Adjoint Operators

The adjoint of an operator is the Hilbert-space operation that moves the operator from one side of an inner product to the other. It is the abstract version of conjugate transposition for matrices, and it is the operation behind Hermitian, unitary, normal, and self-adjoint operators.

In finite-dimensional orthonormal bases, the adjoint is familiar:

$$(A^\dagger)_{mn}=A_{nm}^*.$$

In infinite-dimensional Hilbert spaces, the same idea remains, but unbounded operators require domain care.

Bounded-Operator Definition

Let $A:\mathcal H\to\mathcal H$ be a bounded linear operator on a Hilbert space. The adjoint $A^\dagger$ is the unique bounded operator satisfying

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

for all $\phi,\psi\in\mathcal H$.

The uniqueness and existence follow from the Riesz representation theorem. For each fixed $\phi$, the map

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

is a continuous linear functional of $\psi$. Therefore it is represented by a unique vector, called $A^\dagger\phi$.

For bounded operators, $A^\dagger$ is also bounded and

$$\lVert A^\dagger\rVert_{\mathrm{op}} = \lVert A\rVert_{\mathrm{op}}.$$

Relation to Hermitian Conjugates

In a finite-dimensional orthonormal basis, the abstract adjoint is represented by the conjugate transpose:

$$A^\dagger = (A^*)^T.$$

For example, if

$$A= \begin{pmatrix} 1 & 2+i\\ -i & 3 \end{pmatrix},$$

then

$$A^\dagger = \begin{pmatrix} 1 & i\\ 2-i & 3 \end{pmatrix}.$$

This is often called the Hermitian conjugate of the matrix. Strictly, the adjoint is the operator defined by the inner product; the conjugate transpose is its matrix representation in an orthonormal basis.

In a non-orthonormal basis, taking entrywise conjugate transpose is not by itself the adjoint matrix. The Gram matrix of the basis also enters. This is why the inner product is part of the structure, not decoration.

Basic Algebra Rules

For bounded operators,

$$(A+B)^\dagger=A^\dagger+B^\dagger,$$ $$(cA)^\dagger=c^*A^\dagger,$$ $$(AB)^\dagger=B^\dagger A^\dagger,$$

and

$$(A^\dagger)^\dagger=A.$$

The reversal in $(AB)^\dagger=B^\dagger A^\dagger$ is the same reversal familiar from transposing a product of matrices.

Rank-One Operators

For vectors $u,v\in\mathcal H$, define the rank-one operator

$$A=\lvert u\rangle\langle v\rvert, \qquad A\psi=u\,\langle v\vert\psi\rangle.$$

Then

$$A^\dagger = \lvert v\rangle\langle u\rvert.$$

Indeed,

$$\langle\phi\vert A\psi\rangle = \langle\phi\vert u\rangle \langle v\vert\psi\rangle = \langle v\,\langle u\vert\phi\rangle \vert \psi \rangle = \langle A^\dagger\phi\vert\psi\rangle.$$

This example is the operator form of bra-ket conjugation: taking the adjoint swaps bras and kets and complex-conjugates coefficients.

Multiplication Operators

On $L^2$ spaces, multiplication operators give a useful infinite-dimensional bounded example. If $f$ is essentially bounded and

$$(M_f\psi)(x)=f(x)\psi(x),$$

then

$$M_f^\dagger=M_{f^*}.$$

This follows from

$$\int \phi(x)^* f(x)\psi(x)\,dx = \int \left(f(x)^*\phi(x)\right)^*\psi(x)\,dx.$$

If $f$ is real-valued almost everywhere, then $M_f^\dagger=M_f$.

Hermitian, Unitary, and Normal

The adjoint defines several operator classes used throughout quantum mechanics:

• Hermitian matrices satisfy $A^\dagger=A$ in finite-dimensional orthonormal coordinates.
• Unitary operators satisfy $U^\dagger U=UU^\dagger=I$.
• Normal finite-dimensional operators satisfy $A^\dagger A=AA^\dagger$.
• Self-adjoint unbounded operators satisfy $A=A^\dagger$ including equality of domains.

