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Special Function Theory

We shall now reconsider the eigenvalue problem

$\displaystyle Lu=\lambda u ~~,
$

but we take $ \lambda $ to be a degenerate eigenvalue. This means that we take $ \lambda $ to have more than one eigenvector. These eigenvectors span a subspace, the eigenspace of $ L$ corresponding to $ \lambda $ . This subspace has a basis of eigenvectors, but its choice is not unique.

In spite of this we ask: Is there a way of constructing a basis which is dictated by objective criteria (for our purposes, by geometry and/or physics) and not by subjective preferences?

The answer to this question is ``yes'' whenever one can identify a linear transformation, call it $ T$ , with the following three properties:

(i)
The domain of $ T$ coincides with that of $ L$ ,
(ii)
the transformation $ T$ commutes with $ L$ , i.e.

$\displaystyle TL=LT ~~,
$

and
(iii)
the eigenvalues of $ T$ are non-degenerate.
A transformation with these properties determines a unique eigenbasis for each eigenspace of the original eigenvalue problem. Indeed, let $ u$ be an eigenvector of $ T$ :

$\displaystyle Tu=\tau u ~~.
$

Then

$\displaystyle T(Lu)=L(Tu)=\tau~Lu~~,
$

i.e., $ Lu$ is again an eigenvector of $ T$ corresponding to the same eigenvalue $ \tau$ . The non-degeneracy of $ \tau$ implies that $ Lu$ is a multiple of $ u$ ; in other words,

$\displaystyle Lu=\lambda u ~~.
$

Thus $ u$ is also an eigenvector of $ L$ . Conversely, if $ u$ belongs to the $ \lambda $ -eigenspace of $ L$ , then $ Tu$ also belongs to this subspace. The set of all the eigenvectors of $ T$ which lie in this $ \lambda $ -subspace form a basis for this subspace. This basis is orthonormal if $ T$ and $ L$ are hermitian. The elements of this $ T$ -determined basis are uniquely labelled by the real eigenvalues $ \tau$ and, of course, by the subspace label $ \lambda $ . A set of commuting linear transformations, such as $ L$ and $ T$ , whose eigenvalues uniquely label their common eigenvectors, is called a complete set of commuting operators.

The operator $ T$ is not unique. Suppose there is another hermitian operator, say $ S$ , which together with $ L$ forms another complete set of commuting operators. This means that one now has two orthonormal bases for the $ \lambda $ -eigenspace of $ L$ , one consisting of the eigenvectors of $ T$ , the second consisting of the eigenvectors of $ S$ . Furthermore, these two bases are related by a unitary transformation, i.e. by a rotation in the complex eigenspace of $ L$ .

One of the most far reaching applications of this geometrical framework consists of identifying



Subsections
next up previous contents index
Next: The Helmholtz Equation Up: LINEAR MATHEMATICS IN INFINITE Previous: Fourier Sine Theorem   Contents   Index
Ulrich Gerlach 2010-12-09