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Following this paper, the quantum Fisher information matrix (QFIM) - $\mathcal{F}$ can be calculated as:

$\mathcal{F}_{i, j}(\theta)=4 \operatorname{Re}\left[\left\langle\partial_{i} \psi(\boldsymbol{\theta}) \mid \partial_{j} \psi(\boldsymbol{\theta})\right\rangle-\left\langle\partial_{i} \psi(\boldsymbol{\theta}) \mid \psi(\boldsymbol{\theta})\right\rangle\left\langle\psi(\boldsymbol{\theta}) \mid \partial_{j} \psi(\boldsymbol{\theta})\right\rangle\right]$

$|\psi(\theta)\rangle$ is the current quantum state and $\theta$ is the $N$-dimensional complex vector, that means $\mathcal{F}$ is the $N\times N$ matrix.

The thing that I confused is $\partial_{k} \psi(\boldsymbol{\theta})$ is a scalar so how to calculate $\langle\partial_{i} \psi(\boldsymbol{\theta}) \mid \psi(\boldsymbol{\theta}) \rangle$ and its dagger?

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1 Answer 1

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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 element wise when it acts on a vector.

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  • $\begingroup$ Thanks for your answer, I misunderstand when thinking about $|\psi\rangle$, $|\psi\rangle$, in this case, is a $N$-dimensional vector, and its derivative $\partial\psi(\theta)$ is a $N\times N$ matrix so $\partial_k\psi(\theta)$ is also a $N$-dimensional vector. $\endgroup$
    – Monad
    Nov 18, 2021 at 14:39
  • $\begingroup$ $N\times 1$ matrix, a vector. $\endgroup$
    – narip
    Nov 19, 2021 at 0:10
  • $\begingroup$ shouldn't this be a vector of derivatives, i.e. $$ \left(\begin{array}{c}\partial/\partial \theta_1\\ \partial/\partial\theta_2\\ \vdots \end{array}\right) $$ acting on a scalar function $\psi(\theta)$, resulting in a vector $$ \left(\begin{array}{c}\partial \psi(\theta)/\partial \theta_1\\ \partial\psi(\theta)/\partial\theta_2\\ \vdots \end{array}\right)\quad ? $$ $\endgroup$ May 24 at 20:19
  • $\begingroup$ @ZeroTheHero No, mind that it's $\mathcal F_{i,j} $ instead of the matrix $\mathcal F$ itself. $\endgroup$
    – narip
    May 25 at 0:24
  • $\begingroup$ $\partial_i$ is only partial derivative respect to the $i$-th variable. $\endgroup$
    – narip
    May 25 at 0:29

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