In the adiabatic approximation one looks at the Hamiltonian

$$ H_0 = \sum_{i = 1}^{N_e} \frac{\vec{p}_i^2}{2m_e} + \sum_{i < j} \frac{e^2}{|\vec{r}_i - \vec{r}_j|} + \sum_{k < l} \frac{Z_k Z_l e^2}{|\vec{R}_k - \vec{R}_l|} + \sum_{i, k} \frac{- Z_k e^2}{|\vec{r}_i - \vec{R}_k|}. $$

Now typical statements are

"For a fixed ion configuration $ \{\vec{R}_1,\ldots, \vec{R}_n\}$ let $\Psi_{\alpha} \equiv \Psi_{\alpha}(\vec{r}_1,\ldots,\vec{r}_n,\vec{R}_1,\ldots\vec{R}_n)$ be a solution of the eigenvalue problem $$ H_0 \Psi_{\alpha} = \varepsilon_{\alpha}(\vec{R}_1, \ldots \vec{R}_n) \Psi_{\alpha},$$ where $\alpha$ denotes a complete set of quantum numbers."

What is baffling to me is: How does one know about the existence of such solutions? (Not only, that one finds one solution of that eigenvalue problem, but also a complete set of solutions?!)


Perhaps the answer may lie in properties of Hermitian operators. (The Hamiltonian is Hermitian). For example, the solutions (eigenfunctions) of a Hermitian operator form a complete orthogonal set. Another property of Hermitian matrices is that an $nxn$ Hermitian operator (matrix) has $n$ eigenvalues. So perhaps that answers why the solutions are guaranteed to exist.


So, the Hamiltonian of the question describe a set of atoms and electrons where the atoms are supposed to be fixed in space. There are two answer to this question.

The mathematical answer is: The existence of a complete set of solutions follows from the properties of the Hermitian operators, as stated in the other answer. One can show (it is complicated) that the Hamiltonian can be written as a matrix in the full basis of the Fock space. Therefore the completeness of the set of solutions follows from linear algebra.

The physical answer is: Look around you! Solutions of Hamiltonians like this are all around us. They are called solids. Physically, this guarantees me that solutions exists. Otherwise quantum mechanics would fail spectacularly and it would be not worth considering.

  • $\begingroup$ Why do you even need Fock space here? Nothing more than Hilbert space is needed. And your "physical answer" doesn't really answer anything at all: the Hamiltonians of this form may not necessarily be correct, so this doesn't guarantee anything. $\endgroup$
    – Ruslan
    Jun 3 '20 at 16:40
  • $\begingroup$ @Rusian The Hamiltonian is a multi body Hamiltonian and in this case it is preferable to use the multi body description within the Fock space. Of course also the Hilbert space will do. Regarding the physical answer. Yes, it may be that the Hamiltonian description is not correct. But if there are Hamiltonians like this which do not exhibits a full set of solutions, that would mean that QM is totally failing at the description of reality. Which is not true. $\endgroup$
    – sintetico
    Jun 4 '20 at 0:47

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