Hermitian Operators

A Hermitian operator on a finite-dimensional complex Hilbert space is a linear operator equal to its adjoint. Hermiticity is the condition that makes its quadratic forms real and its spectral data compatible with orthogonal geometry: eigenvalues are real, distinct eigenspaces are orthogonal, and an orthonormal eigenbasis exists.

Why Quantum Mechanics Uses Hermitian Operators

Finite-dimensional Hermitian operators provide the mathematical structure needed for sharp real-valued quantities:

• real spectral values;
• orthogonal eigenspaces;
• orthogonal spectral projectors;
• real expectation values;
• unitary transformations generated by exponentiation.

These properties explain the role of Hermitian Hamiltonians and observable matrices. They do not by themselves specify which physical quantity an operator represents or how an apparatus implements its measurement. That physical layer belongs to Observables.

Definition from the Inner Product

Let $A:\mathcal H\to\mathcal H$ be linear on a finite-dimensional complex Hilbert space. Its adjoint $A^\dagger$ is the unique operator satisfying

$$\langle\phi\vert A\psi\rangle = \langle A^\dagger\phi\vert\psi\rangle$$

for all $\phi,\psi\in\mathcal H$. The operator is Hermitian when

$$A^\dagger=A.$$

Equivalently,

$$\langle\phi\vert A\psi\rangle = \langle A\phi\vert\psi\rangle$$

for every pair of vectors.

The adjoint depends on the inner product. In an orthonormal basis it is represented by the conjugate transpose:

$$(A^\dagger)_{jk}=A_{kj}^*.$$

In a nonorthonormal basis with Gram matrix $G$, the coordinate matrix of the adjoint is instead

$$A^{\dagger_G} = G^{-1}A^\dagger G,$$

where the dagger on the right is the ordinary conjugate transpose of the coordinate matrix. See Adjoint Operators for the basis-independent construction.

Matrix Form

A matrix is Hermitian exactly when

$$A_{jk}=A_{kj}^*.$$

Consequently:

• every diagonal entry is real;
• entries reflected across the diagonal are complex conjugates.

A general two-by-two Hermitian matrix has the form

$$A = \begin{pmatrix} a&z\\ z^*&b \end{pmatrix}, \qquad a,b\in\mathbb R, \quad z\in\mathbb C.$$

Real symmetric matrices are Hermitian, but they are only the real-entry special case. For example,

$$\sigma_y = \begin{pmatrix} 0&-i\\ i&0 \end{pmatrix}$$

is Hermitian and not real.

Real Quadratic Forms

For a Hermitian operator, the quadratic form

$$q_A(\psi) = \langle\psi\vert A\psi\rangle$$

is real for every $\psi$. Indeed,

$$\begin{aligned} q_A(\psi)^* &= \langle A\psi\vert\psi\rangle \\ &= \langle\psi\vert A\psi\rangle. \end{aligned}$$

In finite dimensions, the converse also holds: if $\langle\psi\vert A\psi\rangle$ is real for every $\psi$, then $A$ is Hermitian. Testing only basis vectors is not sufficient, because those tests see diagonal entries but can miss incorrect off-diagonal phases. Superpositions are needed to reconstruct the full sesquilinear form.

The normalized quadratic form is the Rayleigh quotient,

$$\mathcal R_A(\psi) = \frac{ \langle\psi\vert A\psi\rangle }{ \langle\psi\vert\psi\rangle }, \qquad \psi\ne0.$$

For a Hermitian operator it is always real.

Eigenvalues Are Real

Let $Av=\lambda v$ with $v\ne0$. Then

$$\langle v\vert Av\rangle = \lambda\langle v\vert v\rangle.$$

Hermiticity also gives

$$\begin{aligned} \langle v\vert Av\rangle &= \langle Av\vert v\rangle \\ &= \lambda^* \langle v\vert v\rangle. \end{aligned}$$

Since $\langle v\vert v\rangle>0$,

$$\lambda=\lambda^*,$$

so every eigenvalue is real.

