# Is there a concentration inequality for the quantum gate fidelity $F(C,U)$ for a channel $C$ such that $\int dU F(C,U)=X$?

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?

• 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? Commented May 10 at 12:02
• 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 $\epsilon$ Commented May 10 at 12:33
• My point is that we would then have $F(C,U)=X$ for a fixed $C$ and for all $U$. Commented May 10 at 12:59
• 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. Commented May 10 at 13:05
• 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. Commented May 10 at 13:08

## 1 Answer

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.

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