# Tag Info

5

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, with Eq.(5), for the upper bound on the $N-M$ entangled particles, using the convexity asserted in Section II. With respect to the $M$ unentangled particles, ...

5

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 probability distribution: $$L_{\mathrm{classical}}=\frac{\partial [\ln p(\pmb{x}|\theta)]}{\partial \theta},$$ which leads to the relation $$\frac{\partial [ p(\... 4 First of all, here's a short disclaimer: I'm not an in-depth expert in this field, I'm just currently getting in contact with tomography more and more often :) So take the following with a grain of salt. It might be incomplete in the sense that better results have been shown somewhere. We consider the problem of reconstructing a d-dimensional quantum state ... 4 Suppose \lambda_0 = 1 and the rest are 0.$$ 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)^2}{1 + 0} | \langle 0 |A| l \rangle |^2 + 2 \sum_{k\neq 0,l = 0} \frac{(0-1)^2}{0 + 1} | \langle k |A| 0 \rangle |^2\\ = 4 \sum_{l \neq 0} | \langle l |A| 0 \...

4

Here is an approach that requires no specific knowledge about $|\psi\rangle$ whatsoever. In your description you implied that each $H_i$ has the same maximum and minimum eigenvalues $\lambda_m$ and $\lambda_M$ respectively so I will assume this in the derivation. The process of measuring $\langle h \rangle$ empirically can be thought of running a series of ...

4

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 pure state $\rho_\theta=U|\psi\rangle\langle\psi|U^{\dagger}$ with unitary evolution, the quantum Fisher information takes the form $$F_Q[\psi,H]=\langle\psi|H^2|... 4 You are correct that both terms reference the central limit theorem (CLT), which states that[1] ...the average of a large number n of independent measurements (each having standard deviation \Delta \sigma) will converge to a Gaussian distribution with standard deviation \Delta \sigma / \sqrt{n}, so that the error on average scales with n^{-1/2}. ... 3 Let  \rho = \sum_n \rho_n |\psi_n \rangle \langle \psi_n |  be the eigendecomposition of \rho. We will calculate everything in terms of  |\psi_n \rangle basis. Note that  \frac{d \rho_\theta}{d \theta} = i [\rho_\theta, A]  and  \frac{d^2 \rho_\theta}{d \theta^2} = -\big[[\rho_\theta, A], A\big] . Now we write$$ \sqrt{ \sqrt{\rho} \rho_\theta \...

3

You are correct, but the SQL is a local limit, when you already have a very good idea what the value of $\theta$ is, so there is no contradiction. Let's work through it. You measure some relative phase $\Theta$, and you infer that $$\Theta=n\theta+2\pi k,\quad k\in \mathbb{N}.$$ You work out that $$\theta=\frac{\Theta +2\pi k}{n},$$ where $\Theta$ and $n$ ...

1

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 context; they have a deeper origin in the geometric formulation of quantum mechanics. This metric is the geometrized version of the Jordan structure of the ...

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