# Do Hamiltonian operators preserve square integrability? [closed]

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

• Please provide some more details/equations, and point out where you have been stuck in attempting to prove/disprove this, @y255yan – Lelouch Jul 12 at 7:36

## 1 Answer

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.

• 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? – y255yan Jul 12 at 9:10