Eigenstructure and Spectral Decomposition

An eigenvector of a linear operator $A$ is a nonzero ket $|\psi\rangle$ that is only scaled, not rotated, by $A$. The scaling factor $\lambda$ is the corresponding eigenvalue. The equation $A|\psi\rangle = \lambda|\psi\rangle$ is the eigenvalue equation.

Eigenvectors represent special directions in the Hilbert space. In quantum mechanics, they are the states that have definite values for the observable represented by $A$. A measurement of $A$ on an eigenstate $|\lambda\rangle$ always yields the outcome $\lambda$ with certainty.

The set of all eigenvalues of $A$ is called its spectrum. The spectrum may be discrete, continuous, or a combination of both. For finite-dimensional operators, the spectrum consists entirely of eigenvalues; in infinite dimensions, the spectrum can be more subtle.

A subspace of eigenvectors sharing the same eigenvalue is called an eigenspace. The dimension of an eigenspace is the geometric multiplicity of that eigenvalue. The algebraic multiplicity is the number of times the eigenvalue appears as a root of the characteristic polynomial.

For a finite-dimensional operator $A$, the eigenvalue equation can be rewritten $(A - \lambda I)|\psi\rangle = 0$. A nonzero solution exists if and only if $A - \lambda I$ is singular, which means its determinant vanishes: $\det(A - \lambda I) = 0$.

The expression $p(\lambda) = \det(A - \lambda I)$ is a polynomial in $\lambda$ of degree n, called the characteristic polynomial. Its roots are the eigenvalues of $A$. The multiplicity of a root is the algebraic multiplicity of the eigenvalue.

For a 2x2 matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the characteristic polynomial is $\lambda^2 - (a+d)\lambda + (ad-bc) = \lambda^2 - \operatorname{tr}(A)\lambda + \det(A)$. The eigenvalues are $\lambda = (\operatorname{tr} A \pm \sqrt{(\operatorname{tr} A)^2 - 4\det A})/2$.

Once an eigenvalue $\lambda$ is known, the corresponding eigenvectors are found by solving $(A - \lambda I)|\psi\rangle = 0$. The solution space is the eigenspace $E_\lambda$.

Let $A$ be Hermitian and $|\lambda\rangle$ a normalized eigenvector with eigenvalue $\lambda$. Taking the inner product with $\langle\lambda|$ gives $\lambda = \langle\lambda|A|\lambda\rangle$. The complex conjugate of this number is $\langle\lambda|A^\dagger|\lambda\rangle^* = \langle\lambda|A|\lambda\rangle = \lambda$, so $\lambda$ is real.

This reality is essential for observables. Since measurement outcomes are real numbers, every observable must be represented by a Hermitian operator. If $A$ were not Hermitian, its eigenvalues could be complex, and the operator would not correspond to a measurable quantity.

The proof uses only the defining property $A = A^\dagger$ and the conjugate symmetry of the inner product. It applies to finite- and infinite-dimensional Hermitian operators, provided the eigenvectors lie in the domain of $A$.

The converse is not true: an operator can have real eigenvalues without being Hermitian. However, if a diagonalizable operator has a complete orthonormal set of real eigenvalues, then it must be Hermitian. This is the content of the spectral theorem.

Eigenvectors of a Hermitian operator belonging to distinct eigenvalues are orthogonal. To see this, let $A|\lambda\rangle = \lambda|\lambda\rangle$ and $A|\mu\rangle = \mu|\mu\rangle$ with $\lambda \neq \mu$. Then $\mu \langle\lambda|\mu\rangle = \langle\lambda|A|\mu\rangle = \langle\mu|A|\lambda\rangle^* = \lambda^* \langle\mu|\lambda\rangle^* = \lambda \langle\lambda|\mu\rangle$. Since $\lambda \neq \mu$, we conclude $\langle\lambda|\mu\rangle = 0$.

