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Is Hamiltonian operator an operator that preserves the property of square integrability?

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  • $\begingroup$ Please provide some more details/equations, and point out where you have been stuck in attempting to prove/disprove this, @y255yan $\endgroup$ – Lelouch Jul 12 at 7:36
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Whether the question is, if the following expression is finite: $$ \Vert H | \psi \rangle \Vert^2= \langle \psi | H^{\dagger} H | \psi \rangle = \sum_{n, m} c_n^{*} c_m \lambda_n^{*} \lambda_m \langle \psi_n^{*} | \psi_m \rangle = \sum_{n} |c_n|^2|\lambda_n|^2 $$ Where in second equality we have assumed, that $| \psi_m \rangle$ form an orthogonal basis, then it is true, when the spectrum is bounded, or eigenvalues decay sufficiently fast, so that the series (integral in the continuum case) converges.

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  • $\begingroup$ I agree with you. Now I'm pretty confusing about what does it actually mean by saying Hamiltonian operator is a linear operator that acts on Hilbert space? Shouldn't a linear operator on Hilbert space send a square integrable function to a square integrable function? $\endgroup$ – y255yan Jul 12 at 9:10

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