Eigenvalue Problems

An eigenvalue problem asks for nonzero states or functions that are rescaled by an operator:

$$A u=\lambda u.$$

The nonzero solutions $u$ are eigenvectors or eigenfunctions, and the numbers $\lambda$ are eigenvalues. In quantum mechanics, the most important example is the stationary Schrödinger equation

$$H\psi=E\psi,$$

where $H$ is the Hamiltonian, $E$ is an allowed energy, and $\psi$ is an energy eigenstate when it is an acceptable state.

This page owns the mathematical problem type. The finite-dimensional algebraic definition is treated in Eigenvalues and Eigenvectors, while the full self-adjoint-operator structure is treated in Spectral Theorem, Practical Version.

Operator, Domain, and Data

The symbolic equation $A u=\lambda u$ is incomplete until the following are specified:

• the space where $u$ lives;
• the action of $A$;
• the domain on which $A$ is allowed to act;
• any boundary, regularity, decay, or matching conditions.

For a matrix acting on $\mathbb C^n$, this data is usually implicit. For a differential operator, it is not. The expression

$$-\frac{d^2}{dx^2}$$

has different spectra on the real line, on a finite interval with Dirichlet boundary conditions, on a ring with periodic boundary conditions, or on a half-line with a boundary condition at the endpoint.

This is why Boundary Conditions and Domains of Operators are not optional refinements. They are part of the eigenvalue problem.

Finite-Dimensional Model

For an $n\times n$ matrix $A$, the equation

$$Av=\lambda v$$

has a nonzero solution only when

$$\det(A-\lambda I)=0.$$

This characteristic equation gives the eigenvalues. For each eigenvalue, the eigenspace is the null space of $A-\lambda I$:

$$\ker(A-\lambda I) = \{v:(A-\lambda I)v=0\}.$$

If $A$ is Hermitian, its eigenvalues are real and eigenspaces belonging to distinct eigenvalues are orthogonal. When enough eigenvectors exist to form a basis, the operator can be represented diagonally. The finite-dimensional projector form is Spectral Decomposition.

Boundary-Value Eigenproblems

A differential eigenvalue problem combines a differential equation with admissibility conditions. A simple model is

$$-u''(x)=\lambda u(x), \qquad 0<x<L,$$

with

$$u(0)=u(L)=0.$$

The general solution for $\lambda=k^2>0$ is

$$u(x)=A\sin(kx)+B\cos(kx).$$

The first boundary condition gives $B=0$. The second gives

$$\sin(kL)=0,$$

so

$$k=\frac{n\pi}{L}, \qquad \lambda_n=\frac{n^2\pi^2}{L^2}, \qquad n=1,2,3,\ldots.$$

The boundary conditions have selected a discrete set of allowed values. Without those conditions, the differential equation alone would not determine the same spectrum.

For a particle of mass $m$ in an infinite square well, this becomes

$$E_n = \frac{\hbar^2}{2m}\lambda_n = \frac{n^2\pi^2\hbar^2}{2mL^2}.$$

Generalized Eigenvalue Problems

Some problems have the form

$$A u=\lambda B u,$$

where $B$ is not the identity. In Sturm–Liouville theory, the standard form is

$$-\frac{d}{dx} \left[p(x)\frac{du}{dx}\right] +q(x)u = \lambda w(x)u.$$

The weight $w(x)$ changes the inner product used for orthogonality:

$$\langle f,g\rangle_w = \int_a^b f(x)^*g(x)w(x)\,dx.$$

The specialized theory, including weight functions and completeness intuition, is Sturm–Liouville Theory.

Many multidimensional eigenvalue problems first become one-dimensional eigenvalue problems by Separation of Variables. Green Functions package the inverse or resolvent of an eigenvalue problem and make poles, boundary conditions, and response to sources explicit.

Discrete, Continuous, and Mixed Spectra

In finite dimension, eigenvalues form a finite set. In differential-operator problems, several spectral patterns can occur.

A discrete spectrum has isolated eigenvalues with normalizable eigenfunctions. Bound states in an infinite square well or harmonic oscillator are standard examples:

$$H\psi_n=E_n\psi_n, \qquad \lVert\psi_n\rVert=1.$$

A continuous spectrum is not usually described by normalizable eigenfunctions. For a free particle on the line, the formal plane waves

$$e^{ikx}$$

satisfy the eigenvalue equation for kinetic energy, but they are not in $L^2(\mathbb R)$. They are best understood through spectral representations, wave packets, or generalized eigenfunctions.

Many Hamiltonians have mixed spectra: bound states at isolated energies and scattering states in a continuum. Continuous Spectra explains why continuum labels require more care than a large sum over ordinary eigenvectors.

