Hermitian Operators

Of all the operator families in quantum mechanics, Hermitian operators are the ones you measure. Every physical observable — energy, position, momentum, spin component — is represented by a Hermitian operator, and the reason is a single clean theorem: Hermitian operators have real eigenvalues. Measurement outcomes are real numbers, so the operators that produce them must be Hermitian.

Definition

An operator $A$ is Hermitian (or self-adjoint in finite dimensions) if it equals its own adjoint:

In terms of matrix elements in any orthonormal basis, using $(A^\dagger)_{ij} = \overline{A_{ji}}$ from the previous lesson, this reads

Two immediate consequences of this symmetry:

• The diagonal entries are real. Setting $i = j$ gives $A_{ii} = \overline{A_{ii}}$, which forces $A_{ii}$ to be real.
• Off-diagonal entries are conjugate-reflections across the diagonal: $A_{ij}$ and $A_{ji}$ are complex conjugates of one another.

For example,

is Hermitian: the diagonal is real, and $\overline{1+i} = 1-i$ matches the reflected entry. The three Pauli matrices $X, Y, Z$ are all Hermitian, as is the identity.

Hermitian operators have real eigenvalues

This is the property that makes them physical. Suppose $A|\psi\rangle = \lambda|\psi\rangle$ with $|\psi\rangle \neq 0$. Take the inner product with $\langle\psi|$:

Now take the complex conjugate of the left side. Using $\langle\psi|A|\psi\rangle = \overline{\langle\psi|A^\dagger|\psi\rangle}$ and $A^\dagger = A$,

so $\langle\psi|A|\psi\rangle$ is real. Since $\langle\psi|\psi\rangle > 0$ is also real and positive, $\lambda = \langle\psi|A|\psi\rangle / \langle\psi|\psi\rangle$ is real. Every eigenvalue of a Hermitian operator is real.

Eigenvectors for distinct eigenvalues are orthogonal

A second key fact: if $A|\psi\rangle = \lambda|\psi\rangle$ and $A|\phi\rangle = \mu|\phi\rangle$ with $\lambda \neq \mu$ (both real), then $\langle\phi|\psi\rangle = 0$. Compute $\langle\phi|A|\psi\rangle$ two ways:

where the second line uses $A = A^\dagger$ and the reality of $\mu$. Subtracting, $(\lambda - \mu) \langle\phi|\psi\rangle = 0$, and since $\lambda \neq \mu$ we conclude $\langle\phi|\psi\rangle = 0$. Eigenstates with different eigenvalues are automatically orthogonal — the seed of the spectral theorem two lessons ahead.

Hermitian versus symmetric

Over the real numbers, "Hermitian" reduces to symmetric ($A^\top = A$): there is nothing to conjugate. The conjugation only does work for complex matrices. A real symmetric matrix is therefore a special case of a Hermitian matrix, which is why real-valued classical problems (vibrations, networks) inherit the same real-spectrum, orthogonal-eigenvector structure.

Try it

Decide whether the operator $A = \begin{pmatrix} 2 & 1-i \\ 1+i & 3 \end{pmatrix}$ is Hermitian by checking $A_{ij} = \overline{A_{ji}}$ for every entry. Return a boolean.

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