Hermitian Operators

Every physical quantity you can measure — position, momentum, energy, spin — is represented in quantum mechanics by a linear operator acting on the state space. Not every operator is acceptable, however. The special requirement is Hermiticity.

The adjoint of an operator

Given an operator $\hat{A}$ on a complex vector space, its adjoint (also called the Hermitian conjugate) $\hat{A}^\dagger$ is defined by the condition

\langle \psi | \hat{A} \phi \rangle = \langle \hat{A}^\dagger \psi | \phi \rangle

for all states $|\psi\rangle$ and $|\phi\rangle$ . For a finite-dimensional matrix, taking the adjoint means transposing and complex-conjugating every entry:

(A^\dagger)_{ij} = A_{ji}^*.

An operator is Hermitian (or self-adjoint) when $\hat{A}^\dagger = \hat{A}$ , meaning the matrix is unchanged by this operation.

Why observables must be Hermitian

There are two complementary reasons, and together they make Hermiticity unavoidable.

Reason 1 — real expectation values. A measurement result is a real number. The expectation value of $\hat{A}$ in state $|\psi\rangle$ is

\langle \hat{A} \rangle = \langle \psi | \hat{A} | \psi \rangle.

For this to be real we need $\langle \hat{A} \rangle = \langle \hat{A} \rangle^*$ . Working out the complex conjugate of a matrix element:

\langle \psi | \hat{A} | \psi \rangle^* = \langle \psi | \hat{A}^\dagger | \psi \rangle.

Reality therefore demands $\hat{A} = \hat{A}^\dagger$ , i.e. Hermiticity.

Reason 2 — real eigenvalues. The eigenvalues of $\hat{A}$ are the only outcomes a measurement can return (the spectral postulate). If $\hat{A} | a \rangle = a | a \rangle$ , then taking the adjoint of both sides gives

\langle a | \hat{A}^\dagger = a^* \langle a |.

If $\hat{A}$ is Hermitian, $\hat{A}^\dagger = \hat{A}$ , so we can sandwich the eigenvalue equation between $\langle a |$ and $|a\rangle$ :

\langle a | \hat{A} | a \rangle = a \langle a | a \rangle = a, \quad \langle a | \hat{A}^\dagger | a \rangle = a^* \langle a | a \rangle = a^*.

Since both expressions equal the same inner product, $a = a^*$ , which means $a$ is real.

Orthogonal eigenstates

A further key property: eigenstates of a Hermitian operator belonging to different eigenvalues are orthogonal. If $\hat{A}|a\rangle = a|a\rangle$ and $\hat{A}|b\rangle = b|b\rangle$ with $a \neq b$ , then

(a - b)\langle a | b \rangle = 0 \implies \langle a | b \rangle = 0.

This means the eigenstates form an orthonormal basis — exactly the kind of basis needed to expand an arbitrary state and read off measurement probabilities via the Born rule.

A concrete example: the Pauli-Z operator

The Pauli-Z operator, written as a matrix in the $\{|0\rangle, |1\rangle\}$ basis, is

Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.

Because $Z$ is already real and symmetric (equal to its own transpose), $Z^\dagger = Z$ : it is Hermitian. Its eigenvalues are $+1$ and $-1$ , both real, as required. The corresponding eigenstates are $|0\rangle$ and $|1\rangle$ , which are orthogonal.

Expectation value in the state $|{+}\rangle$

The state $|{+}\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ has equal probability $\tfrac{1}{2}$ for each of the two eigenvalues. Using the Born rule the expectation value of $Z$ is

\langle Z \rangle_{|+\rangle} = \tfrac{1}{2}(+1) + \tfrac{1}{2}(-1) = 0.

This result reflects the fact that $|{+}\rangle$ is an eigenstate of $X$ (not $Z$ ), so the spin has no preferred $z$ -component.

Try it

This is a numerical exercise — your code should return a number. Compute the expectation value of the Pauli-Z operator in the state $|{+}\rangle$ .

Run your code to see the quantum state.

The adjoint of an operator

Why observables must be Hermitian

Orthogonal eigenstates

A concrete example: the Pauli-Z operator

Expectation value in the state ∣+⟩|{+}\rangle∣+⟩

Try it

Expectation value in the state $|{+}\rangle$