Orthogonality allows us to choose an orthonormal basis of eigenvectors for any Hermitian operator. In the nondegenerate case, the eigenvectors are automatically orthogonal after normalization. In the degenerate case, we use Gram-Schmidt within each eigenspace to obtain an orthonormal eigenbasis.

An orthonormal eigenbasis is the most convenient basis for representing $A$ because the matrix becomes diagonal. Diagonal matrices are easy to invert, exponentiate, and analyze. Finding this basis is called diagonalization.

The orthogonality property extends to unitary operators as well: eigenvectors belonging to distinct eigenvalues of a unitary operator are orthogonal. The eigenvalues of a unitary operator lie on the unit circle, while those of a Hermitian operator lie on the real line.

The spectral theorem for a finite-dimensional Hermitian operator $A$ states that $A = \sum_i \lambda_i |e_i\rangle\langle e_i|$, where $\{|e_i\rangle\}$ is an orthonormal eigenbasis and $\lambda_i$ are the corresponding real eigenvalues. This is the spectral decomposition of $A$.

The spectral decomposition makes the action of $A$ transparent. Applying it to an eigenstate $|e_j\rangle$ gives $\sum_i \lambda_i |e_i\rangle\langle e_i|e_j\rangle = \lambda_j |e_j\rangle$. For a general state, each component is scaled by its eigenvalue.

The decomposition also expresses $A$ as a sum of orthogonal projectors weighted by eigenvalues: $A = \sum_i \lambda_i P_i$ with $P_i = |e_i\rangle\langle e_i|$. This form is basis-independent in the sense that it depends only on the spectral projectors, not on arbitrary coordinates.

Spectral decomposition is the finite-dimensional prototype of the general spectral theorem for self-adjoint operators on Hilbert space. In infinite dimensions, sums may be replaced by integrals over projection-valued measures, but the conceptual picture remains the same.

Once $A$ is spectrally decomposed, a function $f(A)$ can be defined by applying $f$ to each eigenvalue: $f(A) = \sum_i f(\lambda_i) |e_i\rangle\langle e_i|$. This definition works for any function $f$ defined on the spectrum of $A$.

Important examples include powers $A^k$, exponentials $e^{A}$, and logarithms $\ln A$. For a Hermitian operator, $e^{iA}$ is unitary because $|e^{i\lambda_i}| = 1$. This is the origin of unitary evolution generated by a Hamiltonian: $U = e^{-iHt/\hbar}$.

If $f$ is analytic, $f(A)$ can also be computed via a power series. The spectral definition agrees with the series definition whenever the series converges. The spectral method is often simpler because it avoids infinite sums and gives closed forms.

Functions of noncommuting operators require care. In general, $f(A+B) \neq f(A)+f(B)$, and the Baker-Campbell-Hausdorff formula is needed for exponentials of sums. However, functions of a single operator are unambiguous.

The Pauli-X matrix is $\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$. Find its eigenvalues, normalized eigenvectors, write its spectral decomposition, and use it to compute $e^{i\theta\sigma_x}$.

Step 1 -- Eigenvalues. The characteristic polynomial is $\det(\sigma_x - \lambda I) = \lambda^2 - 1 = 0$, so $\lambda = \pm 1$.

Step 2 -- Eigenvectors. For $\lambda = +1$, solve $\begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}\mathbf{v} = \mathbf{0}$, giving $|+\rangle = (|0\rangle + |1\rangle)/\sqrt{2}$. For $\lambda = -1$, we obtain $|-\rangle = (|0\rangle - |1\rangle)/\sqrt{2}$.

Step 3 -- Spectral decomposition. $\sigma_x = (+1)|+\rangle\langle +| + (-1)|-\rangle\langle -| = |+\rangle\langle +| - |-\rangle\langle -|$.

Step 4 -- Exponential. Since $\sigma_x^2 = I$, $e^{i\theta\sigma_x} = \cos\theta I + i\sin\theta \sigma_x$. The spectral method gives the same result because $e^{i\theta(+1)}|+\rangle\langle +| + e^{i\theta(-1)}|-\rangle\langle -|$ expands to the cosine-sine form.