The proof uses both positive definiteness and equality with the adjoint. A general complex matrix can have complex eigenvalues, and an indefinite Hermitian form would not provide the same Hilbert-space conclusion.

Distinct Eigenspaces Are Orthogonal

Let

$$Au=\mu u, \qquad Av=\lambda v,$$

where $\mu\ne\lambda$. Because both eigenvalues are real,

$$\begin{aligned} \lambda\langle u\vert v\rangle &= \langle u\vert Av\rangle \\ &= \langle Au\vert v\rangle \\ &= \mu\langle u\vert v\rangle. \end{aligned}$$

Therefore

$$(\lambda-\mu) \langle u\vert v\rangle =0,$$

and hence

$$\langle u\vert v\rangle=0.$$

When an eigenvalue is degenerate, vectors within its eigenspace are not automatically orthogonal, but an orthonormal basis can be chosen inside that subspace. The eigenspace and its projector are canonical; the basis chosen within it is not.

Spectral Theorem

Every finite-dimensional Hermitian operator has an orthonormal eigenbasis. In projector form,

$$A = \sum_a aP_a,$$

where $a$ runs over distinct real eigenvalues and $P_a$ projects orthogonally onto the full eigenspace for $a$. The projectors satisfy

$$P_aP_b = \delta_{ab}P_a, \qquad \sum_aP_a=I.$$

This statement combines three results:

1. the matrix is diagonalizable;
2. the diagonalizing basis can be orthonormal;
3. the diagonal entries are real.

The proof and operator-class hierarchy are developed in Normal Operators; the projector form and functional calculus belong to Spectral Decomposition.

Rayleigh Bounds and Variational Meaning

Using the spectral decomposition, write

$$\psi=\sum_aP_a\psi.$$

Then

$$\mathcal R_A(\psi) = \frac{ \sum_a a\lVert P_a\psi\rVert^2 }{ \sum_a \lVert P_a\psi\rVert^2 }.$$

The Rayleigh quotient is therefore a weighted average of eigenvalues. If $a_{\min}$ and $a_{\max}$ are the smallest and largest eigenvalues,

$$a_{\min} \leq \mathcal R_A(\psi) \leq a_{\max}.$$

For normalized vectors,

$$a_{\min} = \min_{\lVert\psi\rVert=1} \langle\psi\vert A\psi\rangle,$$

and similarly the maximum gives $a_{\max}$. Equality at the minimum occurs exactly for vectors in the lowest eigenspace. This is the finite-dimensional core of variational ground-state methods.

Positive Hermitian Operators

A Hermitian operator is positive semidefinite, written

$$A\geq0,$$

when

$$\langle\psi\vert A\psi\rangle\geq0$$

for every $\psi$. By the spectral theorem, this is equivalent to every eigenvalue of $A$ being nonnegative.

For any linear operator $B$,

$$B^\dagger B\geq0$$

because

$$\langle\psi\vert B^\dagger B\psi\rangle = \lVert B\psi\rVert^2 \geq0.$$

Positivity is central to density operators, effects, covariance matrices, and Hamiltonians bounded below. It is stronger than Hermiticity: a Hermitian operator may have negative eigenvalues.

Hermitian and Skew-Hermitian Parts

Every complex matrix $M$ has a unique decomposition

$$M=H+iK,$$

where

$$H = \frac{M+M^\dagger}{2}, \qquad K = \frac{M-M^\dagger}{2i}.$$

Both $H$ and $K$ are Hermitian. They are the operator analogues of the real and imaginary parts of a complex number. The matrix $M$ is normal exactly when $H$ and $K$ commute.

Equivalently, $iK$ is skew-Hermitian:

$$(iK)^\dagger=-iK.$$

Skew-Hermitian operators are the infinitesimal generators of unitary matrices; physicists usually write them as $-i$ times Hermitian generators.

