Wigner’s symmetry representation theorem

Bargmann 2 — from rays to vectors

Chevalier, Proposition 2, step 1: Construct a phase-equivalent map \(A\) on a subset, given \(H,H'\) complex inner product spaces with \(dim H>2\) and a modulus-preserving mapping \(T\) from \(H\) into \(H'\).

Begin by choosing \(\varphi_0 \in H\) with \(\left\Vert\varphi_0\right\Vert=1\). \(A\) will be defined on the subset \(D=(\varphi_0+[\varphi_0]^\perp) \cup [\varphi_0]^\perp\), where the subspace \([\varphi_0]^\perp=\{\varphi:\langle \varphi | \varphi_0 \rangle = 0 \}\) (the hyperplane orthogonal to \([\varphi_0]\)). The subset \(D\) is a disjoint union of its defining components, \(\varphi_0+[\varphi_0]^\perp\), \([\varphi_0]^\perp\) — i.e. \((\varphi_0+[\varphi_0]^\perp) \cap [\varphi_0]^\perp=\varnothing\).

For \(\varphi \in \varphi_0+[\varphi_0]^\perp\), \(A(\varphi)=T(\varphi)/\langle T(\varphi_0)|T(\varphi)\rangle\). We note \(|\langle T(\varphi_0)|T(\varphi)\rangle|=|\langle \varphi_0|\varphi\rangle|=1\), making the maps phase-equivalent on this subset. Further, \(\langle A(\varphi_0) | A(\varphi) \rangle = 1\).

For \(\varphi \in [\varphi_0]^\perp\), we want \(A(\varphi)=A(\varphi_0+\varphi)-A(\varphi_0)\).

Lemma 1 shows that \(T(\varphi_0+\varphi)\) is in the subspace spanned by the orthogonal vectors \(T(\varphi_0),T(\varphi)\). Using the inner product to project: \[|T(\varphi_0+\varphi)\rangle = \frac{|T(\varphi_0)\rangle \langle T(\varphi_0)| T(\varphi_0+\varphi) \rangle}{\left\Vert T(\varphi_0)\right\Vert^2} + \frac{|T(\varphi)\rangle \langle T(\varphi)| T(\varphi_0+\varphi) \rangle}{\left\Vert T(\varphi)\right\Vert^2}\]

Using facts like \(\left\Vert T(\varphi_0)\right\Vert=1\), we can boil down to: \[A(\varphi)=\frac{\langle T(\varphi)| T(\varphi_0+\varphi) \rangle}{\langle T(\varphi_0)| T(\varphi_0+\varphi) \rangle \left\Vert T(\varphi)\right\Vert^2}T(\varphi)\]

As before \(|\langle T(\varphi_0)| T(\varphi_0+\varphi) \rangle|=|\langle \varphi_0| \varphi_0+\varphi \rangle|=1\). Similar reasoning gives \(|\langle T(\varphi)| T(\varphi_0+\varphi) \rangle|=|\langle \varphi| \varphi_0+\varphi \rangle|=\left\Vert \varphi\right\Vert^2=\left\Vert T(\varphi)\right\Vert^2\). Combining these observations we see that \(A\) and \(T\) differ only by a phase factor on \(D\).