Self-Adjoint Operators — Mathematics Knowledge System

Overview

A bounded linear operator T on a Hilbert space H is self-adjoint (or Hermitian) if T = T*, where T* is the Hilbert space adjoint defined by <Tx, y> = <x, T*y> for all x, y in H. Self-adjoint operators have real spectrum, admit a spectral measure decomposition (the spectral theorem), and model observable quantities in quantum mechanics. They are the infinite-dimensional analogue of real symmetric matrices.

Intuition

A matrix is self-adjoint if it equals its conjugate transpose. The Hilbert space analogue replaces the matrix entry condition with an inner product condition: <Tx, y> = <x, Ty> for all x, y. This symmetry forces eigenvalues to be real and eigenvectors for distinct eigenvalues to be orthogonal. In quantum mechanics, self-adjoint operators represent observables whose measured values (spectral values) must be real.

Formal Definition

Definition

Let H be a Hilbert space. The adjoint T* of T in B(H) is the unique operator satisfying <Tx, y> = <x, T*y> for all x, y in H (existence and uniqueness by the Riesz representation theorem applied to y |-> <Tx,y>). T is self-adjoint if T = T*, i.e., <Tx,y> = <x,Ty> for all x,y in H. T is positive (T >= 0) if <Tx,x> >= 0 for all x. Every positive operator is self-adjoint.

\langle Tx, y \rangle = \langle x, T^* y \rangle \quad \forall x, y \in H

Definition of adjoint

T = T^* \iff \langle Tx, y \rangle = \langle x, Ty \rangle \quad \forall x, y \in H

Self-adjointness

\|T\| = \sup_{\|x\|=1} |\langle Tx, x \rangle| \quad (T = T^*)

Norm via quadratic form for self-adjoint T

Notation

Notation	Meaning
$T^*$	Hilbert space adjoint of T
$T = T^*$	T is self-adjoint
$T \ge 0$	T is positive: <Tx,x> >= 0 for all x
$\sigma(T)$	Spectrum of T (real for self-adjoint T)

Theorems

Theorem 5.1: Theorem 5.1

If T in B(H) is self-adjoint, then sigma(T) subset R and ||T|| = sup_{||x||=1} |<Tx,x>|. Moreover, ||T|| = max(|inf sigma(T)|, |sup sigma(T)|) = r(T).

Theorem 5.2: Theorem 5.2

Eigenvectors of a self-adjoint operator corresponding to distinct eigenvalues are orthogonal: if Tu = lambda u and Tv = mu v with lambda != mu (real), then <u,v> = 0.

Theorem 5.3: Theorem 5.3 (Spectral theorem, bounded self-adjoint version)

For every bounded self-adjoint operator T on a Hilbert space H, there exists a unique spectral measure E on the Borel subsets of sigma(T) subset R such that T = integral_{sigma(T)} lambda dE(lambda).

Worked Examples

1
Compute the inner product:
$\langle M_f g, h \rangle = \int_0^1 f(t)g(t)\overline{h(t)}\,dt$
2
Since f is real-valued, f(t) = overline{f(t)}:
$= \int_0^1 g(t)\overline{f(t)h(t)}\,dt = \langle g, M_f h \rangle$
3
So <M_f g, h> = <g, M_f h> for all g, h in L^2[0,1], i.e., M_f* = M_f.

✓ Answer

M_f is self-adjoint for real-valued f because multiplying by a real function commutes with complex conjugation in the inner product.

Practice Problems

Mediumproof writing

Prove that the spectrum of a self-adjoint operator T in B(H) is real.

Mediumfree response

Give an example of a self-adjoint operator on l^2 and compute its spectrum.

Quiz

Summary

T is self-adjoint if T = T*, equivalently <Tx,y> = <x,Ty> for all x,y in H.
Self-adjoint operators have real spectrum: sigma(T) subset R.
Eigenvectors for distinct eigenvalues are orthogonal.
The norm equals the quadratic form norm: ||T|| = sup_{||x||=1}|<Tx,x>|.
The spectral theorem gives T = integral lambda dE(lambda) as an integral over sigma(T).

References

BookReed, M. & Simon, B. — Methods of Modern Mathematical Physics I, Chapter VI
BookRudin, W. — Functional Analysis (2nd ed.), Chapter 12
Websiteen.wikipedia.org