In the question "Derivative of eigenvectors of a matrix with respect to its components", Liviu Nicolaescu has provided an answer valid for a real matrix. As outlined in the following, the same proof applies to Hermitian matrices, but it is incomplete.

Let us consider an Hermitian matrix $H$ ($H^\dagger = H$). Its eigenvectors satisfy

$$(H-\lambda_i) v_i = 0 \quad\text{with}\quad \lambda_i \in \mathbb{R} \quad\text{and}\quad v_j^\dagger v_i=\delta_{ij}.$$

From the derivative of the first relation one gets $$(H - \lambda_i) \dot v_i + (\dot H - \dot \lambda_i) v_i = 0 \quad\rightarrow\quad \dot \lambda_i = v_i^\dagger \dot H v_i.$$

Considering the eigendecomposition of $\dot v_i$ combined with the previous equation, one gets $$\dot v_i = \sum \alpha_{ij}v_j \quad\rightarrow\quad \alpha_{ij} = \frac{v_j^\dagger \dot H v_i}{\lambda_i-\lambda_j}\quad\text{for}\quad i\ne j.$$

From the derivative of $v_i^\dagger v_i=1$, one gets $Re(v_i^\dagger \dot v_i)=Re(\alpha_{ii})=0$.

So far everything is compatible with the case of real matrices, **BUT** what is the value of $Im(\alpha_{ii})$??

From numerical experiments I can surely state that $Im(\alpha_{ii})$ is non-zero and it has an order of magnitude compatible with the values of $\alpha_{ij}$. But I cannot find any way to derive its analytical expression.