Smoothness of the energy levels of a generic Hamiltonian Let us take an Hamiltonian $H(\xi)$ which depends on a set of parameters $\xi$, and assume that the matrix elements $h_{ij}(\xi)$ of the Hamiltonian are smooth complex functions of the parameters $\xi$ (i.e., each $h_{ij}(\xi)$ has continuous derivatives of any order in the parameters $\xi$). Assume also that such parameters are real ($h_{ij}(\xi)$ are complex in general) and each of them can vary on the whole real line $\mathbb{R}$, and that the Hamiltonian has a fixed rank $n$. 
Under these assumptions, the eigenvalues $E_i(\xi)$ of the Hamiltonian $H(\xi)$ can still have a discontinuous derivative in the presence of level crossings.
However, is it always possible to "label" the eigenvalues in such a way that all eigenvalues are smooth in the parameters? (or at least continuous with continuous derivative)
If this is not the case, is it at least the total energy a smooth function of the parameter $\xi$, under certain assumptions? 
One can define the total energy as $E_T(\xi)=\sum_{E_i<E_F} E_i(\xi)$, i.e., the sum of the energy levels below a certain threshold $E_F$ (e.g., Fermi energy in a fermion system), and assume that $E_i(\xi)\neq E_f$ (if any energy level crosses the energy $E_F$ it is clear that the total energy can have a discontinuous derivative).
 A: If we're dealing with a finite-dimensional Hilbert space, then the answer appears to be that you can always find a way to label the eigenvalues such that they are all differentiable (at least).  See this paper and references therein:

A. Parusinski & A. Rainer, "A New Proof of Bronshtein's Theorem".

In particular, Theorem 2.4 of the paper states that if you have a one-parameter $C^p$ family of hyperbolic polynomials, where $p$ is the maximum multiplicity of the roots, then there exists a "differentiable system of the roots", i.e., a way of labelling the roots by differentiable functions of $\xi$.  (A "hyperbolic" polynomial is one whose roots are all real.)  Since the eigenvalues are the roots of the characteristic polynomial of $H$, and you're assuming that the entries of $H$ are $C^\infty$ in $\xi$, the characteristic polynomial is also $C^\infty$ in $\xi$.  This implies that there exists a set of functions $E_i(\xi)$ that is differentiable and which are always equal to the eigenvalues of $H(\xi)$.
In fact, the authors' note about ref. [7] on p. 1 of their paper implies that the functions $E_i(\xi)$ can be chosen to be twice-differentiable as well.  The article is silent, however, on whether the $E_i(\xi)$ can be chosen to be smooth.  The authors do note (near the top of p. 2) that stronger conclusions can be drawn if stronger assumptions are made, and give a list of references;  you might try looking at those papers as well (unfortunately, I don't have access to them here.)
(Tip of the hat to this Math StackExchange answer for pointing me towards Bronshtein's theorem.)
