# Energy eigenfunctions in time-independent perturbation theory

• Chopin
In summary: The cardinality of a countably infinite set is larger than the cardinality of an uncountably infinite set, so there is "enough" quantized functions to completely span the space.In summary, the perturbation theory technique works by forming a new Hamiltonian by applying a perturbation, and attempting to solve for its energy levels. The technique assumes 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. However, if the Hamiltonian is something like a potential well, then it will only have a discrete set of eigenvalues
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?

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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" 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.

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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?

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Chopin said:
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?

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)

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?

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

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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?

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## What are energy eigenfunctions in time-independent perturbation theory?

Energy eigenfunctions are mathematical functions that describe the possible states of a quantum mechanical system. In time-independent perturbation theory, these eigenfunctions are used to calculate the effects of small perturbations on the system's energy levels.

## How are energy eigenfunctions calculated in time-independent perturbation theory?

Energy eigenfunctions are calculated by solving the Schrödinger equation for the unperturbed system and then using this solution to construct a new equation that includes the effect of the perturbation. This new equation can then be solved to determine the energy eigenfunctions for the perturbed system.

## What is the significance of energy eigenfunctions in time-independent perturbation theory?

Energy eigenfunctions are important because they allow us to understand how small perturbations affect the energy levels of a quantum mechanical system. This information is crucial for predicting and explaining the behavior of particles at the atomic and molecular level.

## Can energy eigenfunctions be used to predict the exact energy levels of a perturbed system?

No, energy eigenfunctions can only provide an approximation of the energy levels for a perturbed system. This is because the perturbation is usually too complex to be solved exactly, and therefore, the resulting energy eigenfunctions are also approximate solutions.

## How do energy eigenfunctions relate to the concept of superposition in time-independent perturbation theory?

In time-independent perturbation theory, energy eigenfunctions can be thought of as the "building blocks" of a perturbed system's wavefunction. The superposition of these eigenfunctions allows us to describe the overall state of the system, taking into account the effects of the perturbation.

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