Timeline for Is there a concentration inequality for the quantum gate fidelity $F(C,U)$ for a channel $C$ such that $\int dU F(C,U)=X$?
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 14 at 8:56 | vote | accept | Davide Li Calsi | ||
| May 12 at 14:37 | answer | added | John Watrous | timeline score: 7 | |
| May 12 at 12:14 | comment | added | Davide Li Calsi | @JohnWatrous One can bound $||Tr[U^\dagger K_i|-Tr[V^\dagger K_i|| \leq |Tr[U^\dagger K_i-Tr[V^\dagger K_i|$. Using the inequality $|Tr[AB]| \leq \|A\|_1 \|B\|_\infty$ we further upper bound $ |Tr[U^\dagger K_i-Tr[V^\dagger K_i^\dagger| \leq \|K_i^\dagger\|_1 \|U^\dagger - V^\dagger\|_\infty \leq \sqrt{d} \|K_i^\dagger\|_1 \|U^\dagger - V^\dagger\|_2$. I am basically stuck here. I don't know how to get rid of the $| |Tr[U^\dagger K_i^\dagger]| +|Tr[V^\dagger K_i^\dagger]| |$ P.S. In my previous comment $K_i$ should be $K_i^\dagger$. I fixed this here, but can no longer edit the old comment | |
| May 12 at 12:06 | comment | added | Davide Li Calsi | I did as suggested by Adam Zalcman. By definition $| F_e(N,U) - F_e(N.V)|= Tr[| \psi \rangle \langle \psi | ( I \otimes U^\dagger N | \psi \rangle \langle \psi | - I \otimes V^\dagger N | \psi \rangle \langle \psi | ) ]$. Since $F_e(C) = \frac{1}{d^2} \sum_i |Tr[K_i]|^2$, where $\{K_i\}$ is a Kraus decomposition of channel $C$, one can rewrite the equation as $\frac{1}{d^2} \sum_i | |Tr[U^\dagger K_i|^2 - |Tr[V^\dagger K_i|^2 | = \frac{1}{d^2} \sum_i | |Tr[U^\dagger K_i| - |Tr[V^\dagger K_i| | \cdot | |Tr[U^\dagger K_i| + |Tr[V^\dagger K_i| | $. | |
| May 11 at 18:14 | history | edited | glS♦ | CC BY-SA 4.0 |
edited title
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| May 10 at 17:34 | comment | added | John Watrous | This function seems like it should be 1-Lipschitz. How hard have you tried to bound $L$? | |
| May 10 at 13:41 | comment | added | Tristan Nemoz♦ | Indeed, I should have carried the computations out, my bad! | |
| May 10 at 13:14 | comment | added | Davide Li Calsi | I agree with @AdamZalcman. Using the invariance of the Haar measure we would still have $N(V | 0 \rangle \langle 0 | V^\dagger)$ inside the integral, and there's no way to get rid of it. As for Horodecki's formula, I tried that too, but with little success. The best upper bound that I was able to find with this strategy is $L<\sqrt{d}$, which is not useful since in Levy's lemma the upper bound is $exp(-\frac{d}{L^2} \epsilon \cdot K)$ for a known constant $K$. | |
| May 10 at 13:08 | comment | added | Adam Zalcman | 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. | |
| May 10 at 13:05 | comment | added | Adam Zalcman | 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. | |
| May 10 at 12:59 | comment | added | Tristan Nemoz♦ | My point is that we would then have $F(C,U)=X$ for a fixed $C$ and for all $U$. | |
| May 10 at 12:33 | comment | added | Davide Li Calsi | 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 $\epsilon$ | |
| May 10 at 12:02 | comment | added | Tristan Nemoz♦ | 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? | |
| S May 10 at 11:42 | review | First questions | |||
| May 10 at 13:18 | |||||
| S May 10 at 11:42 | history | asked | Davide Li Calsi | CC BY-SA 4.0 |