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I have two concerns regarding the delta potential hamiltonian:

  1. Is my understanding that in quantum mechanics we use self-adjoint operators. However I cannot figure it out if the hamiltonian that corresponds to a 1D delta potential is a self-adjoint operator.

  2. In quantum mechanics theory of operators when we get a discrete spectrum, (solving the eigen vector equation) we asume it is complete. However, in the case of the delta potential Hamiltonian the discrete part of the spectrum is not complete (it is only one bound state)

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  • $\begingroup$ how is one bound state not complete? $\endgroup$ – ZeroTheHero Mar 26 '18 at 1:29
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In the case of a free particle there is a continuous spectrum, which is complete. With a potential well there can additionally be discrete spectrum corresponding to bound states.

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