2005
DOI: 10.1088/0305-4470/38/22/014
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Inverse spectral problem for quantum graphs

Abstract: The inverse spectral problem for the Laplace operator on a finite metric graph is investigated. It is shown that this problem has a unique solution for graphs with rationally independent edges and without vertices having valence 2. To prove the result, a trace formula connecting the spectrum of the Laplace operator with the set of periodic orbits for the metric graph is established.

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Cited by 137 publications
(142 citation statements)
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“…Actually, it was noted in [28] (see also [17,14]) that the multiplicity of λ = 0 as a root of (2.6) can be different from its multiplicity as an eigenvalue of the Laplace operator, but that for all positive eigenvalues the multiplicities coincide. To avoid this ambiguity we will omit the zero eigenvalues from the spectra that we consider; i.e.…”
Section: 2mentioning
confidence: 99%
“…Actually, it was noted in [28] (see also [17,14]) that the multiplicity of λ = 0 as a root of (2.6) can be different from its multiplicity as an eigenvalue of the Laplace operator, but that for all positive eigenvalues the multiplicities coincide. To avoid this ambiguity we will omit the zero eigenvalues from the spectra that we consider; i.e.…”
Section: 2mentioning
confidence: 99%
“…Topological characteristics of the graph may be reconstructed [19,20]. Assuming that the edge lengths are rationally independent one may even reconstruct the graph using the trace formula [15,18], but under the condition that the potential is zero. Explicit examples of isospectral graphs have been constructed [4,5,15].…”
Section: Spectral Estimates and Inverse Spectral Theorymentioning
confidence: 99%
“…All nonzero eigenvalues of L can be calculated using the vertex and edge scattering matrices [3,5,6,7,10]. The matrix S, and therefore the matrix S φ 1 ,φ 2 = DSD −1 , appearing in the vertex conditions is not only unitary, but also Hermitian.…”
Section: Getting Startedmentioning
confidence: 99%
“…If one of the fluxes is zero, then k = 0 is not a solution to the secular equation. It follows that the so-called algebraic multiplicity 1 m a (0) [10,6] of the zero eigenvalue is zero. Let us turn to calculation of the spectral multiplicity m s (0) -the number of linearly independent solutions to the equation Lψ = 0.…”
Section: Getting Startedmentioning
confidence: 99%