Contents

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# Contents

## Idea

A linear isometry is a linear map

$\phi \,\colon\, \mathscr{H}_1 \longrightarrow \mathscr{H}_2$

between normed vector spaces which preserves the norm:

$\big\Vert \psi(-) \big\Vert_2 \;=\; \big\Vert - \big\Vert_1 \,.$

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

${\big\Vert \psi \big\Vert}_i \,\equiv\, \sqrt{ \left\langle \psi \vert \psi \right\rangle_i }$

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

$\big\langle \phi(-) \vert \phi(-) \big\rangle_2 \;=\; \left\langle - \vert - \right\rangle_1$

or equivalently, in terms of adjoint operators:

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

###### Remark

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

## Properties

###### Proposition

Linear isometries are injective maps.

###### Proof

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

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

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

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

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