10

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 ...


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

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}\...


6

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_\...


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+...


4

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 ...


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

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

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 ...


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

The expressions are equivalent: $$ F(\rho,\sigma) =\operatorname{tr}\sqrt{\sigma^{1/2}\rho\sigma^{1/2}}= \operatorname{tr}\sqrt{\rho^{1/2}\sigma\rho^{1/2}} \\ = \operatorname{tr}|\sqrt\rho\sqrt\sigma| = \operatorname{tr}|\sqrt\sigma\sqrt\rho| = \max_{\psi_\rho,\psi_\sigma}|\langle\psi_\rho|\psi_\sigma\rangle|.$$ See also this question about the symmetry ...


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

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 ...


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}...


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 ┼╗yczkowski's reasoning in derivation of the fidelity formula (section 9.4)). A positive $ N \times N$ matrix $\rho$ ...


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

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 square of density matrix is equal to density matrix itself and equivalent to the pure ...


2

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


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(...


2

The channel $\mathcal{E}$ is explicitly defined in the preceding paragraph as being the depolarising channel. Thus, all you need to calculate is $$ F=\sqrt{\langle 0|\mathcal{E}(|0\rangle\langle 0|)|0\rangle}. $$


2

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 self-adjoint matrices are not diagonalizable in general, so the square root $\sqrt{\rho\sigma}$ can be not well-defined (see edit below). Anyway, $\text{Tr}(\...


2

You could probably reach the same conclusions by identifying that $tr(\rho \sigma)$ is just the expectation value of $\rho$ under the mixed state $\sigma$, but let's do it explicitly: for $\rho = \sum_i p_i |\psi_i\rangle \langle \psi_i|$ and $\sigma = \sum_i q_i |\phi_i\rangle \langle \phi_i|$, we have: $tr(\rho \sigma)$ = $\sum_{ij} p_i q_j tr(|\psi_i\...


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