Hermitian Generators Produce Unitaries

If $H=H^\dagger$, then

$$U(t) = \exp\left( -\frac{iHt}{\hbar} \right)$$

is unitary. Using the power-series or spectral definition of the exponential,

$$U(t)^\dagger = \exp\left( \frac{iHt}{\hbar} \right),$$

so

$$U(t)^\dagger U(t)=I.$$

Hermiticity of the generator therefore produces norm-preserving evolution. The matrix-function construction is Matrix Functions and Exponentials; the physical statement is Unitary Time Evolution.

Worked Example: A Complex Hermitian Matrix

Consider

$$A = \begin{pmatrix} 1&i\\ -i&2 \end{pmatrix}.$$

The diagonal entries are real and the off-diagonal entries are conjugates, so $A^\dagger=A$. Its characteristic polynomial is

$$\begin{aligned} p_A(\lambda) &= \det \begin{pmatrix} \lambda-1&-i\\ i&\lambda-2 \end{pmatrix} \\ &= (\lambda-1)(\lambda-2)-1 \\ &= \lambda^2-3\lambda+1. \end{aligned}$$

The eigenvalues are

$$\lambda_\pm = \frac{3\pm\sqrt5}{2},$$

which are real as the general theorem requires. The Rayleigh quotient of every nonzero vector lies between these values.

For a general matrix

$$\begin{pmatrix} a&z\\ z^*&b \end{pmatrix},$$

the eigenvalues are

$$\begin{aligned} \lambda_\pm &= \frac{a+b}{2} \\ &\quad \pm \sqrt{ \left(\frac{a-b}{2}\right)^2 +\lvert z\rvert^2 }. \end{aligned}$$

The square root is real and nonnegative. Degeneracy occurs only when $a=b$ and $z=0$, in which case the matrix is a scalar multiple of the identity and every nonzero vector is an eigenvector.

Observables and Expectation Values

In finite-dimensional quantum mechanics, Hermitian matrices represent sharp real-valued observables. For a normalized pure state,

$$\langle A\rangle_\psi = \langle\psi\vert A\vert\psi\rangle$$

is real and lies between the smallest and largest eigenvalues. The spectral projectors, together with the Born rule, determine the full outcome distribution. The mean alone does not.

Hermiticity is not a complete theory of measurement:

• the assignment of a matrix to a physical quantity is model-dependent;
• generalized measurements use positive effects that need not be projectors;
• an observable does not by itself specify a unique apparatus or state update;
• infinite-dimensional sharp observables require self-adjoint operators with domains.

See Expectation Values for the probabilistic interpretation.

Infinite-Dimensional Warning

For finite matrices, Hermitian, symmetric, and self-adjoint coincide because every operator and adjoint is defined on the whole finite-dimensional space. For an unbounded differential operator, the formula and its domain must both be specified. A formally symmetric expression may fail to be self-adjoint or may admit several self-adjoint extensions with different spectra.

The physical caveat is Hermitian vs Self-Adjoint Operators. The canonical domain-sensitive mathematics is Symmetric versus Self-Adjoint Operators.

Numerical Practice

Use a Hermitian eigensolver when the matrix is theoretically Hermitian. Such algorithms exploit the structure, return real eigenvalues up to roundoff, and can choose orthonormal eigenvectors.

Before diagonalization, measure the Hermiticity defect, for example

$$\frac{ \lVert A-A^\dagger\rVert }{ \lVert A\rVert }.$$

A tiny defect may come from roundoff and can sometimes be removed by symmetrizing,

$$A \longmapsto \frac{A+A^\dagger}{2},$$

but only when theory says the exact operator is Hermitian. Symmetrizing a genuinely non-Hermitian effective model changes the problem. See Matrix Diagonalization for residual, orthogonality, degeneracy, and conditioning checks.

