Categories: Mathematics.

The **Fredholm alternative** is a theorem regarding equations involving a linear operator \(\hat{L}\) on a Hilbert space, and is useful in the context of multiple-scale perturbation theory. It is an *alternative* because it gives two mutually exclusive options, given here in Dirac notation:

- \(\hat{L} \ket{u} = \ket{f}\) has a unique solution \(\ket{u}\) for every \(\ket{f}\).
- \(\hat{L}^\dagger \ket{w} = 0\) has non-zero solutions. Then regarding \(\hat{L} \ket{u} = \ket{f}\):
- If \(\braket{w}{f} = 0\) for all \(\ket{w}\), then it has infinitely many solutions \(\ket{u}\).
- If \(\braket{w}{f} \neq 0\) for any \(\ket{w}\), then it has no solutions \(\ket{u}\).

Where \(\hat{L}^\dagger\) is the adjoint of \(\hat{L}\). In other words, \(\hat{L} \ket{u} = \ket{f}\) has non-trivial solutions if and only if for all \(\ket{w}\) (including the trivial case \(\ket{w} = 0\)) it holds that \(\braket{w}{f} = 0\).

As a specific example, if \(\hat{L}\) is a matrix and the kets are vectors, this theorem can alternatively be stated as follows using the determinant:

- If \(\mathrm{det}(\hat{L}) \neq 0\), then \(\hat{L} \vec{u} = \vec{f}\) has a unique solution \(\vec{u}\) for every \(\vec{f}\).
- If \(\mathrm{det}(\hat{L}) = 0\), then \(\hat{L}^\dagger \vec{w} = \vec{0}\) has non-zero solutions. Then regarding \(\hat{L} \vec{u} = \vec{f}\):
- If \(\vec{w} \cdot \vec{f} = 0\) for all \(\vec{w}\), then it has infinitely many solutions \(\vec{u}\).
- If \(\vec{w} \cdot \vec{f} \neq 0\) for any \(\vec{w}\), then it has no solutions \(\vec{u}\).

Consequently, the Fredholm alternative is also brought up in the context of eigenvalue problems. Define \(\hat{M} = (\hat{L} - \lambda \hat{I})\), where \(\lambda\) is an eigenvalue of \(\hat{L}\) if and only if \(\mathrm{det}(\hat{M}) = 0\). Then for the equation \(\hat{M} \ket{u} = \ket{f}\), we can say that:

- If \(\lambda\) is
*not*an eigenvalue, then there is a unique solution \(\ket{u}\) for each \(\ket{f}\). - If \(\lambda\) is an eigenvalue, then \(\hat{M}^\dagger \ket{w} = 0\) has non-zero solutions. Then:
- If \(\braket{w}{f} = 0\) for all \(\ket{w}\), then there are infinitely many solutions \(\ket{u}\).
- If \(\braket{w}{f} \neq 0\) for any \(\ket{w}\), then there are no solutions \(\ket{u}\).

- O. Bang,
*Nonlinear mathematical physics: lecture notes*, 2020, unpublished.

© Marcus R.A. Newman, a.k.a. "Prefetch".
Available under CC BY-SA 4.0.