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For a given finite-dimensional complex Hilbert space $\mathcal{H}$ with the inner product $\langle \cdot | \cdot\rangle$, a map $f: \mathcal{H} \to \mathcal{H}$ is said to be an isometric map if $$ ||f(x) - f(y)|| = ||x-y||, \forall x,y \in {\cal H}. $$ Besides, we assume that $f(0)=0$.

A map $g: \mathcal{H} \to \mathcal{H}$ is said to be a symmetry transformation if $$ |\langle \overline{g(x)}|\overline{g(y)}\rangle | = |\langle \bar{x}|\bar{y}\rangle|, \forall x,y \in {\cal H}, $$ where $\bar{x}$ is the unit vector which share the same direction as $x$, etc.

According to Wigner's theorem, every symmetry transformation is either an unitary or anti-unitary. Hence, symmetry transformation is always an isometric map.

Then the question is,

Is every isometric map a symmetry transformation in a finite-dimensional complex Hilbert space?

This is true indeed if ${\cal H}$ is a real Hilbert space.

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Eden Harder
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1 Answers1

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This is not true. Here is an example.

Let's consider a two-dimensional Hilbert space ${\cal H}_2$ and the map $f$: $$ f: (a+bi,c+di) \to (a+bi,c-di). $$ It's easy to verify that $f$ is an isomorphism. Meanwhile, $f$ is not a symmetry transformation since $$ \langle f((1+i,1+i))|f((1-i,1-i))\rangle = \langle (1+i,1-i)|(1-i,1+i) \rangle = 0. $$ $$ \langle (1+i,1+i)|(1-i,1-i) \rangle = 4i. $$

Eden Harder
  • 1,131