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For a fixed quantum channel $N$ and a unitary channel $U$, we define $N$'s gate fidelity as

$$ F(N,U) = \int \langle \psi| U \, N(| \psi \rangle \langle \psi |) \, U^\dagger| \psi \rangle d\mu_H(\psi)$$

where $\mu_H$ is the Haar measure over $d$-dimensional states. Suppose that for some channel $C$ you know that

$$ \int F(C,U) d\mu_H(U) = X $$

where this time $\mu_H$ is the Haar measure over the $d$-dimensional unitary group. Can we say something about the concentration of $F(C,U)$ around the mean $X$? I tried using Levy's lemma to derive a concentration inequality for $f(U) = F(C,U)$, but I was not able to find a suitable upper bound to the Lipschitz constant of $f$. To avoid misunderstandings, by Lipschitz constant I mean the smallest $L$ such that

$$ |f(U) - f(V)| \leq L \|U-V\|_2 $$

Any ideas?

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  • $\begingroup$ Isn't the first integral independent of $U$? Writing $|\psi\rangle$ as $V|0\rangle$ with the integral being over the Haar-random $V$ and then using the invariance of the Haar measure by unitary transformations? That would give you that $F(C, U)$ is constant w.r.t. $U$, so the average is easily computed. Or did I miss something? $\endgroup$
    – Tristan Nemoz
    Commented May 10 at 12:02
  • $\begingroup$ I already know the average $X$ from previous calculations. My goal here is to upper bound the probability that $|F(C,U) - X| > \epsilon$ for a generic $\epsilon$ $\endgroup$ Commented May 10 at 12:33
  • $\begingroup$ My point is that we would then have $F(C,U)=X$ for a fixed $C$ and for all $U$. $\endgroup$
    – Tristan Nemoz
    Commented May 10 at 12:59
  • $\begingroup$ I don't think the first integral is independent of $U$. Invariance of the Haar measure allows us to absorb $U$ into $|\psi\rangle$ at the output of the channel, but it pops back up at the input. Also, $F(U, U)=1$ and $F(D, U)=1/d$ where $D$ is the completely depolarizing channel. $\endgroup$ Commented May 10 at 13:05
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    $\begingroup$ One approach would be to get rid of the first integral using Horodecki's formula \begin{align} F(N, U)=\frac{dF_e(N, U)+1}{d+1} \end{align} where entanglement fidelity is defined as \begin{align} F_e(N, U)=\langle\psi|(I\otimes U^\dagger N)(|\psi\rangle\langle\psi|)|\psi\rangle \end{align} with $|\psi\rangle=\frac{1}{\sqrt{d}}\sum_i|i\rangle|i\rangle$ before trying to find a concentration inequality. $\endgroup$ Commented May 10 at 13:08

1 Answer 1

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The function described in the question is 1-Lipschitz. To argue this, we'll get an inequality in place before we start writing integrals.

If $\vert \gamma\rangle$ and $\vert\delta\rangle$ are unit vectors, then the operator $\vert\gamma\rangle\langle\gamma\vert - \vert\delta\rangle\langle\delta\vert$ has at most two nonzero eigenvalues: $\pm\sqrt{1 - \vert\langle \gamma\vert \delta\rangle\vert^2}$. We can therefore bound the $\infty$-norm of this operator in terms of the Euclidean norm for vectors like so. $$ \bigl\| \vert\gamma\rangle\langle\gamma\vert - \vert\delta\rangle\langle\delta\vert\bigr\|_{\infty} = \sqrt{1 - \vert\langle \gamma\vert \delta\rangle\vert^2} = \sqrt{1 + \vert\langle \gamma\vert \delta\rangle\vert} \sqrt{1 - \vert\langle \gamma\vert \delta\rangle\vert}\\ \leq \sqrt{2} \sqrt{1 - \operatorname{Re}(\langle \gamma\vert \delta\rangle)} = \bigl\| \vert\gamma\rangle - \vert\delta\rangle\bigr\| $$ (You often see a related bound for the trace norm rather than the $\infty$-norm, for which we pick up an additional factor of 2, but we're going to be interested in the $\infty$-norm instead.)

In particular, taking $\vert\gamma\rangle = U \vert\phi\rangle$ and $\vert\delta\rangle = U\vert\phi\rangle$ for any unit vector $\vert\phi\rangle$ gives this:

$$ \bigl\| U \vert\phi\rangle\langle\phi\vert U^{\dagger} - V \vert\phi\rangle\langle\phi\vert V^{\dagger} \bigr\|_{\infty} \leq \;\bigl\| U \vert\phi\rangle - V\vert\phi\rangle\bigr\| \leq \| U - V\|_{\infty}. $$

The $\infty$-norm is (like all norms) convex, so by thinking about the spectral decomposition of a given density operator we see that the same bound works for density operators: $$ \bigl\| U \sigma U^{\dagger} - V \sigma V^{\dagger} \bigr\|_{\infty} \leq \| U - V\|_{\infty}. $$

That's the inequality we needed. Now, observing that $N(\vert\psi\rangle\langle\psi\vert)$ is a density operator for every unit vector $\vert\psi\rangle$, we get what we're after: $$ \begin{aligned} \bigl\vert f(U) - f(V) \bigr\vert & = \;\Biggl\vert \int \langle \psi \vert U N(\vert\psi\rangle\langle\psi\vert) U^{\dagger} - V N(\vert\psi\rangle\langle\psi\vert) V^{\dagger} \vert\psi\rangle\, \mathrm{d}\psi \Biggr\vert\\ & \leq \int \bigl\vert \langle \psi \vert U N(\vert\psi\rangle\langle\psi\vert) U^{\dagger} - V N(\vert\psi\rangle\langle\psi\vert) V^{\dagger} \vert\psi\rangle \bigr\vert\,\mathrm{d}\psi\\ & \leq \int \bigl\| U N(\vert\psi\rangle\langle\psi\vert) U^{\dagger} - V N(\vert\psi\rangle\langle\psi\vert) V^{\dagger} \bigr\|_{\infty}\,\mathrm{d}\psi\\ & \leq \int \| U - V\|_{\infty}\, \mathrm{d}\psi\\ & = \| U - V\|_{\infty} \end{aligned} $$ Of course you can further upper-bound this by $\|U-V\|_2$ if you wish, and we find that the function is 1-Lipschitz as claimed.

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    $\begingroup$ The answer is concise and impeccable. Thanks a lot for the time and effort you must have put into it! $\endgroup$ Commented May 14 at 8:59

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