What does $\langle\partial_i\psi(\theta)|\psi(\theta)\rangle$ mean when implementing the Quantum Fisher information matrix?

Question

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?

Answer 1

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.