Tag Info

Hot answers tagged fidelity

9

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 qubits that lasted for over 39 minutes; these, however, only had an 81% fidelity rate. (This is for qubits used in computation, not memory storage. For memory ...

6

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 does a quantum channel preserve information?" that when comparing fidelity to the trace norm: Using the properties of the trace distance established in the ...

6

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 calculate $\langle\phi|\psi\rangle$, what it really means is $$(\langle\phi|\otimes\mathbb{I})|\psi\rangle,$$ padding everything with identity matrices to ...

5

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 $\frac{1}{\sqrt{2}}\big(|0\rangle + |1\rangle\big)$. or you can say it is the cosine of the smallest angle between two states, also called the cosine similarity

4

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, our analysis of the error used in approximating one unitary by another involves the distance induced by the operator norm: $$\bigl\lVert U - V \bigr\rVert_\... 4 The quantity \text{Tr}(\sqrt{A}\sqrt{B}) that you defined there is actually referred to as the "just-as-good fidelity" (see 1801.02800) because it does have a relationship with the trace distance very similar to the standard fidelity and is therefore "just as good" for quantifying the distinguishability of states. There is no intrinsic reason to prefer the ... 4 (I will give the argument with formulas for now, hopefully I find time for some pictures later.) Let |m\rangle be the (unnormalized) maximally entangled state. Then, a purification of \rho is given by$$ |\rho\rangle_{AB}=(\sqrt{\rho}_A\otimes1\!\!1_B)|m\rangle\ , $$and correspondingly for \sigma -- this can be seen most easily by first tracing the ... 3 I'll provide a slightly different (but of course equivalent) way to prove Uhlmann's theorem, which I personally find more explicit than the standard one, and might help to understand what is going on. I don't know if this qualifies as sufficient "intuition" (it certainly doesn't fully satisfy me), but I at least prefer it to the standard approach with ... 3 Purifications play an important role in the theory of density matrices (or more generally quantum states) because they provide a geometric tool in the explanation and description algebraic relations. (I'll be following here Bengtsson and Życzkowski's reasoning in derivation of the fidelity formula (section 9.4)). A positive  N \times N matrix \rho ... 3 Actually, there should be a minus. There is a mistake in the paper. Wittek uses a minus in his (expensive) book. Indeed say :$$ |\psi\rangle = \frac{1}{\sqrt{2}} (|0,a\rangle + |1,b\rangle)  |\phi\rangle = \frac{1}{\sqrt{Z}} (|a||0\rangle - |b||1\rangle) $$Then :$$ \langle \phi |\psi\rangle = \frac{1}{\sqrt{2Z}} (|a|\langle 0| - |b|\langle 1|) (|...

3

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 question will therefore depend on which kind of fidelity you want. Any measure of fidelity will typically involve comparing the gate that you wanted to the channel ...

3

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 squared distance between two states is commonly defined as $$D(\rho,\sigma)^2\equiv \|\rho-\sigma\|_2^2=1-\text{Tr}(\rho\sigma).$$ This is useful and used for ...

3

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 fundamentally analog machines. what that means practically is that the gates you apply when you run code on a real quantum computer are going to be parameterized by a real ...

2

You are correct in both assumptions. A total phase on a qubit state $|{\psi}>$ is often referred to as the global phase. Any measurement of a quantum state is the expectation value $\lambda_{M}$ of some (Hermitian) observable $M$: $\lambda_{M} = |<\psi|M|\psi>|^{2}$. Because this is invariant to the global phase, there is no physical meaning to ...

2

Qualitatively, fidelity is the measure of the distance between two quantum states. Fidelity equal to 1, means that two states are equal. In the case of a density matrix, fidelity represents the overlap with a reference pure state. Fidelity equal to 1, means that the density matrix is diagonal and equivalent to the pure reference state. Like every distance ...

2

The idea is to use CS inequality in the form $\newcommand{\tr}{\operatorname{Tr}}\lvert \sum_{ij}A_{ij}^* B_{ij}\rvert\le\sqrt{\sum_{ij} \left\lvert A_{ij}\right\rvert^2}\sqrt{\sum_{ij}\left\lvert B_{ij}\right\rvert^2}$, which in matrix formalism reads $\lvert\tr(A^\dagger B)\rvert\le\sqrt{\tr(A^\dagger A})\sqrt{\tr(B^\dagger B)}$. Therefore, \lvert\tr(...

1

The best I have it's this generic answer, which I put here for clarity, hoping for improvements/corrections or even to be superseded by something better: If the limiting factor for fidelity in a given architecture+algorithm are the single-qubit gates, or the two-qubit gates, or the measurement, and if this limiting factor is not optimized in a ZEFOZ point,...

Only top voted, non community-wiki answers of a minimum length are eligible