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

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?

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.
• 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. Nov 18, 2021 at 14:39
• $N\times 1$ matrix, a vector. Nov 19, 2021 at 0:10