Todd Trimble
Some basic Stone-von Neumann spectral theory

Self-adjoint operators on Hilbert space

We follow the treatment given in Reed and Simon.

Spectral theorem

Let T:HHT: H \to H be a self-adjoint operator on a separable Hilbert space, with domain DD. Then there exists a finite measure space (M,μ)(M, \mu), a unitary operator U:HL 2(M,μ)U: H \to L^2(M, \mu), and a real-valued function f:Mf: M \to \mathbb{R} such that

  • ψD\psi \in D if and only if f(Uψ)L 2(M,μ)f \cdot (U\psi) \in L^2(M, \mu), and

  • For ψD\psi \in D, U(T(ψ))=fU(ψ)U(T(\psi)) = f \cdot U(\psi)

The last equation should be read as a diagonalization; it says that

U(T(ψ))(x)=f(x)U(ψ)(x)U(T(\psi))(x) = f(x)U(\psi)(x)

for almost all xMx \in M.


We have yet to give a careful definition of what it means for an operator to be self-adjoint, but in outline, the proof runs roughly as follows. If TT is self-adjoint, then

  1. T+iT + i and TiT - i map DD bijectively onto HH.

  2. The maps T+iT + i and TiT - i are closed, and therefore by the closed graph theorem, their set-theoretic inverses are (closed) bounded maps (T+i) 1(T + i)^{-1}, (Ti) 1(T - i)^{-1} from HH to DHD \subset H.

  3. (T+i) 1(T + i)^{-1} and (Ti) 1(T - i)^{-1} commute, and one is the adjoint of the other. Hence each is (bounded and) normal.

  4. By the spectral theorem for bounded normal operators, there is a measure space (M,μ)(M, \mu), a unitary isomorphism U:HL 2(M,μ)U: H \to L^2(M, \mu), and a measurable, bounded, complex-valued function gg so that U((T+i) 1(ϕ))=gU(ϕ)U ((T+i)^{-1}(\phi)) = g \cdot U(\phi) for all ϕH\phi \in H.

  5. We therefore have a commutative diagram

    H (T+i) 1 D H U U U L 2(M) g D L 2(M)\array{ H & \stackrel{(T+i)^{-1}}{\to} & D & \hookrightarrow & H \\ \mathllap{U} \downarrow & & \downarrow \mathrlap{U} & & \downarrow \mathrlap{U} \\ L^2(M) & \underset{g \cdot -}{\to} & D' & \hookrightarrow & L^2(M) }

    where DD' is the image of the operator given by multiplying by gg. It follows that we have a commutative diagram

    D T+i H U U D 1g L 2(M)\array{ D & \stackrel{T+i}{\to} & H \\ \mathllap{U} \downarrow & & \downarrow \mathrlap{U} \\ D' & \underset{\frac1{g}\cdot -}{\to} & L^2(M) }

    whence, if we put f(x)=1g(x)if(x) = \frac1{g(x)} - i, we obtain U(T(ψ))=fU(ψ)U(T(\psi)) = f \cdot U(\psi).

We now prove these statements.

Self-adjoint operators TT are closed

Recall that a partially defined linear operator T:HHT: H \to H is closed if it has a closed graph in H×HH \times H.

If TT is densely defined, then for each ϕH\phi \in H there is at most one ηH\eta \in H for which

T,ϕ=,η\langle T-, \phi \rangle = \langle -, \eta \rangle

and the set of ϕ\phi for which such η\eta exists defines the domain of the partially defined transpose map T *T^\ast defined by T *(ϕ)=ηT^\ast(\phi) = \eta.


If T:HHT: H \to H is densely defined, then T *T^\ast is closed.

The trick is to exhibit the graph of T *T^\ast as the orthogonal complement of some space. This suffices because orthogonal complements are always closed…

Revised on July 11, 2011 at 17:20:50 by Todd Trimble