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Cayley correspondence between self-adjoint operators and unitaries

Statement

Assume Countable Choice. The map $T \mapsto C_{T} = (T - i) (T + i)^{- 1}$ is a bijection from the set of self-adjoint operators on $H$ onto the set of unitary operators $U$ on $H$ with $\ker (I - U) = {0}$ . The inverse map assigns to such a $U$ the operator $D (T_{U}) = ran (I - U), T_{U} (I - U) y = i (I + U) y (y \in H) .$ For this $T_{U}$ both $T_{U} - i$ and $T_{U} + i$ map $D (T_{U})$ onto $H$ , so $T_{U}$ is self-adjoint, and $C_{T_{U}} = U$ .

Facts & Assumptions

[A1]

For self-adjoint $T$ the Cayley transform $C_{T}$ is unitary, $\ker (I - C_{T}) = {0}$ , and $ran (I - C_{T}) = D (T)$ ; also $∥ (T \mp i) x ∥^{2} = ∥ T x ∥^{2} + ∥ x ∥^{2}$ for $x \in D (T)$ (Cayley transform of a self-adjoint operator, Resolvent of a self-adjoint operator: nonreal resolvents and the estimate).

[A2]

A densely defined symmetric operator $T$ is self-adjoint if $ran (T - i) = ran (T + i) = H$ (Range criterion for self-adjointness).

[A3]

For unitary $U$ one has $U^{*} U = U U^{*} = I$ and $U$ is bijective, with $U^{- 1} = U^{*}$ ; moreover $ran (A)^{⊥} = \ker (A^{*})$ for bounded $A$ , so $ran (I - U)$ is dense exactly when $\ker (I - U^{*}) = {0}$ (Hilbert-adjoint identities, Kernel–range orthogonality for Hilbert adjoints, Orthogonality and the orthogonal complement, Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets).

Proof

technique · direct

Given: A unitary $U$ with $\ker (I - U) = {0}$ , and the map $T \mapsto C_{T}$ of [A1].

1.1

The assignment is well defined: if $y, y^{'} \in H$ satisfy $(I - U) y = (I - U) y^{'}$ , then $y = y^{'}$ because $\ker (I - U) = {0}$ ; hence $D (T_{U}) : = ran (I - U)$ is well defined and $T_{U} (I - U) y : = i (I + U) y$ defines a map on $D (T_{U})$ .

A3given

1.2

$D (T_{U})$ is dense: by [A3], $ran (I - U)^{⊥} = \ker (I - U^{*})$ , and $I - U^{*} = I - U^{- 1}$ has kernel ${0}$ because $(I - U^{- 1}) y = 0$ means $y = U y$ , that is $y \in \ker (I - U) = {0}$ .

A3given

1.3

Conversely $T_{C_{T}} = T$ for every self-adjoint $T$ : by [A1] $ran (I - C_{T}) = D (T)$ and $I - C_{T} = 2 i (T + i)^{- 1}$ , so for $x \in D (T)$ one has $(I - C_{T}) (T + i) x = 2 i x$ and $i (I + C_{T}) (T + i) x = i (2 T x)$ ; hence the inverse construction sends $C_{T}$ to $T$ .

2.1

$T_{U}$ is symmetric: for $y, y^{'} \in H$ , expanding both pairings and using $⟨ U y, U y^{'} ⟩ = ⟨ y, y^{'} ⟩$ , one gets $⟨ T_{U} (I - U) y, (I - U) y^{'} ⟩ = i (⟨ U y, y^{'} ⟩ - ⟨ y, U y^{'} ⟩) = ⟨ (I - U) y, T_{U} (I - U) y^{'} ⟩$ .

A3step 1.1

2.2

Ranges: for every $y \in H$ one has $(T_{U} + i) (I - U) y = 2 i y$ and $(T_{U} - i) (I - U) y = 2 i U y$ ; hence $ran (T_{U} + i) = H$ and, since $U$ is onto by [A3], $ran (T_{U} - i) = H$ .

A1A3step 1.1

3.1

By steps 1.2, 2.1 and 2.2 the operator $T_{U}$ is densely defined, symmetric, and has both ranges equal to $H$ , so $T_{U}$ is self-adjoint by [A2].

A2step 1.2step 2.1step 2.2

4.1

$C_{T_{U}} = U$ : by [A1] applied to the self-adjoint $T_{U}$ and by step 2.2 one has $(T_{U} + i)^{- 1} (2 i y) = (I - U) y$ , so $C_{T_{U}} (2 i y) = (T_{U} - i) (I - U) y = 2 i U y$ ; since $y \mapsto 2 i y$ is onto $H$ , $C_{T_{U}} = U$ .

A1step 2.2step 3.1

5.1

By steps 1.3 and 4.1 the two constructions are mutually inverse, and every assignment above is a bijection by construction; hence $T \mapsto C_{T}$ is a bijection onto the stated class.

step 1.3step 3.1step 4.1∎

Depends on

Used by

Dependency tree · two levels

37 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Gerald Teschl, Mathematical Methods in Quantum Mechanics, second edition (standard reference, not scraped)
Theo Buehler and Dietmar A. Salamon, Functional Analysis (standard reference, not scraped)