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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 $$\|\rho_A - \sigma_A\|_p = \varepsilon$$ and $$\operatorname{F}(\rho_A,\sigma_A) = \bigl\|\sqrt{\rho_A}\sqrt{\sigma_A}\bigr\|_1 = \delta,$$ where $\varepsilon$ is small and $... 7 Recall the law of cosines for two unit vectors$\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. Similarly, you'll recall the definition of the inner product, $$\langle \mathbf u|\mathbf {v}\rangle = \cos\theta.$$ So, $$\|\mathbf{u}-\mathbf{v}\| = \sqrt{2}\... 7 When you ask about an 'ideal' fidelity measure, it assumes that there is one measure which inherently is the most meaningful or truest measure. But this isn't really the case. For unitary operators, our analysis of the error used in approximating one unitary by another involves the distance induced by the operator norm:$$ \bigl\lVert U - V \bigr\rVert_\... 7 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 ... 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 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 overcome the decay, we need gates with fidelity so close to one that the decay is negligible over the course of quantum algorithms we intend to run. As you ... 6 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|K_0|\psi\rangle\langle\psi|K_0^\dagger|\psi\rangle + \langle\psi|K_1|\psi\rangle\langle\psi|K_1^\dagger|\psi\rangle d\psi\\ & =\frac{1}{4\pi}\int_0^\pi\... 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 5 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 ... 5 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 & \frac{1-\varepsilon}{2} \end{pmatrix},\quad \rho_1 = \begin{pmatrix} \frac{1-\varepsilon}{2} & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & \frac{1+... 5 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}$of$\sigma_A$could happen to be optimal for the right choice of$\rho_{AR}$. For example, if we suppose that$\sigma_{AR}$is any given extension of$\sigma_A$, and we ... 5 Recall that for any Hermitian operator$A$and any unit vector$|\psi\rangle$the real number$\langle \psi|A|\psi\rangle$, known as the Rayleigh quotient, is bounded by the largest eigenvalue$\lambda_{max}$of$A$$$\langle \psi|A|\psi\rangle \le \lambda_{max}.$$ Moreover, the maximum is achieved when$|\psi\rangle$is the unit norm eigenvector of$A$... 4 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 ... 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 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 ... 4 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 : $$\langle \phi |\psi\rangle = \frac{1}{\sqrt{2Z}} (|a|\langle 0| - |b|\langle 1|) (|... 4 Answer: Fidelity of 0.9999 at 1.08 seconds in 2013: http://science.sciencemag.org/content/342/6160/830.full?ijkey=uhZaDNPnwgTdA More details: The T_2 was 180 minutes, or 3 hours. What about the 81% that Heather mentioned?: The fidelity of 81% that Heather quotes, was actually referring to something else. In the same paper they wanted to show that they ... 4 (I will give the argument with formulas for now, hopefully I find time for some pictures later.) Let |m\rangle be the (unnormalized) maximally entangled state. Then, a purification of \rho is given by$$ |\rho\rangle_{AB}=(\sqrt{\rho}_A\otimes1\!\!1_B)|m\rangle\ , $$and correspondingly for \sigma -- this can be seen most easily by first tracing the ... 4 Okay, this is a rather subtle situation, but I think I've figured it out. The key is to be very careful about which mathematical results about Hermitian operators do and do not hold for generic operators. Let H represent an arbitrary Hermitian matrix, N an arbitrary normal one, D be a generic diagonalizable matrix, and M an arbitrary matrix, all ... 4 Both definitions are used and authors usually make it clear which one they mean. Wikipedia also points this out under the Alternative Defintion section. 4 Usually, error rate for a qubit is defined as probability of undesired change in the qubit state (see for example this paper). Then we have state fidelity, which is a measure of the difference between the state we have and the state we would like to have, for any (single or multi qubit) quantum system. Quantum state tomography is a means to characterize the ... 4 In general, it would seem no. The quantity$$ \mathrm{Tr}[(\rho - \sigma)|\psi\rangle\langle\psi|] $$is only concerned with the distance between \rho and \sigma on the subspace \mathrm{span}(|\psi\rangle). For example, we know we can decompose the Hilbert space as \mathcal{H} = \mathrm{span}(|\psi\rangle) \oplus \mathrm{span}(|\psi\rangle)^{\perp}. ... 4 Here's a concrete example for a single qubit. We can always change the basis to have |\psi\rangle=|0\rangle. Let us further suppose that \langle0|\rho|0\rangle=0, so that$$\rho=\begin{pmatrix}0&0\\0&1\end{pmatrix}.$$The requirement \operatorname{Tr}[(\sigma-\rho)|\psi\rangle\!\langle\psi|]=\langle\psi|\sigma-\rho|\psi\rangle=\epsilon then ... 4 For any quantum error correcting code, it is possible to construct a channel which introduces errors that the code cannot correct. However, the key point is that such channels are highly adversarial and not at all representative of any physically reasonable error mechanism. An easy way to construct such adversarial noise is to build it from the logical ... 3 No, you wouldn't find 0.9 again. To make the partial trace calculation simpler you can note that the state |\psi'\rangle is separable under the bipartition a_1b_1 | a_2 b_2, i.e. |\psi'\rangle = |00\rangle \otimes (\sqrt{a} |00\rangle + \sqrt{1-a} |11\rangle). So irrespective of the value of a we have \operatorname{Tr}_{a_2b_2}[|\psi'\rangle\... 3 Prohibited device Such a circuit C would enable faster-than-light communication and therefore does not exist. Suppose Alice and Bob share a Bell pair |\psi\rangle = \frac{1}{\sqrt{2}}(|01\rangle - |10\rangle) and Alice has a classical bit b \in \{0, 1\} she wishes to communicate to Bob. She proceeds as follows. If b=0 Alice measures her half of |\... 3 If these are qubit states, the formula in your question simplifies dramatically to$$F'(\rho,\pi)=\sqrt{\text{tr}(\rho \pi) + 2 \sqrt{\text{det}(\rho) \text{det}(\pi)}}.$$If you consider the components of the vectors, \vec s = (s_1, s_2, s_3) and \vec r = (r_1, r_2, r_3), this can be expressed simply as$$F'(\rho, \pi) = \frac{1}{\sqrt{2}} \left[1+ \... 3 I'll try to address the problem from the Riemannian geometry point of view. In this approach, the distances are identified as length of geodesics of Riemannian metrics on spaces of quantum states. The advantage of this approach lies in the fact that the Riemannian distances automatically satisfy the metric axioms of positivity, symmetry and the triangle ... 3 Short version Consider the two following observations: Given a state$\rho$, the problem of finding purifications of$\rho$is equivalent to that of finding matrices$A$such that$\rho=AA^\dagger$. The purifications of$\rho$are then the vectorisations of these$A$(i.e. a vector$\Psi$is a purification of$\rho$iff its matrix of coefficients$\Psi_{ij}...