Symmetric versus Self-Adjoint Operators

Symmetric and self-adjoint are the same idea for finite-dimensional Hermitian matrices, but they are not the same condition for unbounded operators. The difference is not philosophical. It is a domain issue, and it can change spectra, boundary conditions, and time evolution.

The short version is:

Symmetric means the inner-product identity holds on the chosen domain. Self-adjoint means the operator equals its adjoint, including equality of domains.

Every self-adjoint operator is symmetric. Not every symmetric operator is self-adjoint.

This Toolkit page is the compact prerequisite contrast. The full operator-theoretic owner, including deficiency indices, extension tests, resolvent consequences, and links to spectral calculus and dynamics, is Self-Adjoint Operators.

Setup

Let $A$ be a densely defined linear operator on a Hilbert space:

$$A:D(A)\subseteq\mathcal H\to\mathcal H.$$

The adjoint $A^\dagger$ is defined on those $\phi\in\mathcal H$ for which $\langle\phi\vert A\psi\rangle$ can be represented as an inner product with every $\psi\in D(A)$. See Adjoint Operators for the construction.

The important point is that $D(A^\dagger)$ may differ from $D(A)$.

Symmetric Operators

The operator $A$ is symmetric if

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

for all $\phi,\psi\in D(A)$.

Equivalently,

$$D(A)\subseteq D(A^\dagger), \qquad A^\dagger\psi=A\psi \quad \text{for all }\psi\in D(A).$$

This condition is often what physicists check by integration by parts. It guarantees, for example, that expectation values on the domain are real:

$$\langle\psi\vert A\psi\rangle = \langle A\psi\vert\psi\rangle = \langle\psi\vert A\psi\rangle^*.$$

But symmetry alone does not guarantee the full spectral theorem or unitary time evolution.

Self-Adjoint Operators

The operator $A$ is self-adjoint if

$$A=A^\dagger$$

as operators. For unbounded operators, this includes both the formula and the domain:

$$D(A)=D(A^\dagger), \qquad A\psi=A^\dagger\psi \quad \text{on that common domain}.$$

The domain equality is the condition finite-dimensional notation hides.

Finite-Dimensional Case

In a finite-dimensional Hilbert space, every linear operator is bounded and defined on all of $\mathcal H$. The adjoint is also defined on all of $\mathcal H$. Therefore the domain issue disappears.

That is why a finite matrix is Hermitian exactly when

$$A^\dagger=A.$$

The finite-dimensional matrix facts are developed in Hermitian Operators. This page explains why that intuition must be refined in wave mechanics.

Momentum on an Interval

Let $\mathcal H=L^2([0,L])$ and consider

$$P=-i\hbar\frac{d}{dx}.$$

For sufficiently regular $\phi$ and $\psi$,

$$\langle\phi\vert P\psi\rangle - \langle P\phi\vert\psi\rangle = -i\hbar \left[ \phi(x)^*\psi(x) \right]_0^L.$$

On the domain

$$D_0(P) = \{\psi\in H^1([0,L]):\psi(0)=\psi(L)=0\},$$

the boundary term vanishes for all allowed $\phi,\psi$. Thus the operator is symmetric on this domain.

However, it is not self-adjoint. One can show that the adjoint has a larger domain, essentially $H^1([0,L])$ with no endpoint condition. Therefore

$$D_0(P)\ne D(P^\dagger).$$

The same formal differential expression becomes self-adjoint on phase-twisted domains

$$D_\theta(P) = \{\psi\in H^1([0,L]):\psi(L)=e^{i\theta}\psi(0)\}, \qquad \theta\in[0,2\pi).$$

Different $\theta$ define different self-adjoint momentum operators. The boundary condition is part of the operator.

Why Self-Adjointness Matters

Self-adjoint operators support the standard sharp-observable machinery:

• their spectra are real;
• they admit a spectral theorem, including continuous spectra;
• functions of the operator can be defined by spectral calculus;
• self-adjoint Hamiltonians generate unitary time evolution.

Symmetric operators may have real expectation values on a chosen domain, but that is not enough for the full measurement and dynamics framework. For the physics-facing warning, see Hermitian vs Self-Adjoint Operators.

