Wigner’s symmetry representation theorem

Bargmann 4 — additivity on \([\varphi_0]^\perp\)

Chevalier, Proposition 2, step 3: The restriction of \(A\) to \([\varphi_0]^\perp\) is additive.

First we consider orthogonal unit vectors \(\varphi_1,\varphi_2 \in [\varphi_0]^\perp\). We want to show that the double prime coefficients \(A(\alpha_1\varphi_1+\alpha_2\varphi_2)=\alpha_1''A(\varphi_1)+\alpha_2''A(\varphi_2)\) are the same as for \(A(\alpha_1\varphi_1)=\alpha_1'A(\varphi_1)\) and \(A(\alpha_2\varphi_2)=\alpha_2'A(\varphi_2)\). As above, we consider the two forms for \(|\langle A(\varphi_0+\alpha_1\varphi_1)|A(\varphi_0+\alpha_1\varphi_1+\alpha_2\varphi_2)\rangle|\): this gives \(|1+\alpha_1'^*\alpha_1''|=|1+\alpha_1^*\alpha_1|\). Since \(\alpha_1^*\alpha_1\) is real positive, we have that \(\alpha_1'^*\alpha_1''\) is real positive (so in Lemma 3, the two alternative results are in fact the same), and further \(\alpha_1'^*\alpha_1''=\alpha_1^*\alpha_1\). Since \(|\alpha_1|=|\alpha_1'|=|\alpha_1''|\), we deduce \(\alpha_1''=\alpha_1'\). Similarly, \(\alpha_2''=\alpha_2'\). Hence \(A(\alpha_1\varphi_1+\alpha_2\varphi_2)=A(\alpha_1\varphi_1)+A(\alpha_2\varphi_2)\).

Now consider vectors \(\psi_1,\psi_2\) in the subspace spanned by \(\varphi_1,\varphi_2\). We have \(\psi_1=\alpha_1\varphi_1+\alpha_2\varphi_2\) and \(\psi_2=\beta_1\varphi_1+\beta_2\varphi_2\). Further, \(A(\psi_1+\psi_2)=(\alpha_1+\beta_1)'A(\varphi_1)+(\alpha_2+\beta_2)'A(\varphi_2)\), \(A(\psi_1)=\alpha_1'A(\varphi_1)+\alpha_2'A(\varphi_2)\), and \(A(\psi_2)=\beta_1'A(\varphi_1)+\beta_2'A(\varphi_2)\). But since the prime operation is linear or antilinear on rays, we have \((\alpha_1+\beta_1)'=\alpha_1'+\beta_1'\) and \((\alpha_2+\beta_2)'=\alpha_2'+\beta_2'.\) Combining the pieces: \(A(\psi_1+\psi_2)=A(\psi_1)+A(\psi_2)\). Since the subspace spanned by two arbitrary vectors is either a ray or contains two orthogonal vectors, additivity results in either case.