14 votes
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What are min and max overlaps of a maximally entangled state with a separable state?

The minimum overlap is zero and the maximum overlap is $\frac{1}{d}$. The overlap is a linear function of $\rho$ and the set $S$ of separable states is convex, so the overlap is both minimized and ...
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11 votes
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What is the longest time a qubit has survived with 0.9999 fidelity?

Well, for the longest coherence time ever, I'm finding this Science from 2013 entitled Room-Temperature Quantum Bit Storage Exceeding 39 Minutes Using Ionized Donors in Silicon-28, which indicates ...
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  • 3,281
10 votes
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What is a stabilizer state?

Let $\mathcal{G}_n$ denote the Pauli group on $n$ qubits. An $n$-qubit state $|\psi\rangle$ is called a stabilizer state if there exists a subgroup $S \subset \mathcal{G}_n$ such that $|S|=2^n$ and $A|...
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8 votes
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What is the difference between the "Fubini-Study distances" $\arccos|\langle\psi|\phi\rangle|$ and $\sqrt{1-|\langle\psi|\phi\rangle|}$?

Recall the law of cosines for two unit vectors $\mathbf{u}$ and $\mathbf{v}$ in $\mathbb R^2$: $$ \|\mathbf{u}-\mathbf{v}\|^2 = 2-2\cos\theta, $$ where $\theta$ is the angle between the vectors. ...
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8 votes
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Closeness of purifications of states

No dimension-independent bound is possible. Consider states $\rho_A$ and $\sigma_A$ that are close in $p$-norm (for $p>1$) but have relatively low fidelity. Specifically, assume $$ \|\rho_A - \...
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  • 4,413
8 votes
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How to calculate the average fidelity of an amplitude damping channel

An elementary method is to simply carry out the integration $$ \begin{align} \overline{F} &= \int\langle\psi|\mathcal{N_\gamma}(|\psi\rangle\langle\psi|)|\psi\rangle d\psi\\ &=\int\langle\psi|...
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7 votes
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What would be an ideal fidelity measure to determine the closeness between two non unitary matrices?

When you ask about an 'ideal' fidelity measure, it assumes that there is one measure which inherently is the most meaningful or truest measure. But this isn't really the case. For unitary operators, ...
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7 votes

How can I calculate the inner product of two quantum registers of different sizes?

I guess you're looking at equations (130) and (131)? So, here, you have $|\psi\rangle=(|0\rangle|a\rangle+|1\rangle|b\rangle)/\sqrt{2}$ and $|\phi\rangle=|a| |0\rangle+|b| |1\rangle$. When it says to ...
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6 votes

Purpose of using Fidelity in Randomised Benchmarking

Nielsen and Chuang in their book "Quantum Computation and Quantum Information" have section (Chapter 9) on distance measures for quantum information. Surprisingly they say in Section 9.3 " How well ...
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  • 581
6 votes
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What does fidelity mean?

It might be worth mentioning the physical motivation for these definitions and the concept of fidelity itself. Unlike the classical computers we all know and love, quantum computers are ...
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  • 176
6 votes
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How to limit the error probability in large scale quantum computers

It is true that fidelity decays exponentially in the course of quantum computation. This is indeed a major limitation of NISQ computers that imposes a stringent "depth budget". In order to ...
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6 votes

If $\rho,\sigma$ are classical-quantum states, can the fidelity $F(\rho,\sigma)$ be expressed in terms of $F(\rho_i,\sigma_i)$?

Observe that, for any collection of matrices $A_i$, we have $$\sqrt{\sum_i |i\rangle\!\langle i|\otimes A_i} = \sum_i |i\rangle\!\langle i|\otimes \sqrt{A_i}, \\ {\rm Tr}\left(\sum_i |i\rangle\!\...
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6 votes
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Approximate Cloning

Given the constraints you impose (why those are appropriate constraints is perhaps another discussion), I think you're over-complicating things. Without loss of generality, you can assume $$ |\alpha_0\...
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5 votes

What does fidelity mean?

Simply it is the distance (similarity measure) between two quantum states, for example the fidelity between $|0\rangle$ and $|1\rangle$ is less than the fidelity between $|0\rangle$ and $\frac{1}{\...
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  • 473
5 votes

Are there disadvantages in using the inner product between states instead of the fidelity?