Self-Adjointness and Reality of Spectrum

In quantum mechanics, observables are represented by self-adjoint operators. For eigenvalue problems this matters because self-adjointness gives:

• real spectral values;
• orthogonality of eigenfunctions associated with distinct discrete eigenvalues;
• unitary time evolution for Hamiltonians;
• the projection-valued spectral structure used for probabilities.

Formal symmetry is not enough in infinite-dimensional problems. Boundary conditions can make a differential expression self-adjoint, fail to make it self-adjoint, or define different self-adjoint operators with different spectra. The distinction is developed in Symmetric versus Self-Adjoint Operators.

Degeneracy

An eigenvalue is degenerate when the eigenspace has dimension greater than one. In finite dimensions, degeneracy means

$$\dim\ker(A-\lambda I)>1.$$

In quantum mechanics, degeneracy often reflects symmetry. For example, rotational symmetry can make several angular-momentum states share one energy. Degenerate eigenspaces should be treated as subspaces, not as a single preferred eigenvector. The correct measurement object is usually the projector onto the whole eigenspace.

Solving Strategy

For a practical eigenvalue problem:

1. Specify the space, operator, domain, and boundary conditions.
2. Solve the differential or algebraic equation for a general spectral parameter.
3. Impose boundary, matching, regularity, and normalizability conditions.
4. Identify the allowed spectral values and eigenspaces.
5. Normalize normalizable eigenfunctions using the correct inner product.
6. Decide whether the spectrum is discrete, continuous, or mixed.
7. Use the spectral decomposition or spectral theorem appropriate to the problem.

For numerical work, a discretized differential operator turns the problem into a matrix eigenvalue problem, but the discretization should preserve the essential boundary conditions and self-adjoint structure. See Matrix Diagonalization.

Common Mistakes

• Solving only the differential equation and treating every value of $\lambda$ as allowed.
• Forgetting that the zero function is not an eigenfunction.
• Confusing a formal plane wave with a normalizable eigenstate.
• Applying finite-dimensional diagonalization intuition to continuous spectra without the spectral theorem.
• Changing boundary conditions without recognizing that the operator and spectrum have changed.
• Ignoring degeneracy and summing over arbitrary eigenvectors when a projector onto the eigenspace is the invariant object.
• Treating self-adjointness as a technicality rather than part of the physical definition of an observable.

Cross-Links

• Eigenvalues and Eigenvectors
• Time-Independent Schrödinger Equation
• Boundary Conditions
• Domains of Operators
• Spectral Decomposition
• Spectral Theorem, Practical Version
• Continuous Spectra
• Sturm–Liouville Theory
• Separation of Variables
• Green Functions

References

• G. Teschl, Mathematical Methods in Quantum Mechanics, 2nd ed., American Mathematical Society, 2014.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• G. B. Arfken, H. J. Weber, and F. E. Harris, Mathematical Methods for Physicists, 7th ed., Academic Press, 2013.
• S. Axler, Linear Algebra Done Right, 3rd ed., Springer, 2015.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.

Exercises

1. For

$$A= \begin{pmatrix} 1 & 0\\ 0 & 4 \end{pmatrix},$$

find the eigenvalues and eigenspaces.

Solution

The eigenvalues are $1$ and $4$. The eigenspace for $\lambda=1$ is the span of $(1,0)^T$, and the eigenspace for $\lambda=4$ is the span of $(0,1)^T$.

2. Solve the eigenvalue problem

$$-u''=\lambda u, \qquad u(0)=u(L)=0.$$

Why is $\lambda=0$ not an allowed eigenvalue?

Solution

For $\lambda=k^2>0$, the solution satisfying $u(0)=0$ is $u(x)=A\sin(kx)$. The condition $u(L)=0$ requires $kL=n\pi$, so

$$\lambda_n=\frac{n^2\pi^2}{L^2}, \qquad n=1,2,3,\ldots.$$

For $\lambda=0$, the equation gives $u(x)=Ax+B$. The two boundary conditions force $A=B=0$, leaving only the zero function, which is not an eigenfunction.

3. Explain why changing from Dirichlet to periodic boundary conditions changes the eigenvalue problem, even if the differential expression is still $-d^2/dx^2$.

Solution

The operator is not just the differential expression. Its domain includes the boundary conditions. Dirichlet conditions require $u(0)=u(L)=0$, while periodic conditions identify endpoint values and derivatives. These domains define different self-adjoint operators and generally produce different spectra.

4. Why are free-particle plane waves not ordinary normalized eigenvectors in $L^2(\mathbb R)$?

Solution

The function $e^{ikx}$ has constant modulus, so

$$\int_{-\infty}^{\infty} \lvert e^{ikx}\rvert^2\,dx = \infty.$$

It is not square-integrable. Plane waves are useful as generalized eigenfunctions or as components in wave-packet and spectral representations, not as ordinary normalized Hilbert-space vectors.