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Let $\Gamma$ be a metric graph with finitely many edges. Consider the operator H acting as $\frac{-d^2}{dx_e^2}$ on each edge $e$, with the domain consisting of functions that belong to $H^2(e)$ on each edge $e$ and satisfy the boundary condition $A_vF + B_vF' = 0$ at each vertex. Here $\{A_v, B_v :v \in V\}$ is a collection of matrices of sizes $d_v \times d_v$ (Here $d_v$ is the number of edges that are connected to v) such that each matrix $(A_v \; B_v)$ has the maximal rank. In order for H to be self-adjoint, the following condition at each vertex is necessary and sufficient:

$\textrm{the matrix }A_vB_v^*\textrm{ is self-adjoint.}$

This is a theorem from: P. Kuchment. Quantum Graphs I. Some Basic Structures. Texas A\& M University.

He says that proof can be found in V. Kostrykin and R. Schrader. Kirchhoff's rule in quantum wires but does someone know more direct proof? In the second paper they use maximal isotropic subspaces but I would like not to use that.

Thank you very much.

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  • $\begingroup$ These things usually boil down to exercises in (symplectic) linear algebra. The operator will be self-adjoint precisely if it is symmetric and a restriction of the maximal operator of the right dimension, as indicated by the deficiency indices. $\endgroup$
    – Christian Remling
    Commented Apr 13, 2015 at 21:12

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A more direct proof can be found in the book "Introduction to Quantum Graphs", http://www.ams.org/bookstore-getitem/item=SURV-186, Theorem 1.4.4

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