Hermitian Matrices

The conjugate transpose

Before defining Hermitian matrices, recall the conjugate transpose (also called the adjoint or dagger) of a matrix. Given a matrix $A$ with entries $a_{ij}$ , its conjugate transpose $A^\dagger$ is the matrix whose $(i,j)$ entry is $\overline{a_{ji}}$ — the complex conjugate of the entry at position $(j,i)$ in the original. In two steps: transpose the matrix (swap rows and columns), then take the complex conjugate of every entry.

For a concrete $2 \times 2$ example,

A = \begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix} \implies A^\dagger = \begin{pmatrix} \overline{\alpha} & \overline{\gamma} \\ \overline{\beta} & \overline{\delta} \end{pmatrix}.

Definition of a Hermitian matrix

A square matrix $A$ is Hermitian (sometimes called self-adjoint) if it equals its own conjugate transpose:

A = A^\dagger.

Entry by entry this means $a_{ij} = \overline{a_{ji}}$ for every pair of indices $i$ and $j$ . Two immediate consequences follow from this single equation:

Diagonal entries are real. Setting $i = j$ gives $a_{ii} = \overline{a_{ii}}$ , which forces $a_{ii} \in \mathbb{R}$ .
Off-diagonal entries come in conjugate pairs. The entry above the diagonal at position $(i,j)$ is the complex conjugate of the entry below the diagonal at $(j,i)$ .

A simple $2 \times 2$ Hermitian matrix therefore has the form

H = \begin{pmatrix} r & z \\ \overline{z} & s \end{pmatrix}, \quad r, s \in \mathbb{R},\; z \in \mathbb{C}.

Why Hermitian matrices matter in quantum mechanics

In quantum mechanics every observable — every measurable physical quantity such as energy, momentum, or spin — is represented by a Hermitian operator. Three reasons make Hermitian operators the right choice:

Real eigenvalues. If $H$ is Hermitian and $H |v\rangle = \lambda |v\rangle$ , then $\lambda \in \mathbb{R}$ . The proof is a one-liner using the inner product:

\lambda \langle v | v \rangle = \langle v | H | v \rangle = \langle v | H^\dagger | v \rangle = \overline{\langle v | H | v \rangle} = \overline{\lambda} \langle v | v \rangle.

Since $\langle v | v \rangle \neq 0$ , it follows that $\lambda = \overline{\lambda}$ , so $\lambda$ is real. This is physically necessary: measurement outcomes must be real numbers.

Orthogonal eigenstates. Eigenvectors of a Hermitian matrix that correspond to different eigenvalues are orthogonal. This allows the eigenstates to form a complete orthonormal basis for the Hilbert space — the set of possible outcomes of a measurement forms a well-defined sample space.

Spectral theorem. Every Hermitian matrix on a finite-dimensional space is diagonalisable by a unitary matrix: $H = U D U^\dagger$ , where $D$ is real diagonal. This guarantees that the eigenstates of any observable provide a complete description of the system.

A worked example

Consider the matrix

A = \begin{pmatrix} 3 & 1+2i \\ 1-2i & 5 \end{pmatrix}.

Diagonal entries: $a_{11} = 3 \in \mathbb{R}$ and $a_{22} = 5 \in \mathbb{R}$ . Both are real.

Off-diagonal pair: $a_{12} = 1+2i$ and $a_{21} = 1-2i$ . The complex conjugate of $a_{21}$ is $\overline{1-2i} = 1+2i$ , which equals $a_{12}$ . The condition $a_{12} = \overline{a_{21}}$ is satisfied.

Because every condition holds, $A$ is Hermitian: $A = A^\dagger$ .

As a check, the eigenvalues of $A$ must be real. The characteristic polynomial is

\det(A - \lambda I) = (3-\lambda)(5-\lambda) - (1+2i)(1-2i) = \lambda^2 - 8\lambda + 15 - 5 = \lambda^2 - 8\lambda + 10.

The discriminant is $64 - 40 = 24 \geq 0$ , so both eigenvalues are real: $\lambda = 4 \pm \sqrt{6}$ .

Non-example

The matrix

B = \begin{pmatrix} 2i & 1 \\ 1 & 3 \end{pmatrix}

is not Hermitian because $b_{11} = 2i$ is not real. Observables are never represented by such matrices.

Try it

This is a logical exercise — your code should return a boolean, not a circuit. The matrix $A$ from the worked example above has entries $a_{11} = 3$ , $a_{12} = 1+2i$ , $a_{21} = 1-2i$ , $a_{22} = 5$ . Is $A$ Hermitian?

Run your code to see the quantum state.