Common Mistakes

• Checking only whether diagonal entries are real. Off-diagonal entries must be conjugates across the diagonal.
• Using a transpose instead of a conjugate transpose. The distinction is essential for complex matrices.
• Taking the conjugate transpose in a nonorthonormal basis without the Gram matrix. The adjoint is defined by the inner product.
• Treating Hermitian as synonymous with real symmetric. Complex Hermitian matrices are standard in quantum mechanics.
• Assuming Hermiticity means positivity. Hermitian eigenvalues are real but may be negative.
• Assuming every operator with real eigenvalues is Hermitian. Nonnormal matrices can have real spectra.
• Calling a formal differential expression Hermitian without a domain. Infinite-dimensional observables require self-adjointness.
• Inferring a complete measurement model from one matrix. Outcome probabilities and state updates require the quantum measurement framework.

Exercises

1. Find the conditions on $a,b,c,d\in\mathbb C$ for

$$M = \begin{pmatrix} a&b\\ c&d \end{pmatrix}$$

to be Hermitian.

Solution

The adjoint is

$$M^\dagger = \begin{pmatrix} a^*&c^*\\ b^*&d^* \end{pmatrix}.$$

Thus $M^\dagger=M$ exactly when

$$a,d\in\mathbb R, \qquad c=b^*.$$

2. Let $A$ be Hermitian and let $u,v$ be eigenvectors with eigenvalues $\mu$ and $\lambda$. Derive both the reality of $\lambda$ and the orthogonality of $u$ and $v$ when $\mu\ne\lambda$.

Solution

For $v\ne0$,

$$\begin{aligned} \lambda\lVert v\rVert^2 &= \langle v\vert Av\rangle \\ &= \langle Av\vert v\rangle \\ &= \lambda^*\lVert v\rVert^2, \end{aligned}$$

so $\lambda=\lambda^*$. The same holds for $\mu$. Then

$$\begin{aligned} \lambda\langle u\vert v\rangle &= \langle u\vert Av\rangle \\ &= \langle Au\vert v\rangle \\ &= \mu\langle u\vert v\rangle. \end{aligned}$$

If $\lambda\ne\mu$, the overlap must vanish.

3. For

$$A = \begin{pmatrix} 1&i\\ -i&2 \end{pmatrix},$$

give the sharpest possible interval containing $\langle\psi\vert A\vert\psi\rangle$ for every normalized $\psi$.

Solution

The characteristic polynomial is

$$\lambda^2-3\lambda+1,$$

so the eigenvalues are

$$\lambda_\pm = \frac{3\pm\sqrt5}{2}.$$

The Rayleigh quotient of a Hermitian matrix lies between its extreme eigenvalues. Therefore

$$\frac{3-\sqrt5}{2} \leq \langle\psi\vert A\vert\psi\rangle \leq \frac{3+\sqrt5}{2}.$$

The bounds are sharp because they are attained by normalized eigenvectors.

4. Let $H=H^\dagger$ and

$$U(t)=e^{-iHt/\hbar}.$$

Show that $U(t)$ is unitary and state which step would fail if $H$ were not Hermitian.

Solution

Taking the adjoint gives

$$\begin{aligned} U(t)^\dagger &= \left( e^{-iHt/\hbar} \right)^\dagger \\ &= e^{iH^\dagger t/\hbar} \\ &= e^{iHt/\hbar}. \end{aligned}$$

Since both exponentials are functions of the same matrix $H$, they commute and

$$U(t)^\dagger U(t) = e^{iHt/\hbar}e^{-iHt/\hbar} =I.$$

If $H\ne H^\dagger$, the adjoint exponential contains $H^\dagger$ rather than $H$, and the cancellation does not generally occur.

References

• S. Axler, Linear Algebra Done Right, 3rd ed., Springer, 2015.
• R. A. Horn and C. R. Johnson, Matrix Analysis, 2nd ed., Cambridge University Press, 2013.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Volume I: Functional Analysis, revised and enlarged ed., Academic Press, 1980.