The finite-dimensional Hermitian matrix story is developed in Hermitian Operators. The domain-sensitive mathematical distinction is Symmetric versus Self-Adjoint Operators, and the physical warning is Hermitian vs Self-Adjoint Operators.

Practical Warning for Unbounded Operators

For a densely defined unbounded operator

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

the adjoint is not automatically defined on all of $\mathcal H$. A vector $\phi$ lies in $D(A^\dagger)$ when there exists some $\eta\in\mathcal H$ such that

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

for every $\psi\in D(A)$. Then $A^\dagger\phi=\eta$.

The domain $D(A^\dagger)$ can be larger than, smaller than, or different from the original domain, depending on the operator. This is why unbounded self-adjointness cannot be checked by looking only at a formal expression.

Formal Adjoints and Boundary Terms

Differential expressions often have a formal adjoint obtained by integration by parts. For example,

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

is formally its own adjoint. On an interval, however,

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

The boundary term must vanish on the chosen domain for the operator to be symmetric. Self-adjointness requires the stronger statement that the operator and its adjoint have the same domain. Boundary conditions are therefore not cosmetic; they decide the adjoint domain.

For the domain viewpoint, see Domains of Operators.

Common Mistakes

• Treating the adjoint as merely “transpose and conjugate” without checking the basis is orthonormal.
• Forgetting the order reversal in $(AB)^\dagger=B^\dagger A^\dagger$.
• Forgetting the complex conjugate in $(cA)^\dagger=c^*A^\dagger$.
• Calling a differential expression self-adjoint because its formal adjoint looks the same.
• Ignoring the domain $D(A^\dagger)$ for unbounded operators.
• Confusing Hermitian matrices with self-adjoint unbounded operators.

Cross-Links

• Inner Products
• Bounded Operators
• Domains of Operators
• Unbounded Operators
• Symmetric versus Self-Adjoint Operators
• Hermitian Operators
• Unitary Operators
• Normal Operators
• Hermitian vs Self-Adjoint Operators

References

• J. B. Conway, A Course in Functional Analysis, 2nd ed., Springer, 1990.
• 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.
• P. R. Halmos, Introduction to Hilbert Space and the Theory of Spectral Multiplicity, 2nd ed., Chelsea, 1957.

Exercises

1. Find the adjoint of

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

Solution

Conjugate and transpose:

$$A^\dagger = \begin{pmatrix} 1-i & -3i\\ 2 & -1 \end{pmatrix}.$$

2. Show that $\left(\lvert u\rangle\langle v\rvert\right)^\dagger=\lvert v\rangle\langle u\rvert$.

Solution

For all $\phi,\psi$,

$$\langle\phi\vert \left(\lvert u\rangle\langle v\rvert\right) \psi\rangle = \langle\phi\vert u\rangle \langle v\vert\psi\rangle.$$

The operator $\lvert v\rangle\langle u\rvert$ satisfies

$$\langle \left(\lvert v\rangle\langle u\rvert\right)\phi \vert \psi\rangle = \langle v\,\langle u\vert\phi\rangle\vert\psi\rangle = \langle\phi\vert u\rangle \langle v\vert\psi\rangle.$$

Thus it is the adjoint.

3. Let $(M_f\psi)(x)=f(x)\psi(x)$ on an $L^2$ space, with $f$ bounded. Show that $M_f^\dagger=M_{f^*}$.

Solution

Compute

$$\langle\phi\vert M_f\psi\rangle = \int \phi(x)^*f(x)\psi(x)\,dx = \int \left(f(x)^*\phi(x)\right)^*\psi(x)\,dx = \langle M_{f^*}\phi\vert\psi\rangle.$$

Therefore $M_f^\dagger=M_{f^*}$.

4. Why does a vanishing formal boundary term not automatically prove self-adjointness?

Solution

A vanishing boundary term on a chosen domain shows symmetry on that domain. Self-adjointness also requires equality with the adjoint operator, including equality of domains:

$$A=A^\dagger, \qquad D(A)=D(A^\dagger).$$

For unbounded differential operators, the adjoint may have a different domain even when the formal differential expression is the same.