- #1

gaiussheh

- 12

- 1

- TL;DR Summary
- What is the theorem that clearly states the completeness of eigenfunctions?

I know that completeness (roughly) implies that (almost) all functions can be decomposed into a sum of eigenfunctions of a Hermitian operator. ##\psi=\sum_n \alpha_n \psi_n##. Clearly, there have to be some restrictions on the function itself, and the operator as well. But what is that?

My question arises from the fact that the eigenfunctions of a particle in a hard box model (i.e., infinite potential well) can not be used to express wavefunction that is non-zero outside of the potential well. Surely, it is because the Hamiltonian is only defined within the well, and so are the eigenfunctions. Still, if we extend the definitions to ##(-\infty,+\infty)##, like what textbooks do (i.e. define ##V[x]=+\infty ## for ##x## outside the box), it seems to be a contradiction of the completeness theorem, as the function is defined on the whole axis, and the boundary conditions are met as ##\psi -\rightarrow 0## as ##x\rightarrow \infty##, so there must be some condition that is not met. (Is it because you can't really define an operator that "equals to" ##\infty##?) Nevertheless, can you expand an arbitrary wave function using the basis of hydrogen atom orbitals? There seems to be nothing wrong with the Hamiltonian itself apart from ##x=0## is not continuous.

My question arises from the fact that the eigenfunctions of a particle in a hard box model (i.e., infinite potential well) can not be used to express wavefunction that is non-zero outside of the potential well. Surely, it is because the Hamiltonian is only defined within the well, and so are the eigenfunctions. Still, if we extend the definitions to ##(-\infty,+\infty)##, like what textbooks do (i.e. define ##V[x]=+\infty ## for ##x## outside the box), it seems to be a contradiction of the completeness theorem, as the function is defined on the whole axis, and the boundary conditions are met as ##\psi -\rightarrow 0## as ##x\rightarrow \infty##, so there must be some condition that is not met. (Is it because you can't really define an operator that "equals to" ##\infty##?) Nevertheless, can you expand an arbitrary wave function using the basis of hydrogen atom orbitals? There seems to be nothing wrong with the Hamiltonian itself apart from ##x=0## is not continuous.