The quantity $\text{Tr}(\sqrt{A}\sqrt{B})$ that you defined there is actually referred to as the "just-as-good fidelity" (see 1801.02800) because it does have a relationship with the trace distance ...
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  • 81
5 votes
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Quantum circuit for computing fidelity

Prohibited device Such a circuit $C$ would enable faster-than-light communication and therefore does not exist. Suppose Alice and Bob share a Bell pair $|\psi\rangle = \frac{1}{\sqrt{2}}(|01\rangle - |...
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5 votes
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Can the fidelity $F(\rho,\sigma)$ be computed knowing only $\rho - \sigma$?

The answer is no, as the following counter-example reveals. Let $\varepsilon\in(0,1)$ and define $$ \rho_0 = \begin{pmatrix} \frac{1+\varepsilon}{2} & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 &...
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  • 4,413
5 votes
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Fidelity of extensions of states

Let's start with the second question. There is nothing special about an extension $\sigma_{AR}^{\ast}$ that allows it to be optimal for the right-hand side of (1); any extension $\sigma_{AR}$ of $\...
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  • 4,413
5 votes
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How to find the distance between a given $\rho$ and the nearest pure state(s)?

Recall that for any Hermitian operator $A$ and any unit vector $|\psi\rangle$ the real number $\langle \psi|A|\psi\rangle$, known as the Rayleigh quotient, is bounded by the largest eigenvalue $\...
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5 votes
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Prove the fidelity can be written in terms of Pauli expectation values as ${\rm tr}(\rho\sigma)=\sum_k \chi_\rho(k)\chi_\sigma(\rho)$

Background If $v_1, v_2, \dots, v_n$ is an orthonormal basis in the inner product space $V$, then any vector $u\in V$ can be expressed as a linear combination $$ u = \alpha_1 v_1 + \alpha_2 v_2 + \...
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5 votes
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What use cases are there for 127 qubit QPUs?

I think that the main reason behind is to tackle technical difficulties connected with building huge number of qubits. Having hundred of qubits brings about issues with interconnection, connections to ...
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4 votes
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Intuitive role of the polar decomposition in proof of Uhlmann's theorem for fidelity

(I will give the argument with formulas for now, hopefully I find time for some pictures later.) Let $|m\rangle$ be the (unnormalized) maximally entangled state. Then, a purification of $\rho$ is ...
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4 votes

Are there disadvantages in using the inner product between states instead of the fidelity?

A few thoughts: It mostly depends on what you are trying to quantify. The inner product of states, $\text{Tr}(\rho\sigma)$, is used to quantify the distance in state space. More precisely, the ...
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4 votes
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Proving the inequality $|\mathrm{tr}(AU)|\le \mathrm{tr}|A|$ in Uhlmann's theorem

I'll give a couple of methods to do this: (Using matrix inequalities) The idea is to use CS inequality in the form $$\newcommand{\tr}{\operatorname{Tr}}\lvert \sum_{ij}A_{ij}^* B_{ij}\rvert\le\sqrt{\...
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4 votes

How can I calculate the inner product of two quantum registers of different sizes?

Actually, there should be a minus. There is a mistake in the paper. Wittek uses a minus in his (expensive) book. Indeed say : $$ |\psi\rangle = \frac{1}{\sqrt{2}} (|0,a\rangle + |1,b\rangle) $$ $$ |...
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  • 4,544
4 votes

What is the longest time a qubit has survived with 0.9999 fidelity?

Answer: Fidelity of 0.9999 at 1.08 seconds in 2013: http://science.sciencemag.org/content/342/6160/830.full?ijkey=uhZaDNPnwgTdA More details: The $T_2$ was 180 minutes, or 3 hours. What about the ...
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  • 11.9k
4 votes
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How to calculate the fidelity of a certain gate of a IBMQ device in Qiskit using randomized benchmarking/tomography?

Fidelity is a single-number measure of how good a gate is. Since there are many ways that a gate can go wrong, there are multiple ways that the fidelity can be defined. The exact answer to your ...
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4 votes

Can we combine the square roots inside the definition of the fidelity?

Okay, this is a rather subtle situation, but I think I've figured it out. The key is to be very careful about which mathematical results about Hermitian operators do and do not hold for generic ...
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  • 2,057
4 votes
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Can we combine the square roots inside the definition of the fidelity?

Square root is not differentiable at 0, so that cyclic property cannot be applied While $\rho\sigma$ has the same non-negative eigenvalues as $\sqrt\rho\sigma\sqrt\rho$, it's not self-adjoint. Non ...
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  • 5,423
4 votes
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Is the quantum state fidelity defined as $F(\rho, \sigma)=\text{tr}\sqrt{\rho^{1/2}\sigma\rho^{1/2}}$ or its square?

Both definitions are used and authors usually make it clear which one they mean. Wikipedia also points this out under the Alternative Defintion section.
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