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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).

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    $\begingroup$ Relevant and potentially a duplicate: Are Born-Oppenheimer energies analytic functions of nuclear positions? $\endgroup$
    – Emilio Pisanty
    Commented Jul 22, 2015 at 14:42
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    $\begingroup$ Just to clarify, are you only interested in smoothness or does analyticity matter? If the latter, is $\xi$ a purely real parameter or is it allowed to be complex? Note that in the complex case, the eigenstate labels are not uniquely+continuously assignable due to branch cuts, cf. Pechukas et al., Analytic structure of the eigenvalue problem as used in semiclassical theory of electronically inelastic collisions, J. Chem. Phys. 64, 1099 (1976). $\endgroup$
    – Emilio Pisanty
    Commented Jul 22, 2015 at 14:46
  • $\begingroup$ The parameters are real but the matrix elements are not. For example, a matrix element can be $a+i b$ where $a$ and $b$ are real, or one can have a matrix element $a e^{\imath \phi}$ where $a$ and $\phi$ are real. I think smoothness will suffice, since in general physical properties of a system can be calculated as derivatives of the energy with respect to the parameters. $\endgroup$
    – sintetico
    Commented Jul 22, 2015 at 15:01
  • $\begingroup$ Are you considering a finite- or infinite-dimensional system? Smoothness properties are a good bit harder in the latter. $\endgroup$
    – Emilio Pisanty
    Commented Jul 22, 2015 at 15:08
  • $\begingroup$ Let us assume the simple case, discrete energy levels. The rank $n$ of the matrix is fixed. Is it this the finite case, or you refer to something different? $\endgroup$
    – sintetico
    Commented Jul 22, 2015 at 15:11

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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.)

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  • $\begingroup$ So, the hypothesis of having a maximum multiplicity $p$ of the roots does exclude the infinite case? If one has an infinite set of eigenvalues in an interval, one can in principle have a singular point where one has an infinite degeneracy, ie, an infinite number of energy levels which are degenerate and therefore the multiplicity of the characteristic polynomial is infinite. In this case one cannot apply the theorem. Am I right? $\endgroup$
    – sintetico
    Commented Jul 22, 2015 at 15:53
  • $\begingroup$ That would be my reading of it as well. You would need to prove that you don't have infinite degeneracy for any values of $\xi$ to apply this particular theorem. This doesn't necessarily mean that a smooth system of roots doesn't exist in such a case, of course; just that you can't use this theorem to prove it. $\endgroup$
    – Michael Seifert
    Commented Jul 22, 2015 at 16:18
  • $\begingroup$ I agree. Anyway, the theorem is valid for a polynomial, and in the infinite case (continuum spectrum) the characteristic "polynomial" is not a polynomial in the strict sense...so I am not sure that in the infinite case one can assume the differentiability of the eigenvalues just by excluding the infinite degeneracy case... $\endgroup$
    – sintetico
    Commented Jul 22, 2015 at 18:00

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