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Completeness of the eigenfunctions (Which vectorspace?)

  1. Mar 16, 2014 #1
    Completeness of the eigenfunctions (Which vectorspace??)

    Once again in need of brain power from the interwebz :)

    So I get that the eigenfunctions to the hamiltonoperator forms a complete set, but I'm unsure now as to which vectorspace it is?

    And we're talking the one-dimensional case!

    My first guess would be that it is all the functions that are summable? But that doesn't completely satisfy me, since I know from Fourier Series and Fourier Transform know that any summable function can be written as an infinite sum of complex exponentials, and the eigenfunctions to the hamiltonian are a finite set?

    Maybe someone can give me the hints of what I need to look at, to get a full understanding?

    Right now I'm working with perturbation theory, and I'm trying to reconcile myself with the fact that the the first order correction to the wavefunction can be written as a linear combination of the solutions (eigenfunctions) to the unperturbed hamiltonoperator.

    Thank you.
     
  2. jcsd
  3. Mar 16, 2014 #2
    The energy eigenfunctions form a basis for the vector space of the wave functions. Wave functions are in general complex and they are quadratically integrable.
     
  4. Mar 16, 2014 #3

    WannabeNewton

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    No they aren't, at least not in general. If we consider an infinite square well with the usual Hamiltonian ##\hat{H} = -\frac{\hbar^2}{2m}\nabla^2 + V(x)## then the eigenfunctions ##|n \rangle## will range over all ##n \in \mathbb{N}## with periodic boundary conditions at the ends of the well resulting in ##\langle x |n \rangle \propto \sin(\frac{n\pi x}{a})##. Here we have a countably infinite set of energy eigenfunctions. For a free particle, ##\hat{H} = -\frac{\hbar^2}{2m}\nabla^2##, we have the eigenfunctions ##\langle x | \psi_k \rangle \propto e^{ikx}## ranging over all real ##k## so we have a continuous energy spectrum.

    On the other hand, say we have a spinning electron at rest (meaning it is in a momentum eigenstate and we boost to the frame in which the momentum eigenvalue vanishes) in a uniform magnetic field ##\vec{B} = B \hat{z}## relative to the axes of a coordinate system and we are interested only in the spin evolution of the electron. Then the Hamiltonian ##\hat{H} = \propto B S_z## corresponds to that of the spin component of our system. It is time-independent and its eigenfunctions are just those of ##S_z## i.e. the two-component spinors. So here the eigenfunctions constitute a finite complete set.

    In most cases the underlying vector space will just be the Hilbert space of square-integrable functions. However this isn't always true e.g. in the case of the free particle the space is actually an associated Rigged Hilbert space.
     
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