Completeness of the eigenfunctions (Which vectorspace?)

In summary, the completeness of the eigenfunctions of the Hamiltonian operator means that they can form a complete set in the vector space of wave functions. This vector space is usually the Hilbert space of square-integrable functions, but in some cases it may be a Rigged Hilbert space. The eigenfunctions can either be a finite or infinite set, depending on the specific system.
  • #1
Runei
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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.
 
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  • #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.
 
  • #3
Runei said:
...and the eigenfunctions to the hamiltonian are a finite set?

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.
 

FAQ: Completeness of the eigenfunctions (Which vectorspace?)

1. What is meant by the completeness of eigenfunctions?

The completeness of eigenfunctions refers to the ability of a set of eigenfunctions to fully describe a vector space. This means that any function within the vector space can be expressed as a linear combination of the eigenfunctions.

2. Which vector space is being referred to in the completeness of eigenfunctions?

In the context of quantum mechanics, the vector space being referred to is the space of all possible states of a physical system. This can also be referred to as the Hilbert space.

3. How is the completeness of eigenfunctions related to the eigenvalue equation?

The completeness of eigenfunctions is closely related to the eigenvalue equation, which states that a linear operator acting on an eigenfunction yields a scalar multiple of that eigenfunction. The eigenfunctions form a basis for the vector space, allowing for the completeness of the space.

4. Why is the completeness of eigenfunctions important in quantum mechanics?

The completeness of eigenfunctions is important in quantum mechanics because it allows for the mathematical representation of physical systems and their properties. It also allows for the calculation of probabilities and predictions of the behavior of quantum systems.

5. Can a set of eigenfunctions be complete for one vector space but not for another?

Yes, a set of eigenfunctions can be complete for one vector space but not for another. This is because the completeness of eigenfunctions is dependent on the specific vector space and the properties of the physical system being studied. The eigenfunctions must be chosen carefully to ensure completeness for a particular vector space.

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