Formal Hermiticity

Physicists often say that a differential operator is “Hermitian” after checking that integration by parts gives no leftover boundary term. This can mean one of several things:

• the formal differential expression is equal to its formal adjoint;
• the operator is symmetric on a specified domain;
• the operator is truly self-adjoint;
• the operator is essentially self-adjoint on a smaller dense core.

These are not interchangeable. The safest language is:

• use Hermitian for finite-dimensional matrices when no domain issue is present;
• use symmetric for the inner-product identity on a specified domain;
• use self-adjoint when the adjoint operator has the same domain and action.

Essentially Self-Adjoint Operators

A symmetric operator can sometimes have a unique self-adjoint closure. Such an operator is called essentially self-adjoint on its starting domain.

This is common in mathematical physics. One first defines a differential operator on a convenient dense core, such as smooth compactly supported functions, and then proves that its closure is self-adjoint. When this works, the simple domain is a safe calculation domain because it determines a unique self-adjoint operator.

When it does not work, there may be several self-adjoint extensions or none. Boundary conditions often classify the possible extensions.

Boundary Conditions and Extensions

For second-order Hamiltonians, boundary conditions can choose among self-adjoint realizations. Dirichlet, Neumann, periodic, and phase-twisted conditions may all make sense for related differential expressions, but they describe different operators and can have different spectra.

This is why a phrase such as “the Hamiltonian $-\hbar^2 d^2/(2m\,dx^2)$ on an interval” is incomplete. One must also specify the domain. The practical differential-equation page is Boundary Conditions, and the domain language is Domains of Operators.

Common Mistakes

• Assuming symmetric automatically means self-adjoint.
• Treating a formal integration-by-parts identity as a complete operator proof.
• Forgetting that $D(A)$ and $D(A^\dagger)$ can differ.
• Saying “Hermitian” without specifying whether the setting is finite-dimensional or unbounded.
• Ignoring boundary conditions when defining momentum or Hamiltonian operators.
• Assuming every symmetric operator has a unique self-adjoint extension.
• Applying finite-dimensional spectral intuition before checking self-adjointness.

Cross-Links

• Domains of Operators
• Adjoint Operators
• Unbounded Operators
• Bounded Operators
• Spectral Theorem, Practical Version
• Hermitian Operators
• Boundary Conditions
• Hermitian vs Self-Adjoint Operators
• Propagators and Boundary Conditions
• Discrete and Continuous Spectra

References

• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Volume I: Functional Analysis, Academic Press, 1980.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• G. Teschl, Mathematical Methods in Quantum Mechanics, 2nd ed., American Mathematical Society, 2014.
• G. Bonneau, J. Faraut, and G. Valent, “Self-adjoint extensions of operators and the teaching of quantum mechanics,” American Journal of Physics 69, 322-331, 2001.
• J. von Neumann, Mathematical Foundations of Quantum Mechanics, Princeton University Press, 1955.

Exercises

1. In finite dimension, show that the symmetric inner-product identity implies $A^\dagger=A$.

Solution

If

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

for all vectors $\phi,\psi$, then by definition of the adjoint,

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

for all $\phi,\psi$. Therefore $(A^\dagger-A)\phi$ is orthogonal to every $\psi$, so $(A^\dagger-A)\phi=0$ for all $\phi$. Hence $A^\dagger=A$.

2. For $P=-i\hbar d/dx$ on $[0,L]$, show that Dirichlet boundary conditions make $P$ symmetric.

Solution

The boundary form is

$$-i\hbar \left[ \phi(x)^*\psi(x) \right]_0^L.$$

If $\phi(0)=\phi(L)=0$ and $\psi(0)=\psi(L)=0$, then both endpoint products vanish. Therefore

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

on that domain.

3. Why does the previous exercise not prove self-adjointness?

Solution

It proves the inner-product identity only for vectors in the chosen Dirichlet domain. Self-adjointness also requires the adjoint domain to be the same. For this momentum example, the adjoint domain is larger, so the Dirichlet-domain operator is symmetric but not self-adjoint.

4. Explain why self-adjointness, not mere symmetry, is the natural condition for a Hamiltonian.

Solution

A closed-system Hamiltonian should generate unitary time evolution. The theorem behind this statement uses self-adjointness of $H$, not merely symmetry on a convenient domain. Symmetry gives real expectation values on that domain, but self-adjointness supplies the spectral calculus needed to define $e^{-iHt/\hbar}$ as a unitary group.