- #1

Chopin

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I've been working my way through some basic quantum mechanics, and have gotten up to perturbation theory. It basically makes sense to me, but there's one thing that bothers me, and I was wondering if somebody could shed some light on it.

The essential idea behind perturbation theory is that we start with a basic Hamiltonian [tex]\hat{H}^0[/tex], and solve for its energy levels:

[tex]\hat{H}^0_n\Psi^0_n = E^0_n\Psi^0_n[/tex]

We then form a new Hamiltonian by applying a perturbation, to form [tex]\hat{H} = \hat{H}^0 + \lambda\hat{H}'}[/tex], and attempt to solve for its energy levels.

It looks as though the technique attempts to do this by assuming that the eigenfunctions of the full Hamiltonian, [tex]\Psi[/tex] can be expressed as a linear combination of the set of base eigenfunctions, [tex]\Psi^0_n[/tex], and it's here that I have a question. That assumption is true if the eigenfunctions of [tex]\hat{H}^0[/tex] form a complete basis, but is that always the case? If [tex]\hat{H}^0[/tex] is something like a potential well, then it will only have a discrete set of eigenvalues, not a continuous range. Is it still possible for a linear combination of these discrete eigenfunctions to completely span the state space?

The essential idea behind perturbation theory is that we start with a basic Hamiltonian [tex]\hat{H}^0[/tex], and solve for its energy levels:

[tex]\hat{H}^0_n\Psi^0_n = E^0_n\Psi^0_n[/tex]

We then form a new Hamiltonian by applying a perturbation, to form [tex]\hat{H} = \hat{H}^0 + \lambda\hat{H}'}[/tex], and attempt to solve for its energy levels.

It looks as though the technique attempts to do this by assuming that the eigenfunctions of the full Hamiltonian, [tex]\Psi[/tex] can be expressed as a linear combination of the set of base eigenfunctions, [tex]\Psi^0_n[/tex], and it's here that I have a question. That assumption is true if the eigenfunctions of [tex]\hat{H}^0[/tex] form a complete basis, but is that always the case? If [tex]\hat{H}^0[/tex] is something like a potential well, then it will only have a discrete set of eigenvalues, not a continuous range. Is it still possible for a linear combination of these discrete eigenfunctions to completely span the state space?

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