# Tag Info

14

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 maximized at extreme points. Extreme points of $S$ are the states of the form$^1$ $\rho = \overline\sigma\otimes\tau$. The reason we choose to define $\sigma$ as ...

11

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 ...

9

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 $A|\psi\rangle = |\psi\rangle$ for every $A\in S$. For example, $(|00\rangle+|11\rangle)/\sqrt2$ is a stabilizer state, because it is a $+1$ eigenstate of the ...

8

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, $$\|\mathbf{u}-\mathbf{v}\| = \sqrt{2}\... 8 No dimension-independent bound is possible. Consider states \rho_A and \sigma_A that are close in p-norm (for p>1) but have relatively low fidelity. Specifically, assume$$ \|\rho_A - \sigma_A\|_p = \varepsilon $$and$$ \operatorname{F}(\rho_A,\sigma_A) = \bigl\|\sqrt{\rho_A}\sqrt{\sigma_A}\bigr\|_1 = \delta, $$where \varepsilon is small and ... 7 When you ask about an 'ideal' fidelity measure, it assumes that there is one measure which inherently is the most meaningful or truest measure. But this isn't really the case. For unitary operators, our analysis of the error used in approximating one unitary by another involves the distance induced by the operator norm:$$ \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\... 6 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\!\langle i|\otimes A_i\right) = \sum_i {\rm Tr}(A_i).$$It immediately follows that \|\sqrt\rho\sqrt\sigma\|_1\equiv {\rm Tr}|\sqrt\rho\sqrt\sigma| can be written as ... 6 Given the constraints you impose (why those are appropriate constraints is perhaps another discussion), I think you're over-complicating things. Without loss of generality, you can assume$$ |\alpha_0\rangle=|0\rangle,\qquad |\alpha_1\rangle=\cos\theta|0\rangle+\sin\theta|1\rangle $$and clearly$$ U|\alpha_0\rangle|0\rangle=a_0|00\rangle+a_1|\alpha_1\...

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 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 ...

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

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}. $$Moreover, the maximum is achieved when |\psi\rangle is the unit norm eigenvector of A ... 5 Background If v_1, v_2, \dots, v_n is an orthonormal basis in the inner product space V, then any vector u\in V can be expressed as a linear combination$$ u = \alpha_1 v_1 + \alpha_2 v_2 + \dots + \alpha_n v_n.\tag1 $$Moreover, the coefficients can be computed using \alpha_k=\langle v_k, u\rangle, as can be seen by applying \langle v_k, .\rangle ... 4 Fidelity is a single-number measure of how good a gate is. Since there are many ways that a gate can go wrong, there are multiple ways that the fidelity can be defined. The exact answer to your question will therefore depend on which kind of fidelity you want. Any measure of fidelity will typically involve comparing the gate that you wanted to the channel ... 4 Actually, there should be a minus. There is a mistake in the paper. Wittek uses a minus in his (expensive) book. Indeed say :$$ |\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

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

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 $|\... 4 (I will give the argument with formulas for now, hopefully I find time for some pictures later.) Let$|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 ...

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