i.e. a completely positive, trace-preserving, linear (CPTP) map then $\|\Omega\|_\diamond=1$, so we have

Corollary. If $\Phi$ and $\Psi$ are CPTP maps and $\|\Phi-\Psi\|_\diamond\leqslant\varepsilon$, then $\|\Phi^{\otimes t}-\Psi^{\otimes t}\|_\diamond\leqslant t\varepsilon$ for all $t\in\mathbb{N}$.

However, if we drop the assumption that $m\leqslant 1$ then there are counterexamples to $(1)$. Choose any quantum channel $\Psi$ and set $\Phi:=b\Psi$ for $b:=1+\varepsilon>1$. Then $\|\Phi-\Psi\|_\diamond=\varepsilon$, but $$ \|\Phi^{\otimes t}-\Psi^{\otimes t}\|_\diamond=\|b^t\Psi^{\otimes t}-\Psi^{\otimes t}\|_\diamond=b^t-1>t\epsilon.\tag7 $$

The corollary has important practical implications. Namely, it makes it easy and convenient to obtain fairly tight bounds on error rates of quantum circuits given the knowledge of error rates of component gates. This highly desirable property does not hold for many popular quantities used for describing gate error rates, such as average gate fidelity.

  • 1
    $\begingroup$ Wish I could put another +1 for the "too good to be true" and the simplicity of the proof. Thanks! $\endgroup$
    – Tristan Nemoz
    Commented Jan 18, 2023 at 19:16

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