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How to show two self-adjoint operators (unbounded) on a Hilbert space with the same eigenvalues and eigenfunctions are the same.

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  • $\begingroup$ In general, they are not, since the eigenfunctions may or may not form a basis for the Hilbert space. $\endgroup$ – Jas Ter May 8 '15 at 10:59
  • $\begingroup$ suppose they form a basis, can we show they are same? I don't know if how to do that since they may be unbounded… $\endgroup$ – tomography May 8 '15 at 11:01
  • $\begingroup$ Are you assuming that the eigenfunctions have the same eigenvalues, or just that the sets of eigenvalues and eigenfunctions are the same? $\endgroup$ – Dom May 8 '15 at 11:04
  • $\begingroup$ same eigenvalue and same eigenspace if have multiplicity more than one. $\endgroup$ – tomography May 8 '15 at 11:05
  • $\begingroup$ @changer: you probably mean that the eigenfunctions form an orthonormal base of the Hilbert space, or the linear span of eigenfunctions is dense. $\endgroup$ – Yurii Savchuk May 8 '15 at 11:11
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That's not true. Consider the operators $T,S \colon L^2[0,1] \to L^2[0,1]$, given by $$ Tx(t) = tx(t), \quad Sx(t) = t^2x(t). $$ Both are self-adjoint, and both have no eigenvalues. But $S \ne T$, as $S1 \ne T1$.

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