Derivative of eigenvalues with respect to parameter

Question

By trying to find precise ways to calculate the derivative of numerical Hermitian matrices, I've recently stumbled upon this post in Math Stack Exchange. From the first answer on that post we get an expression for the derivative of the eigenvalues with respect to the matrix entries. From that we get that for a Hermitian matrix $\mathbf{H}$ parametrized by a real quantity $\varphi$, its eigenvalues $E_p(\varphi)$ and the unitary matrix that diagonalizes it $\mathbf{U}$, the following identity holds: \begin{align} \frac{\partial E_p}{\partial\varphi} &= \sum_{ij} \frac{\partial E_p}{\partial \mathbf{H}_{ij}} \frac{\partial \mathbf{H}_{ij}}{\partial \varphi} \\ &= \left[ \mathbf{U}\frac{\partial \mathbf{H}}{\partial \varphi}\mathbf{U}^{\dagger} \right]_{pp}. \end{align}

I was not able to find this information anywhere else and the references presented to the original expression in that post have a bit more complicated math that I find hard to follow.

My main question regarding it is:

Does this expression generalize to higher order derivatives? That is, does the following expression hold?

\begin{align} \frac{\partial^n E_p}{\partial\varphi^n} &\stackrel{?}{=} \left[ \mathbf{U}\frac{\partial^n \mathbf{H}}{\partial \varphi^n}\mathbf{U}^{\dagger} \right]_{pp} \end{align}

Answer 1

This is standard Rayleigh-Schrödinger perturbation theory, but described a bit differently from how it is usually done in quantum mechanics books. Here (assuming a non-degenerate spectrum) is the quickest route to the usual treatment:

Suppose that we have an eigenvector $|n\rangle$ with eigenvalue $E_n$ $$ H|n\rangle= E_n |n\rangle. $$ Then differentiate with respect to the parameter to get $$ dH |n\rangle+ Hd|n\rangle= dE_n |n\rangle +E_n d|n\rangle $$ Find matrix elements by applying $\langle m|$ to this get $$ \langle m|dH |n\rangle+ \langle m| Hd|n\rangle= dE_n \langle m|n\rangle +E_n \langle m|d|n\rangle. $$

First, take $m=n$ so $$ \langle n|dH |n\rangle+ \langle n| Hd|n\rangle= dE_n \langle n|n\rangle +E_n \langle n|d|n\rangle. $$ or $$ \langle n|dH |n\rangle+ E_n\langle n| d|n\rangle= dE_n \langle n|n\rangle +E_n \langle n|d|n\rangle, $$ so $$ \langle n|dH |n\rangle= dE_n \langle n|n\rangle. $$ This is your (and Feynman-Hellmann's) equation. Now let $m\ne n$ and assume that $d|n\rangle$ is orthogonal to $|n\rangle$ (This does not preserve normalization but this will not matter)

Now we get $$ \langle m|dH |n\rangle+ E_m \langle m| d|n\rangle= +E_n \langle m|d|n\rangle, $$ so $$ \langle m| d|n\rangle = \frac{\langle m|dH |n\rangle}{E_n-E_m}. $$ Now you know how the derivative $$ dE_n\equiv \frac{dE_n}{d\lambda}$$ of the eigenvalue and $$ d|n\rangle\equiv \frac{d}{d\lambda}|n\rangle\; $$ of the eigenvector. You can use these formulas to compute the second derivative of $E_n$ and so on. the answers rapidly get much more complicated (see the section in the cited Wikipedia article called "Second-order and higher-order corrections").

Answer 2

During the last days I have found out that my initial guess was incomplete and I'm posting here to share what I have found.

The terms with derivatives of $\mathbf{U}$ can contribute non-trivially to the expression for derivatives of order higher than one. I have calculated an expression for the second-order case. The derivation is too long for this post but final result is \begin{align} \frac{\partial^2 \epsilon_p}{\partial \varphi^2} &= \left( \mathbf{U} \frac{\partial^2 \mathbf{H}}{\partial \varphi^2} \mathbf{U}^{\dagger} \right)_{pp} + 2 \left( \mathbf{U} \frac{\partial \mathbf{H}}{\partial \varphi} \mathbf{U}^{\dagger} \mathbf{A} \right)_{pp}, \end{align} where $\mathbf{A}$ is defined by \begin{align} \mathbf{A}_{ab} &= \begin{cases} 0, \quad a=b \\ (\epsilon_b - \epsilon_a)^{-1} \left( \mathbf{U} \frac{\partial \mathbf{H}}{\partial \varphi} \mathbf{U}^{\dagger} \right)_{ab}, \quad a \neq b. \end{cases} \end{align}

I'm still looking for possible errors in the derivation and I have no idea if this method will extend for higher order derivatives. Feedback on both is welcome.