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Hermitian Matrices and Normal Operators

Source lecture(s): PHY621 Ch. 2

Intuition

Physics needs operators whose eigenvalues are real (they are measured quantities) and whose eigenvectors form a complete orthogonal set (you want to expand any state in them). Hermiticity delivers both, and it is not a coincidence — it is the reason quantum mechanics insists observables be Hermitian.

\[A = A^\dagger \quad\text{i.e.}\quad A_{ij} = A_{ji}^*\]

Two theorems, both short

Eigenvalues are real. For \(A\mathbf{v} = \lambda\mathbf{v}\) with \(\mathbf{v}\neq0\):

\[\lambda\langle\mathbf{v},\mathbf{v}\rangle = \langle\mathbf{v},A\mathbf{v}\rangle = \langle A\mathbf{v},\mathbf{v}\rangle = \bar\lambda\langle\mathbf{v},\mathbf{v}\rangle\]

Since \(\langle\mathbf{v},\mathbf{v}\rangle > 0\), \(\lambda = \bar\lambda\). ∎

Eigenvectors for distinct eigenvalues are orthogonal. For \(A\mathbf{u} = \lambda\mathbf{u}\), \(A\mathbf{v} = \mu\mathbf{v}\) with \(\lambda\neq\mu\):

\[\lambda\langle\mathbf{v},\mathbf{u}\rangle = \langle\mathbf{v},A\mathbf{u}\rangle = \langle A\mathbf{v},\mathbf{u}\rangle = \mu\langle\mathbf{v},\mathbf{u}\rangle\]

so \((\lambda-\mu)\langle\mathbf{v},\mathbf{u}\rangle = 0\), forcing \(\langle\mathbf{v},\mathbf{u}\rangle = 0\). ∎

Both proofs are the same move — push \(A\) across the inner product — which is what "self-adjoint" means and why the inner product is doing the real work here.

Degeneracy and Gram–Schmidt

If two eigenvalues coincide, the argument above says nothing. But the eigenvectors of a repeated eigenvalue span a subspace, and any basis of that subspace consists of eigenvectors — so you may simply choose an orthogonal one by Gram–Schmidt. Orthogonality is always achievable, never automatic within a degenerate subspace.

Physically, degeneracy usually signals a symmetry, and the freedom to rotate within the eigenspace is the freedom to label states by that symmetry's quantum numbers.

Completeness and spectral decomposition

Together these give the spectral theorem: a Hermitian \(A\) on an \(n\)-dimensional space has an orthonormal eigenbasis, and

\[A = \sum_n \lambda_n|\mathbf{v}_n\rangle\langle\mathbf{v}_n|, \qquad \hat 1 = \sum_n |\mathbf{v}_n\rangle\langle\mathbf{v}_n|\]

The operator is its eigenvalues, weighted by projectors. Everything follows: functions of operators are defined by \(f(A) = \sum f(\lambda_n)|\mathbf{v}_n\rangle\langle\mathbf{v}_n|\), so \(e^{iAt}\), \(A^{-1}\) and \(\sqrt{A}\) all become elementary once you have diagonalised.

The wider family: normal matrices

Hermitian is sufficient for a complete orthonormal eigenbasis but not necessary. The precise condition is normality, \(AA^\dagger = A^\dagger A\):

Type Condition Eigenvalues Physics
Hermitian \(A = A^\dagger\) real observables
Anti-Hermitian \(A = -A^\dagger\) pure imaginary generators of unitary evolution
Unitary \(A^\dagger = A^{-1}\) on the unit circle symmetries, time evolution
Normal \(AA^\dagger = A^\dagger A\) any complex all of the above

All are normal, hence all are unitarily diagonalisable. A matrix that is not normal may fail to be diagonalisable at all — see diagonalization.

Common mistakes

  • Confusing Hermitian with symmetric. They coincide only for real matrices. A complex symmetric matrix (\(A = A^{\mathsf T}\), no conjugate) is generally not Hermitian and can have complex eigenvalues.
  • Expecting orthogonality within a degenerate subspace automatically. You must construct it.
  • Assuming real eigenvalues imply Hermitian. The converse is false — a non-normal matrix can have real eigenvalues without orthogonal eigenvectors.
  • Forgetting which inner product. "Self-adjoint" is relative to an inner product; in the normal-modes problem the relevant one is mass-weighted, and the matrix that matters is \(A^{-1/2}BA^{-1/2}\).

Knowledge graph position

Prerequisites: inner product spaces, eigenvalues. Leads to: diagonalization, normal modes, Sturm–Liouville theory, quantum mechanics.

Quiz

Q1 (conceptual). Why does Hermiticity guarantee real eigenvalues and orthogonal eigenvectors with essentially the same proof?

Answer

Both follow from being able to move \(A\) across the inner product: \(\langle\mathbf{v},A\mathbf{u}\rangle = \langle A\mathbf{v},\mathbf{u}\rangle\). Setting \(\mathbf{u} = \mathbf{v}\) gives \(\lambda = \bar\lambda\); taking two different eigenvectors gives \((\lambda-\mu)\langle\mathbf{v},\mathbf{u}\rangle = 0\). One property, two corollaries.

Q2 (computational). Is \(A = \begin{pmatrix}1 & i\\ i & 1\end{pmatrix}\) Hermitian? Symmetric?

Answer

Symmetric (\(A = A^{\mathsf T}\)) but not Hermitian: \(A^\dagger = \begin{pmatrix}1&-i\\-i&1\end{pmatrix} \neq A\). Its eigenvalues are \(1\pm i\) — complex, as non-Hermitian permits. A clean demonstration that symmetry is not enough over \(\mathbb{C}\).

Q3 (MCQ). The property that guarantees a complete orthonormal eigenbasis is:

  • (a) being symmetric
  • (b) being normal, \(AA^\dagger = A^\dagger A\)
  • (c) having distinct eigenvalues
  • (d) being invertible
Answer

(b). Hermitian, anti-Hermitian and unitary are all special cases of normal. Distinct eigenvalues (c) guarantee independence of eigenvectors but not orthogonality unless the matrix is also normal.