# Energy eigenfunctions in time-independent perturbation theory

1. Sep 26, 2010

### Chopin

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 $$\hat{H}^0$$, and solve for its energy levels:

$$\hat{H}^0_n\Psi^0_n = E^0_n\Psi^0_n$$

We then form a new Hamiltonian by applying a perturbation, to form $$\hat{H} = \hat{H}^0 + \lambda\hat{H}'}$$, 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, $$\Psi$$ can be expressed as a linear combination of the set of base eigenfunctions, $$\Psi^0_n$$, and it's here that I have a question. That assumption is true if the eigenfunctions of $$\hat{H}^0$$ form a complete basis, but is that always the case? If $$\hat{H}^0$$ 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?

Last edited: Sep 27, 2010
2. Sep 27, 2010

### alxm

Yes, the eigenfunctions always form a complete orthonormal set.

If you take the 1D infinite potential well as an example, you'll recall the time-independent energy eigenfunctions are:
$$\psi_n = \sin(\frac{nx\pi}{L})$$ (or cos, depending on if you had an even or odd potential)

Well, can you represent any well-behaved (non-divergent, piecewise-continuous, etc) function with a linear combination of these?
Sure - that's just a http://en.wikipedia.org/wiki/Fourier_series" [Broken] expansion, isn't it?

Now, naturally that equality only holds if you have an expansion with an infinite number of terms.
But if your perturbation is small, the difference in the eigenfunctions will be small, and so the series will hopefully converge relatively quickly.

Last edited by a moderator: May 4, 2017
3. Sep 27, 2010

### Chopin

Sure, but that only works for functions that are periodic in $$L$$. What if my perturbation widens the potential well? The functions will now be periodic in $$L+\Delta L$$, so can the original set of eigenfunctions still represent these?

In free space, we need a complete continuum of energy or momentum eigenstates to span the space, so how can a discrete subset of these suffice to do the same job? Or does the combination of base eigenfunctions only approximate the full eigenfunctions?

Last edited: Sep 27, 2010
4. Sep 27, 2010

### alxm

It works for any periodic function; a Fourier series expansion doesn't require the function to be periodic on the same interval as the basis. Your expansion coefficients for any odd function $$f(x)$$ are:

$$c_n = \frac{2}{L}\int_0^L f(x)\sin(\frac{nx\pi}{L})dx$$

This is just your basic Fourier sine series. Now if you interpret that in a QM framework, your new wave function can be written as an expansion of its eigenfunctions:
$$\psi' = \sum_n c_n\sin(\frac{nx\pi}{L+\Delta L})$$

If the box expands instantaneously, the wave-function is (at that point in time) equal to $$\psi_0$$, the wave-function in the original box, and your Fourier series coefficients:
$$c_n = \frac{2}{L+\Delta L}\int_0^{L+\Delta L} \psi_0(x)\sin(\frac{nx\pi}{L+\Delta L})dx$$

Are just the populations of your new states, and equal to the overlap of the original wave function and the new states.
(Edit: Oops, you were asking about the opposite representation. Well, it's still the same math, point is just that the function need not be periodic on the interval you're integrating over)

A plane wave solution isn't 'well-behaved'; as it's not integrable. Nor is it a valid wave function.

Last edited: Sep 27, 2010
5. Sep 27, 2010

### Chopin

Ok, that makes sense. But that's for an infinite potential well, where the eigenfunctions just happen to form a nice set of Fourier components. What if the Hamiltonian is something more complicated, like a Coloumb potential or something like that? Now the eigenfunctions are more irregular. Will they still span the space? Or more generally, will the eigenfunctions of any Hermitian operator form a complete basis for the state space?

I guess what makes me uncomfortable about this is the fact that if you take, say, the x-representation, you can span the space with delta functions, but you need an uncountably infinite number to do so. If we move from that to quantized energy levels, that's only a countably infinite set. I know infinities are weird, but since the cardinality of a countably infinite set is smaller than that of an uncountably infinite set, intuitively it seems like there wouldn't be "enough" quantized functions to completely span the space, like if you tried to span a three dimensional vector space with only two vectors. Presumably there's something wrong with my reasoning here--can you point out what it is?

Last edited: Sep 27, 2010