nLab linear isometry




A linear isometry is a linear map

ϕ: 1 2 \phi \,\colon\, \mathscr{H}_1 \longrightarrow \mathscr{H}_2

between normed vector spaces which preserves the norm:

ψ() 2= 1. \big\Vert \psi(-) \big\Vert_2 \;=\; \big\Vert - \big\Vert_1 \,.

In particular, when the norm is induced by (Hermitian) inner products | i{\langle \cdot \vert \cdot \rangle}_{i} (as on Hilbert spaces)

ψ iψ|ψ i {\big\Vert \psi \big\Vert}_i \,\equiv\, \sqrt{ \left\langle \psi \vert \psi \right\rangle_i }

then a linear isometry is an isometry in that it preserves these inner products:

ϕ()|ϕ() 2=| 1 \big\langle \phi(-) \vert \phi(-) \big\rangle_2 \;=\; \left\langle - \vert - \right\rangle_1

or equivalently, in terms of adjoint operators:

(1)ϕ ϕ=Id 1. \phi^\dagger \cdot \phi \,=\, Id_{\mathscr{H}_1} \,.


(relation to unitary operators)
If in addition to (1) also the reverse condition ϕϕ =Id 2\phi \cdot \phi^\dagger \,=\, Id_{\mathscr{H}_2} holds, then ϕ\phi is called a unitary operator, which is the case iff ϕ\phi is surjective map.



Linear isometries are injective maps.


For |ψ,|ψ:\left\vert \psi \right\rangle, \left\vert \psi' \right\rangle \,\colon\, \mathscr{H} we need to show that

ϕ|ψ=ϕ|ψ|ψ=|ψ. \phi\left\vert \psi \right\rangle \;=\; \phi\left\vert \psi' \right\rangle \;\;\;\; \Rightarrow \;\;\;\; \left\vert \psi \right\rangle \;=\; \left\vert \psi' \right\rangle \mathrlap{\,.}

But by non-degeneracy of the norm this is equivalent to showing

ϕ(|ψ|ψ)=0|ψ|ψ=0. { \Big\Vert \phi \big( \left\vert \psi \right\rangle - \left\vert \psi' \right\rangle \big) \Big\Vert } \,=\, 0 \;\;\;\;\;\;\; \Rightarrow \;\;\;\;\;\;\; { \Big\Vert \left\vert \psi \right\rangle - \left\vert \psi' \right\rangle \Big\Vert } \,=\, 0 \mathrlap{\,.}

But since ϕ\phi is an isometry, here the left hand side already coincides with the right hand side and hence certainly implies it.


See also:

Last revised on September 26, 2023 at 06:03:04. See the history of this page for a list of all contributions to it.