In the question "[Derivative of eigenvectors of a matrix with respect to its components](https://mathoverflow.net/questions/229425)", [Liviu Nicolaescu](https://mathoverflow.net/a/229467/175010) 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](https://mathoverflow.net/a/229467/175010), **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.



Derivative of eigenvectors of an Hermitian matrix
This is first order perturbation theory, see [Wikipedia](https://en.wikipedia.org/wiki/Perturbation_theory_(quantum_mechanics)#First_order_corrections). As explained there (search for "*Since the overall phase is not determined in quantum mechanics*"), you need to use the freedom you have to multiply the eigenstate by a complex phase vector, to ensure that $\alpha\_{ii}$ is real and hence zero.


More precisely, the phase ambiguity is resolved by demanding that the inner product $\langle v\_i^{(0)}|v\_i^{(n)}\rangle$ of the zeroth order eigenstate $v\_i^{(0)}$ and the $n$-th order correction $v\_i^{(n)}$ is real for each $n$. To first order this implies $\langle v\_i^{(0)}|v\_i^{(1)}\rangle=0$, but in higher order this overlap is nonzero. For example, $\langle v\_i^{(0)}|v\_i^{(2)}\rangle=-\tfrac{1}{2}\langle v\_i^{(1)}|v\_i^{(1)}\rangle$. [These notes](https://web.pa.msu.edu/people/mmoore/TIPT.pdf) gives a clear derivation.


