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8 votes
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What is the difference between "Shot-Noise-Limit" and "Standard Quantum Limit"?

You are correct that both terms reference the central limit theorem (CLT), which states that[1] ...the average of a large number $n$ of independent measurements (each having standard deviation $\...
ryanhill1's user avatar
  • 2,503
7 votes
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Does the symmetric logarithmic derivative operator have a geometric interpretation?

First, the classical correspondence, explaining why the SLD should be present. The Fisher information is the expectation value of the score, where the score is the logarithmic derivative of the ...
Quantum Mechanic's user avatar
6 votes
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Why is the quantum Fisher information for pure states $F_Q[\rho,A]=4(\Delta A)^2$?

Suppose $\lambda_0 = 1$ and the rest are $0$. $$ F_Q [\rho,A] = 2 \sum_{k,l} \frac{(\lambda_k-\lambda_l)^2}{\lambda_k + \lambda_l} | \langle k |A| l \rangle |^2\\ = 2 \sum_{k=0,l \neq 0} \frac{(1-0)^...
AHusain's user avatar
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6 votes
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Unit vanishes in the Quantum Cramer-Rao Bound?

You are correct: the units must indeed match. If we take a standard evolution with unitary $U=\exp(-i H \theta)$, then the units of $H$ and $\theta$ must match such that $H\theta$ is unitless. For a ...
Quantum Mechanic's user avatar
6 votes
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Understanding the $M$ upper bound in the paper: "Multipartite entanglement and high-precision metrology"

Your conclusion appears correct to me. It seems that Eq.(23), modified with your proposed change to the RHS, can be verified by combining Eq.(3), for the upper bound on the $M$ unentangled particles, ...
Jonathan Trousdale's user avatar
4 votes

How is the connection between Bures fidelity and quantum Fisher information derived?

Let $ \rho = \sum_n \rho_n |\psi_n \rangle \langle \psi_n | $ be the eigendecomposition of $\rho$. We will calculate everything in terms of $ |\psi_n \rangle$ basis. Note that $ \frac{d \rho_\theta}{d ...
tsgeorgios's user avatar
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4 votes
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Stabilizer state QFI lower limit query

The state $\psi$ (this is denoting the density matrix, even though it's a pure state) can be described as a sum of all the products of the stabilizers. We are promised that $X_i$ is not in the ...
DaftWullie's user avatar
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4 votes
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How to derive the quantum Fisher information from the relative entropy?

Expressing the derivative $\partial_i\rho$ in terms of its eigenvalues and eigenvectors will show us that these two are not equal. I will assume a full-rank density matrix $\rho$ to streamline the ...
Quantum Mechanic's user avatar
4 votes
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Is Quantum Cramer-Rao bound for single parameter always attainable?

The answer given in the literature is always yes, this is guaranteed to be possible for single-parameter estimation. What you are noticing eventually gives rise to some cool things that I'll mention ...
Quantum Mechanic's user avatar
3 votes
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Paris 2009 paper on Quantum Estimation. From eq. 12 to eq. 16

The first question is correct: consider unitary evolution with generator $H_\lambda$ such that $|\partial_\lambda \psi_n\rangle=iH_\lambda |\psi_n\rangle$ for all $n$. The states $|\psi_n\rangle$ need ...
Quantum Mechanic's user avatar
3 votes
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Problems trying to plot the classical Fisher information with Pennylane

There seems to be a bug in classical_fisher in combination with np.linalg.norm. I opened an issue here, should be resolved soon (...
Korbinian's user avatar
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3 votes
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How to compute the SLDs for pure single-qubit states?

These results are specific to qubits and your direct verification is the best way to do it. You can see this because they required the property of Pauli matrices that $\{\pmb{v}\cdot\pmb{\sigma},\pmb{...
Quantum Mechanic's user avatar
3 votes
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How to compute the QFI of a thermal state?

QFI must always be computed with respect to a parameter "$\theta$". Perhaps it is the temperature that you want to use here, or $\beta$? Regardless, we can put the state into your desired ...
Quantum Mechanic's user avatar
3 votes
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What does $\langle\partial_i\psi(\theta)|\psi(\theta)\rangle$ mean when implementing the Quantum Fisher information matrix?

In fact, $\partial_{k} \psi(\boldsymbol{\theta})$ is a vector. For example, $\frac{\mathrm{d}}{dx} \begin{pmatrix}x\\x^2\end{pmatrix}=\begin{pmatrix}1\\2x\end{pmatrix}$ , i.e., the derivative is ...
narip's user avatar
  • 3,009
2 votes

Does the symmetric logarithmic derivative operator have a geometric interpretation?

Although, the Bures metric, the Fisher tensor and the symmetric logarithmic derivative appear mainly in quantum estimation theory, and even though the original discovery by Helstrom was in this ...
David Bar Moshe's user avatar
2 votes
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Quantum computation of classical Fisher information

Section 3 of this paper talks about strategies to obtain the classical Fisher information in a NISQ computing setting. More concretely, this recent preprint details how to measure it in practice with ...
David Wierichs's user avatar
2 votes
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Fisher information from likelihood function for discrete quantum case

This is correct. If you retained each of the measurement results as opposed to the total number of times each result occured, you would neglect the multinomial factor. In addition, it is worth ...
Quantum Mechanic's user avatar
2 votes
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What are the antisymmetric terms in $\sigma_{mn}$ in the expression for the Fisher information?

Antisymmetric means that $\sigma_{nm}=-\sigma_{mn}$. Since the sum ranges over all values of $m$ and $n$, adding an antisymmetric term adds something proportional to $$|\langle \psi_m^\lambda|\...
Quantum Mechanic's user avatar
1 vote

Generalizing error propagation formula to multi-parameters

Short answer: It always behooves oneself to learn the classical version of a theory before the quantum one. Many many people (including but certainly not limited to OP) could have their problems ...
Quantum Mechanic's user avatar
1 vote

In what limit does the estimator sample variance converge to the Cramer-Rao bound?

The answer is the same as for the other question of yours: no, the efficiency is asymptotic in $\nu$ not in $\mu$. The MLE is not efficient (in general) for any finite finite number of observations. ...
glS's user avatar
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1 vote
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Create qnode with density matrix on pennylane

My previous answer mistakenly read quantum_fisher instead of classical_fisher. It is actually possible to compute the latter ...
Tristan Nemoz's user avatar
  • 6,897
1 vote

How is the quantum geometric tensor derived?

One expands the pure state to leading order, as in the paper, with $$|\psi(\lambda_\mu+d\lambda_\mu)\rangle=|\psi(\lambda_\mu)\rangle+d\lambda_\mu|\partial_\mu \psi(\lambda_\mu)\rangle.$$ For multiple ...
Quantum Mechanic's user avatar

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