Wigner’s symmetry representation theorem

Uhlhorn 3 — normalizing \(\theta\)

Normalize \(\theta\) to give an isometry \(\Theta:H\rightarrow H'\).

Define \(\delta:H-{0}\rightarrow \mathbb R\) from \(\Vert\theta(x)\Vert=\delta(x)\Vert x\Vert\).

For non-orthogonal \(x,y\), \(y\) is orthogonal to \(z=\langle y|y\rangle x-y\langle y|x\rangle\). Orthogonality is preserved by \(\theta\) so \(0=\langle \theta(y)|\theta(z)\rangle=\phi(\Vert y\Vert^2)\langle \theta(y)|\theta(x)\rangle-\phi(\langle y|x\rangle)\Vert \theta(y)\Vert^2\). Similarly, \(\phi(\Vert x\Vert^2)\langle \theta(x)|\theta(y)\rangle-\phi(\langle x|y\rangle)\Vert \theta(x)\Vert^2=0\). From \(\phi(\overline \alpha)=\overline{\phi(\alpha)}\), the reals are mapped onto themselves (Yale, theorem 3, pdf), and the action on the complex numbers is either the identity or conjugation. Hence \[\begin{align}\langle \theta(y)|\theta(x)\rangle&=\frac{\phi(\langle y|x\rangle)\Vert \theta(y)\Vert^2}{\phi(\Vert y\Vert^2)}=\frac{\phi(\langle y|x\rangle)\Vert \theta(y)\Vert^2}{\Vert y\Vert^2}=\phi(\langle y|x\rangle)\delta(y)^2\\&=\overline{\phi(\langle x|y\rangle)}\delta(x)^2=\phi(\langle y|x\rangle)\delta(x)^2\end{align}.\] Hence \(\delta(x)=\delta(y)\). The value of \(\delta\) is independent of the vector. If we define \(\Theta=\theta/\delta\), one has \(\langle \Theta(y)|\Theta(x)\rangle=\phi(\langle y|x\rangle)\). Since \(\phi\) is the identity or complex conjugate \(\Theta\) is an (anti)isometry. It \(T\) is onto, \(\Theta\) is also onto (and hence (anti)unitary), since \(\Theta(\alpha x)=\phi(\alpha)\Theta(x)\), so rays are mapped onto rays. Hence:

PROPOSITION 8 ([Uhlhorn, 1962, Lemma 3.4, Theorems 4.1 and 4.2]). Let \(H,H'\) be complex Hilbert spaces with \(\mathrm{dim} H\ge 3\). If \(T:[H]\rightarrow[H']\) is an \(\perp\)-symmetry then there exists a unitary or antiunitary operator \(\Theta:H\rightarrow H'\) which induces \(T\). The operator \(\Theta\) is determined by \(T\) up to a complex factor of modulus 1.